Wireless & Mobile Communications Pune University MCQs
Wireless & Mobile Communications Pune University MCQs
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Paging System”.
1. Which of the following is not a standard used for paging system?
a) POCSAG
b) ERMES
c) IS-95
d) FLEX
Answer: c
Explanation: IS-95 is a standard used for cellular system which is based on code division multiple access . POCSAG , ERMES and FLEX are the protocols used in paging system.
2. Paging system uses which mode of transmission?
a) Full duplex
b) Simplex
c) Half Duplex
d) Duplex
Answer: b
Explanation: Simplex systems communicate in only one way. For the paging systems, messages received are not acknowledged and thus they use simplex mode of transmission.
3. The information sent by paging system is known as a ___________
a) Note
b) Line
c) Message
d) Page
Answer: d
Explanation: Page is concise information sent by a paging system to the subscribers of entire service area.
4. Which type of message cannot be sent with the help of paging system?
a) Alphanumeric message
b) Video message
c) Voice message
d) Numeric message
Answer: b
Explanation: The message sent by a paging system can be numeric, alphanumeric or voice depending upon the type of service.
5. What is a paging access number?
a) An e mail id
b) A username
c) A toll free telephone number
d) A registration number
Answer: c
Explanation: Paging access number is a telephone number which is used to send the information to the subscriber.
6. Which type of transmission technique is employed by paging system?
a) Simulcasting
b) Multicasting
c) Unicasting
d) Hybrid
Answer: a
Explanation: Simulcast is a reliable technique used by paging system by transmitting the same paging signal from multiple paging transmitters at approximately equal times.
7. Which of the following is not the property of paging system?
a) Asymmetric communication
b) Light weight
c) High cost
d) Wide area coverage
Answer: c
Explanation: One of the main reasons for the increased users of paging system was its less cost. But as the prices of cellular system declined, the users of paging system also decreased.
8. Which of the following properties describes the transmitters and receivers in paging system?
a) High complexity and high power transmitter, high complexity and high power receivers
b) Low complexity and low power transmitter, low complexity and low power receivers
c) Low complexity and low power transmitter, high complexity and high power receivers
d) High complexity and high power transmitter, low complexity and low power receivers
Answer: d
Explanation: High power of transmitters helps paging signal to easily penetrate the building and low power receivers allow long usage time and light weight batteries.
9. What is a pager in the paging system?
a) A transmitter
b) A receiver
c) A transceiver
d) An equalizer
Answer: b
Explanation: A pager is a wireless device which receives the page, i.e. numeric, alphanumeric or voice message sent by the transmitter.
10. Who introduced the paging system for the first time?
a) Al Gross
b) Teri Pall
c) Alexander Graham Bell
d) Martin Cooper
Answer: a
Explanation: AL Gross introduced the first paging system during world war in 1949.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Cordless Telephone Systems”.
1. Which of the following is a protocol used for cordless telephone system?
a) PACS
b) ERMES
c) IS-95
d) FLEX
Answer: a
Explanation: PACS is a protocol used for cordless telephone system. ERMES and FLEX are used by paging system. And IS-95 is used for cellular system.
2. In which frequency range do the cordless phones mostly work?
a) 43-50 MHz
b) 88-108 MHz
c) 540-1600 KHz
d) 200-540 KHz
Answer: a
Explanation: Cordless phones mostly operate in the frequency range of 43-50 MHz. The frequency range of FM is 88-108 MHz and for AM is 540-1600 KHz.
3. Which of the following is the drawback for cordless telephones?
a) Wireless technology
b) Limited coverage area
c) Mobile
d) Security
Answer: b
Explanation: Cordless telephone systems are wireless, but they have a disadvantage of limited coverage area with cell size of approx. 300 m. Being digital, they have very less chance of eaves dropping.
4. Which of the following is a fully digital cordless system?
a) CT0
b) CT1
c) CT1+
d) DECT
Answer: d
Explanation: Digital enhanced cordless telecommunication is a fully digital system established in 1991. CT0, Ct1, Ct1+ were analog systems established in 1980, 1984 and 1987 respectively.
5. Which of the following is an example of local wireless system?
a) GSM
b) Cordless telephone system
c) UMTS
d) EDGE
Answer: b
Explanation: GSM, UMTS and EDGE covers worldwide area whereas cordless system has vey less coverage area.
6. Which of the following is not a standard for cordless telephony?
a) CT-2
b) DECT
c) UMTS
d) PHS
Answer: c
Explanation: CT-2 is a cordless telephone standard used in Europe and Asia. DECT and PHS are cordless telephone standards used in Europe and Japan respectively. UMTS is a cellular system standard.
7. What is the range of cell diameter of DECT?
a) 300 m
b) 2 km
c) 10 km
d) 70 km
Answer: a
Explanation: DECT has the cell diameter of 300 m from the base station while GSM is designed for outdoor use with a cell diameter of 70 km.
8. Which of the following standard of cordless telephone system is also approved as a 3G standard?
a) PHS
b) PACS
c) DECT
d) CT2
Answer: c
Explanation: DECT fulfills the IMT 2000 requirements and is approved as a 3G standard by ITU .
9. Cordless telephone system will not work under which of the following criteria?
a) Within a home
b) Within a building
c) Within campus
d) Within a city
Answer: d
Explanation: Cordless telephones are used in homes, in offices, on campus, at trade shows. But due to their limited coverage they cannot cover a whole city.
10. Which of the following is not an application of DECT?
a) Multimedia processing
b) Cordless private branch exchange
c) Wireless local loop
d) Home cordless phones
Answer: a
Explanation: Users in a neighborhood served by a telephone company wired local loop can be connected by a cordless phone that exchanges signals with a neighborhood antenna.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Cellular Telephone Systems”.
1. Which of the following is not a characteristic of cellular telephone system?
a) Accommodate a large number of users
b) Large geographic area
c) Limited frequency spectrum
d) Large frequency spectrum
Answer: d
Explanation: Cellular systems accommodate a large number of users within a limited frequency spectrum over a large geographic area.
2. What is the responsibility of MSC in cellular telephone system?
a) Connection of mobile to base stations
b) Connection of mobile to PSTN
c) Connection of base station to PSTN
d) Connection of base station to MSC
Answer: b
Explanation: Mobile Switching Center is responsible for connecting all mobiles to the PSTN in a cellular system.
3. Who has the responsibility of billing and system maintenance function in cellular system?
a) Base Station
b) PSTN
c) MSC
d) Mobile system
Answer: c
Explanation: Mobile switching center accommodates 100,000 subscribers and 5,000 simultaneous conversations at a time and handles all billing and system maintenance functions.
4. What is the function of FVC ?
a) Voice transmission from base station to mobiles
b) Voice transmission from mobile to base station
c) Initiating mobile calls
d) Broadcast all traffic request for all mobile
Answer: a
Explanation: FVC and RVC are responsible for voice transmission. FVC is used for voice transmission from base station to mobile and RVC is used for voice transmission from mobile to base station.
5. Which two channels are responsible for initiating mobile calls?
a) FVC and FCC
b) FVC and RVC
c) FCC and RCC
d) FCC and RVC
Answer: c
Explanation: FCC and RCC are control channels responsible for initiating mobile calls.
6. Of the total channels present in the cellular system, what is the percentage of voice and control channels?
a) 95% voice channels, 5% control channels
b) 5% voice channels, 95% control channels
c) 50% voice channels, 50% control channels
d) 25% voice channels, 75% control channels
Answer: a
Explanation: In each cellular system, control channels are 5% of the total channels available and remaining 95% are dedicated to voice and data traffic.
7. What is MIN?
a) Subscriber’s telephone number
b) Paging message
c) Traffic request number
d) Mobile Internet
Answer: a
Explanation: MIN is a 10 digit unique number which represents the telephone number of subscriber.
8. What is transmitted along with the call initiation request during the origin of call by a mobile?
a) MIN
b) ESN
c) ESN and SCM
d) MIN, ESN and SCM
Answer: d
Explanation: When a mobile originates the call, it sends the MIN , ESN and SCM along with the call initiation request.
9. What does SCM indicates?
a) Maximum receiver power level for a particular user
b) Maximum transmitter power level for a particular user
c) Minimum receiver power level for a particular user
d) Minimum transmitter power level for a particular user
Answer: b
Explanation: SCM indicates the maximum transmitter power level for a particular user.
10. What is the shape of the cell present in the cellular system?
a) Circular
b) Square
c) Hexagonal
d) Triangular
Answer: c
Explanation: The shape of the cell present in the cellular network is hexagonal since it can cover the entire geographical area without any gap and overlapping.
11. Why the size of the cell is kept small in cellular network?
a) Increase capacity
b) Decrease capacity
c) Increased size of base station electronics
d) Slow process of handoffs
Answer: a
Explanation: The size of the cells in cellular network is kept small because of the need of high capacity in areas with high user density and reduced size and cost of base station electronics.
12. What is handoff?
a) Forward channel
b) Switching technique
c) Roamer
d) Guard channel
Answer: b
Explanation: Handoff is a switching technique which refers to the process of transferring an active call or data session from one cell in a cellular network to another.
13. Which one is not an advantage of using frequency reuse?
a) Increased capacity
b) Limited spectrum is required
c) Same spectrum may be allocated to other network
d) Number of base stations is reduced
Answer: d
Explanation: Frequency reuse is a technique of reusing frequencies and channels within a cellular system to improve capacity and spectral efficiency.
14. The process of transferring a mobile station from one base station to another is ____________
a) MSC
b) Roamer
c) Handoff
d) Forward channel
Answer: c
Explanation: Handoff is the process of changing the channel associated with current connection while a call is in progress.
15. The interference between the neighbouring base stations is avoided by ____________
a) Assigning different group of channels
b) Using transmitters with different power level
c) Using different antennas
d) Using different base stations
Answer: a
Explanation: The interference between the neighbouring base stations is avoided by assigning different group of channels and reusing the same channel after a certain amount of distance.
This set of Wireless & Mobile Communications Interview Questions and Answers focuses on “Second Generation Cellular Networks”.
1. Which of the following multiple access techniques are used by second generation cellular systems?
a) FDMA/FDD and TDMA/FDD
b) TDMA/FDD and CDMA/FDD
c) FDMA/FDD and CDMA/FDD
d) FDMA/FDD only
Answer: b
Explanation: First generation cellular system used FDMA/FDD techniques. Second generation standards uses TDMA/FDD and CDMA/FDD multiple access techniques. 2G networks are digital.
2. Which one is not a TDMA standard of second generation networks?
a) GSM
b) IS-136
c) AMPS
d) PDC
Answer: c
Explanation: GSM , IS-136 and PDC are the three most popular TDMA standards of second generation. AMPS is a first generation standard.
3. Which of the following is a CDMA standard of second generation network?
a) IS-95
b) IS-136
c) ETACS
d) EDGE
Answer: a
Explanation: Interim Standard 95 is the most popular CDMA standard of second generation networks. IS-136 is a TDMA standard of 2G. EDGE is a standard of 2.5G and ETACS is a 1G standard.
4. Popular 2G CDMA standard IS-95 is also known as ______________
a) CdmaOne
b) CdmaTwo
c) IS-136
d) IS-95B
Answer: a
Explanation: The popular 2G CDMA standard, Interim Standard is also known as CdmaOne. The 2.5G CDMA standard, IS-95B is called CdmaTwo. And IS-136 is a TDMA standard for 2G.
5. How many users or voice channels are supported for each 200 KHz channel in GSM?
a) Eight
b) Three
c) Sixty four
d) Twelve
Answer: a
Explanation: GSM is a circuit switched system that divides each 200 KHz channel into eight 25 KHz time slots, i.e. each radio channel is divided into eight voice channels.
6. How many voice channels are supported for each 30 KHz radio channel in IS-136?
a) Eight
b) Thirty
c) Three
d) Sixteen
Answer: c
Explanation: Interim Standard 136 was popularly known as North American Digital Cellular system. It divides each 30 KHz radio channel into three time slots, each of 10 KHz.
7. How many users are supported in IS-95 for each 1.25 MHz?
a) Eight
b) Sixty four
c) Sixteen
d) Twenty five
Answer: b
Explanation: IS-95 supports upto 64 users which are orthogonally coded and simultaneously transmitted on each 1.25 MHz. The services of IS-95 standard are short messaging service, slotted paging, over-the-air activation, enhanced mobile station identities etc.
8. Which modulation technique is used by GSM?
a) GMSK
b) BPSK
c) QPSK
d) GFSK
Answer: a
Explanation: GSM uses a form of modulation known as GMSK . It is a form of modulation with no phase discontinuities and provides data transmission with efficient spectrum usage.
9. IS-95 uses which modulation technique?
a) GMSK
b) BPSK
c) QAM
d) AFSK
Answer: b
Explanation: IS- 95 uses BPSK with quadrature spreading. It is regarded as one of the most robust digital modulation technique and is used for long distance wireless communication.
10. IS-136 uses which modulation technique?
a) π/4 DQPSK
b) BPSK
c) GMSK
d) AFSK
Answer: a
Explanation: IS-136 uses π/4 DQPSK modulation technique. This technique allows a bit rate of 48.6 Kbit/s with 30 KHz channel spacing which gives a bandwidth efficiency of 1.62 bit/s/Hz.
11. Which is one of the disadvantages of 2G standards?
a) Short Messaging Service
b) Digital modulation
c) Limited capacity
d) Limited Internet Browsing
Answer: d
Explanation: 2G technologies use circuit switched data modems that limits data users to a single circuit switched voice channel. The advantages of 2G network are that they are digital in nature and supports SMS service.
12. GSM was earlier also known as _____________
a) Group System Mobile
b) Global Special Meaning
c) Group Special Mobile
d) Global Special Mobile
Answer: c
Explanation: GSM was earlier known as Group Special Mobile. As it became more global, the meaning of acronym was changed to Global System for Mobile.
13. 2G CDMA standard, IS-95, was proposed by which company?
a) Nippon Telephone and Telegraph
b) Qualcomm
c) Bellcore and Motorola
d) AT&T Bell Laboratories
Answer: b
Explanation: IS-95 was proposed by Qualcomm in early 1990s. Later it was adopted as a standard by Telecommunications Industry Association in TIA/EIA/IS-95 release published in 1995.
14. Which one of the following 2G standard is used in Japan?
a) IS-136
b) GSM
c) PDC
d) AMPS
Answer: c
Explanation: PDC was standardized by Japanese Ministry of Posts and Telecommunication in 1991. It is similar to IS- 136, but with 25 KHz voice channels to be compatible with the Japanese analog channels.
15. The 2G GSM technology uses a carrier separation of _____________
a) 1.25 MHz
b) 200 KHz
c) 30 KHz
d) 300 KHz
Answer: b
Explanation: The Global System for Mobile uses a carrier separation of 200 KHz, each channel supporting upto eight users.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “2.5G”.
1. What is the name of the web browsing format language supported by 2.5G technology?
a) Wireless Application Protocol
b) Hypertext Markup Language
c) Extensible Markup Language
d) Hypertext Transfer Protocol
Answer: a
Explanation: 2.5G technology supports a new web browsing format language, which is called Wireless Application Protocol . It allows standard web pages to be viewed in a compressed format specifically designed for small, portable hand held wireless devices.
2. What is the name of the internet microbrowser technology used by NTT DoCoMo in Japan?
a) Wireless Application Protocol
b) I-mode
c).W-mode
d) Hypertext Markup Language
Answer: b
Explanation: I-mode is a wireless data service and Internet microbrowser technology introduced by NTT DoCoMo on its PDC network in 1998. It is currently used by other wireless services throughout the world.
3. 2.5G upgrade path for a particular wireless carrier does not match the original 2G technology choice made earlier by the same carrier.
a) True
b) False
Answer: b
Explanation: As 2.5G is the upgradation of 2G technology, 2.5G upgradation path must match the original 2G technology. For example, 2.5G upgrade solution designed for GSM must dovetail with original GSM interface so that change of hardware is not required.
4. Which of the following is not a TDMA standard of 2.5G network?
a) HSCSD
b) GPRS
c) EDGE
d) GSM
Answer: d
Explanation: GSM is a TDMA standard for 2G network. HSCSD , GPRS and EDGE are TDMA standards of 2.5G technology.
5. Which of the following is a 2.5G CDMA standard?
a) IS-95
b) Cdma2000
c) IS-95B
d) CdmaOne
Answer: c
Explanation: IS-95B is code division multiple access standard for 2.5G. It is an upgradation of IS- 95 which is a second generation standard of CDMA.
6. HSCSD supports which 2G standard?
a) GSM
b) IS-136
c) GSM and IS-136
d) PDC
Answer: a
Explanation: High Speed Circuit Switched Data supports the Global system for Mobile standard. It only requires a software upgrade at the base station.
7. How does HSCSD differs from the GSM to obtain higher speed data rate?
a) By allowing single user to use one specific time slot
b) By allowing single user to use consecutive user time slots
c) By using 8-PSK modulation technique
d) By allowing multiple users to use individual time slot
Answer: b
Explanation: HSCSD allows individual data users to use consecutive time slots in order to obtain higher speed data access on the GSM network. In case of GSM, it limits each user to use only one specific time slot.
8. GPRS and EDGE supports which 2G standard?
a) GSM only
b) IS-136 only
c) GSM and IS-136 both
d) PDC
Answer: c
Explanation: GPRS network provides a packet network on dedicated GSM or IS-136 radio channels. EDGE is also developed keeping in desire both GSM and IS-136 operators.
9. How is HSCSD different from GPRS?
a) Infrastructure
b) Multiple Access Scheme
c) Modulation technique
d) Switching Technique
Answer: d
Explanation: GPRS is a packet based network. HSCSD dedicates circuit switched channels to specific users whereas GPRS supports many more users, but in a bursty manner.
10. What changes GPRS need to acquire while upgrading itself from GSM?
a) A whole new base station
b) New transceiver at base station
c) New channel cards
d) New packet overlay including routers and gateways
Answer: d
Explanation: GPRS requires a GSM operator to install new routers and Internet gateways at the base station along with new software upgrade. New base station RF hardware is not required.
11. Which new modulation technique is used by EDGE?
a) BPSK
b) 8- PSK
c) DQPSK
d) AFSK
Answer: b
Explanation: EDGE uses a new digital modulation format, 8- PSK . It is used in addition to GSM’s standard GMSK modulation.
12. Various air interface formats used by EDGE are also known as ___________
a) Modulation and coding schemes
b) Coding schemes
c) Modulating air interface
d) Air interface coding schemes
Answer: a
Explanation: EDGE allows nine different air interface formats known as multiple modulation and coding schemes . Each MCS state may use either GMSK or 8- PSK modulation for network access, depending upon the instantaneous demands of the network and the operating conditions.
13. EDGE is sometimes also referred as ____________
a) HSCSD
b) 3GPP
c) EGPRS
d) EGSCSD
Answer: c
Explanation: EDGE is sometimes also referred as Enhanced GPRS . It is an enhancement of a GSM network in which EDGE is introduced on top of the General Packet Radio Service . It is used to transfer data in a packet switched mode on various time slots.
14. What is one disadvantage of EDGE in comparison to HSCSD and GPRS?
a) Low data rates
b) Small coverage range
c) Low speed
d) No advancement
Answer: b
Explanation: Because of the higher data rates and relaxed error control covering in many of the selectable air interface formats, the coverage range is smaller in EDGE than in HSCSD or GPRS.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Third Generation Wireless Networks”.
1. Which of the following is not a characteristic of 3G network?
a) Communication over VoIP
b) Unparalleled network capacity
c) Multi-megabit Internet access
d) LTE based network
Answer: d
Explanation: Multi-megabit Internet access, communication using Voice over internet Protocol , voice activated calls, unparalleled network capacity are some of the characteristics of 3G network. 3G systems promise unparalleled wireless access which is not possible in 2G systems. LTE is a standard of 4G systems.
2. What is the term used by ITU for a set of global standards of 3G systems?
a) IMT 2000
b) GSM
c) CDMA
d) EDGE
Answer: a
Explanation: International Telecommunications Union used the term IMT-2000 in 1998. It is used for a set of global standards for third generation mobile telecoms services and equipment.
3. Which of the following leads to evolution of 3G networks in CDMA systems?
a) IS-95
b) IS-95B
c) CdmaOne
d) Cdma2000
Answer: d
Explanation: 3G evolution of CDMA system leads to cdma2000. It is based on the fundamentals of IS-95 and IS-95B. IS-95 is a 2G standard for CDMA systems. IS-95B is a CDMA system for 2.5G networks.
4. Which of the following leads to the 3G evolution of GSM, IS-136 and PDC systems?
a) W-CDMA
b) GPRS
c) EDGE
d) HSCSD
Answer: a
Explanation: The 3G evolution for GSM, IS-136 and PDC systems leads to W-CDMA . It is based on the network fundamentals of GSM, as well as merged versions of GSM and IS-136 through EDGE. GPRS, EDGE and HSCSD are 2.5G networks.
5. What is 3GPP?
a) Project based on W-CDMA
b) Project based on cdma2000
c) Project based on 2G standards
d) Project based on 2.5G standards
Answer: a
Explanation: 3GPP is a 3G Partnership Project for Wideband CDMA standards based on backward compatibility with GSM and IS-136. The project was established in December 1998. Its initial scope was to make a globally applicable third generation mobile phone system.
6. What is 3GPP2?
a) Project based on W-CDMA
b) Project based on cdma2000
c) Project based on 2G standards
d) Project based on 2.5G standards
Answer: b
Explanation: 3GPP2 is a 3G Partnership Project for Cdma2000 standards based on backward compatibility with earlier CdmaOne 2G CDMA technology. It was initiated by IMT-2000 to cover high speed, broadband and Internet Protocol based mobile systems. It mainly focuses on North American and Asian regions.
7. Which of the following is not a standard of 3G?
a) UMTS
b) Cdma2000
c) TD-SCDMA
d) LTE
Answer: d
Explanation: UMTS , TD-SCDMA , Cdma2000 are the standards defined for 3G networks. LTE is a 4G standard for high speed wireless communication.
8. Which of the following 3G standard is used in Japan?
a) Cdma2000
b) TD-SCDMA
c) UMTS
d) UTRA
Answer: c
Explanation: Japan uses UMTS standard for its 3G network. The standards used are UMTS 800, UMTS 900, UMTS 1500, UMTS 1700 and UMTS 2100. They are standardized by ARIB .
9. What does the number 2000 in IMT-2000 signifies?
a) Year
b) Number of subscribers per cell
c) Number of cells
d) Area
Answer: a
Explanation: The International Telecommunication Union defined the third generation of mobile telephony standards, IMT-2000 to facilitate growth, increase bandwidth, and support more diverse applications. The number 2000 in IMT-2000 indicates the start of the system and the spectrum used .
10. Which of the following is not an application of third generation network?
a) Global Positioning System
b) Video conferencing
c) Mobile TV
d) Downloading rate upto 1 Gbps
Answer: d
Explanation: 3G applications include GPS , MMS , video conferencing, location based services, video on demand, wireless voice telephony and high data rates with peak downloading rate of 100 Mbps. For 4G networks, the peak downloading rate is 1 Gbps.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “3G W-CDMA ”.
1. What is the full form of UMTS?
a) Universal Mobile Telephone System
b) Ubiquitous Mobile Telephone System
c) Ubiquitous Mobile Telemetry System
d) Universal Machine Telemedicine System
Answer: a
Explanation: UMTS is a visionary air interface standard that was introduced in 1996. European carriers, manufacturers, and government regulators collectively developed the early version of UMTS as an open air interface standard for third generation wireless telecommunication.
3. UMTS use which multiple access technique?
a) CDMA
b) TDMA
c) FDMA
d) SDMA
Answer: a
Explanation: Although UMTS is designed to operate on evolved GSM core networks, it uses code division multiple access for its air interface. The majority of the 3G systems in operation employ CDMA, while the rest use TDMA. CDMA allows various users to share a channel at the same time, while TDMA allows users to share the same channel by chopping it into different time slots.
3. UMTS does not has backward compatibility with ____________
a) GSM
b) IS-136
c) IS-95
d) GPRS
Answer: c
Explanation: UMTS assures backward compatibility with the second generation GSM, IS-136 and PDC TDMA technologies. It is also compatible with all 2.5G TDMA techniques like GPRS and EDGE. But it does not provide compatibility to CDMA technologies of 2G and 2.5 G. IS-95 is a CDMA standard of 2G.
4. UMTS is also known as _____________
a) IS-95
b) GPRS
c) CdmaOne
d) W-CDMA
Answer: d
Explanation: UMTS uses Wideband CDMA to carry the radio transmissions. Therefore, it is also referred as W-CDMA. W-CDMA offers greater spectral efficiency and bandwidth to mobile network operators.
5. What is the chip rate of W-CDMA?
a) 1.2288 Mcps
b) 3.84 Mcps
c) 270.833 Ksps
d) 100 Mcps
Answer: b
Explanation: W-CDMA uses a chip rate of 3.84 Mcps. Chip rate is the product of symbol rate and spreading factor. If the symbol rate is 960 Kbps and spreading factor is 4 for W-CDMA, then the chip rate is 3.84 Mcps. The chip rate for Cdma2000 and GSM are 1.2288 Mcps and 27.0833 Ksps respectively.
6. W-CDMA works in FDD mode only.
a) True
b) False
Answer: b
Explanation: W-CDMA works in both FDD and TDD mode. W-CDMA developed for wide area cellular coverage uses FDD. And TDD is used by W-CDMA for indoor cordless type applications.
7. How much packet data rate per user is supported by W-CDMA if the user is stationary?
a) 2.048 Kbps
b) 100 Mbps
c) 2.048 Mbps
d) 1 Gbps
Answer: c
Explanation: If the user is stationary, W-CDMA supports packet data rates upto 2.048Mbps per user. Thus, it allows high quality data, multimedia, streaming audio video and broadcast type services to consumers. Future version of WCDMA will support stationary user data rates in excess of 8Mbps.
8. What is the minimum spectrum allocation required by W-CDMA?
a) 5 MHz
b) 20MHz
c) 1.25 MHz
d) 200 KHz
Answer: a
Explanation: W-CDMA/UMTS requires a minimum spectrum allocation of 5 MHz. Using this bandwidth, it has the capacity to carry over 100 simultaneous voice calls. It is able to carry data at speeds up to 2 Mbps in its original format. 20 MHz is the bandwidth defined for LTE. CdmaOne uses a bandwidth of 1.25 MHz. GSM’s bandwidth is 200 KHz.
9. W-CDMA requires a complete change of RF equipment at each base station.
a) True
b) False
Answer: a
Explanation: W-CDMA is designed to provide backward compatibility and interoperability for all GSM, IS-136/PDC, GPRS and EDGE equipment. But due to a wider air interface bandwidth of W-CDMA, it requires a complete change of RF-equipment at each base station.
10. How much increase in spectral efficiency is provided by W-CDMA in comparison to GSM?
a) Two times
b) Three times
c) No increase
d) Six times
Answer: d
Explanation: W-CDMA can provide at least six times an increase in spectral efficiency over GSM at system level. Such a wider bandwidth is chosen to higher data rates as low as 8 kbps to as high as 2 Mbps on a single 5 MHz W-CDMA radio channel.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “3G Cdma2000”.
1. Which of the following has no backward compatibility with 3G Cdma2000?
a) IS-95
b) GPRS
c) IS-95A
d) IS-95B
Answer: b
Explanation: 3G Cdma2000 is based on the original IS-95 and IS-95A CDMA standards, as well as the 2.5G IS-95B air interface. While upgrading, Cdma2000 maintains backward compatibility with existing IS-95, IS-95A and IS-95B equipments. Thus, Cdma2000 allow wireless carriers to introduce a family of new high data rate Internet access capabilities within existing systems.
2. 2G and 2.5G CDMA operators may selectively introduce 3G capabilities at each cell without changing entire base stations and reallocate spectrums.
a) True
b) False
Answer: a
Explanation: Cdma2000 does not require change in entire base station or additional RF equipment. All the changes are made in software or in baseband hardware.
3. Which of the following the first 3G CDMA air interface?
a) IS-95
b) IS-95B
c) Cdma2000 1xRTT
d) CdmaOne
Answer: c
Explanation: Cdma2000 1xRTT is the first 3G air interface. Here, RTT stands for Radio Transmission Technology and 1x indicates that the bandwidth is one times that of the original CdmaOne channel. It is modulated on a single carrier.
4. Within ITU IMT-2000 body, Cdma2000 1xRTT is also known as ____________
a) Cdma2000 1xEV-DO
b) Cdma2000 1xEV-DV
c) IS-95B
d) G3G-MC-CDMA-1X
Answer: d
Explanation: Cdma2000 1xRTT is also known as G3G-MC-CDMA-1X. The initials MC stand for multicarrier. RTT stand for Radio Transmission Technology, a language suggested by IMT-2000 body. Usually, MC and RTT are omitted for convenience.
5. How many users are supported by Cdma2000 1X in comparison to 2G CDMA standard?
a) Half
b) Twice
c) Six times
d) Ten times
Answer: b
Explanation: Cdma2000 support up to twice as many users as the 2G CDMA standard. It also provides the two times the standby time for longer lasting battery life.
6. Cdma2000 works in TDD mode only.
a) True
b) False
Answer: b
Explanation: Cdma2000 works in both FDD and TDD mode. Cdma2000 developed for wide area cellular coverage uses FDD. And TDD is used by Cdma2000 for indoor cordless type applications.
7. Which of the following is not a characteristic of Cdma2000?
a) Adaptable baseband signalling rates
b) Adaptable baseband chipping rates
c) Multicarrier technologies
d) OFDMA
Answer: d
Explanation: Cdma2000 does not use OFDMA technique. OFDMA is used by 4G networks. Cdma2000 uses rapidly adaptable baseband signalling rates and chipping rates for each user. It also provides multi-level keying within same framework.
8. Cdma2000 1xEV was developed by ________
a) Motorola
b) AT&T Laboratories
c) Qualcomm
d) NTT
Answer: c
Explanation: Cdma2000 is an evolutionary advancement for CDMA. It was developed by Qualcomm Inc. It is a proprietary high data rate packet standard that can be overlaid upon existing IS-95, IS-95B and Cdma2000 networks.
9. How is bandwidth increased in Cdma2000?
a) Clubbing adjacent radio channels
b) Changing the hardware of base stations
c) Change of spectrum
d) Change of RF equipment
Answer: a
Explanation: The increase in the bandwidth is achieved through clubbing the adjacent radio channels of CdmaOne and using multicarrier technologies. For example, Cdma2000 3xRTT uses these technologies by combining three adjacent 1.25 MHz bandwidth of CdmaOne channels.
10. What are the two options provided by Cdma2000 1xEV?
a) Cdma2000 1xRTT. Cdma2000 3xRTT
b) Cdma2000 1xEV-DO, Cdma2000 1xEV-DV
c) Cdma2000 1xEV-DT, Cdma2000 1xEV-DO
d) Cdma2000 1xRTT, Cdma2000 1xEV-DV
Answer: b
Explanation: Cdma2000 1xEV is an evolutionary upgrade for Cdma2000. It provides two options, for accessing only data and for both data & voice .
11. Which of the following is not backward compatible with Cdma2000?
a) Cdma2000 1xRTT
b) Cdma2000 3xRTT
c) Cdma2000 1xEV-DO
d) Cdma2000 1xEV-DT
Answer: c
Explanation: Cdma2000 1xEV-DO option provides data rates of about 2.4 Mbps and supports data traffic only. No voice communication is supported. It relaxes the minimum latency requirement. But this mode is not backward compatible with Cdma2000.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Wireless Local Area Networks ”.
1. What is the full form of WLAN?
a) Wide Local Area Network
b) Wireless Local Area Network
c) Wireless Land Access Network
d) Wireless Local Area Node
Answer: b
Explanation: WLAN stands for Wireless Local Area Network. Wireless networks is increasingly used as a replacement for wires within homes, buildings, and office settings through the deployment of wireless local area networks .
2. WLANs use high power levels and generally require a license for spectrum use.
a) True
b) False
Answer: b
Explanation: WLANs use low power and generally do not require a license for spectrum. They provide ad hoc high data transmission rate connections deployed by individuals. In the late 1980s, FCC provided licence free bands for low power spread spectrum devices in ISM band, which is used by WLAN.
3. What is the name of 300 MHz of unlicensed spectrum allocated by FCC in ISM band?
a) UNII
b) Unlicensed PCS
c) Millimetre wave
d) Bluetooth
Answer: a
Explanation: FCC allocated 300 MHz of unlicensed spectrum in the ISM bands. This allocation is called the Unlicensed National Information Infrastructure band. It was allocated for the express purpose of supporting low power license free spread spectrum data communication.
4. Which of the following specifies a set of media access control and physical layer specifications for implementing WLANs?
a) IEEE 802.16
b) IEEE 802.3
c) IEEE 802.11
d) IEEE 802.15
Answer: c
Explanation: IEEE 802.11 is a set of media access control and physical layer specification for implementing WLAN computer communication. It was founded in 1987 to begin standardization of spread spectrum WLANs for use in the ISM bands.
5. Which of the following is not a standard of WLAN?
a) HIPER-LAN
b) HIPERLAN/2
c) IEEE 802.11b
d) AMPS
Answer: d
Explanation: AMPS is a standard of first generation network. HIPERLAN is a WLAN standard developed in Europe in mid 1990s. HIPERLAN/2 is also developed in Europe that provides upto 54 Mbps of user data.
6. Which of the following is the 802.11 High Rate Standard?
a) IEEE 802.15
b) IEEE 802.15.4
c) IEEE 802.11g
d) IEEE 802.11b
Answer: d
Explanation: IEEE 802.11b was a high rate standard approved in 1999. It provided new data rate capabilities of 11 Mbps, 5.5 Mbps in addition to the original 2 Mbps and 1 Mbps user rates of IEEE 802.11.
7. Which of the following spread spectrum techniques were used in the original IEEE 802.11 standard?
a) FHSS and DSSS
b) THSS and FHSS
c) THSS and DSSS
d) Hybrid technique
Answer: a
Explanation: Original IEEE 802.11 used both the approaches of FHSS and DSSS . But from late 2001s, only DSSS modems are used within IEEE 802.11.
8. Which of the following WLAN standard has been named Wi-Fi?
a) IEEE 802.6
b) IEEE 802.15.4
c) DSSS IEEE 802.11b
d) IEEE 802.11g
Answer: c
Explanation: The DSSS IEEE 802.11b standard has been named Wi-Fi by the Wireless Ethernet Compatibility Alliance. It is a group that promotes adoption of 802.11 DSSS WLAN.
9. Which of the following is developing CCK-OFDM?
a) IEEE 802.11a
b) IEEE 802.11b
c) IEEE 802.15.4
d) IEEE 802.11g
Answer: d
Explanation: IEEE 802.11g is developing CCK-OFDM standards. It will support roaming capabilities and dual band use for public WLAN networks. It also has backward compatibility with 802.11b technologies.
10. What is the data rate of HomeRF 2.0?
a) 10 Mbps
b) 54 Mbps
c) 200 Mbps
d) 1 Mbps
Answer: a
Explanation: HomeRF 2.0 has data rate of the order of 10 Mbps. The FHSS proponents of IEEE 802.11 have formed the HomeRF standard that supports the frequency hopping equipment. In 2001, HomeRF developed a 10 Mbps FHSS standard called HomeRF 2.0.
11. HIPER-LAN stands for ____________
a) High Precision Radio Local Area Network
b) High Performance Radio Local Area Network
c) High Precision Radio Land Area Network
d) Huge Performance Radio Link Access Node
Answer: b
Explanation: HIPER-LAN stands for High Performance Radio Local Area Network. It was developed in Europe in mid 1990s. It was intended to provide individual wireless LANs for computer communication.
12. What is the range of asynchronous user data rates provided by HIPER-LAN?
a) 1-100 Mbps
b) 50-100 Mbps
c) 1-20 Mbps
d) 500 Mbps to 1 Gbps
Answer: c
Explanation: HIPER-LAN provides asynchronous user data rates of between 1 to 20 Mbps, as well as time bounded messaging of rates of 64 kbps to 2.048 Mbps. It uses 5.2 GHz and 17.1 GHz frequency bands.
13. What is the name of the European WLAN standard that provides user data rate upto 54 Mbps?
a) UNII
b) WISP
c) MMAC
d) HIPERLAN/2
Answer: d
Explanation: HIPERLAN/2 has emerged as the next generation European WLAN standard. It provides upto 54 Mbps of user data to a variety of networks. The networks includes the ATM backbone, IP based networks and the UMTS network.
14. What is WISP?
a) Wideband Internet Service Protocol
b) Wireless Internet Service Provider
c) Wireless Instantaneous Source Provider
d) Wideband Internet Source Protocol
Answer: b
Explanation: WISP is wireless Internet Service Provider used to explore public LANs . It builds a nationwide infrastructure of WLAN access points in selected hotels, restaurants or airports. It then charges a monthly subscription fee to users who wish to have always on Internet access in those selected locations.
15. The price of WLAN hardware is more than 3G telephones and fixed wireless equipment.
a) True
b) False
Answer: b
Explanation: As, WLAN could be used to provide access for the last 100 meters into homes and businesses. Therefore, the price of WLAN hardware is far below 3G telephones and fixed wireless equipment.
This set of Wireless & Mobile Communications Questions and Answers for Freshers focuses on “Bluetooth and Personal Area Networks ”.
1. Which of the following is not an open standard?
a) Bluetooth
b) WWW
c) HTML
d) VPN
Answer: d
Explanation: An open standard is a standard that is publicly available. It has various rights to use associated with it. Bluetooth is an open standard that has been embraced by over thousand manufacturers of electronic appliances. VPN is a private network.
2. What is the nominal range of Bluetooth?
a) 1 Km
b) 10 m
c) 1 m
d) 10 Km
Answer: b
Explanation: The power of the transmitter governs the range over which a Bluetooth device can operate. Generally, Bluetooth devices are said to fall into one of three classes. The most common kind is class 2, and it operates in 10 m range.
3. Bluetooth standard is named after ___________
a) King Ronaldo Bluetooth
b) Pope Vincent Bluetooth
c) King Herald Bluetooth
d) Pope Francis Bluetooth
Answer: c
Explanation: Bluetooth standard is named after King Harald Bluetooth. He was the 10th century Viking who united Denmark and Norway. The Bluetooth standard aims to unify the connectivity chores of appliances within the personal workspace of an individual.
4. Bluetooth operates in which band?
a) Ka Band
b) L Band
c) Ku Band
d) 2.4 GHz ISM Band
Answer: d
Explanation: Bluetooth is a radio technology operating in 2.4 GHz frequency band. Bluetooth is best suited for low-bandwidth applications like transferring sound data with telephones or byte data with hand-held computers or keyboard.
5. Which of the following scheme is used by Bluetooth?
a) Frequency hopping TDD scheme
b) Frequency hopping FDD scheme
c) DSSS TDD scheme
d) DSSS FDD scheme
Answer: a
Explanation: Frequency hopping TDD scheme is used by Bluetooth. Frequency hopping provides a form of multiple access among co-located devices in different piconets.
6. What is the range of time slot in Bluetooth?
a) 120 milliseconds
b) 625 microseconds
c) 577 microseconds
d) 5.7 seconds
Answer: b
Explanation: Bluetooth uses a timeslot of 625 microseconds. A data channel hops randomly 1600 times per second between 79 RF channels. Thus, each channel is divided into time slots 625 microseconds.
7. Which modulation scheme is used by Bluetooth?
a) DQPSK
b) MSK
c) GFSK
d) BPSK
Answer: c
Explanation: Bluetooth uses GFSK . When GFSK is used for Bluetooth modulation, the frequency of the carrier is shifted to carry the modulation. By doing this the Bluetooth modulation achieves a bandwidth of 1 MHz with stringent filter requirements to prevent interference on other channels.
8. What is the channel symbol rate in Bluetooth for each user?
a) 270.833 Kbps
b) 1 Gbps
c) 100 Mbps
d) 1 Mbps
Answer: d
Explanation: Each user in Bluetooth uses a radio channel symbol rate of 1 Mbps using GFSK modulation. The frequency hopping scheme of each Bluetooth user is determined from a cyclic code with a length of 10 27 -1.
9. What is the raw channel bit error rate of Bluetooth?
a) 10 -3
b) 10 -10
c) 10 3
d) 10 -1
Answer: a
Explanation: Bluetooth has a bit error rate of 10 -3 . Bluetooth uses a number of forward error control coding and automatic repeat request schemes to achieve this bit rate.
10. Which of the following standard committee specifies Bluetooth and other Personal Area Networks ?
a) IEEE 802.11b
b) IEEE 802.15
c) IEEE 802.11g
d) IEEE 802.16
Answer: b
Explanation: IEEE 802.15 standards committee is formed to provide an international forum for developing Bluetooth and other PANs. PANs are used to interconnect pocket PCs, personal digital assistants , cell phones, light projectors and other appliances.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Frequency Reuse”.
1. Cellular concept replaces many low power transmitters to a single high power transmitter.
a) True
b) False
Answer: b
Explanation: Cellular concept is a system level idea that replaces a single high power transmitter to many low power transmitters. High power transmitters lead to large cell, and thus it was impossible to use the same frequencies throughout the systems. But, it is possible with low power transmitter.
2. Why neighbouring stations are assigned different group of channels in cellular system?
a) To minimize interference
b) To minimize area
c) To maximize throughput
d) To maximize capacity of each cell
Answer: a
Explanation: Neighbouring base stations are assigned different group of channels. It minimizes the interference between base stations and the users under their control.
3. What is a cell in cellular system?
a) A group of cells
b) A group of subscribers
c) A small geographical area
d) A large group of mobile systems
Answer: c
Explanation: Cell is a small geographic area in a cellular system. Each cellular base station within a cell is allocated a group of radio channels that could be used in another cell.
4. What is frequency reuse?
a) Process of selecting and allocating channels
b) Process of selection of mobile users
c) Process of selecting frequency of mobile equipment
d) Process of selection of number of cells
Answer: a
Explanation: Frequency reuse is the process of using the same radio frequencies on radio transmitter sites within a geographic area. They are separated by sufficient distance to cause minimal interference with each other.
5. Which of the following is a universally adopted shape of cell?
a) Square
b) Circle
c) Triangle
d) Hexagon
Answer: d
Explanation: Hexagonal cell shape is a simplistic model of radio coverage for each base station. It has been universally adopted since the hexagon permits easy and manageable analysis of a cellular system.
6. Actual radio coverage of a cell is called __________
a) Fingerprint
b) Footprint
c) Imprint
d) Matrix
Answer: b
Explanation: Actual radio coverage of a cell is known as the footprint. It is determined from field measurements or propagation prediction models. Although the real footprint is amorphous in nature, a regular cell shape is needed for systematic system design.
7. Why the shape of cell is not circle?
a) Omni directionality
b) Small area
c) Overlapping regions or gaps are left
d) Complex design
Answer: c
Explanation: Circle is the first natural choice to represent the coverage area of a base station. But while adopting this shape, adjacent cells cannot be overlaid upon a map without leaving gaps or creating overlapping regions.
8. What is the main reason to adopt hexagon shape in comparison to square and triangle?
a) Largest area
b) Simple design
c) Small area
d) Single directional
Answer: a
Explanation: For a given distance between the center of a polygon and its farthest perimeter points, the hexagon has the largest area. Thus, by using the hexagon geometry, the fewest number of cells can cover a geographic region.
9. Which type of antenna is used for center excited cells?
a) Dipole antenna
b) Grid antenna
c) Sectored antenna
d) Omnidirectional antenna
Answer: d
Explanation: For center excited cells, base station transmitters are used at the center of cell. To cover the whole cell, omnidirectional antenna is the best choice for base station transmitters.
10. Which type of antenna is used for edge excited cells?
a) Omnidirectional antenna
b) Grid antenna
c) Sectored directional antenna
d) Dipole antenna
Answer: c
Explanation: For edge excited cell, mostly base station transmitters are placed on three of the six cell vertices. To cover the assigned portion of a cell, sectored directional antenna is the best choice.
11. For a cellular system, if there are N cells and each cell is allocated k channel. What is the total number of available radio channels, S?
a) S=k*N
b) S=k/N
c) S=N/k
d) S=k N
Answer: a
Explanation: If there is a cellular system with total of S duplex channels. Each cell is allocated a group of k channels and there are total N cells in the system, S channels are divide among N cells into unique and disjoint channel groups. Therefore, total number of radio channel is the product of total number of cells in the system and number of channel allocated to each cell .
12. What is a cluster in a cellular system?
a) Group of frequencies
b) Group of cells
c) Group of subscribers
d) Group of mobile systems
Answer: b
Explanation: Cluster is group of N cells. These cells use the complete set of frequency available for the cellular system at that location.
13. What is a frequency reuse factor for N number of cells in a system?
a) N
b) N 2
c) 2*N
d) 1/N
Answer: d
Explanation: The frequency reuse factor is defined as 1 over the number of cells in the cluster of the system . It is given by 1/N since each cell within a cluster is only assigned 1/N of the total available channels in the system.
14. Capacity of a cellular system is directly proportional to __________
a) Number of cells
b) Number of times a cluster is replicated
c) Number of Base stations
d) Number of users
Answer: b
Explanation: The capacity of a cellular system is directly proportional to the number of times a cluster is replicated in a fixed area. If the cluster size N is reduced while the cell size is kept constant, more clusters are required to cover a given area, and hence more capacity is achieved.
15. A spectrum of 30 MHz is allocated to a cellular system which uses two 25 KHz simplex channels to provide full duplex voice channels. What is the number of channels available per cell for 4 cell reuse factor?
a) 150 channels
b) 600 channels
c) 50 channels
d) 85 channels
Answer: a
Explanation: Total bandwidth is 30 MHz. And the channel bandwidth is 50 KHz/duplex channel . Therefore, total available channels are 600 channels . For 4 cell reuse factor, total number of channels available per cell will be 150 channels .
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Channel Assignment Strategies”.
1. Which of the following is not an objective for channel assignment strategies?
a) Efficient utilization of spectrum
b) Increase of capacity
c) Minimize the interference
d) Maximize the interference
Answer: d
Explanation: The objective of channel assignment strategy is to utilize the spectrum efficiently. And for efficient utilization, a frequency reuse scheme consistent with the objective of increasing capacity and minimizing interference is required.
2. The choice of channel assignment strategy does not impact the performance of the system.
a) True
b) False
Answer: b
Explanation: The choice of channel assignment strategy impacts the performance of the system. Particularly as to how calls are managed, when a mobile user is handed off from one cell to another.
3. In fixed channel assignment strategy, each cell is allocated a predetermined set of _______
a) Voice channels
b) Control channels
c) Frequency
d) base stations
Answer: a
Explanation: In a foxed channel strategy, each cell is allocated a predetermined set of voice channels. Any call attempt within the cell can only be served by the unused channels in that particular cell.
4. What happen to a call in fixed channel strategy, if all the channels in a cell are occupied?
a) Queued
b) Cross talk
c) Blocked
d) Delayed
Answer: c
Explanation: As any call attempt within a cell can be served by unused channels in fixed channel strategy. If all the channels in that cell are occupied, the call is blocked and subscriber does not receive any service.
5. What is a borrowing strategy in fixed channel assignments?
a) Borrowing channels from neighbouring cell
b) Borrowing channels from neighbouring cluster
c) Borrowing channels from same cell
d) Borrowing channels from other base station in same cell
Answer: a
Explanation: In borrowing strategy, a cell is allowed to borrow channels from a neighbouring cell if all of its own channels are already occupied. The MSC supervises such borrowing procedure and ensures that the borrowing of channel does not interfere with any call in progress.
6. In dynamic channel assignment strategy, voice channels are allocated to different cells permanently.
a) True
b) False
Answer: b
Explanation: In a dynamic channel strategy, voice channels are not allocated to different cells permanently. Instead, serving base station requests a channel from MSC each time a cell request is made.
7. In dynamic channel assignment strategy, base station requests channel from ____________
a) MSC
b) Neighbouring cell
c) Neighbouring cluster
d) Neighbouring base station
Answer: a
Explanation: Each time a call request is made, the serving base station requests a channel from the MSC. The switch then allocates a channel to the requested cell following an algorithm that takes into account the likelihood of future blocking within the cell.
8. Dynamic channel assignment reduces the likelihood of blocking in comparison to fixed channel assignment.
a) True
b) False
Answer: a
Explanation: Dynamic channel assignment reduces the likelihood of blocking. Accordingly, the MSC only allocates a given frequency if that frequency is not presently in use in the cell or any other cell which falls within the minimum restricted distance of frequency reuse.
9. RSSI stands for ________
a) Received Signal Strength Indicator
b) Restricted Signal Strength Indicator
c) Radio Signal Strength Indication
d) Restricted System Software Indicator
Answer: a
Explanation: Received signal strength indicator is a measurement of the power present in a received radio signal. RSSI is usually invisible to a user of a receiving device.
10. What is the drawback of dynamic channel assignment?
a) Decrease channel utilization
b) Increase probability of blocked call
c) Cross talk
d) Increase storage and computational load on system
Answer: d
Explanation: Dynamic channel assignment requires the MSC to collect real time data on channel occupancy, traffic distribution and RSSI of all channels on continuous basis. This increases the storage and computational load on the system but provides the advantage of increased channel utilization and decreased probability of blocked call.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Handoff Strategies”.
1. What is the condition for handoff?
a) A mobile moves into a different cell while in conversation
b) A mobile remains in the same cell while in conversation
c) A mobile moves to different cell when idle
d) A mobile remains in the same cell and is idle
Answer: a
Explanation: Handoff occurs when a mobile moves into a different cell while a conversation is in progress. The MSC automatically transfers the call to a new channel belonging to the new base station.
2. Handoff does not require voice and control channel to be allocated to channels associated with the new base station.
a) True
b) False
Answer: b
Explanation: Handoff operation involves identifying a new base station. It also requires that the voice and control signal be allocated to channels associated with the new base station.
3. The time over which a call can be maintained within a cell without handoff is called _________
a) Run time
b) Peak time
c) Dwell time
d) Cell time
Answer: c
Explanation: The time over which a call is maintained within a cell without handoff is called as dwell time. Dwell time vary depending on speed of user and type of radio coverage.
4. Dwell time does not depend on which of the following factor?
a) Propagation
b) Interference
c) Distance between subscriber and base station
d) Mobile station
Answer: d
Explanation: Dwell time of a particular user is governed by a number of factors. They include propagation, interference, distance between the subscriber and the base station, and other time varying effects.
5. Which of the following is associated with the handoff in first generation analog cellular systems?
a) Locator receiver
b) MAHO
c) Cell dragging
d) Breathing cell
Answer: a
Explanation: Locator receiver is a spare receiver in each base station. It is used to scan and determine signal strengths of mobile users which are in neighbouring cells.
6. MAHO stands for ______
a) MSC assisted handoff
b) Mobile assisted handoff
c) Machine assisted handoff
d) Man assisted handoff
Answer: b
Explanation: MAHO stands for mobile assisted handoff. In 2G systems, handoff decisions are mobile assisted. In MAHO, every mobile station measure the received power from surrounding base station and continuously reports the results to serving base station.
7. A handoff is initiated when the power received from the base station of a neighbouring cell falls behind the power received from the current base station by certain level.
a) True
b) False
Answer: b
Explanation: MAHO measures the power received from the surrounding base station. And a handoff is initiated when the power received from the base station of a neighbouring cell begins to exceed the power received from current base station.
8. What is the condition for intersystem interference?
a) Mobile moves from one cell to another cell
b) Mobile remains in the same cell
c) Mobile moves from one cellular system to another cellular system
d) Mobile remains in the same cluster
Answer: c
Explanation: An intersystem handoff is initiated when a mobile moves from one cellular system to another during a course of a call. An MSC engages in an intersystem interference when a mobile becomes weak in a given cell and MSC cannot find another cell to which call can be transferred.
9. What is the disadvantage of guard channel?
a) Efficient utilization of spectrum
b) Cross talk
c) Near far effect
d) Reduce total carried traffic
Answer: d
Explanation: Guard channel is a concept for handling priority in handoff. Here, a fraction of the total available channels in a cell is reserved exclusively for handoff requests from ongoing calls. This method has the disadvantage of reducing the total carried traffic, as fewer channels are allocated to originating calls.
10. Which of the following priority handoff method decrease the probability of forced termination of a call due to lack of available channels?
a) Queuing
b) Guard channel
c) Cell dragging
d) Near far effect
Answer: a
Explanation: Queuing of handoff requests is a method to decrease the probability of forced termination of a call due to lack of available channels. Queuing of handoff is possible due to the fact that there is a finite time interval between the time the received signal level drops below the handoff threshold and the time the call is terminated.
11. Umbrella cell approach is possible by using _________
a) Antenna of same heights
b) Antenna of different heights
c) Different voice channels
d) Different control channels
Answer: b
Explanation: Umbrella cell approach is possible by using different antenna heights and different power levels. By using this approach, it is possible to provide large and small cells which are co-located at a single location.
12. Cell dragging is a problem occur due to __________
a) Pedestrian users
b) Stationary users
c) High speed mobile systems
d) Base stations having same frequency
Answer: a
Explanation: Cell dragging is a practical handoff problem in microcell system. It results from pedestrian users that provide a very strong signal to the base station.
13. What was the typical handoff time in first generation analog cellular systems?
a) 1 second
b) 10 seconds
c) 1 minute
d) 10 milliseconds
Answer: b
Explanation: In first generation analog cellular system, the typical time to make a handoff once the signal level is below the threshold, is about 10 seconds. This requires the value for threshold to be 6 dB to 12 dB.
14. How much time it takes for handoff in digital cellular systems like GSM?
a) 1 second
b) 10 seconds
c) 1 minute
d) 10 milliseconds
Answer: a
Explanation: In digital cellular systems, the mobile assista with the handoff procedure by determining the best candidate. Once the decision is made, it typically requires 1 to 2 seconds for handoff.
15. Soft handoff is also known as _________
a) MAHO
b) Hand over
c) Break before make
d) Make before break
Answer: d
Explanation: Soft handoff is one in which the channel in the source cell is retained and used for a while in parallel with the channel in the target cell. In this case, the connection with the receiver target is established before the connection to the source is broken, hence this handover is called make-before-break.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Interference and System Capacity”.
1. Which of the following is not a source of interference?
a) Base station in a different cluster
b) Another mobile in same cell
c) A call in progress in neighbouring cell
d) Any BS operating on same frequency
Answer: a
Explanation: Interference is a major limiting factor in the performance of cellular radio systems. Sources of interference includes another mobile in the same cell, a call in progress in neighbouring cell, other base stations operating in the same frequency band, or any non-cellular system which inadvertently leaks energy into the cellular frequency band.
2. Interference on voice channels causes _______
a) Blocked calls
b) Cross talk
c) Queuing
d) Missed calls
Answer: b
Explanation: Interference on voice channels causes crass talk. Here, the subscriber hears interference in the background due to an undesired transmission.
3. Interference in control channel leads to ________
a) Cross talk
b) Queuing
c) Blocked calls
d) Voice traffic
Answer: c
Explanation: On control channels, interference leads to missed and blocked calls. This happens due to errors in the digital signalling.
4. Interference is more severe in rural areas.
a) True
b) False
Answer: a
Explanation: Interference is more severe in rural areas. It happens due to the greater RF noise floor and the large number of base stations and mobiles.
5. What are co-channel cells?
a) Cells having different base stations
b) Cells using different frequency
c) Cells using adjacent frequency
d) Cells using same frequency
Answer: d
Explanation: Due to frequency reuse concept, there are several cells that use the same set of frequencies. These cells are called co-channel cells. And the interference between these cells is called co-channel interference.
6. Co-channel interference is a function of _________
a) Radius of cell
b) Transmitted power
c) Received power
d) Frequency of mobile user
Answer: a
Explanation: This is the case when the size of each cell is approximately the same and the base stations transmit the same power. Co-channel interference ratio is independent of the transmitted power and becomes a function of the radius of the cell ® and the distance between centers of the nearest co channel cell .
7. Co-channel reuse ratio is define by _________
a) Q=D*R
b) Q=D/R
c) Q=D^R
d) Q=1/R
Answer: b
Explanation: Co-channel reuse ratio is defined by Q=D/R. By increasing the ratio of D/R, the spatial separation between co-channel cells relative to the coverage distance of a cell is increased. Thus, interference I reduced from improved isolation of RF energy from the co-channel cells.
8. Co-channel ratio in terms of cluster size is defined as _________
a) \ N
c) 3N
d) √N
Answer: a
Explanation: Co-channel reuse is defined using \(\sqrt{}\). A small value of Q provides larger capacity since the cluster size N is small. However, a large value of Q improves the transmission quality, due to smaller level of co-channel interference.
9. What is the cluster size for CDMA?
a) N=10
b) N=100
c) N=1
d) N=50
Answer: c
Explanation: CDMA systems have a cluster size of N=1.Therefore, frequency reuse is not as difficult as for TDMA or first generation cellular systems.
10. What is breathing cell effect?
a) Fixed coverage region
b) Dynamic and time varying coverage region
c) Large coverage region
d) Very small coverage region
Answer: b
Explanation: Breathing cell is a concept used by CDMA systems. They had a dynamic, time varying coverage region which varies depending on the instantaneous number of users on the CDMA radio channel.
11. Adjacent channel interference occurs due to _______
a) Power transmitted by Base station
b) MSCs
c) Same frequency of mobile users
d) Imperfect receiver filters
Answer: d
Explanation: Interference resulting from signals which are adjacent in frequency to the desired signal is called adjacent channel interference. It results from imperfect receiver filters which allow nearby frequencies to leak into the passband.
12. Which of the following problem occur due to adjacent channel interference?
a) Blocked calls
b) Cross talk
c) Near-far effect
d) Missed calls
Answer: c
Explanation: One of the main problems with adjacent channel interference is the near-far effect. It occurs when a mobile close to a base station transmits on a channel close to one being used by a weak mobile.
13. In near-far effect, a nearby transmitter captures the __________
a) Receiver of the subscriber
b) Transmitter of the subscriber
c) Nearby MSC
d) Neighbouring base station
Answer: a
Explanation: Near-far effect occurs if an adjacent channel user is transmitting in very close range to a subscriber’s receiver while the receiver attempts to receive a base station on the desired channel. In this effect, a nearby transmitter captures the receiver of the subscriber.
14. Adjacent channel interference can be minimized through _______
a) Changing frequency of base stations
b) Careful filtering and channel assignments
c) Increasing number of base stations
d) Increasing number of control channels
Answer: b
Explanation: Adjacent channel assignment can be minimized through careful filtering and channel assignments. Each cell is given only a fraction of the available channels, a cell need not be assigned channels which are all adjacent in frequency.
15. In dynamic channel assignment, any channel which is being used in one cell can be reassigned simultaneously to another cell in the system at a reasonable distance.
a) True
b) False
Answer: a
Explanation: Dynamic channel assignment is one well known solution to the micro cellular channel assignment problem. The dynamic nature of the strategy permits adaptation to spatial and traffic variations while the distribution of control reduces the required computational load.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Trunking and Grade of Service”.
1. What is the concept for accommodating a large number of users in a limited radio spectrum?
a) Grade of service
b) Trunking
c) Multiplexing
d) Multitasking
Answer: b
Explanation: Cellular radio systems rely on trunking to accommodate a large number of users in a limited radio spectrum. The concept of trunking allows a large number of users to share the relatively small number of channels in a cell by providing access to each user, on demand.
2. On termination of call, the occupied channel is not returned to the pool of available channels in trunking.
a) True
b) False
Answer: b
Explanation: In a trunked radio system, each user is allocated a channel on a per call basis. Upon termination of the call, the previously occupied channel is immediately returned to the pool of available channels. It is a method for a system to provide network access to many clients by sharing a set of lines or frequencies instead of providing them individually.
3. In trunking system, when the channel is already in use, the call is blocked or queued.
a) True
b) False
Answer: a
Explanation: In a trunked mobile radio system when a particular user requests service, there is a possibility that all the channels are already in use. Then the user is blocked, or denied access to the system. Sometimes, a queue may be used to hold the requesting users until a channel becomes available.
4. Who developed the fundamental of trunking theory?
a) Newton
b) Ohm
c) Erlang
d) Einstein
Answer: c
Explanation: The fundamentals of trunking theory were developed by Erlang. He was a Danish mathematician. He embarked on the study of how a large population could be accommodated by a limited number of servers in late 19th century.
5. What is the unit for the measure of traffic intensity?
a) Meters
b) Henry
c) Ohm
d) Erlang
Answer: d
Explanation: The measure of traffic intensity is given by Erlang. It is defined as the ratio of the time during which a facility is cumulatively occupied to the time this facility is available for occupancy. Telecommunication operators are vitally interested in traffic intensity as it dictates the amount of equipment they must supply.
6. One Erlang represents _________
a) One call- hour per hour
b) One call-minute per hour
c) One call- hour per minute
d) Many calls- hour per hour
Answer: a
Explanation: One Erlang represents the amount of traffic intensity carried by a channel that is completely occupied . For example, a radio channel that is occupied for 30 minutes during an hour carries 0.5 Erlangs of traffic.
7. What is the measure of the ability of user to access a trunked system during the busiest hour?
a) Trunking
b) Grade of Service
c) Multiplexing
d) Sectoring
Answer: b
Explanation: The grade of service is a measure of the ability of a user to access a trunked system during the busiest hour. The busy hour is based upon customer demand at the busiest hour during a week, month or a year.
8. GOS is typically given as a likelihood that a ________
a) Call is in progress
b) Channels are busy
c) Call is blocked
d) Channel are free
Answer: c
Explanation: GOS is typically given as the likelihood that a call is blocked or the likelihood of a call experiencing a delay greater than a certain queuing time. It is the wireless designer’s job to estimate the maximum required capacity used to allocate the proper number of channels in order to meet GOS.
9. The time requires to allocate a trunked radio channel to a requesting user is called _______
a) Dwell time
b) Holding time
c) Run time
d) Set up Time
Answer: d
Explanation: When any user makes a call request, the time required to allocate a trunked channel to a user is known as set up time. TETREA call set up time is 0.3 seconds. GSM uses a call set up time of several seconds.
10. Average duration of a typical call is called ________
a) Holding time
b) Dwell time
c) Set up time
d) Run time
Answer: a
Explanation: Average time of a typical call is called holding time. It is denoted by H . It is used to measure the traffic intensity per user. The time over which a call may be maintained within a cell, without handoff is called dwell time.
11. The average number of call requests per unit time is also known as ________
a) Request rate
b) Load
c) Grade o Service
d) Traffic intensity
Answer: a
Explanation: Request rate is the average number of call requests per unit time. It is denoted by λ. Unit for request rate is second-1. It can also be defined as the ratio of traffic intensity of each user and the holding time.
12. Traffic intensity offered by each user is the product of __________
a) Set up time and holding time
b) Call request rate and holding time
c) Load and holding time
d) Call request rate and set up time
Answer: b
Explanation: The traffic intensity offered by each user is equal to the call request rate multiplied by the holding time. Each user generates a traffic intensity A=λH Erlang. Here, H is the average duration of a call and λ is the average number of call requests per unit time for each user.
13. AMPS cellular system is designed for a GOS of _____ blocking.
a) 10%
b) 50 %
c) 2%
d) 1%
Answer: c
Explanation: The AMPS cellular system is designed for a GOS of 2% blocking. This implies that the channel allocations for cell sites are designed so that 2 out of 100 calls will be blocked due to channel occupancy during the busiest hour.
14. Blocked calls cleared formula is also known as _______ formula.
a) Erlang C
b) Erlang A
c) Erlang D
d) Erlang B
Answer: d
Explanation: Erlang B formula is also known as the blocked calls cleared formula. The Erlang B formula determines the probability that a call is blocked. And, it is a measure of the GOS for a trunked system which provides no queuing for blocked calls.
15. Blocked calls delayed formula is also known as _______
a) Erlang A
b) Erlang B
c) Erlang C
d) Erlang D
Answer: c
Explanation: Erlang C is also known as Blocked Calls Delayed. In this trunked system, a queue is provided to hold calls which are blocked. If a channel is not available immediately, the call request may be delayed until a channel becomes available.
This set of Wireless & Mobile Communications Interview Questions and Answers for freshers focuses on “Improving Coverage and Capacity in Cellular Systems”.
1. Which of the following techniques do not help in expanding the capacity of cellular system?
a) Sectoring
b) Scattering
c) Splitting
d) Microcell zone concept
Answer: b
Explanation: As the demand for wireless service increases, the number of channels assigned to a call eventually becomes insufficient to support the required number of user. Techniques such as cell splitting, sectoring and coverage zone approaches are used in practice to expand the capacity of cellular system.
2. ________ uses directional antennas to control interference.
a) Sectoring
b) Cell splitting
c) Repeaters
d) Micro cell zone concept
Answer: a
Explanation: Sectoring uses directional antenna to further control the interference and frequency reuse of channels. By decreasing the cell radius R and keeping the co-channel reuse ratio D/R unchanged, cell splitting increases the number of channels per unit area.
3. _______ allows an orderly growth of cellular system.
a) Sectoring
b) Scattering
c) Cell splitting
d) Micro cell zone technique
Answer: c
Explanation: Cell splitting allows an orderly growth of cellular system. By defining new cells which have a smaller radius than the original cells, capacity increases due to additional number of channels per unit area.
4. Which of the following technology distributes the coverage of the cell and extends the cell boundary to hard-to-reach places?
a) Cell splitting
b) Scattering
c) Sectoring
d) Micro cell zone concept
Answer: d
Explanation: Micro cell zone concept distributes the coverage of the cell and extends the cell boundary to hard-to reach places. It is the solution for the problem of increased number of handoffs when sectoring which results in an increase of load on switching.
5. Which of the following increases the number of base stations in order to increase capacity?
a) Cell splitting
b) Sectoring
c) Repeaters
d) Micro cell zone concept
Answer: a
Explanation: Cell splitting increases the number of base stations in order to increase capacity. Whereas, sectoring and zone microcells rely on base station antenna placements to improve capacity by reducing co-channel interference.
6. Which of the following trunking inefficiencies?
a) Cell splitting
b) Micro cell zone technique
c) Sectoring
d) Repeaters
Answer: c
Explanation: Sectored cells experience trunking inefficiencies. Cell splitting and zone micro cell techniques do not suffer the trunking inefficiencies experienced by sectored cells. They enable the base station to oversee all handoff chores related to microcells, thus reducing the computational load at MSC.
7. The process of subdividing a congested cell into smaller cells is called _______
a) Cell splitting
b) Sectoring
c) Micro cell technique
d) Repeaters
Answer: a
Explanation: Cell splitting is the process of subdividing a congested cell into smaller cells. Each small cell has its own base station and a there is a corresponding reduction in antenna height and transmitter power.
8. Cell splitting increases the capacity of a cellular system since it increases the number of times ________ are reused.
a) Cells
b) Channels
c) Transmitters
d) Mobile stations
Answer: b
Explanation: Cell splitting increases the capacity of a cellular system since it increases the number of times channels are reused. But it has a limitation that handoffs are more frequent and channel assignments become more difficult.
9. Cell splitting do not maintain the minimum c-channel reuse ratio.
a) True
b) False
Answer: b
Explanation: Cell splitting allows a system to grow by replacing large cells with smaller cells. It does not upset the channel allocation scheme required to maintain the minimum co channel reuse ratio Q between co-channel cells.
10. Which of the following technique is used to limit radio coverage of newly formed microcells?
a) Sectoring
b) Splitting
c) Antenna downtilting
d) Scattering
Answer: c
Explanation: Antenna downtilting deliberately focuses radiated energy from the base station toward the ground . It is often used to limit the radio coverage of newly formed microcells.
11. Sectoring increases SIR .
a) True
b) False
Answer: a
Explanation: Sectoring increases SIR so that cluster size may be reduced. SIR is improved using directional antenna. And then capacity improvement is achieved by reducing the number of cells in a cluster, thus increasing the frequency reuse.
12. Which of the following has range extension capability?
a) Sectoring
b) Repeaters
c) Scattering
d) Micro cell zone concept
Answer: b
Explanation: Wireless operator needs to provide dedicated coverage for hard-to-reach areas, such as within buildings, or in valleys or tunnels. Radio transmitters used to provide such range extension capabilities are called as repeaters. They are bidirectional in nature.
13. Repeaters has one drawback of reradiating ____________
a) Frequency
b) Channels
c) Power
d) Repeater noise and interference
Answer: d
Explanation: Upon receiving signals from a base station forward link, the repeater amplifies and reradiates the base station signals to the specific coverage region. Unfortunately, the received noise and interference is also reradiated by repeater on both the forward and reverse link.
14. Which of the following is not an advantage of micro cell zone technique?
a) Reduced co channel interference
b) Improved signal quality
c) Increase in capacity
d) Increasing number of base stations
Answer: d
Explanation: The advantage of the zone cell technique is that co-channel interference in the cellular system is reduced since a large central base station is replaced by several low powered transmitters on the edges of the cell. Thus, signal quality is reduced and it leads to an increase in capacity.
15. In a micro cell zone concept, when a mobile travels from one zone to another within the cell, it retains the same _________
a) Power level
b) Base station
c) Channel
d) Receiver
Answer: c
Explanation: As a mobile travels from one zone to another within the cell, it retains the same channel. Thus, unlike in sectoring, a handoff is not required at the MSC when the mobile travels between zones within the cell. The channels are re used in co channel cells in a normal fashion.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Free Space Propagation Model”.
1. The mechanism behind electromagnetic wave propagation cannot be attributed to ___________
a) Reflection
b) Diffraction
c) Scattering
d) Sectoring
Answer: d
Explanation: The mechanisms behind electromagnetic wave propagation are diverse. They can be greatly attributed to reflection, diffraction and scattering. Due to multiple reflections from various objects, the electromagnetic waves travel along different paths of varying lengths.
2. The propagation model that estimates radio coverage of a transmitter is called ___________
a) Large scale propagation model
b) Small scale propagation model
c) Fading model
d) Okumura model
Answer: a
Explanation: Large scale propagation model are useful in estimating the radio coverage area of a transmitter. They can predict the mean signal strength for an arbitrary transmitter-receiver separation distance. They characterize signal strength over large T-R separation distances.
3. Propagation model that characterize rapid fluctuation is called _________
a) Hata model
b) Fading model
c) Large scale propagation model
d) Okumura model
Answer: b
Explanation: Fading models characterize the rapid fluctuations of the received signal strength over very short travel distance or shot time durations .
4. Small scale propagation model is also known as _________
a) Fading model
b) Micro scale propagation model
c) Okumura model
d) Hata model
Answer: a
Explanation: Small scale propagation model is also called fading model. Fading model characterize the rapid fluctuations of the received signal strength over very short distance of a few wavelengths or short time duration. The propagation models are used to estimate the performance of wireless channels.
5. Free space propagation model is to predict ______
a) Received signal strength
b) Transmitted power
c) Gain of transmitter
d) Gain of receiver
Answer: a
Explanation: Free space propagation model predicts the received signal strength when there is an unobstructed line of sight path between transmitter and receiver. It assumes the ideal propagation condition that the environment is empty between the transmitter and receiver.
6. Which of the following do not undergo free space propagation?
a) Satellite communication system
b) Microwave line of sight radio links
c) Wireless line of sight radio links
d) Wired telephone systems
Answer: d
Explanation: EM signals when traveling through wireless channels experience fading effects due to various effects. But in some cases the transmission is with no obstruction and direct line of sight such as in satellite communication, microwave and wireless line of sight radio links.
7. The free space model predicts that received signal decays as a function of _________
a) Gain of transmitter antenna
b) T-R separation
c) Power of transmitter antenna
d) Effective aperture of the antenna
Answer: b
Explanation: As with most large scale radio wave propagation models, the free space model predicts that received signal decays as a function of the T-R separation distance raised to some power. Often it is given as a function of negative square root of the distance.
8. Relation between gain and effective aperture is given by ______
a) G=(4πA e )/λ 2
b) G=(4π λ 2 )/A e
c) G=4Ď€A e
d) G=A e /λ 2
Answer: a
Explanation: The gain of the antenna is proportional to effective aperture area. Therefore, antennas with large effective apertures are high gain antennas and have small angular beam widths. Most of their power is radiated in a narrow beam in one direction, and little in other directions.
9. Relation between wavelength and carrier frequency is _________
a) λ=c/f
b) λ=c*f
c) λ=f/c
d) λ=1/f
Answer: a
Explanation: Wavelength is inversely proportional to carrier frequency. For electromagnetic radiation in free space, wavelength is a ratio of speed of light and carrier frequency . Speed of light is 3*10 8 m/s. The unit for wavelength is meters.
10. Which of the following antenna radiates power with unit gain uniformly in all directions?
a) Directional antenna
b) Dipole antenna
c) Isotropic antenna
d) Loop antenna
Answer: c
Explanation: Isotropic antenna radiates the power with unit gain uniformly in all directions. It is an ideal antenna. From practical point of view, there is no actual isotropic antenna. But, an isotropic antenna is often used as a reference antenna for the antenna gain.
11. EIRP is abbreviated as __________
a) Effective isotropic radiated power
b) Effective isotropic radio power
c) Effective and immediate radiated power
d) Effective and immediate ratio of power
Answer: a
Explanation: EIRP stands for Effective Isotropic Radiated Power. It is the amount of power that a theoretical isotropic antenna would emit to produce the peak power density observed in the direction of maximum antenna gain. EIRP also takes into account the losses in transmission line and connectors and includes the gain of the antenna.
12. Path loss in free space model is defined as difference of ________
a) Effective transmitted power and gain
b) Effective received power and distance between T-R
c) Gain and received power
d) Effective transmitter power and receiver power
Answer: d
Explanation: I Path loss is defined as difference of effective transmitter power and receiver power. Free-space path loss is the loss in signal strength of an electromagnetic wave that would result from a line-of-sight path through free space, with no obstacles nearby to cause reflection or diffraction.
13. Far field region is also known as _________
a) Near field region
b) Fraunhofer region
c) Erlang region
d) Fresnel region
Answer: b
Explanation: The far field is the region far from the antenna. In this region, the radiation pattern does not change shape with distance. Also, this region is dominated by radiated fields, with the E- and H-fields orthogonal to each other and the direction of propagation as with plane waves.
14. Fraunhofer distance is given by _____
a) 2D2/λ
b) 2D/λ
c) D/λ
d) 2D/λ2
Answer: a
Explanation: Fraunhofer distance, also known as far field distance is inversely proportional to wavelength. It depends on the largest physical dimension of the antennal . This distance basically denotes the boundary between far field and near field region.
15. Which of the following is called an ideal antenna?
a) Dipole antenna
b) Directional antenna
c) Isotropic antenna
d) Loop antenna
Answer: c
Explanation: Isotropic antenna is an ideal antenna that directs the power uniformly in all directions. It is a theoretical point source of electromagnetic. It is practically not possible. It is mainly used as a hypothetical antenna to measure the gain.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Reflection”.
1. Which of the following mechanism do not impact propagation in mobile communication system?
a) Reflection
b) Diffraction
c) Scattering
d) Refraction
Answer: d
Explanation: Reflection, diffraction and scattering are the three basic propagation mechanism which impact propagation in mobile communication system. Large scale propagation model and small scale fading and multipath propagation are described by the physics of reflection, diffraction and scattering.
2. What is the dimension of object as compared to wavelength of propagating wave when reflection occurs?
a) Large
b) Small
c) Same
d) Very small
Answer: a
Explanation: Reflection occurs when a propagating electromagnetic wave impinges upon an object which has very large dimensions when compared to the wavelength of the propagation wave. Reflection occurs from the surface of the Earth and from buildings and walls.
3. When does the wave propagating from one medium to another gets partially reflection and partially transmitted?
a) Both mediums have same electrical properties
b) Both mediums have different electrical properties
c) Both mediums have same magnetic properties
d) Both mediums have different magnetic properties
Answer: b
Explanation: When a radio wave propagating in one medium impinges upon another medium having different electrical properties. The wave is partially reflected and partially transmitted.
4. What is the case of reflection, in course of second medium being a perfect dielectric?
a) Loss of energy during absorption
b) Total energy reflected back to first medium
c) No loss of energy in absorption
d) Total energy transmitted into second medium
Answer: c
Explanation: If the plane wave is incident on a perfect dielectric, part of the energy is transmitted into the second medium and part of the energy is reflected back into the first medium. There is no loss of energy in absorption.
5. What is the case of reflection, in course of second medium being a perfect conductor?
a) Loss of energy during absorption
b) Total energy reflected back to first medium
c) Partly transmission and reflection
d) Total energy transmitted into second medium
Answer: b
Explanation: If the second medium is perfect conductor, then all incident energy is reflected back into the second medium. There is no loss of energy during absorption.
6. Which of the following relates the incident and reflected & transmitted wave?
a) Fresnel transmission coefficient
b) Scattering coefficient
c) Diffraction coefficients
d) Fresnel reflection coefficient
Answer: d
Explanation: The electric field intensity of the reflected and transmitted waves may be related to the incident waves in the medium of origin through the Fresnel reflection coefficient. It is equal to the ratio of the amplitude of the reflected wave to the incident wave, with each expressed as phasors.
7. Reflection coefficient is not a function of __________
a) Material property
b) Diffraction loss
c) Wave polarization
d) Angle of incidence
Answer: b
Explanation: The reflection coefficient is a function of the material properties, and generally depends upon the wave polarization, angle of incidence and frequency of propagating waves. It is a parameter that describes how much of an electromagnetic wave is reflected by an impedance discontinuity in the transmission medium.
8. Polarized wave can be mathematically represented as sum of ________
a) Four orthogonal components
b) Two spatially adjacent components
c) Two spatially orthogonal components
d) Six orthogonal components
Answer: c
Explanation: A polarized wave may be mathematically represented as sum of two spatially orthogonal components. For an arbitrary polarization, super position may be used to compute the reflected fields from a reflecting surface.
9. The plane of incidence contains only incident rays.
a) True
b) False
Answer: b
Explanation: The plane of incidence is defined as the plane containing the incident, reflected and transmitted waves. The incident light is polarized with its electric field perpendicular to the plane containing the incident, reflected, and refracted rays.
10. Permittivity and conductivity are insensitive to ______ for a good conductor.
a) Operating frequency
b) Polarization density
c) Electric field
d) Property of material
Answer: a
Explanation: The terms permittivity and conductivity are insensitive to operating frequency when the material is a good conductor. In the case of conductors, it is evident that electric field inside a conductor is zero. That is because free charges reside only on the surface of conductor and not inside.
11. Velocity of electromagnetic wave can be given by _______
a) 1/√
b) ÎĽ/∈
c) 1/
d) ÎĽ∈
Answer: a
Explanation: For a medium with permittivity, ∈ and permeability, ÎĽ the velocity of electromagnetic wave is given by 1/√. It is also known as phase velocity. The velocity of light is given by 3*10 8 m/s.
12. The boundary condition at the surface of incidence obeys ________
a) Kepler’s law
b) Gauss law
c) Faraday law
d) Snell’s law
Answer: d
Explanation: Snell’s law is also known as Snell–Descartes law or the law of refraction. It gives a formula to describe the relationship between the angles of incidence and refraction, when referring to light or other waves passing through a boundary between two different isotropic media, such as water, glass, or air.
13. The angle at which no reflection occurs in the medium of origin is called _________
a) Azimuth angle
b) Elevation angle
c) Brewster angle
d) Inclination angle
Answer: c
Explanation: The Brewster angle is the angle at which no reflection occurs in the medium of origin. It occurs when the incident angle is such that the reflection coefficient is equal to zero. The critical Brewster’s angles for diamond, glass and water are 67.5°, 57° and 53° respectively.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Diffraction”.
1. Diffraction occurs when radio path between Tx. And Rx. Is obstructed by ____________
a) Surface having sharp irregularities
b) Smooth irregularities
c) Rough surface
d) All types of surfaces
Answer: a
Explanation: Diffraction occurs when radio path between transmitter and receiver is obstructed by a surface that has sharp irregularities . The secondary waves resulting from the obstructing surface are present throughout the space and even behind the obstacle.
2. At high frequencies, diffraction does not depends on ___________
a) Geometry of the object
b) Distance between Tx and Rx
c) Amplitude of incident wave
d) Polarization of incident wave
Answer: b
Explanation: At high frequency, diffraction depends on the geometry of the object, as well as the amplitude, phase, and polarization of the incident wave at the point of diffraction. It gives rise to a bending of waves even when line of sight does not exist between transmitter and receiver.
3. Diffraction allows radio signals to propagate around ________
a) Continuous surface
b) Smooth surface
c) Curved surface of Earth
d) Does not allow propagation
Answer: c
Explanation: Diffraction allows radio signals to propagate around the curved surface of the Earth. Signals can propagate beyond the horizon and to propagate behind obstruction. It is the slight bending of light as it passes around the edge of an object.
4. Which principle explains the phenomenon of diffraction?
a) Principle of Simultaneity
b) Pascal’s Principle
c) Archimedes’ Principle
d) Huygen’s principle
Answer: d
Explanation: The phenomenon of diffraction can be explained by Huygen’s principle. It states that all points on a wavefront can be considered as point sources for the production of secondary wavelets. And these wavelets combine to produce a new wavefront in direction of propagation.
5. Diffraction is caused by propagation of secondary wavelets into _______
a) Bright region
b) Shadowed region
c) Smooth region
d) Large region
Answer: b
Explanation: Diffraction is caused due to propagation of secondary wavelets into a shadowed region. The field strength in the shadowed region is the vector sum of the electric field components of all the secondary wavelets in the space around the obstacle.
6. Difference between the direct path and the diffracted path is called _______
a) Average length
b) Radio path length
c) Excess path length
d) Wavelength
Answer: c
Explanation: Excess path length denoted by ∆, is the difference between the direct path and the diffracted path. It is calculated with the help of Fresnel zone geometry.
7. The phase difference between a direct line of sight path and diffracted path is function of _______
a) Height and position of obstruction
b) Only height
c) Operating frequency
d) Polarization
Answer: a
Explanation: The phase difference between a direct line of sight path and diffracted path is a function of height and position of the diffraction. It is also a function of transmitter and receiver location.
8. Which of the following explains the concept of diffraction loss?
a) Principle of Simultaneity
b) Pascal’s Principle
c) Fresnel zone
d) Archimedes’ Principle
Answer: c
Explanation: The concept of diffraction loss is a function of the path difference around an obstruction. It can be explained by Fresnel zones. Fresnel zones represent successive regions where secondary waves have a path length from Tx to Rx which are nλ/2 greater than total path length.
9. In mobile communication system, diffraction loss occurs due to ______
a) Dielectric medium
b) Obstruction
c) Electric field
d) Operating frequency
Answer: b
Explanation: Diffraction loss occurs from the blockage of secondary waves such that only a portion of the energy is diffracted around an obstacle. An obstruction causes a blockage of energy from source some of the Fresnel zones, allowing only some of the transmitted energy to reach the receiver.
10. For predicting the field strength in a given service area, it is essential to estimate ______
a) Polarization
b) Magnetic field
c) Height of transmitter
d) Signal attenuation
Answer: d
Explanation: Estimating the signal attenuation caused by diffraction of radio waves over hills and buildings is essential in predicting the field strength in a given service area. In practice, prediction is a process of theoretical approximation modified by necessary empirical corrections.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Scattering”.
1. Scattering occurs when medium consists of objects with dimensions _______ compared to the wavelength.
a) Same
b) Small
c) Large
d) Very large
Answer: b
Explanation: Scattering occurs when the medium through which the wave travels consists of objects with dimensions that are small compared to the wavelength. But the number of obstacles per unit volume is large.
2. Scattered waves are produced at ________
a) Rough surface
b) Shadowed region
c) Smooth surface
d) Horizon
Answer: a
Explanation: Scattered waves are produced by rough surfaces, small objects or by other irregularities in the channel. In practice, foliage, street signs, and lamp posts induce scattering in a mobile communication system.
3. The actual received signal is ______ than what is predicted by reflection and diffraction model.
a) Weaker
b) Equal
c) Stronger
d) Very weak
Answer: c
Explanation: The actual received signal in a mobile radio environment is often stronger than what is predicted by reflection and diffraction model alone. This is because when a radio wave incidence upon the rough surface, reflected energy is spread out in all directions.
4. Scattered energy in all directions provides _________ at a receiver.
a) Channels
b) Loss of signal
c) No energy
d) Additional radio energy
Answer: d
Explanation: Objects such as lamp posts and trees tend to scatter energy in all directions. They provide additional radio energy at a receiver. Scattering may also refer to particle-particle collisions between molecules, atoms, electrons, photons and other particles.
5. Surface roughness are often tested using __________
a) Rayleigh criterion
b) Lawson criterion
c) Barkhausen stability criterion
d) Nyquist criterion
Answer: a
Explanation: Rough surface is often tested using a Rayleigh criterion. It defines the critical height of surface protuberances for a given angle of incidence. The Rayleigh criterion is the criterion for the minimum resolvable detail. The imaging process is said to be diffraction-limited when the first diffraction minimum of the image of one source point coincides with the maximum of another.
6. A surface is considered rough if protuberance is ________ than critical height.
a) Equal
b) Less
c) Greater
d) No relation
Answer: c
Explanation: A surface is considered rough if its minimum to maximum protuberance is greater than the critical height calculated using Rayleigh criterion. It is considered smooth if protuberance is less than critical height.
7. RCS of scattering object is defined as the ratio of _______
a) Power density of signal scattered to power density of radio wave incident
b) Power density of radio wave incident to power density of signal scattered
c) Power density of incident waves to power density of reflected wave
d) Power density of reflected wave to power density of incident waves
Answer: a
Explanation: The radar cross section of a scattering object is defined as the ratio of the power density of the signal scattered in the direction of the receiver to the power density of the radio wave incident upon the scattering object. It has unit of square meters.
8. Which equation is used to calculate the received power due to scattering for urban mobile radio system?
a) Laplace equation
b) Bistatic radar equation
c) Poisson’s equation
d) Maxwell equation
Answer: b
Explanation: For urban mobile radio systems, models based on bistatic radar equation is used to compute the received power due to scattering in the far field. This equation describes the propagation of wave in free space which impinges on a scattering object and then reradiated in the direction of receiver.
9. In ionosphere propagation, waves arriving at the receiving antenna using the phenomenon of _______
a) Scattering
b) Refraction
c) Diffraction
d) Radiation
Answer: a
Explanation: Scattering is a general physical process where some forms of radiation, such as light, sound, or moving particles, are forced to deviate from a straight trajectory by one or more paths. It is due to localized non-uniformities in the medium through which they pass.
10. Power density is basically termed as ________ power per unit area.
a) Reflected
b) Refracted
c) Radiated
d) Diffracted
Answer: c
Explanation: Power density is the amount of power per unit volume. It is also termed as radiated power per unit area. In energy transformers including batteries, fuel cells, motors, etc., power density refers to a volume. It is then also called volume power density.
This set of Wireless & Mobile Communications Questions and Answers for Experienced people focuses on “Practical Link Budget Design Using Path Loss Models”.
1. Empirical approach is based on fitting curve or analytical expressions.
a) True
b) False
Answer: a
Explanation: The empirical approach is based on fitting curves or analytical expressions that recreate a set of measured data. This has the advantage of implicitly taking into account all propagation factors, both known and unknown through actual field measurements.
2. Which of the following is not a practical path loss estimation technique?
a) Log distance path loss model
b) Log normal shadowing
c) Determination of percentage of coverage area
d) Hata model
Answer: d
Explanation: Log normal shadowing, log distance path loss model and determination of percentage of coverage area are practical pat loss estimation techniques. Hata model is only valid for exterior environment.
3. Average received signal power decreases __________ with distance.
a) Exponentially
b) Logarithmically
c) Two times
d) Four times
Answer: b
Explanation: Both theoretical and measurement based propagation models indicate that average received signal power decreases logarithmically with distance. It is valid for both outdoor and indoor channels.
4. What does path loss exponent indicates?
a) Rate at which path loss decreases with distance
b) Rate at which path loss increases with distance
c) Rate at which path loss decreases with power density
d) Rate at which path loss increases with power density
Answer: b
Explanation: The average large scale path loss for an arbitrary T-R separation is expressed as a function of distance by using a path loss exponent, n. It indicates the rate at which the path loss increases with distance.
5. The reference distance should not be in the far field of the antenna.
a) True
b) False
Answer: b
Explanation: It is important to select a free space reference distance that is appropriate for the propagation environment. The reference distance should always be in the far field of the antenna so that near field effects do not alter the reference path loss.
6. Which distribution describes the shadowing effect?
a) Log normal distribution
b) Nakagami distribution
c) Cauchy distribution
d) Rayleigh distribution
Answer: a
Explanation: Log normal distribution describes the random shadowing effects. It occurs over a large number of measurement locations which have the same T-R separation, but have different clutter on the propagation path.
7. Log normal shadowing is a phenomenon that occurs with same T-R separation having same level clutter on the propagation path.
a) True
b) False
Answer: b
Explanation: Log normal shadowing occurs over a large number of measurement locations which have the same T-R separation, but have different levels of clutter on the propagation path. It follows log normal distribution.
8. Log normal shadowing implies that measured signal levels at specific T-R separation have ______ distribution when signal levels have values in dB units.
a) Rayleigh
b) Gamma
c) Gaussian
d) Nakagami
Answer: c
Explanation: Log normal shadowing implies that measured signal levels at a specific T-R separation have Gaussian distribution. It is about the distance dependent mean of 4.68 where the signal levels have values in dB units.
9. A link budget is accounting of all __________
a) Gain and losses from the transmitter
b) Power transmitted by transmitter
c) Power received by receiver
d) Power transmitted and received
Answer: a
Explanation: A link budget is accounting of all of the gains and losses from the transmitter, through the medium to the receiver in a telecommunication system. It accounts for the attenuation of the transmitted signal due to propagation, as well as the antenna gains and miscellaneous losses.
10. Antenna’s efficiency is given by the ratio of __________
a) Losses
b) Physical aperture to effective aperture
c) Signal power to noise power
d) Effective aperture to physical aperture
Answer: d
Explanation: The larger the antenna aperture the larger is the resulting signal power density in the desired direction. The ratio of effective aperture to the physical aperture is the antenna’s efficiency.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Outdoor Propagation Models”.
1. Which of the following is not an outdoor propagation model?
a) Longley-Rice model
b) Ericson Multiple Breakpoint Model
c) Hata model
d) Okumura model
Answer: b
Explanation: Ericson multiple breakpoint model is an indoor propagation model. Longley-Rice, Hata and Okumura model are outdoor propagation models. Most of these models are based on a systematic interpretation of measurement data obtained in the service area.
2. Longley –Rice model is applicable to _________
a) Point to point communication
b) All to all communication
c) Point to multipoint communication
d) Multipoint microwave distribution sstem
Answer: a
Explanation: The Longley-Rice model is applicable to point-to-point communication systems in the frequency range from 40 MHz to 100 GHz. They are applicable for different kinds of terrain. Terrain profile may vary from a simple curved Earth profile to a highly mountainous profile.
3. Longley-Rice prediction model is also referred as _________
a) Okumura model
b) Hata model
c) ITS irregular terrain model
d) Bertoni model
Answer: c
Explanation: The Longley Rice prediction model is also referred to as ITS irregular terrain model. The model is based on electromagnetic theory and on statistical analyses of both terrain features and radio measurements. It predicts the median attenuation of a radio signal as a function of distance and the variability of the signal in time and in space.
4. The extra term for additional attenuation due to urban clutter near the receiving antenna is called __________
a) Power factor
b) Urban gain
c) Clutter factor
d) Urban factor
Answer: d
Explanation: The urban factor is derived by comparing the predictions by the original Longley –Rice model with those obtained by Okumura. It deals with radio propagation in urban areas and is relevant to mobile radio.
5. Longley Rice model’s merit is to provide corrections due to environmental factors.
a) True
b) False
Answer: b
Explanation: One shortcoming of the Longley –Rice model is that it does not provide a way of determining corrections due to environmental factors. It does not consider correction factors to account for the effects of buildings and foliage. Multipath is also not considered.
6. Which method is used by Edwards and Durkin algorithm to calculate the loss associated with diffraction edges?
a) Epstein and Peterson method
b) Interpolation method
c) Knife edge diffraction method
d) Fresnel- Kirchoff method
Answer: a
Explanation: The Edwards and Durkin algorithm uses Epstein and Peterson method to calculate the loss associated with two diffraction edges. It is the sum of two attenuations. First is loss at second diffraction edge caused by first diffraction edge. And second is the loss at receiver caused by second diffraction edge.
7. Durkin’s model can read digital elevation map.
a) True
b) False
Answer: a
Explanation: Durkin’s model is very attractive because it can read in a digital elevation map and perform a site specific propagation computation on the elevation data. It can produce a signal strength contour that is reported to be good within a few dB.
8. Which of the most widely used model for signal prediction in urban areas?
a) Ericsson Multiple Breakpoint Model
b) Log distance path loss model
c) Okumura model
d) Attenuation factor model
Answer: c
Explanation: Okumura’s model is one of the most widely used models for signal prediction in urban areas. This model is applicable for frequencies in the range 150 MHz to 1920 MHz .
9. Okumura model is applicable for distances of _________
a) 1 m to 10 m
b) 1 km to 100 km
c) 100 km to 1000 km
d) 10 km to 10000 km
Answer: b
Explanation: Okumura’s model is applicable for distances of 1 km to 100 km. It can be used for base station antenna heights ranging from 30 m to 1000 m. Okumura developed a set of curves giving the median attenuation relative to free space in an urban area.
10. Okumura model is considered to be complex in predicting path loss.
a) True
b) False
Answer: b
Explanation: Okumura’s model is considered to be among the simplest and best in terms of accuracy in path loss prediction for mature cellular and land mobile radio system. It is very practical and has become a standard for system planning in modern land mobile system in Japan.
11. Which of the following is the major disadvantage of the Okumura model?
a) Complex
b) Inaccurate
c) Not practical
d) Slow response to rapid change in terrain
Answer: d
Explanation: The major disadvantage with the model is its slow response to rapid changes in terrain. Therefore the model is fairly good in urban and suburban areas, but not as good in rural areas. Common standard deviations between predicted and measured path loss values are 10 dB to 14 dB.
12. The Hata model is empirical formulation of which model?
a) Okumura model
b) Longley- Rice model
c) Durkin’s model
d) Walfisch and Bertoni model
Answer: a
Explanation: The Hata model is an empirical formulation of the graphical path loss data provided by Okumura. It is valid from 150 MHz to 1500 MHz. Hata presented the urban area propagation loss as a standard formulation. It supplied correct Equations for application to other situations.
13. Hata model is well suited for _________
a) Personal communication system
b) Large cell mobile radio system
c) Small cell mobile radio system
d) Every mobile radio system
Answer: b
Explanation: Hata model is well suited for large cell mobile radio systems. But it is not well suited for personal communication system which have cells on the order of 1 km radius. Hata model does not have any path specific corrections which are available in Okumura model.
14. Which of the following considers the impact of rooftops and building?
a) Okumura model
b) Hata model
c) Walfisch and Bertoni model
d) Longley- Rice model
Answer: c
Explanation: The impact of rooftops and building height is considered by Walfisch and Bertoni model. It uses diffraction to predict average signal strength at street level. It considers path loss to be a product of three factors.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Indoor Propagation Models”.
1. The variability of the environment is slower for a smaller range of T-R separation distances in indoor models.
a) True
b) False
Answer: b
Explanation: The indoor radio channel differs from the traditional mobile radio channel. The distances covered are much smaller and variability of the environment is much greater for a much smaller range of T-R separation distances.
2. Propagation within building is not influenced by _________
a) Layout of the building
b) Construction materials
c) Building type
d) Trees outside the building
Answer: d
Explanation: It has been observed that propagation within buildings is strongly influenced by specific features. These features are layout of the building, the construction materials, and the building type.
3. Smaller propagation distances make it more difficult to insure far-field radiation for all receiver location and types of antenna.
a) True
b) False
Answer: a
Explanation: Smaller propagation distances make it more difficult to insure far field radiation for all receiver location and types of antenna. The condition is very variable for smaller propagation distances.
4. What is hard partition?
a) Partition as part of the building
b) Partition that can be moved
c) Partition not touching ceiling
d) Partition between different floors
Answer: a
Explanation: Partitions that are formed as part of the building structure are called hard partitions. Partitions vary widely in their electrical and physical characteristics. Thus, it makes difficult in applying general models to specific indoor installation.
5. Partitions that can be moved are called _______
a) Soft partitions
b) Hard partitions
c) Disk partition
d) Dynamic partition
Answer: a
Explanation: Partitions that may be moved are called soft partitions. They do not span to the ceiling. Office buildings often have large open areas which are constructed by using moveable office partitions. Thus, space can be reconfigured easily.
6. Losses between the floors of the building can be determined using ________
a) Internal dimensions
b) Material used to create antenna
c) External dimension
d) Line of sight path
Answer: c
Explanation: The losses between floors of a building are determined by external dimensions as well as materials of the building. It is also determined using type of construction used to create the floors and external surroundings.
7. Technique of drawing a single ray between the transmitter and receiver is called ______
a) Secondary ray tracing
b) Primary ray tracing
c) Line of sight
d) Straight line tracing
Answer: b
Explanation: PAF represents a specific obstruction encountered by a ray drawn between the transmitter and receiver in 3-D. This technique of drawing a single ray between transmitter and receiver is called primary ray tracing.
8. ________ is a process of converting plain text into cipher text.
a) Authentication
b) Decryption
c) Encryption
d) Compression
Answer: c
Explanation: Encryption is the most effective way to achieve data security. It is the process of encoding a message in such a way that only authorized parties can access it. Encryption does not itself prevent interference but denies the intelligible content to a would-be interceptor.
9. _______ reduces the cell size to increase capacity.
a) Intelligent cell approach
b) Microcell approach
c) Top down approach
d) Bottom up approach
Answer: b
Explanation: The microcell zone concept is used in cellular systems to specifically increase the capacity and coverage in cellular systems. A handoff is not required at MSC when the mobile travels between zones within a cell or when a mobile travels from one zone to another within the cell.
10. _________ configuration describes a desktop in an office.
a) Mobile and wired
b) Fixed and wired
c) Fixed and wireless
d) Mobile and wireless
Answer: b
Explanation: Fixed and wired configuration describes a desktop in an office. The device use fixed networks for performance reasons. Neither weight nor power consumption of the devices allows for mobile usage.
This set of Basic Wireless & Mobile Communications Questions and Answers focuses on “Small- Scale Multipath Propagation”.
1. Small scale fading describes the _________ fluctuations of the amplitude, phases of a signal.
a) Rapid
b) Slow
c) Instantaneous
d) Different
Answer: a
Explanation: Small scale fading or simply fading, is used to describe the rapid fluctuations of amplitudes, phases, or multipath delays of a radio signal over a short period of time or travel distance. It ignores the large scale path loss.
2. Fading is caused by interference.
a) True
b) False
Answer: a
Explanation: Fading is caused by interference. It is caused by interference between two or more versions of the transmitted signal which arrive at the receiver at slightly different times.
3. Which of the following is not an effect caused by multipath in radio channel?
a) Rapid changes in signal strength
b) Random frequency modulation
c) Power of base station
d) Time dispersion
Answer: c
Explanation: Rapid changes in signal strength over a small travel distance are caused due to multipath. It causes random frequency modulation due to varying Doppler shifts on different multipath signals. Time dispersion is also caused by multipath propagation delays.
4. In urban areas, fading occurs due to height of mobile antenna ________ than height of surrounding structure.
a) Same
b) Smaller
c) Greater
d) Very larger
Answer: b
Explanation: In urban areas, fading occurs because height of the mobile antenna is below the height of surrounding structures. Therefore, there is no single line of sight path to the base station.
5. Fading does not occur when mobile receiver is stationary.
a) True
b) False
Answer: b
Explanation: The received signal may fade even when the mobile receiver is stationary. It is due to the movement of surrounding objects in the radio channel. The multipath components combine vectorially at the receiver antenna and cause signal to distort or fade.
6. Apparent shift in frequency in multipath wave is caused due to relative motion between________
a) Base station and MSC
b) Mobile and surrounding objects
c) Mobile and MSC
d) Mobile and base station
Answer: d
Explanation: Due to relative motion between mobile and base station, each multipath wave experiences an apparent shift in frequency. This shift in received signal frequency due to motion called Doppler shift.
7. Doppler shift is directly proportional to __________
a) Velocity
b) Height of antenna
c) Power of receiving antenna
d) Power of transmitter
Answer: a
Explanation: The shift in received signal frequency due to motion is called Doppler shift. It is directly proportional to the velocity and direction of motion of mobile with respect to the direction of arrival of the received multipath wave.
8. Which of the following factor does not influence small scale fading?
a) Multipath propagation
b) Power density of base station
c) Speed of mobile
d) Speed of surrounding objects
Answer: b
Explanation: Many physical factors in radio channel influence small scale fading. Multipath propagation, speed of mobile, speed of surrounding objects, transmission bandwidth of the signal influences small scale fading in a large way.
9. Signal will distort if transmitted signal bandwidth is greater than bandwidth of __________
a) Receiver
b) Radio channel
c) Multipath channel
d) Transceiver
Answer: c
Explanation: Received signal will be distorted if transmitted signal bandwidth is greater than bandwidth of multipath channel. But received signal strength will not fade much over a local area.
10. What is a measure of the maximum frequency difference for which signals are strongly correlated in amplitude?
a) Coherence bandwidth
b) Narrow bandwidth
c) Incoherent bandwidth
d) Wide bandwidth
Answer: a
Explanation: The bandwidth of the channel can be quantified by the coherence bandwidth. It is related to the specific multipath structure of channel. It is a measure of maximum frequency difference for which signals are strongly correlated in amplitude.
11. The Doppler shift for mobile moving with constant velocity, v is given by _______
a) /λ
b) v/λ
c) v*cos θ
d) v*λ
Answer: a
Explanation: Doppler shift is given by /λ. This formula relates the Doppler shift to the mobile velocity and spatial angle between the direction of motion of mobile and the direction of arrival of the wave.
12. Doppler shift is positive if mobile is moving away from direction of arrival of the wave.
a) True
b) False
Answer: b
Explanation: Doppler shift is positive if the mobile is moving toward the direction of arrival of the wave, as the apparent received frequency is increased. And if the mobile is moving away from the direction of arrival of the wave, Doppler shift is negative.
This set of Wireless & Mobile Communications Interview Questions and Answers for Experienced people focuses on “Impulse Response Model of a Multipath Channel”.
1. Small scale variations of a mobile radio signal are directly related to _______
a) Impulse response of mobile radio channel
b) Impulse response of base station
c) Frequency response of antenna
d) Frequency response of base station
Answer: a
Explanation: The small scale variations of a mobile radio signal can be directly related to the impulse response of mobile radio channel. he impulse response is a wideband channel characterization and contains all information necessary to simulate or analyze any type of radio transmission through the channel.
2. Impulse response is a narrowband characterization.
a) True
b) False
Answer: b
Explanation: Impulse response is a wideband channel characterization. It contains all information necessary to simulate or analyse any type of radio transmission through the channel.
3. Mobile radio channel can be modelled as a ______ filter.
a) Non-linear
b) Low-pass
c) Linear
d) Bandpass
Answer: c
Explanation: Mobile radio channel may be modelled as a linear filter with a time varying impulse response. The time variation is due to receiver motion in space. The filtering nature is caused by summation of amplitudes and delays of multiple arriving waves at any instant of time.
4. Impulse response does not play any role in characterization of the channel.
a) True
b) False
Answer: b
Explanation: Impulse response is a useful characterization of channel. It may be used to predict and compare the performance of different mobile communication systems and transmission bandwidth for a particular mobile channel condition.
5. Received signal can be expressed as ______ of transmitted signal with channel impulse response.
a) Addition
b) Subtraction
c) Division
d) Convolution
Answer: d
Explanation: The received signal y can be expressed as a convolution of transmitted signal x with impulse response of mobile radio channel. The variable t represents the time variations due to motion and the channel impulse multipath delay for fixed value of t.
6. Discretization of multipath delay axis of impulse response into equal time delay segments is called __________
a) Excess delay bins
b) Delay bins
c) Discrete bins
d) Digital bins
Answer: a
Explanation: It is useful to discretize the multipath delay axis of the impulse response into equal time delay segments called excess delay bins. The technique of quantizing the delay bins determines the time delay resolution of the channel model.
7. Small scale received power is ________ of average powers received in each multipath component.
a) Log
b) Exponential
c) Multiplication
d) Sum
Answer: d
Explanation: The average small received power is the sum of the average powers received in each multipath component. This is the case if transmitted signal is able to resolve the multipath.
8. The received power of a wideband signal fluctuates significantly when a receiver is moved about a local area.
a) True
b) False
Answer: b
Explanation: Amplitude of individual multipath components does not fluctuate widely in local area. Therefore, the received power of a wideband signal does not fluctuate significantly when a receiver is moved about a local area.
9. Average power for a CW signal is _______ to average received power for a wideband signal in small scale region.
a) Equivalent
b) Two times
c) Four times
d) Ten times
Answer: a
Explanation: Average power for a CW signal is equivalent to average received power for a wideband signal in small scale region. This can occur when either the multipath phases are identically and independently distributed or when path amplitudes are uncorrelated.
10. The received local ensemble average power of wideband and narrowband signals are ________
a) Different
b) Equivalent
c) Not dependent
d) Double
Answer: b
Explanation: The received local ensemble average power of wideband and narrowband signals is equivalent. When the transmitted signal has bandwidth greater than bandwidth of the channel, the received power varies very little. However, if transmitted signal has very narrow bandwidth, large fluctuation occurs at receiver.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Small- Scale Multipath Measurements”.
1. Which of the following is not a small scale multipath measurement technique?
a) Indirect pulse measurement
b) Direct pulse measurement
c) Spread spectrum sliding correlator measurement
d) Swept frequency measurement
Answer: a
Explanation: Because of the importance of multipath structure in determining the small scale fading effects, a number of wideband channel sounding techniques have been developed. These are direct pulse measurement, spread spectrum sliding correlator and swept frequency measurement.
2. Direct RF pulse systems are complex.
a) True
b) False
Answer: b
Explanation: Direct RF pulse systems have a very attractive system. They lack complexity because of the use of off the shelf equipment. This system can also provide a local average power delay profile.
3. What is the main disadvantage of RF pulse system?
a) Complexity
b) Not real time
c) Interference and noise
d) Simplicity
Answer: c
Explanation: The main problem with RF pulse system is that it is subject to interference and noise. This is due to wide passband filter required for multipath time resolution.
4. Why is the phase of individual multipath components are not received in RF Pulse system?
a) Due to use of duplexer
b) Due to use of ADC
c) Due to use of flip flops
d) Due to use of envelope detector
Answer: d
Explanation: Another disadvantage of RF pulse system is that phase of individual multipath components are not received due to use of an envelope detector. However, use of a coherent detector permits measurement of the multipath phase using this technique.
5. Spread spectrum sliding correlator has better ________ in comparison to RF pulse system.
a) Dynamic range
b) Frequency
c) Power density
d) Structure
Answer: a
Explanation: While the probing signal may be wideband, it is possible to detect the transmitted signal using a narrowband receiver preceded by a wideband mixer. Thus, it improves the dynamic range of the system in comparison to RF pulse system.
6. If a PN sequence has chip duration of T c , then chip rate is given by ___________
a) T c
b) 1/T c
c) 2 T c
d) T c 2
Answer: b
Explanation: In a spread spectrum channel sounder, a carrier signal is spread over a large bandwidth. It is done by mixing it with a binary pseudo number sequence having a chip duration of Tc and chip rate of 1/T c .
7. The maximal length of PN sequence, the sequence length for n number of shift register is ______
a) 2 n
b) 1/2 n
c) 2 n – 1
d) 2 n + 1
Answer: c
Explanation: For a maximal length pseudo number sequence, the sequence length is 2 n – 1. Here, n is the number of shift registers in the sequence register.
8. The ratio between transmitter chip clock rate and the difference between the transmitter chip clock and difference between transmitter and receiver chip clock rates is called ___________
a) Slide factor
b) Chip factor
c) Reuse factor
d) Shape factor
Answer: a
Explanation: The slide factor is defined as ratio between transmitter chip clock rate and the difference between the transmitter chip clock and difference between transmitter and receiver chip clock rates. It is unit less quantity.
9. Which of the following is not an advantage of spread spectrum channel sounding system?
a) Rejection of passband noise
b) Real time
c) Coverage range improvement
d) Less transmitter power required
Answer: b
Explanation: One of the main advantage of spread spectrum channel sounding system is ability to reject passsband noise. Thus, it improves the coverage range for a given transmitter power. Also, required transmitter powers can be considerably lower compared to direct pulse systems due to inherent processing gain.
10. Which of the following is a disadvantage of spread spectrum channel sounding system?
a) Passband noise
b) Less coverage area
c) Large transmitter power required
d) Not in real time
Answer: d
Explanation: A disadvantage of spread spectrum system as compared to direct RF pulse system, is that measurements are not made in real time. But they are compiled as the PN codes slide past one another.
11. There is no relationship between time domain and frequency domain techniques.
a) True
b) False
Answer: b
Explanation: There is a dual relationship between time domain and frequency domain techniques. Therefore, it is possible to measure the channel impulse response in the frequency domain. Frequency domain channel sounder is used for measuring channel impulse response.
12. Transmissivity is a _________ domain response.
a) Time
b) Frequency
c) Time-frequency
d) Spatial
Answer: b
Explanation: Transmissivity response is a frequency domain response. It is a frequency domain representation of channel impulse response. This response is then converted to the time domain using inverse discrete Fourier transform.
13. Frequency domain channel sounding technique do not require hard wired synchronization between transmitter and receiver.
a) True
b) False
Answer: b
Explanation: Frequency domain channel sounding system requires careful calibration and hardwired synchronization between transmitter and receiver. However, this technique works well and indirectly provides amplitude and phase information in the time domain.
14. Frequency domain channel sounding system is _______ in nature.
a) Real time
b) Digital
c) Non real time
d) Analog
Answer: c
Explanation: Frequency domain channel sounding system makes measurements of non real time nature. For time varying channels, the channel frequency response can change rapidly giving an erroneous impulse response measurement.
15. Faster sweep time can be accomplished by ________
a) Increasing frequency steps
b) Increasing time slots
c) Reducing time slots
d) Reducing frequency steps
Answer: d
Explanation: A faster sweep time can be accomplished by reducing the number of frequency steps. But this sacrifices the time resolution and excess delay range in the time domain.
This set of Wireless & Mobile Communications Problems focuses on “Parameters of Mobile Multipath Channels”.
1. Power delay profile is represented as plots of __________ with respect to fixed time delay reference.
a) Relative received power
b) Frequency
c) Transmitted power
d) Relative power
Answer: a
Explanation: Power delay profiles are generally represented as plots of relative received power as a function of excess delay with respect to a fixed time delay reference. They are calculated by averaging instantaneous power delay profile measurements over a local area.
2. Which of the following is not a multipath channel parameter that can be determined from power delay profile?
a) Mean excess delay
b) RMS delay spread
c) Excess delay spread
d) Doppler spread
Answer: d
Explanation: The mean excess delay, excess delay spread and rms delay spread are some multipath channel parameters. They can be determined from a power delay profile. Doppler spread is a measure of spectral broadening caused by time rate of change of mobile radio channel.
3. The time dispersive properties of wideband multipath channel are quantified by ______ and _______
a) Mean excess delay, rms delay spread
b) Doppler spread, rms delay spread
c) Doppler spread, coherence time
d) Mean excess delay, Doppler spread
Answer: a
Explanation: The time dispersive properties of wide band multipath channels are most commonly quantified by their mean excess delay and rms delay spread. Coherence time characterizes the time varying nature of frequency dispersiveness of mobile radio channel in time domain.
4. _______ is the square root of the second central moment of the power delay profile.
a) Mean excess delay
b) Rms delay spread
c) Excess delay spread
d) Coherence time
Answer: b
Explanation: The rms delay spread is the square root of the second central moment of the power delay profile. Many measurements are made at many local areas in order to determine a statistical range of multipath channel parameters for a mobile communication system over a large scale area.
5. Which of the following is the first moment of the power delay profile?
a) Rms delay spread
b) Excess delay spread
c) Mean excess delay
d) Doppler spread
Answer: c
Explanation: Mean excess delay is the first moment of the power delay profile. It is defined from a single power delay profile which is temporal or spatial average of consecutive impulse response measurements collected and averaged over a local area.
6. What is the order of typical values of rms delay spread in outdoor mobile radio channels?
a) Microseconds
b) Nanoseconds
c) Seconds
d) Minutes
Answer: a
Explanation: Typical values of rms delay spread are on the order of microseconds in outdoor mobile radio channels. For indoor mobile radio channels, they are of the order of nanoseconds.
7. Power delay profile and magnitude frequency response of a mobile radio channel are related through _______
a) Laplace Transform
b) Fourier Transform
c) S Transform
d) Wavelet Transform
Answer: b
Explanation: Power delay profile and magnitude frequency response of a mobile radio channel are related through the Fourier transform. Therefore, it is possible to obtain an equivalent description of the channel in the frequency domain using its frequency response characteristics.
8. ______ and coherence bandwidth are inversely proportional to one another.
a) Rms delay spread
b) Mean excess delay
c) Excess delay spread
d) Doppler spread
Answer: a
Explanation: The rms delay spread and coherence bandwidth are inversely proportional to one another. However, their exact relationship is a function of the exact multipath structure.
9. Coherence bandwidth is a statistical measure of range of frequencies over which channel is considered _______
a) Time dispersive
b) Frequency selective
c) Time variant
d) Flat
Answer: d
Explanation: Coherence bandwidth is a statistical measure of range of frequencies over which channel is considered flat. A flat channel is a channel which passes all spectral components with approximately equal gain and linear phase.
10. Which of the following describes time varying nature of the channel in a small scale region?
a) Delay spread and coherence time
b) Coherence bandwidth and delay spread
c) Doppler spread and coherence time
d) Delay spread and doppler spread
Answer: c
Explanation: Delay spread and coherence bandwidth describes the time dispersive nature of the channel in a local area. They do not give information about the time varying nature of the channel caused by either relative motion between the mobile and base station, or by movement of objects in the channel.
11. Doppler spread is a range of frequencies over which received Doppler spread is _______
a) Zero
b) Non zero
c) Infinite
d) One
Answer: b
Explanation: Doppler spread is a range of frequencies over which received Doppler spectrum is essentially non-zero. It is a measure of spectral broadening caused by time rate of change of mobile radio channel.
12. _______ is a statistical measure of time duration over which channel impulse response is invariant.
a) Coherence time
b) Doppler spread
c) Mean excess delay
d) Rms delay spread
Answer: a
Explanation: Coherence time is actually a statistical measure of the time duration over which the channel impulse response is essentially invariant. It quantifies the similarity of the channel response at varying times.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Types of Small- Scale Fading”.
1. Which of the following is not a channel parameter?
a) Bandwidth
b) Coherence time
c) Rms delay spread
d) Doppler spread
Answer: a
Explanation: Channel parameters are coherence bandwidth, rms delay spread and Doppler spread. Signal parameters are bandwidth and symbol period. Different types of fading occur depending on the relation between channel and signal parameters.
2. ______ leads to time dispersion and frequency selective fading.
a) Doppler spread
b) Multipath delay spread
c) Time dispersive parameters
d) Frequency delay spread
Answer: b
Explanation: Multipath delay spread leads to time dispersion and frequency selective fading. Doppler spread leads to frequency dispersion and time selective fading. These two propagation mechanism are independent of each other.
3. Which of the following s not a characteristic of flat fading?
a) Mobile radio channel has constant gain
b) Linear phase response
c) Non linear phase response
d) Bandwidth is greater than the bandwidth of transmitted signal
Answer: c
Explanation: The received signal will undergo flat fading if mobile radio channel has a constant gain and linear phase response over a bandwidth which is greater than the bandwidth of transmitted signal. It is the most common type of fading.
4. Spectral characteristics of the channel changes with time in flat fading.
a) True
b) False
Answer: b
Explanation: In flat fading, the multipath structure of the channel is such that spectral characteristics of the transmitted signal are preserved at the receiver. But, the strength of the received signal changes with time due to fluctuations in the gain of channel caused by multipath.
5. Flat fading channel is also known as _______________
a) Amplitude varying channel
b) Wideband channel
c) Phase varying channel
d) Frequency varying channel
Answer: a
Explanation: Flat fading channel is also known as amplitude varying channel. They are also sometimes referred to as narrowband channel. In flat fading channel, the bandwidth of the applied signal is narrow as compared to the channel flat fading bandwidth.
6. In a frequency selective fading, mobile radio channel possess a constant gain and a linear phase over bandwidth smaller than bandwidth of transmitted signal.
a) True
b) False
Answer: a
Explanation: If the channel possesses a constant gain and linear phase response over a bandwidth that is smaller than the bandwidth of transmitted signal, then the channel creates frequency selective fading on the received signal. Frequency selective fading channel are much more difficult to model than flat fading channel.
7. Frequency selective fading channels are also known as ________
a) Narrowband channel
b) Wideband channel
c) Amplitude varying channel
d) Phase varying channel
Answer: b
Explanation: Frequency selective fading channels are also known as wideband channels. The bandwidth of the signal is wider than the bandwidth of channel impulse response.
8. Frequency selective fading does not induce intersymbol interference.
a) True
b) False
Answer: b
Explanation: Frequency selective fading is due to time dispersion of the transmitted symbols within the channel. Thus, the channel induces intersymbol interference . The channel have certain frequency components in the received signal spectrum that have greater gain than others.
9. For fast fading channel, the coherence time of the channel is smaller than _______ of transmitted signal.
a) Doppler spread
b) Bandwidth
c) Symbol period
d) Coherence bandwidth
Answer: c
Explanation: In fast fading channel, the channel impulse response changes rapidly within the symbol duration. Thus, the coherence time of the channel is much smaller than the symbol period of the transmitted signal.
10. In slow fading channel, Doppler spread of the channel is much less than the ________ of baseband signal.
a) Symbol period
b) Phase
c) Coherence time
d) Bandwidth
Answer: d
Explanation: Slow fading channel may be assumed to be static over one or several reciprocal bandwidth intervals. In the frequency domain, this implies that the Doppler spread of the channel is much less than the bandwidth of the baseband signals.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Rayleigh and Ricean Distribution”.
1. Which of the following distribution is used for describing statistical time varying nature of received envelope of multipath component?
a) Log normal distribution
b) Levy distribution
c) Rayleigh distribution
d) Gaussian distribution
Answer: c
Explanation: Rayleigh distribution is the most common distribution for statistical modelling. It is used to describe the statistical time varying nature of the received envelope of a flat fading signal. It also describes the envelope of an individual multipath component.
2. Envelope of the sum of two quadrature Gaussian noise signal obeys _________ distribution.
a) Rayleigh
b) Inverse Gaussian
c) Nakagami
d) Gamma
Answer: a
Explanation: It is well known that the envelope of the sum of two quadrature Gaussian noise signal obeys Rayleigh distribution. This fading distribution could be applied to any scenario where there is no line of sight path between transmitter and receiver antennas.
3. For a Rayleigh fading signal, mean and median differ by _______
a) 2 dB
b) 10 dB
c) 0.55 dB
d) 100 dB
Answer: c
Explanation: The mean and median differ by only 0.55 dB in a Rayleigh fading signal. The differences between the rms values and the other two values are higher.
4. It is easy to compare different fading distributions using mean values instead of median values.
a) True
b) False
Answer: b
Explanation: By using median values instead of mean values, it is easy to compare different fading distributions which may have widely varying means. Median is used in practice since fading data is measured in the field and a particular distribution cannot be assumed.
5. For a nonfading signal component present, the small scale fading envelope distribution is ____________
a) Rayleigh
b) Gaussian
c) Log normal
d) Ricean
Answer: d
Explanation: The small scale fading envelope is Ricean when there is a dominant stationary signal component, such as line of sight propagation path. In such a situation, random multipath components arriving at different angles are superimposed on a stationary dominant signal.
6. Ricean distribution degenerates to ________ distribution when the dominant component fades away.
a) Log normal
b) Gamma
c) Rayleigh
d) Gaussian
Answer: c
Explanation: Ricean distribution degenerates to Rayleigh distribution when the dominant component fades away. As the dominant signal becomes weaker, the composite signal resembles a noise signal which has an envelope that is Rayleigh.
7. The envelope of a bandpass noise is __________
a) Uniformly distributed
b) Rayleigh
c) Ricean
d) Gaussian
Answer: b
Explanation: The envelope of only bandpass noise is Rayleigh distribution. Rayleigh distribution is a continuous probability density function for positive random variables.
8. The envelope of a sinusoid plus bandpass noise has __________
a) Uniformly distributed
b) Rayleigh
c) Ricean
d) Gaussian
Answer: c
Explanation: The envelope of a sinusoid plus bandpass noise has Ricean distribution. In probability theory, Ricean distribution is the probability distribution which has magnitude of a circular bivariate normal random variable with potentially non-zero mean.
9. What do you call an attenuation that occurs over many different wavelengths of the carrier?
a) Rayleigh fading
b) Ricean fading
c) Wavelength fading
d) Slow fading
Answer: d
Explanation: Slow fading does not vary quickly with the frequency. It originates due to effect of mobility. Slow fading is the result of signal path change due to shadowing and obstructions such as tree or buildings etc.
10. Which of the reception problems below that is not due to multipath?
a) Delayed spreading
b) Rayleigh fading
c) Random Doppler shift
d) Slow fading
Answer: d
Explanation: Slow fading arises when the coherence time of the channel is large relative to the delay requirement of the application. Slow fading is caused by events such as shadowing, where a large obstruction such as a hill or large building obscures the main signal path between the transmitter and the receiver.
This set of Wireless & Mobile Communications test focuses on “Statistical Models for Multipath Fading Channels”.
1. Which of the following is not a statistical models for multipath fading channels?
a) Clarke’s model for flat fading
b) Saleh and Valenzuela indoor statistical model
c) Two ray Rayleigh fading model
d) Faraday model
Answer: d
Explanation: Several multipath models have been suggested to explain the observed statistical nature of a mobile channel. Clarke’s model for flat fading, Saleh and Valenzuela indoor statistical model, two ray Rayleigh fading model are some of the statistical model for multipath fading channels.
2. Who presented the first statistical model for multipath fading channel?
a) Ossana
b) Rayleigh
c) Newton
d) Faraday
Answer: a
Explanation: The first statistical model was presented by Ossana. It was based on interference of wave incident and reflected from the flat sides of randomly located buildings. Ossana model predicts flat fading power spectra that were in agreement with measurements in suburban areas.
3. Clarke’s model assumes a horizontal polarized antenna.
a) True
b) False
Answer: b
Explanation: Clarke developed a model where the statistical characteristics of the electromagnetic fields of the received signal at the mobile are deduced from scattering. Clarke’s model assumes a fixed transmitter with a vertically polarized antenna.
4. A wave that is incident on mobile does not undergo Doppler shift.
a) True
b) False
Answer: b
Explanation: Every wave that is incident on the mobile undergoes a Doppler shift. It is due to the motion of the receiver and it arrives at the receiver at the same time. Therefore, there is no excess delay due to multipath for any of the waves.
5. Which of the following is an important statistics of a Rayleigh fading useful for designing error control codes and diversity schemes?
a) Mobile speed
b) Doppler frequency
c) Level crossing rate
d) Power density
Answer: c
Explanation: The level crossing rate and average fade duration of a Rayleigh fading signal are important statistics. They are useful for designing error control codes and diversity schemes to be used in mobile communication.
6. The level crossing rate is defined as expected rate at which _______ fading envelope crosses a specified level.
a) Rayleigh
b) Saleh
c) Vanezuela
d) Faraday
Answer: a
Explanation: The LCR is defined as the expected rate at which the Rayleigh fading envelop is normalized to the local rms signal level. And it crosses a specified level in a positive going direction.
7. Level crossing rate is a function of _______
a) Power transmitted by base station
b) Power density of receiver
c) Mobile speed
d) Bit error rate
Answer: c
Explanation: The level crossing rate is a function of mobile speed. Because, it is possible to relate the time rate of change of the received signal of the signal level and velocity of the mobile.
8. Clarke’s model considers the multipath time delay.
a) True
b) False
Answer: b
Explanation: Clarke’s model and the statistics for Rayleigh fading are for flat fading conditions. They do not consider multipath time delay.
9. Saleh and Venezuela reported the results of ______ propagation measurements.
a) Indoor
b) Outdoor
c) Air
d) High frequency
Answer: a
Explanation: Saleh and Venezuela reported the results of indoor propagation measurements. The measurements were made between two vertically polarized omnidirectional antennas located on the same floor of a medium sized office building.
10. Saleh and Venezuela show that indoor channel is _______ time varying.
a) Not
b) Very slow
c) Fast
d) Very fast
Answer: b
Explanation: The results obtained by Saleh and Venezuela shows that the indoor channel is quasi static or very slow time varying. It shows that the statistics of the channel impulse response are independent of transmitting and receiving antenna polarization.
11. What is the full form of SIRCIM?
a) Simulation of Indoor Radio Channel Impulse response Model
b) Statistical Indoor Radio Channel for Impulse Model
c) Statistical Impulse Radio Channel for Indoor Model
d) Simulation of Impulse Radio Channel for Indoor Model
Answer: a
Explanation: SIRCIM stands for Simulation of Indoor Radio Channel Impulse response Model. SIRCIM generates realistic samples of small scale indoor channel impulse response measurements.
This set of Wireless & Mobile Communications Assessment Questions and Answers focuses on “Multipath Shape Factors for Small – Scale Fading”.
1. The term small scale fading describes the slow fluctuations of received power level due to changes in receiver position.
a) True
b) False
Answer: b
Explanation: The term small scale fading describes the rapid fluctuations of received power level due to changes in receiver position. This effect is due to the constructive and destructive interference of the numerous multipath waves that impinge upon a wireless receiver.
2. The rapid fluctuations due to small scale fading affect the _________ design.
a) Receiver
b) Transmitter
c) MSC
d) Base station
Answer: a
Explanation: The rapid fluctuation due to small scale fading affect every aspect of receiver design such as dynamic range, equalization, diversity, modulation scheme and channel error correction coding. These fluctuations are a function of direction of travel as related to the angle of arrival of multipath delay.
3. An approximately omnidirectional channel model accurately describes fading statistics if directional antennas are employed at the receiver.
a) True
b) False
Answer: a
Explanation: An approximately omnidirectional channel model accurately describes fading statistics if directional or smart antenna systems are employed at the receiver. Unfortunately, recent measurements and models have shown that the arriving multipath in a local area bears little resemblance to the omnidirectional propagation.
4. Which of the following is not a principle shape factor?
a) Angular spread
b) Angular constriction
c) Azimuthal direction of maximum fading
d) Angle of arrival
Answer: d
Explanation: Three principle shape factors are angular spread, angular constriction, and azimuthal direction of maximum fading. They are exactly related to the average rate at which a received signal fades.
5. Angular spread is a measure of how multipath concentrates about __________
a) Angle of arrival
b) Transmitted power
c) Single azimuthal direction of arrival
d) Received power
Answer: c
Explanation: The shape factor, angular spread is a measure of how multipath concentrates about a single azimuthal direction of arrival. It has several advantages like it is invariant under changes in transmitted power.
6. Angular constriction is a measure of how multipath concentrates about ______ azimuthal direction.
a) Single
b) Two
c) Three
d) Four
Answer: b
Explanation: The shape factor, angular constriction is a measure of how multipath concentrates about two azimuthal directions. The measure for angular constriction is invariant under changes in transmitted power.
7. Shape factor, azimuthal direction of maximum fading is a directional parameter.
a) True
b) False
Answer: a
Explanation: Azimuthal direction of maximum fading is the third shape factor which is a directional or orientation parameter. Its value corresponds to the direction in which s mobile user would move in order to experience the maximum fading rate in the local area.
8. _______ of a stationary process is actually the variance of the rate of change.
a) Mean
b) Mean square
c) Mean squared derivative
d) Mean squared integral
Answer: c
Explanation: Mean squared derivative of a stationary process is actually the variance of the rate of change. The mean derivative of the stationary process is zero; the mean squared derivative is the simplest statistics that measures the fading rate of a channel.
9. Complex received voltage is a summation of __________ that have impinged upon receiver antenna.
a) Multipath waves
b) Waves
c) Power density
d) Single path waves
Answer: a
Explanation: Complex received voltage is a baseband representation. It us a summation of numerous multipath waves that have impinged upon the receiver antenna and have excited a complex voltage component at the input of a receiver.
10. Which of the following is equal to received power?
a) Square of complex voltage
b) Complex voltage
c) Magnitude of complex voltage
d) Magnitude squared of complex voltage
Answer: d
Explanation: The received power is equal to the magnitude squared of complex voltage. The mathematical operation of taking the squared magnitude of a complex quantity is a nonlinear operation.
11. Which of the following is equal to received envelope?
a) Square of complex voltage
b) Complex voltage
c) Magnitude of complex voltage
d) Magnitude squared of complex voltage
Answer: c
Explanation: The received envelope is equal to the magnitude of the complex voltage. The channel is assumed to be Rayleigh fading in order to calculate the mean squared fading rate.
12. Which of the following describes the average fading rate within a local area?
a) Angular spread
b) Angular constriction
c) Azimuthal direction of maximum fading
d) Angle of arrival
Answer: a
Explanation: Angular spread describes the average fading rate within a local area. The average of the two fading rate variances, regardless of the orientation of the measurement, is always given by averaging the variances observed over two perpendicular directions within the local area.
13. Angular constriction affects the average fading rate within local area.
a) True
b) False
Answer: b
Explanation: Angular constriction does not affect the average fading rate within local area. It describes the variability of fading rates taken along different azimuthal directions.
This set of Wireless & Mobile Communications Questions and Answers for Aptitude test focuses on “Frequency Modulation Vs. Amplitude Modulation”.
1. Which is the process of encoding information from a message source in suitable manner for transmission?
a) Modulation
b) Demodulation
c) Encryption
d) Decryption
Answer: a
Explanation: Modulation is the process of encoding information from a message source in suitable manner for transmission. It translates the baseband message signal to a bandpass signal at frequencies that are very higher compared to the baseband frequency.
2. The bandpass signal is called the modulating signal.
a) True
b) False
Answer: b
Explanation: The bandpass signal is called the modulated signal and the baseband message signal is called the modulating signal. Modulation is done by varying the amplitude, phase or frequency of a high frequency carrier in accordance with the amplitude of the message signal.
3. AM signal have all information in ______ of the carrier.
a) Amplitude
b) Magnitude
c) Frequency
d) Power
Answer: a
Explanation: AM signals have all information in the amplitude of the carrier. Because, AM signal superimpose the exact relative amplitudes of the modulating signal onto the carrier.
4. In frequency modulation, the amplitude of modulated carrier signal is ______and its frequency is ______ by the modulating message signal.
a) Constant, constant
b) Varied, constant
c) Constant, varied
d) Varied, varied
Answer: c
Explanation: In frequency modulation , the amplitude of modulated carrier signal is kept constant while its frequency is varied by the modulating message signal. FM is the most popular analog modulation technique used in mobile communication.
5. FM signals have all their information in ______ of the carrier.
a) Amplitude
b) Magnitude
c) Frequency
d) Power
Answer: c
Explanation: FM signal have all their information in the phase or frequency of the carrier. This provides a nonlinear and very rapid improvement in reception quality once a certain minimum received signal level, called FM threshold is achieved.
6. Amplitude modulation has ______ relationship between the quality of the received signal and the power of the received signal.
a) No
b) Non linear
c) Constant
d) Linear
Answer: d
Explanation: In amplitude modulation schemes, there is a linear relationship between the quality of the received signal and the power of the received signal. AM signals superimpose the exact relative amplitudes of the modulating signal onto the carrier.
7. FM has _______ noise immunity when compared to amplitude modulation.
a) Same
b) Less
c) No
d) Better
Answer: d
Explanation: FM offers many advantages over amplitude modulation which makes it a better choice for many mobile radio applications. Frequency modulation has better noise immunity when compared to amplitude modulation.
8. FM signals are _____ susceptible to atmospheric and impulse noise as compared to AM.
a) Largely
b) Less
c) Not
d) Better
Answer: b
Explanation: FM signals are represented as frequency variations rather than amplitude variations. Therefore, FM signals are less susceptible to atmospheric and impulse noise.
9. Burst noise affects FM systems.
a) True
b) False
Answer: b
Explanation: If the FM received signal is above the FM threshold, burst noise does not affect FM system performance as much as AM systems. It is due to the fact that message amplitude variations do not carry information in FM.
10. AM has a modulation index.
a) True
b) False
Answer: b
Explanation: Unlike AM, FM has a modulation index. Hence, bandwidth occupancy can be varied to obtain greater signal to noise performance. Thus, it is possible to tradeoff bandwidth occupancy for improved noise performance.
11. Which of the following is an advantage of AM systems in comparison to FM sysytems?
a) Occupy less bandwidth
b) Superior performance in fading
c) Better noise immunity
d) Not susceptible to impulse noise
Answer: a
Explanation: AM signals are able to occupy less bandwidth as compared to FM signals, since the transmission system is linear. Small scale fading causes rapid fluctuations in the received signal. Thus, FM offers superior qualitative performance in fading when compared to AM.
12. An FM signal is a ______ envelope signal.
a) Varied
b) Small
c) Large
d) Constant
Answer: d
Explanation: An FM signal is a constant envelope signal. It is due to the fact that the envelope of the carrier does not change with changes in the modulating signal. Hence, the transmitted power of an FM signal is constant regardless of the amplitude of the message signal.
13. FM uses class______ amplifiers and AM uses class ______ amplifiers.
a) C, C
b) A, C
c) C, A
d) AB, C
Answer: c
Explanation: The constant envelope of the transmitted signal allows efficient class C amplifiers. However in AM, it is critical to maintain linearity between the applied message and amplitude of transmitted signal. Therefore, class A or AB are used for AM systems.
14. AM exhibits capture effect characteristics.
a) True
b) False
Answer: b
Explanation: AM systems do not exhibit capture effect characteristics because all the interferers are received at once and must be discriminated after the modulation process. FM exhibits this characteristic. Capture effect is a direct result of rapid non-linear improvement in received quality for an increase in receives power.
15. Which of the following is a drawback of FM systems?
a) Burst noise
b) Susceptible to atmospheric noise
c) Wider frequency band
d) Poor performance in fading
Answer: c
Explanation: FM systems have many advantages over AM systems. They also have certain disadvantages. FM systems require a wider frequency band in transmitting media in order to obtain the advantages of reduced noise and capture effect. It is generally several times as large as that needed for AM.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Amplitude Modulation”.
1. Carrier signal in modulation technique is _______ signal.
a) High frequency
b) Low frequency
c) High amplitude
d) Low amplitude
Answer: a
Explanation: Carrier signal in modulation technique is a high frequency signal. In amplitude modulation, the amplitude of a high frequency carrier signal is varied in accordance to the instantaneous amplitude of the modulating signal.
2. Modulation index of an AM signal is ratio of __________ to the _______
a) Peak carrier amplitude, Peak message signal amplitude
b) Peak message signal amplitude, Peak carrier amplitude
c) Carrier signal frequency, Message signal frequency
d) Message signal frequency, Carrier signal frequency
Answer: b
Explanation: The modulation index k of an AM signal is defined as the ratio of the peak message signal amplitude to the peak carrier amplitude. The modulation index is often expressed as a percentage. It is also called percentage modulation.
3. If the peak message signal amplitude is half the peak amplitude of the carrier signal, the signal is _____ modulated.
a) 100%
b) 2%
c) 50%
d) 70%
Answer: c
Explanation: The modulation is also expressed in percentage. It is also called percentage modulation. The signal is said to be 50% modulated if the peak message signal amplitude is half the peak amplitude of the carrier signal.
4. A percentage of modulation greater than ___________ will distort the message signal.
a) 10%
b) 25%
c) 50%
d) 100%
Answer: d
Explanation: A percentage of modulation greater than 100% will distort the message signal if detected by an envelope detector. In this case the lower excursion of the signal will drive the carrier amplitude below zero, making it negative .
5. The RF bandwidth of AM is ____________ the maximum frequency contained in the modulating message signal.
a) Equal
b) Two times
c) Four times
d) Ten times
Answer: b
Explanation: The RF bandwidth of an AM signal is equal to B AM =2f m . It is double the maximum frequency contained in the modulating message signal. AM spectrum consists of an impulse at the carrier frequency and two sidebands which replicate the message spectrum.
6. Single sideband AM systems occupy same bandwidth as of conventional AM systems.
a) True
b) False
Answer: b
Explanation: Single sideband AM systems transmit only one of the sidebands about the carrier. Hence, they occupy only half the bandwidth of conventional AM systems.
7. How is the performance of SSB AM systems in fading channels?
a) Poor
b) Best
c) Good
d) Average
Answer: a
Explanation: SSB systems have the advantage of being very bandwidth efficient. But their performance in fading channels is very poor. For proper detection, the frequency of the oscillator at the product detector mixer in the receiver must be same as that of the incoming carrier frequency.
8. Which of the following is a disadvantage of tone-in-band SSB system?
a) High bandwidth
b) Bad adjacent channel protection
c) Effects of multipath
d) Generation and reception of signal is complicated
Answer: d
Explanation: Tone-in-band SSB systems has the advantage of maintaining the low bandwidth property of the SSB signals, while at the same time providing good adjacent channel protection. The tone in band system employs feedforward automatic gain and frequency control to mitigate the effects of multipath induced fading.
9. FFSR in AM systems stands for ________
a) Feedforward signal regeneration
b) Feedbackward signal regeneration
c) Feedbackward system restoration
d) Feedforward system restoration
Answer: a
Explanation: FFSR stands for Feedforward signal regeneration. If the pilot tone and the information bearing signal undergo correlated fading, it is possible at the receiver to counteract the effects of fading through signal processing based on tracking of pilot tone. This process is called FFSR.
10. AM demodulation technique can be divided into _____ and _____ demodulation.
a) Direct, indirect
b) Slope detector, zero crossing
c) Coherent, noncoherent
d) Quadrature detection, coherent detection
Answer: c
Explanation: AM demodulation techniques may be broadly divide into two main categories. They are called coherent and noncoherent demodulation. They are differentiated by the knowledge of transmitted carrier frequency and phase at the receiver.
11. Non coherent detection requires the knowledge of transmitted carrier frequency and phase at the receiver.
a) True
b) False
Answer: b
Explanation: Non coherent detection does require the knowledge of phase information. However, coherent detection requires knowledge of the transmitted carrier frequency and phase at the receiver.
12. A product detector in AM systems is also called ___________
a) Envelope detector
b) Differentiator
c) Integrator
d) Phase detector
Answer: d
Explanation: A product detector is also called a phase detector. It forms a coherent demodulator for AM signals. It is a down converter circuit which converts the input bandpass signal to a baseband signal.
13. AM system use only product detector for demodulation. They never use envelope detectors.
a) True
b) False
Answer: b
Explanation: AM systems can use either product detector or envelope detector for demodulation. As a rule, envelope detectors are useful when input signal power is at least 10dB greater than noise power, whereas product detectors are able to process the AM signals with input signal to noise ratios well below 0 dB.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Angle modulation”.
1. FM is a part of general class of modulation known as ______
a) Angle modulation
b) Phase modulation
c) Amplitude modulation
d) Frequency modulation
Answer: a
Explanation: FM is a part of general class of modulation known as angle modulation. Angle modulation varies a sinusoidal carrier signal in such a way that the angle of the carrier is varied according to the amplitude of the modulating baseband signal.
2. FM is called constant envelope because ______ of carrier wave is kept constant.
a) Frequency
b) Amplitude
c) Phase
d) Angle
Answer: b
Explanation: FM is called the constant envelope because amplitude of the carrier wave is kept constant. It is duo to the fact that the envelope of the carrier does not change with changes in the modulating signal.
3. Which of the following are two most important classes of angle modulation?
a) Amplitude modulation, frequency modulation
b) Amplitude modulation, phase modulation
c) Frequency modulation, phase modulation
d) Single sideband amplitude modulation, phase modulation
Answer: c
Explanation: The two most important classes of angle modulation are frequency modulation and phase modulation. They provide the ways in which phase of a carrier signal may be varied in accordance with the baseband signal.
4. Frequency modulated signal is regarded as the phase modulated signal in which the modulating wave is differentiated before modulation.
a) True
b) False
Answer: b
Explanation: Frequency modulated signal is regarded as the phase modulated signal in which the modulating wave is integrated before modulation. This means that an FM signal can be generated by first integrating the message signal and then using the result as an input to a phase modulator.
5. Frequency modulation index defines the relationship between the ______ and bandwidth of transmitted signal.
a) Frequency of message signal
b) Amplitude of message signal
c) Amplitude of carrier signal
d) Frequency of carrier signal
Answer: b
Explanation: The frequency modulation index defines the relationship between the message amplitude and the bandwidth of the transmitted signal. If the modulating signal is a low pass signal, maximum bandwidth of the modulating signal is equal to the highest frequency component present in the modulating signal.
6. FM bandwidth is approximated using _______ rule.
a) Carson’s
b) Faraday’s
c) Maxwell’s
d) Armstrong’s
Answer: a
Explanation: The approximation of bandwidth is done using Carson’s rule. Carson’s bandwidth rule defines the approximate bandwidth requirements of communications system components for a carrier signal that is frequency modulated by a continuous or broad spectrum of frequencies rather than a single frequency.
7. Which of the following are two methods for generating FM signal?
a) Coherent method, noncoherent method
b) Product detector, envelope detector
c) Direct method, indirect method
d) Slope detector, Zero crossing detector
Answer: c
Explanation: Direct method and indirect method are the methods used for generating FM signals. These methods are differentiated by the variation of the carrier frequency.
8. In indirect method, the carrier frequency is directly varied in accordance with the input modulating signal.
a) True
b) False
Answer: b
Explanation: The above is the case for direct method. In the indirect method, a narrowband FM signal is generated using a balanced modulator, and frequency multiplication is used to increase both the frequency deviation and the carrier frequency to the required level.
9. Which of the following is used to vary the frequency of the carrier frequency in accordance with the baseband signal amplitude variations in direct method of FM generation?
a) Integrator
b) Envelope detector
c) Multivibrator
d) Voltage controlled oscillators
Answer: d
Explanation: In direct method, VCOs are used to vary the frequency of the carrier signal in accordance with the baseband signal amplitude variations. These oscillators use devices with reactance that can be varied by the application of a voltage.
10. Frequency demodulator is a frequency to amplitude converter circuit.
a) True
b) False
Answer: a
Explanation: Frequency demodulator produces an output voltage with instantaneous amplitude that is directly proportional to the instantaneous frequency of the input FM signal. Thus, frequency demodulator is a frequency to amplitude converter circuit.
11. Which of the following is not a technique for FM demodulation?
a) Slope detection
b) Zero crossing detection
c) Product detector
d) Phase locked discriminator
Answer: c
Explanation: Various techniques such as slope detection, zero crossing detection, phase locked discrimination and quadrature detection are used to demodulate FM. Product detector is used for demodulating AM signals.
12. Which of the following FM demodulator is sometimes known as pulse averaging discriminator?
a) Slope detection
b) Zero crossing detection
c) Quadrature detection
d) Phase locked discriminator
Answer: b
Explanation: Zero crossing detector is sometimes known as pulse averaging discriminator. The rationale behind this technique is to use the output of the zero crossing detector to generate a pulse train with an average value that is proportional to frequency of the input signal.
13. PLL in FM detection stands for ______
a) Phase locked loop
b) Programmable logic loop
c) Phase locked logic
d) Programmable locked loop
Answer: a
Explanation: PLL stands for phase locked loop. The PLL is a closed loop control system which can track the variations in the received signal phase and frequency.
14. In angle modulation, signal to noise ratio before detection is a function of ______
a) Modulation index
b) Input signal to noise ratio
c) Maximum frequency of the message
d) IF filter bandwidth
Answer: d
Explanation: In angle modulation systems, the signal to noise ratio before detection is the function of the receiver IF filter bandwidth, received carrier power, and received interference. However, signal to noise ratio after detection is a function of maximum frequency of the message, input signal to noise ratio and modulation index.
15. FM can improve the receiver performance through adjustment of transmitted power.
a) True
b) False
Answer: b
Explanation: FM can improve receiver performance through adjustment of the modulation index at the transmitter, and not the transmitted power. This is not the case in AM since linear modulation techniques do not trade bandwidth for SNR.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Digital Modulation”.
1. Modern mobile communication systems use analog modulation techniques.
a) True
b) False
Answer: b
Explanation: Modern mobile communication systems use digital modulation techniques. Advancements in VLSI and digital signal processing technology have made digital modulation more cost effective than analog transmission systems.
2. Which of the following is not an advantage of digital modulation?
a) Greater noise immunity
b) Greater security
c) Easier multiplexing
d) Less bandwidth requirement
Answer: d
Explanation: Digital modulation offer many advantages over analog modulation. Some advantages include greater noise immunity and robustness. They provide easier multiplexing of various forms of information and greater security.
3. A desirable modulation scheme provides _________ bit error rates at __________ received signal to noise ratios.
a) Low, low
b) Low, high
c) High, high
d) High, low
Answer: a
Explanation: A desirable modulation scheme provides low bit error rates at low received signal to noise ratios. They perform well in multipath and fading conditions, occupies a minimum bandwidth and is easy and cost effective to implement.
4. The performance of modulation scheme is not measured in terms of __________
a) Power efficiency
b) Bandwidth efficiency
c) Cost and complexity
d) Transmitted power
Answer: d
Explanation: The performance of modulation scheme is often measured in terms of its power efficiency and bandwidth efficiency. Other factors also affect the choice of modulation scheme, such as cost and complexity of the subscriber receiver and modulation which is simple to detect.
5. In digital communication system, in order to increase noise immunity, it is necessary to increase _________
a) Signal power
b) Signal amplitude
c) Signal frequency
d) Signal magnitude
Answer: a
Explanation: In digital communication system, in order to increase noise immunity, it is necessary to increase signal power. However, the amount by which the signal power should be increased to obtain a certain level of fidelity depends on the particular type of modulation employed.
6. Which of the following is the ratio of signal energy per bit to noise power spectral density?
a) Bandwidth efficiency
b) Spectral density
c) Power efficiency
d) Power density
Answer: c
Explanation: Power efficiency is often expressed as the ratio of signal energy per bit to noise power spectral density required at the receiver input for a certain probability of error. Power efficiency is a measure of how favourably the trade-off between fidelity and signal power is made.
7. Increasing the data rate implies the increase in pulse width of digital symbol.
a) True
b) False
Answer: b
Explanation: There is an unavoidable relationship between data rate and bandwidth occupancy. Increasing the data rate implies decreasing the pulse width of a digital symbol, which increases the bandwidth of the signal.
8. Which of the following is the ratio of the throughput data rate per Hertz?
a) Bandwidth efficiency
b) Spectral density
c) Power efficiency
d) Power density
Answer: a
Explanation: Bandwidth efficiency reflects how efficiently the allocated bandwidth is utilized. It is defined as the ratio of throughput data rate per Hertz in a given bandwidth. It describes the ability of a modulation scheme to accommodate data within a limited bandwidth.
9. Which of the following is defined as the range of frequencies over which the signal has a non zero power spectral density?
a) Null to null bandwidth
b) Half power bandwidth
c) 3 dB bandwidth
d) Absolute bandwidth
Answer: d
Explanation: The absolute bandwidth is defined as the range of frequencies over which the signal has a non-zero power spectral density. For symbols represented as rectangular baseband pulses, the PSD profile extends over an infinite range of frequencies, and has an absolute bandwidth of infinity.
10. _______ is equal to width of main spectral lobe.
a) Null to null bandwidth
b) Half power bandwidth
c) 3 dB bandwidth
d) Absolute bandwidth
Answer: a
Explanation: Null to null bandwidth is a simpler and more widely accepted measure of bandwidth. It is equal to the width of main spectral lobe.
11. Half power bandwidth is also called ______
a) Absolute bandwidth
b) Null to null bandwidth
c) 3 dB bandwidth
d) Zero dB bandwidth
Answer: c
Explanation: Half power bandwidth is also called the 3 dB bandwidth. It is defined as the interval between frequencies at which the PSD has dropped to half power, or 3 dB below the peak value.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Pulse Shaping Techniques”.
1. Intersymbol interference leads to ________ probability of the receiver for making an error in detecting the symbols.
a) Increased
b) Decreased
c) Zero
d) One
Answer: a
Explanation: ISI leads to increased probability of the receiver making an error in detecting a symbol. When rectangular pulses are passed through a bandlimited channel, the pulses will spread in time, and the pulse for each symbol will smear into the time intervals of succeeding symbols.
2. ISI is ________ by increasing channel bandwidth.
a) Maximized
b) Minimized
c) Zero
d) Infinite
Answer: b
Explanation: Increasing channel bandwidth is one of the method to minimize intersymbol interference. But mobile communication systems use minimal bandwidth, thus other methods to reduce ISI are desirable.
3. Why is pulse shaping technique used?
a) To increase ISI
b) To increase spectral width of modulated signal
c) To reduce ISI
d) To reduce power spectral density
Answer: c
Explanation: Pulse shaping techniques reduces the intersymbol interference. They are also used to reduce the spectral width of the modulated digital signal.
4. Who was the first to solve the problem of ISI?
a) Manchester
b) Faraday
c) Graham Bell
d) Nyquist
Answer: d
Explanation: Nyquist was the first to solve the problem of ISI. He overcome the problem of ISI while keeping the transmission bandwidth low. He observed that ISI can be completely nullified if at every instant, the response due to all symbols except the current symbol is equal to zero.
5. According to Nyquist, the impulse response of the overall communication system should have ______ decay with _______ magnitude for sample values not equal to zero.
a) Fast, small
b) Slow, small
c) Slow, large
d) Fast. Large
Answer: a
Explanation: According to Nyquist, the impulse response of the overall communication system should have fast decay with small magnitude for sample values not equal to zero. If the channel is ideal then it should be possible to realize approximate shaping filters at both transmitter and receiver.
6. Raised cosine filter does not satisfy Nyquist criteria.
a) True
b) False
Answer: b
Explanation: Raised cosine filter is the most popular pulse shaping filter used in mobile communication. It belongs to the class of filters that satisfy Nyquist criterion.
7. As the roll off factor in raised cosine rolloff filter __________ the occupied bandwidth ________
a) Increases, decreases
b) Decreases, constant
c) Increases, increases
d) Decreases, increases
Answer: c
Explanation: As the rolloff factor increases, the bandwidth of the filter also increases and the time sidelobe levels decrease in adjacent symbol slots. Thus, it implies that increasing rolloff factor decreases the sensitivity to timing jitter but increases the occupied bandwidth.
8. Gaussian pulse shaping filter follows Nyquist criterion.
a) True
b) False
Answer: b
Explanation: Gaussian pulse shaping filter uses non Nyquist technique. It is effective when used in conjunction with minimum shift keying modulation, or other modulation which is well suited for power efficient nonlinear amplifiers.
9. Gaussian filter has zero crossings at adjacent symbol peaks.
a) True
b) False
Answer: b
Explanation: Nyquist filters have zero crossings at adjacent symbol peaks and a truncated transfer function. Gaussian filter does not follow Nyquist criterion and has a smooth transfer function with no zero crossings.
10. Which of the following is true for a Gaussian filter?
a) Large bandwidth
b) Minimum ISI
c) High overshoot
d) Sharp cut off
Answer: d
Explanation: The Gaussian filter has a narrow absolute bandwidth, and has a sharp cut off, low overshoot and pulse area preservation properties. This makes it attractive for use in mobile communication that uses nonlinear RF amplifiers.
11. Gaussian pulse shaping filter reduces the spectral occupancy and ISI.
a) True
b) False
Answer: b
Explanation: Gaussian pulse shaping does not satisfy Nyquist criterion for ISI cancellation. Thus, it reduces the spectral occupancy but there is degradation in the performance due to increased ISI.
12. Gaussian pulses are used when cost and power efficiency are major factors.
a) True
b) False
Answer: a
Explanation: Gaussian pulses are used when cost and power efficiency are major factors. But the bit error rates due to ISI are deemed to be lower than what is nominally required. Thus, there is a trade-off between desired RF bandwidth and irreducible error due to ISI.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Linear Modulation Techniques”.
1. In linear modulation technique __________ of transmitted signal varies linearly with modulating digital signal.
a) Amplitude
b) Frequency
c) Phase
d) Angle
Answer: a
Explanation: In linear modulation technique, the amplitude of transmitted signal varies linearly with modulating digital signal. It is a form of digital modulation technique.
2. Linear modulation techniques are not bandwidth efficient.
a) True
b) False
Answer: b
Explanation: Linear modulation techniques are bandwidth efficient. They are used in wireless communication systems when there is an increasing demand to accommodate more and more users within a limited spectrum.
3. Which of the following is not a linear modulation technique?
a) OQPSK
b) π/4 QPSK
c) FSK
d) BPSK
Answer: c
Explanation: OQPSK, π/4 QPSK and BPSK are the most popular linear modulation techniques. They have very good spectral efficiency. However, FSK is an non-linear modulation technique.
4. In BPSK, the ________ of constant amplitude carrier signal is switched between two values according to the two possible values.
a) Amplitude
b) Phase
c) Frequency
d) Angle
Answer: b
Explanation: In binary phase shift keying , the phase of a constant amplitude carrier signal is switched between two possible values m1 and m2. These two values corresponds to binary 1 and 0 respectively.
5. By applying cos, BPSK signal is equivalent to ________
a) Double sideband suppressed carrier amplitude modulated waveform
b) Single sideband suppressed carrier amplitude modulated waveform
c) Frequency modulated waveform
d) SSB amplitude waveform
Answer: a
Explanation: The BPSK signal is equivalent to a double sideband suppressed carrier amplitude modulated waveform, where cos is applied as the carrier. Hence, a BPSK signal can be generated using a balanced modulator.
6. BPSK uses non-coherent demodulator.
a) True
b) False
Answer: b
Explanation: BPSK uses coherent or synchronous demodulation. It requires the information about the phase and frequency of the carrier be available at the receiver.
7. DPSK uses coherent form of PSK.
a) True
b) False
Answer: b
Explanation: Differential phase shift keying uses noncoherent form of phase shift keying. Noncoherent form avoids the need for a coherent reference signal at the receiver. Noncoherent receivers are also easy and cheap to build.
8. In DPSK system, input signal is differentially encoded and then modulated using a ________ modulator.
a) Amplitude
b) Frequency
c) BPSK
d) QPSK
Answer: c
Explanation: In DPSK system, input binary sequence is first differentially encoded and then modulated using a BPSK modulator. The differentially encoded sequence is generated from input binary sequence by complimenting their modulo-2 sum.
9. The energy efficiency of DPSK is _________ to coherent PSK.
a) Superior
b) Same
c) Zero
d) Inferior
Answer: d
Explanation: The energy efficiency of DPSK is inferior to that of coherent PSK by about 3 dB. But, it has an advantage of reduced receiver complexity.
10. QPSK has ________ the bandwidth efficiency of BPSK.
a) Twice
b) Same
c) Half
d) Four times
Answer: a
Explanation: Quadrature phase shift keying has twice the bandwidth of BPSK. It is because two bits are transmitted in a single modulation symbol. The phase of the carrier takes on one of the four equally spaced values, where each value of phase corresponds to a unique pair of message bit.
11. QPSK provides twice the bandwidth efficiency and _______ energy efficiency as compared to BPSK.
a) Twice
b) Half
c) Same
d) Four times
Answer: c
Explanation: The bit error probability of QPSK is identical to BPSK but twice as much data can be sent in the same bandwidth. Thus, when compared to BPSK, QPSK provides twice the spectral efficiency with exactly the same efficiency.
12. What is the full form of OQPSK?
a) Optical Quadrature Phase Shift Keying
b) Orthogonal Quadrature Pulse Shift Keying
c) Orthogonal Quadrature Phase Shift Keying
d) Offset Quadrature Phase Shift Keying
Answer: d
Explanation: OQPSK stands for offset quadrature phase shift keying. It is a modified form of QPSK which is less susceptible to deleterious effects and supports more efficient amplification. OQPSK is sometimes also called staggered QPSK.
13. The bandwidth of OQPSK is _______ to QPSK.
a) Identical
b) Twice
c) Half
d) Four times
Answer: a
Explanation: The spectrum of an OQPSK signal is identical to that of QPSK signal. Hence, both signals occupy the same bandwidth. The staggered alignment of the even and odd bit streams in OQPSK signal does not change the nature of spectrum.
14. QPSK signals perform better than OQPSK in the presence of phase jitter.
a) True
b) False
Answer: b
Explanation: OQPSK signal perform better than QPSK in the presence of phase jitter. It is due to the presence of noisy reference signal at the receiver.
15. Which of the following is not a detection technique used for detection of π/4 QPSK signals?
a) Baseband differential detection
b) IF differential detection
c) FM discriminator detection
d) Envelope detection
Answer: d
Explanation: There are various types of detection techniques used for the detection of π/4 QPSK signals. They include baseband differential detection, IF differential detection and FM discriminator detection.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Constant Envelope Modulation”.
1. In non-linear modulation, the amplitude of the carrier varies with the variation of modulating signal.
a) True
b) False
Answer: b
Explanation: In non-linear modulation, the amplitude of the carrier is constant regardless of the variation in the modulating signal. Many practical mobile radio communication systems use these types of nonlinear modulation methods.
2. In constant envelope family of modulation, class C amplifiers introduces degradation in spectrum occupancy.
a) True
b) False
Answer: b
Explanation: The constant envelope family of modulation has an advantage of satisfying various conditions. In this, power efficient Class C amplifiers can be used without introducing degradation in the spectrum occupancy of the transmitted signal.
3. Constant envelope modulation techniques occupy ______ bandwidth than linear modulation schemes.
a) Larger
b) Smaller
c) Same
d) Twice
Answer: a
Explanation: Constant envelope modulation technique occupies a larger bandwidth than linear modulation technique. It is one of the disadvantage of constant envelope modulation. It is not well suited where bandwidth efficiency is more important than power efficiency.
4. In BFSK __________ of constant amplitude carrier signal is switched between two values.
a) Phase
b) Angle
c) Frequency
d) Amplitude
Answer: c
Explanation: In BFSK, the frequency of a constant amplitude carrier signal is switched between two values according to the two possible message states. These states are called high and low tunes, which corresponds to binary 1 or 0.
5. MSK stands for ________
a) Maximum shift keying
b) Minimum shift keying
c) Minimum space keying
d) Maximum space keying
Answer: b
Explanation: MSK stands for minimum shift keying. It is a special type of continuous phase shift keying. It is form of digital modulation technique that was developed in 1950s.
6. What is the modulation index of MSK?
a) 0.1
b) 1
c) 0.5
d) 0
Answer: c
Explanation: Minimum shift keying is a special type of CPFSK. Its peak frequency deviation is equal to ¼ the bit rate. In other words, MSK is continuous phase FSK with a modulation index of 0.5.
7. The modulation index of an FSK signal is similar to modulation index of ________
a) Amplitude modulation
b) Phase modulation
c) QPSK
d) Frequency modulation
Answer: d
Explanation: The modulation index of an FSK signal is similar to FM modulation index. It is defined by /Rb. Here ∆F is the peak RF frequency and Rb is the bit rate.
8. The name minimum phase shift keying implies minimum _________
a) Frequency separation
b) Amplitude separation
c) Phase change
d) Amplitude deviation
Answer: a
Explanation: The name minimum phase shift keying implies minimum frequency separation, i.e. the bandwidth that allows orthogonal detection. A modulation index of 0.5 corresponds to the minimum frequency spacing that allows two FSK signals to be coherently orthogonal.
9. MSK is sometimes also referred as _________
a) Slow FSK
b) Fast FSK
c) Slow PSK
d) Fast PSK
Answer: b
Explanation: Minimum shift keying is sometimes also referred as fast FSK. It is so called because frequency spacing used is only half as much as that used in conventional noncoherent frequency shift keying.
10. Which of the following is not a property of MSK?
a) Variable envelope
b) Spectral efficiency
c) Good BER performance
d) Self synchronizing capability
Answer: a
Explanation: MSK has a constant envelope. It is a spectrally efficient scheme. It possesses properties such as constant envelope, spectral efficiency, good BER performance and self-synchronizing capability.
11. MSK is a special form of OQPSK.
a) True
b) False
Answer: a
Explanation: Yes, MSK can be thought of as a special form of offset quadrature phase shift keying. The condition is that baseband rectangular pulses are replaced with half sinusoidal pulses.
12. GMSK is a ________ of MSK.
a) Integral
b) Opposite
c) Derivative
d) Similar
Answer: c
Explanation: Gaussian minimum phase shift keying is a simple binary modulation scheme. It is viewed as a derivative of MSK. GMSK considerably reduces the sidelobe levels in the transmitted spectrum.
13. Which of the following holds true for GMSK?
a) Minimum ISI
b) Minimum error rate
c) Good spectral efficiency
d) Variable envelope property
Answer: c
Explanation: GMSK sacrifices the irreducible error rate caused by partial response signalling in exchange for extremely good spectral efficiency and constant envelope properties. And the premodulation Gaussian filtering introduces ISI in the transmitted signal.
14. MSK has complex demodulation and synchronization circuits.
a) True
b) False
Answer: b
Explanation: MSK has simple demodulation and synchronization circuits. It has various other advantages like continuous phase property makes it highly desirable for highly reactive loads. Due to these advantages, MSK is a popular modulation scheme for mobile radio communication.
This set of Wireless & Mobile Communications Quiz focuses on “Combined Linear and Constant Envelope Modulation Techniques”.
1. Which of the following is a combined linear and constant envelope technique?
a) MPSK
b) PSK
c) BPSK
d) QPSK
Answer: a
Explanation: M-ary phase shift keying is a combined linear and constant envelope technique. It is a part of M-ary modulation techniques. These modern modulation techniques exploit the fact that digital baseband data may be sent by varying both the envelope and phase of an RF carrier.
2. In an M-ary signalling scheme two or more bits are grouped together to form a _______
a) Chip
b) Symbol
c) Byte
d) Pattern
Answer: b
Explanation: In an M-ary signalling scheme two or more bits are grouped together to form symbols. And one of the M possible signals is transmitted during each symbol period of duration Ts.
3. The number of possible signal in M-ary signalling is given by M and M = __________ where n is an integer.
a) n
b) 2n
c) 2 n
d) n 2
Answer: c
Explanation: Two or more bits are grouped to form a symbol in M-ary modulation. And the number of possible symbols should be equal to 2 n , where n is an integer.
4. M-ary signalling techniques are not sensitive to timing jitters.
a) True
b) False
Answer: b
Explanation: Timing errors increase when smaller distances between signals in the constellation diagram are used. M-ary signalling techniques are attractive for use in bandlimited channel, but are limited in their applications due to sensitivity in timing jitter.
5. M-ary modulation schemes have very good power efficiency.
a) True
b) False
Answer: b
Explanation: M-ary modulation schemes have poor power efficiency, but they have a better bandwidth efficiency. An 8-PSK system requires a bandwidth that is 3 times smaller than a BPSK system, whereas its BER performance is very worse since signals are packed more closely in the signal constellation.
6. In M-ary PSK, the carrier ___________ takes one of M possible values.
a) Amplitude
b) Frequency
c) Angle
d) Phase
Answer: d
Explanation: In an M-ary PSK, the carrier phase takes one of the M possible values. The possible values of phase are θ i =2/M, where i=1,2,…….,M.
7. The constellation of M-ary PSK is ____________ dimensional.
a) One
b) Does not exist
c) Two
d) Three
Answer: c
Explanation: The constellation of M-ary PSK is two dimensional. It is due to the presence of two basis signals. And the M-ary message points are equally spaced on a circle.
8. What is the radius of the circle in M-ary PSK on which message points are equaly spaced?
a) √E s
b) √E b
c) E b
d) E s
Answer: a
Explanation: The M-ary message points are equally spaced on a circle of radius √E s centred at the origin. Here E s is energy per symbol. Thus, MPSK is a constant envelope signal when no pulse shaping is used.
9. As the value of M _________ the bandwidth efficiency ________
a) Increases, same.
b) Increases, decreases
c) Increases, increases
d) Decreases, same
Answer: c
Explanation: The first null bandwidth of M-ary PSK signals decrease as M increases while Rb is held constant. Therefore, as the value of M increases, the bandwidth efficiency also increases.
10. The power efficiency of the M ary PSK decreases because of the _____
a) Freely packed constellation
b) Increment of bandwidth efficiency
c) Fixed null bandwidth
d) Densely packed constellation
Answer: d
Explanation: Bandwidth efficiency increases as the value of M increases. But at the same time, increasing M implies that the constellation is more densely packed. Hence the power efficiency or noise tolerance is decreased.
11. In QAM, the amplitude is _______ and phase is _______
a) Varied, constant
b) Varied, varied
c) Constant, varied
d) Constant, constant
Answer: b
Explanation: Quadrature amplitude modulation is obtained by allowing the amplitude to also vary with the phase. Thus, the constellation consists of square lattice of signal points.
12. M-ary QAM signal have constant energy per symbol.
a) True
b) False
Answer: b
Explanation: M-ary QAM does not have constant energy per symbol. It also does not have constant distance between possible symbol states. It reasons that particular values of M-ary QAM signal will be detected with higher probability than others.
13. In comparison to M-ary PSK, M-ary QAM bandwidth efficiency is _____ and power efficiency is ________
a) Identical, superior
b) Less, superior
c) Identical, identical
d) Superior, superior
Answer: a
Explanation: The power spectrum and bandwidth efficiency of QAM modulator is identical to M-ary PSK modulation. But, in terms of power efficiency QAM is superior to M-ary PSK.
14. The bandwidth efficiency of an M-ary FSK signal ________ with ________ in M.
a) Constant, increase
b) Increases, increase
c) Decreases, increase
d) Decreases, decrease
Answer: c
Explanation: The bandwidth efficiency of an M-ary FSK signal decreases with increase in M. Therefore, unlike M-PSK signals, M-FSK signals are bandwidth inefficient.
15. Power efficiency of M-ary FSK increases, since _________
a) Constellation is densely packed
b) M signals are non-orthogonal
c) Fixed null bandwidth
d) M-signals are orthogonal
Answer: d
Explanation: In M-ary FSK, all the M signals are orthogonal and there is no crowding in the signal space. Hence, power efficiency of M-ary FSK increases with M.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Spread Spectrum Modulation Techniques”.
1. The transmission bandwidth of spread spectrum techniques is equal to the minimum required signal bandwidth.
a) True
b) False
Answer: b
Explanation: Spread spectrum techniques employ a transmission bandwidth that is several orders of magnitude greater than the minimum required signal bandwidth. On the other hand, primary objective of all the modulation schemes is to minimize the required transmission bandwidth.
2. Why spread spectrum technique is inefficient for a single user?
a) Large transmission bandwidth
b) Small transmission bandwidth
c) Fixed transmission bandwidth
d) Fixed null bandwidth
Answer: a
Explanation: Spread spectrum systems are bandwidth inefficient for single users. But in spread spectrum systems, many users can simultaneously use the same bandwidth without significantly interfering with one another. It is one of the advantages of spread spectrum.
3. Which of the following is not a property of spread spectrum techniques?
a) Interference rejection capability
b) Multipath fading
c) Frequency planning elimination
d) Multiple user, multiple access interface
Answer: b
Explanation: Resistance to multipath fading is one of the fundamental reasons for considering spread spectrum systems for wireless communication. Since spread spectrum signals have uniform energy over a very large bandwidth, at any given time only a small portion of the spectrum will undergo fading.
4. Which of the following is not a characteristic of PN sequence?
a) Nearly equal number of 0s and1s
b) Low correlation between shifted version of sequence
c) Non deterministic
d) Low cross-correlation between any two sequences
Answer: c
Explanation: Pseudo-Noise sequences are deterministic in nature. Certain characteristics of PN sequence are nearly equal number of 0s and 1s, very low correlation between shifted versions of the sequence, very low cross correlation between any two sequences.
5. PN sequence can be generated using sequential logic circuits.
a) True
b) False
Answer: a
Explanation: PN sequence is usually generated using sequential logic circuits. When the feedback logic consists of exclusive OR gates, the shift register is called a linear PN sequence generator.
6. The period of a PN sequence produced by a linear m stage shift register cannot exceed _____ symbols.
a) 2m
b) m
c) 2 m
d) 2 m -1
Answer: d
Explanation: There are exactly 2 m -1 non-zero states for an m stage feedback shift register. Thus, the period of a PN sequence produced by a linear m stage shift register cannot exceed 2 m -1.
7. DSSS system spreads the baseband signal by ________ the baseband pulses with a pseudo noise sequence.
a) Adding
b) Subtracting
c) Multiplying
d) Dividing
Answer: c
Explanation: A direct sequence spread spectrum system spreads the baseband data by directly multiplying the baseband data pulses with a pseudo noise sequence. And the pseudo noise sequence is produced by a pseudo noise code generator.
8. Frequency hopping involves a periodic change of transmission _______
a) Signal
b) Frequency
c) Phase
d) Amplitude
Answer: b
Explanation: Frequency hopping involves a periodic change of transmission frequency. A frequency hopping signal is regarded as a sequence of modulated data bursts with time varying, pseudo random carrier frequencies.
9. What is the set of possible carrier frequencies in FH-SS?
a) Hopset
b) Hop
c) Chips
d) Symbols
Answer: a
Explanation: The set of possible carrier frequencies in FH-SS is called hopset. Hopping occurs of a frequency band that includes a number of channels. Each channel is defined as a spectral region with central frequency in the hopset.
10. The bandwidth of the channel used in the hopset is called _________
a) Hopping bandwidth
b) Total hopping bandwidth
c) Instantaneous bandwidth
d) 3 dB bandwidth
Answer: c
Explanation: The bandwidth of a channel used in the hopset is called the instantaneous bandwidth. And the bandwidth of the spectrum over which the hopping occurs is called total hopping bandwidth.
11. The processing gain of FH systems is given by ratio of _______
a) Hopping bandwidth and hopping period
b) Instantaneous bandwidth and hopping duration
c) 3 dB bandwidth and bit rate
d) Total hopping bandwidth and instantaneous bandwidth
Answer: d
Explanation: The processing gain of frequency hopping systems is given by B ss /B. Here, B ss and B denote the total hopping bandwidth and instantaneous bandwidth respectively.
12. FH systems do not have collisions.
a) True
b) False
Answer: b
Explanation: It is possible to have collisions in an FH system where an undesired user transmits in the same channel at the same time as the desired user. Whenever an undesired signal occupies a particular hopping channel in FH, the noise and interference in the channel are translated in frequency.
13. In fast frequency hopping, hopping rate is less than the information symbol rate.
a) True
b) False
Answer: b
Explanation: Fast frequency hopping occurs if there is more than one frequency hop during each transmitted symbol. Thus, in fast frequency hopping the hopping rate equals or exceeds the information symbol rate.
This set of Wireless & Mobile Communications MCQs focuses on “Modulation Performance in Fading and Multipath Channels”.
1. Bit error rate provides the information about the type of error.
a) True
b) False
Answer: b
Explanation: Bit error rate does not provide information about the type of error. However, bit error rate evaluation gives a good indication of the performance of a particular modulation scheme.
2. Which of the following is specified by a specific number of bit errors occurring in a given transmission?
a) Bit error rate
b) Equally likely event
c) Outage event
d) Exhaustive events
Answer: c
Explanation: Outage event is specified by a specific number of bit errors occurring in a given transmission. Evaluating the probability of outage is one of the means to judge the effectiveness of the signalling scheme in a mobile radio channel.
3. Irreducible BER floor is created in frequency selective channels due to ____________
a) Intersymbol interference
b) Random spectral spreading
c) Time varying Doppler spread
d) Blind speed
Answer: a
Explanation: Frequency selective fading is caused due to multipath delay spread which causes intersymbol interference. It results in an irreducible BER floor for mobile systems.
4. Irreducible BER floor is created in non frequency selective channels due to ____________
a) Intersymbol interference
b) Multipath time delay
c) Time varying Doppler spread
d) Blind speed
Answer: c
Explanation: Even if a mobile channel is not frequency selective, the tie varying Doppler spread due to motion creates an irreducible BER floor. It is caused due random spectral spreading.
5. The performance of BPSK is best is term of BER because _______
a) Symbol offset interference does not exist
b) Existence of cross rail interference
c) No multipath delay
d) Doppler spread
Answer: a
Explanation: BER performance of BPSK is best among all the modulation schemes compared. This is because symbol offset interference does not exist in BPSK. Symbol offset interference is also called cross rail interference due to the fact that the eye diagram has multiple rails.
6. High capacity mobile systems are interference limited.
a) True
b) False
Answer: a
Explanation: High capacity mobile systems are interference limited, but they are not noise limited. It was clearly seen that when carrier to interference ratio is large, the errors are primarily due to fading, and interference has very little effect. However, as C/I drops below a certain level, interference dominates the link performance.
7. Which of the following do not impact bit error rate in mobile communication systems?
a) Mobile velocity
b) Channel delay spread
c) Modulation format
d) Base station
Answer: d
Explanation: The mobile velocity, channel delay spread, interference levels and modulation formats all independently impact the raw bit error rate in mobile communication systems. And simulation is a powerful way to design or predict the performance of wireless communication links.
8. Coherence time refers to ____________
a) Time required attaining a call with the busy base station
b) Time required for synchronization between the transmitter and the receiver
c) Minimum time for change in magnitude and phase of the channel
d) None of the mentioned
Answer: b
Explanation: Coherence time is the time required for synchronization between the transmitter and receiver. It is the over which a propagating wave is said to be coherent.
9. Doppler spread refers to _________
a) Signal fading due to Doppler shift in the channel
b) Temporary failure of message transfer
c) Large coherence time of the channel as compared to the delay constraints
d) All of the mentioned
Answer: a
Explanation: Doppler spread refers to signal fading due to Doppler shift in the channel. It is a measure of spectral broadening caused by time rate of change of the mobile radio channel.
10. A rake receiver uses multiple ______
a) Delay circuits
b) Correlators
c) Detectors
d) Flip flops
Answer: b
Explanation: A rake receiver uses multiple correlators to separately detect multiple strongest components. It is designed to counter the effects of multipath fading.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Fundamentals of Equalization”.
1. Which of the following is not used to improve received signal quality over small scale times and distance?
a) Modulation
b) Equalization
c) Diversity
d) Channel coding
Answer: a
Explanation: Equalization, diversity and channel coding are the three techniques which are used to improve received signal quality and link performance over small scale times and distance. But, the approach, cost, complexity and effectiveness varies in wireless communication system.
2. Equalization is used to compensate __________
a) Peak signal to noise ratio
b) Intersymbol interference
c) Channel fading
d) Noises present in the signal
Answer: b
Explanation: Equalization compensate the intersymbol interference created by multipath within time dispersive channels. An equalizer within a receiver compensates for the average range of the expected channel amplitude and delay characteristics.
3. Training and tracking are the operating modes of _________
a) Diversity techniques
b) Channel coding techniques
c) Equalization techniques
d) Demodulation techniques
Answer: c
Explanation: General operating modes of an adaptive equalizer includes training and tracking. A known fixed length training sequence is sent by the transmitter so that the receiver’s equalizer may adapt to a proper setting for minimum bit error rate detection.
4. An equalizer is said to be converged when it is properly _______
a) Trained
b) Tracked
c) Installed
d) Used
Answer: a
Explanation: When an equalizer has been properly trained, it is said to have converged. Equalizers are commonly used in digital communication systems where user data is segmented into short time blocks or time slots.
5. Time for convergence of an equalizer is not a function of _______
a) Equalizer algorithm
b) Equalizer structure
c) Time rate of change of multipath radio channel
d) Transmitter characteristics
Answer: d
Explanation: The timespan over which an equalizer converges is a function of the equalizer algorithm, the equalizer structure and the time rate of change of the multipath radio channel. Equalizers require proper retraining in order to maintain effective ISI cancellation.
6. Equalizer is usually implemented in __________
a) Transmitter
b) Baseband or at IF in a receiver
c) Radio channel
d) Modulator stage
Answer: b
Explanation: An equalizer is implemented at baseband or at IF in a receiver. Since the baseband complex envelope expression can be used to represent bandpass waveform, the channel response, the demodulated signal and adaptive equalizer algorithm are usually simulated and implemented at baseband.
7. Equalizer is ________ of the channel.
a) Opposite
b) Same characteristics
c) Inverse filter
d) Add on
Answer: c
Explanation: An equalizer is actually an inverse filter of the channel. The goal of equalization is to satisfy that the combination of the transmitter, channel and receiver be an all pass channel.
8. ______ controls the adaptive algorithm in an equalizer.
a) Error signal
b) Transmitted signal
c) Received signal
d) Channel impulse response
Answer: a
Explanation: The adaptive algorithm is controlled by the error signal. The error signal is derived by comparing the output of the equalizer and some signal which is either an exact scaled replica of the transmitted signal or represents a property of transmitted signal.
9. The adaptive algorithms in equalizer that do not require training sequence are called ________
a) Linear adaptive algorithms
b) Blind algorithms
c) Non-linear adaptive algorithms
d) Spatially adaptive algorithms
Answer: b
Explanation: Blind algorithms exploit the characteristics of the transmitted signal and do not require training sequence. These type of algorithm are able to acquire equalization through property restoral techniques of transmitted signal.
10. Which of the following is a blind algorithm?
a) Linear adaptive algorithms
b) Constant modulus algorithm
c) Non-linear adaptive algorithms
d) Spatially adaptive algorithms
Answer: b
Explanation: Blind algorithm technique uses algorithms such as the constant modulus algorithm and the spectral coherence restoral algorithm . CMA is used for constant envelope modulation and forces the equalizer weights to maintain a constant envelope on the received signal.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Equalization Techniques”.
1. Equalization techniques can be categorised into _______ and ______ techniques.
a) Linear, non linear
b) Active, passive
c) Direct, indirect
d) Slow, fast
Answer: a
Explanation: Equalization techniques can be classified into linear and non linear techniques. These categories are determined from how the output of an adaptive equalizer is used for subsequent control of the equalizer.
2. Equalization is linear if an analog signal is fed back to change the subsequent outputs of the equalizer.
a) True
b) False
Answer: b
Explanation: If the analog signal is not used in the feedback path to adapt the equalizer, the equalization is linear. On the other hand, if it is fed back to change the subsequent outputs of the equalizer, the equalization is non-linear.
3. In the context of equalizers, LTE stands for ________
a) Long transversal equalizer
b) Least time-varying equalizer
c) Linear transversal equalizer
d) Linear time-varying equalizer
Answer: c
Explanation: The most common equalizer structure used for equalization is linear transveral equalizer . It is made up of tapped delay lines, with the tappings speed a symbol period apart.
4. Which of the following is not a characteristic of FIR filter?
a) Many zeroes
b) Poles only at z=0
c) Transfer function is a polynomial of z-1
d) Many poles
Answer: d
Explanation: Finite impulse response filter has many zeroes but poles only at z=0. The transfer function of the filter is a polynomial of z-1. It is also referred to as transversal filter.
5. Which of the following is not an advantage of lattice equalizer?
a) Simple structure
b) Numerical stability
c) Faster convergence
d) Dynamic assignment
Answer: a
Explanation: The structure of lattice equalizer is more complicated than a linear transversal equalizer. But, numerical stability and faster convergence are two advantage of laatic equalizer. Also, its unique structure allows dynamic assignment of the most effective length of lattice equalizer.
6. Non-linear equalizers are used in applications where channel distortion is not severe.
a) True
b) False
Answer: b
Explanation: Non-linear equalizers are used in applications where the channel distortion is too severe for a linear equalizer to handle. They are most commonly used in practical wireless communication.
7. Which of the following is not a non-linear equalization technique?
a) Decision feedback equalization
b) Maximum likelihood symbol detection
c) Minimum square error detection
d) Maximum likelihood sequence detection
Answer: c
Explanation: Decision feedback equalization, maximum likelihood symbol detection and maximum likelihood sequence detection offers non-linear equalization. They offer improvements over linear equalization techniques and are used in most 2G and 3G systems.
8. For a distorted channel, LTE performance is superior to DFE.
a) True
b) False
Answer: b
Explanation: LTE is well behaved when the channel spectrum is comparatively flat. But if the channel is severely distorted or exhibits null in the spectrum, the performance of LTE deteriorates and MSE of DFE is better than LTE.
9. Which of the following does not hold true for MLSE?
a) Minimizes probability of sequence error
b) Require knowledge of channel characteristics
c) Requires the statistical distribution of noise
d) Operates on continuous time signal
Answer: d
Explanation: Matched filter operates on the continuous time signal, whereas maximum likelihood sequence estimation equalizer and channel estimator rely on discretized samples. MLSE is optimal in the sense that it minimizes the probability of sequence error.
10. MLSE decodes each received signal by itself.
a) True
b) False
Answer: b
Explanation: Rather than decoding each received signal by itself, MLSE tests all possible data sequences by using a channel impulse response simulator within the algorithm. It chooses the data sequence with maximum probability as the output.
This set of Wireless & Mobile Communications Questions and Answers for Campus interviews focuses on “Algorithms for Adaptive Equalization”.
1. Which of the following factor could not determine the performance of algorithm?
a) Structural properties
b) Rate of convergence
c) Computational complexity
d) Numerical properties
Answer: a
Explanation: The performance of an algorithm is determined by various factors. These factors are rate of convergence, computational complexity and numerical properties. The performance of algorithm does not depend on structural properties.
2. Rate of convergence is defined by __________ of algorithm.
a) Time span
b) Number of iterations
c) Accuracy
d) Complexity
Answer: b
Explanation: Rate of convergence is required as number of iterations required for the algorithm to converge close enough to the optimum solution. It enables the algorithm to track statistical variations when operating in non stationary environment.
3. Computational complexity is a measure of ________
a) Time
b) Number of iterations
c) Number of operations
d) Accuracy
Answer: c
Explanation: Computational complexity is the number of operations required to make one complete iteration of the algorithm. It helps in comparing the performance with other algorithms.
4. Choice of equalizer structure and its algorithm is not dependent on ________
a) Cost of computing platform
b) Power budget
c) Radio propagation characteristics
d) Statistical distribution of transmitted power
Answer: d
Explanation: The cost of the computing platform, the power budget and the radio propagation characteristics dominate the choice of an equalizer structure and its algorithm. Battery drain at the subscriber unit is also a paramount consideration.
5. Coherence time is dependent on the choice of the algorithm and corresponding rate of convergence.
a) True
b) False
Answer: a
Explanation: The choice of algorithm and its corresponding rate of convergence depends on the channel data rate and coherence time. The speed of the mobile unit determines the channel fading rate and the Doppler spread, which is directly related to coherence time of the channel.
6. Which of the following is not an algorithm for equalizer?
a) Zero forcing algorithm
b) Least mean square algorithm
c) Recursive least square algorithm
d) Mean square error algorithm
Answer: d
Explanation: Three classic equalizer algorithm are zero forcing algorithm, least mean squares algorithm and recursive least squares algorithm. They offer fundamental insight into algorithm design and operation.
7. Which of the following is a drawback of zero forcing algorithm?
a) Long training sequence
b) Amplification of noise
c) Not suitable for static channels
d) Non zero ISI
Answer: b
Explanation: The zero forcing algorithm has the disadvantage that the inverse filter may excessively amplify noise at frequencies where the folded channel spectrum has high attenuation.
8. Zero forcing algorithm performs well for wireless links.
a) True
b) False
Answer: b
Explanation: ZF is not often used in wireless links as it neglects the effect of noise altogether. However, it performs well for static channels with high SNR, such as local wired telephone links.
9. LMS equalizer minimizes __________
a) Computational complexity
b) Cost
c) Mean square error
d) Power density of output signal
Answer: c
Explanation: LMS equalizer is a robust equalizer. It is used to minimize mean square error between the desired equalizer output and the actual equalizer output.
10. For N symbol inputs, LMS algorithm requires ______ operations per iterations.
a) 2N
b) N+1
c) 2N+1
d) N 2
Answer: c
Explanation: The LMS algorithm is the simplest algorithm. For N symbol inputs, it requires only 2N+1 operations per iteration.
11. Stochastic gradient algorithm is also called ________
a) Zero forcing algorithm
b) Least mean square algorithm
c) Recursive least square algorithm
d) Mean square error algorithm
Answer: b
Explanation: The minimization of the MSE is carried out recursively, and it can be performed by the use of stochastic gradient algorithm. This more commonly called the least mean square algorithm.
12. Convergence rate of LMS is fast.
a) True
b) False
Answer: b
Explanation: The convergence rate of the LMS algorithm is slow. It is slow due to the fact that it uses only one parameter i.e. step size that control the adaptation rate.
13. Which of the following does not hold true for RLS algorithms?
a) Complex
b) Adaptive signal processing
c) Slow convergence rate
d) Powerful
Answer: c
Explanation: Recursive least square algorithm uses fast convergence rate as opposed to LMS algorithms. They are powerful, albeit complex, adaptive signal processing techniques which significantly improves the convergence of adaptive equalizer.
14. Which of the following algorithm uses simple programming?
a) LMS Gradient DFE
b) FTF algorithm
c) Fast Kalman DFE
d) Gradient Lattice DFE
Answer: a
Explanation: Advantages of LMS gradient DFE algorithm are low computational complexity and simple programming. While fast tranversal filter algorithm, Fast Kalman DFE and gradient lattice DFE uses complex programming.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Diversity Techniques”.
1. Diversity requires a training sequence.
a) True
b) False
Answer: b
Explanation: Unlike equalization, diversity requires no training overhead since a training sequence is not required by the transmitter. Diversity is a powerful communication receiver technique that provides wireless link improvement at a relatively low cost.
2. Diversity decisions are made by ____________
a) Receiver
b) Transmitter
c) Channel
d) Adaptive algorithms
Answer: a
Explanation: In virtually all applications, diversity decisions are made by the receiver and are unknown to the transmitter. Diversity exploits the random nature of radio propagation by finding independent signal paths for communication.
3. Small scale fades are characterized by ____________ amplitude fluctuations.
a) Large
b) Small
c) Rapid
d) Slow
Answer: c
Explanation: Small scale fades are characterized by deep and rapid fluctuations. They occur as the mobile system moves over distances of just a few wavelengths. These fades are caused by multiple reflections from the surrounding in the vicinity of the mobile.
4. ____________ is used to prevent deep fade for rapidly varying channel.
a) Modulation
b) Demodulation
c) Macroscopic diversity technique
d) Microscopic diversity technique
Answer: d
Explanation: In order to prevent deep fades from occurring, microscopic diversity techniques can exploit the rapidly changing signal. By selecting the best signal at all times, a receiver can mitigate small scale fading effects.
5. Large scale fading can be mitigated with the help of _________
a) Modulation
b) Demodulation
c) Macroscopic diversity technique
d) Microscopic diversity technique
Answer: c
Explanation: Large scale fading is mitigated with macroscopic diversity techniques. It is done by selecting a base station which is not shadowed when others are, the mobile can improve substantially the average signal to noise ratio.
6. Space diversity s also known as ________
a) Antenna diversity
b) Time diversity
c) Frequency diversity
d) Polarization diversity
Answer: a
Explanation: Space diversity is also known as antenna diversity. It is one of the popular forms of diversity used in wireless communications. Signals received from the spatially separated antenna on the mobile would have essentially uncorrelated envelopes for antenna separation.
7. Which of the following is not a category of space diversity technique?
a) Selection diversity
b) Time diversity
c) Feedback diversity
d) Equal gain diversity
Answer: b
Explanation: Space diversity reception methods can be classified into four categories. They are selection diversity, feedback diversity, maximal ratio combining and equal gain diversity.
8. In selection diversity, the gain of each diversity branch provides different SNR.
a) True
b) False
Answer: b
Explanation: Selection diversity uses m demodulators to provide m diversity branches. Their gain is adjusted to provide the same average SNR for each branch.
9. Polarization diversity uses the ________ as the diversity element.
a) Modulation index
b) Carrier frequency
c) Reflection coefficient
d) Coherence time
Answer: c
Explanation: Decorrelation of the signal in each polarization is caused by multiple reflections in the channel between mobile and base station antenna. Reflection coefficient for each polarization is different, which results in different amplitudes and phases for each reflection.
10. Which of the factor does not determine the correlation coefficient?
a) Polarization angle
b) Cross polarization discrimination
c) Offset angle from the main beam direction
d) Coherence time
Answer: d
Explanation: The correlation coefficient is determined by three factors, polarization angle, offset angle from the main beam direction of the diversity antenna, and the cross polarization discrimination. The correlation coefficient generally becomes higher as offset angle becomes large.
11. Frequency diversity is implemented by transmitting information on more than one ___________
a) Carrier frequency
b) Amplitude
c) Phase
d) Modulation scheme
Answer: a
Explanation: Frequency diversity is implemented by transmitting information on more than one carrier frequency. Frequency diversity is often employed in microwave line of sight links which carry several channels in frequency division multiplex mode.
12. Frequency diversity uses ________ as a diversity element.
a) Correlation coefficient
b) Coherence time
c) Coherence bandwidth
d) SNR
Answer: c
Explanation: The rationale behind the frequency diversity is that frequencies separated by more than the coherence bandwidth of the channel will be uncorrelated. Thus, they will not experience the same fade.
13. Frequency diversity is good for low traffic conditions.
a) True
b) False
Answer: b
Explanation: Frequency diversity is not good for low traffic conditions. This technique has a disadvantage that it not only requires spare bandwidth but also requires that there be as many receivers as there are channels used for frequency diversity. However, for critical traffic, the expense may be justified.
14. Time diversity repeatedly transmits information at time spacings that exceed ___________
a) Coherence bandwidth
b) Dwell time
c) Run time
d) Coherence time
Answer: d
Explanation: Time diversity repeatedly transmits information at time spacings that exceed coherence time of the channel. Thus, multiple repetitions of the signal will be received with independent fading conditions, thereby providing for diversity.
15. In maximal ratio combining, the output SNR is equal to __________
a) Mean of all individual SNRs
b) Maximum of all SNRs
c) Sum of individual SNR
d) Minimum of all SNRs
Answer: c
Explanation: Maximal ratio combining produces an output SNR equal to the sum of the individual SNRs. Thus, it has the advantage of producing an output with an acceptable SNR even when none of the individual signals are themselves acceptable.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Rake Receiver”.
1. In CDMA spread spectrum systems, chip rate is less than the bandwidth of the channel.
a) True
b) False
Answer: b
Explanation: In CDMA spread spectrum systems, the chip rate is typically much greater than the flat fading bandwidth of the channel. Whereas conventional modulation techniques require an equalizer to undo intersymbol interference between adjacent channels.
2. A RAKE receiver collects the __________ versions of the original signal.
a) Time shifted
b) Amplitude shifted
c) Frequency shifted
d) Phase shifted
Answer: a
Explanation: RAKE receiver attempts to collect the time shifted versions of the original signal. It is due to the fact that there is useful information present in the multipath components.
3. RAKE receiver uses separate _________ to provide the time shifted version of the signal.
a) IF receiver
b) Equalizer
c) Correlation receiver
d) Channel
Answer: c
Explanation: RAKE receiver uses separate correlation receivers to provide the time shifted version of the original signal for each of the multipath signal. CDMA receivers combine these time shifted versions of the original signal to transmission in order to improve the signal to noise ratio of the receiver.
4. Each correlation receiver in RAKE receiver is adjusted in ____________
a) Frequency shift
b) Amplitude change
c) Phase shift
d) Time delay
Answer: d
Explanation: Each correlation receiver may be adjusted in time delay, so that a microprocessor controller can cause different correlation receivers to search in different time windows for significant multipath.
5. The range of time delays that a particular correlator can search is called ________
a) Search window
b) Sliding window
c) Time span
d) Dwell time
Answer: a
Explanation: The range of time delays that a particular correlator an search is called a search window. RAKE receiver attempts to collect the time shifted version of the original signal by providing a separate correlation receiver for each of the multipath signal.
6. RAKE receiver is used for _______ technique.
a) CDMA
b) TDMA
c) FDMA
d) OFDM
Answer: a
Explanation: RAKE receiver is essentially a diversity receiver which is used specifically for CDMA. It uses the fact that the multipath components are practically uncorrelated from one another when their relative propagation delays exceed a chip period.
7. A RAKE receiver uses __________ to separately detect the M strongest signals.
a) Single correlator
b) Multiple correlator
c) Single IF receiver
d) Multiple IF receivers
Answer: b
Explanation: A RAKE receiver uses multiple correlators to separately detect the M strongest multipath components. Demodulation and bit decisions are then based on the weighted ouputs of the M correlators.
8. In a RAKE receiver, if the output from one correlator is corrupted by fading, all the other correlator’s output are also corrupted.
a) True
b) False
Answer: b
Explanation: In a RAKE receiver, if the output from one correlator is corrupted by fading, the others may not be. And the corrupted signal may be discounted through weighting process.
9. A RAKE receiver uses __________
a) Equalization
b) Channel coding
c) Diversity
d) Encryption
Answer: c
Explanation: RAKE receiver is a diversity receiver. Diversity is provided by the fact that the multipath components are practically uncorrelated from one another when their relative propagation delays exceed chip period.
10. Interleaving is used to obtain ___________ diversity.
a) Time
b) Frequency
c) Polarization
d) Antenna
Answer: a
Explanation: Interleaving is used to obtain time diversity in a digital communication system without adding any overhead. It provides rapid proliferation of digital speech coders which transform analog voices into efficient digital messages.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Block Codes and Finite Fields”.
1. In block codes, parity bits are ___________ to block of messages.
a) Added
b) Subtracted
c) Multiplied
d) Divided
Answer: a
Explanation: In block codes, parity bits are added to the block of message bits to make codewords or code blocks. They enable a limited number of errors to be detected and corrected without retransmission.
2. How many redundant bits are added in block codes for k information bits and n code bits?
a) n+k
b) n-k
c) k 2
d) n 2
Answer: b
Explanation: In block encoder, k information bits are encoded into n code bits. A total of n-k redundant bits are added to the k information bits for the purpose of detecting and correcting errors.
3. For block codes, rate of the code is defined as __________
a) n 2 /k
b) k 2 /n
c) n/k
d) k/n
Answer: c
Explanation: The block code is referred to as an code, and the rate of the code is defined as R=n/k. It is equal to the rate of information divided by the raw channel rate.
4. The ability of the block code to correct errors is a function of __________
a) Number of parity bits
b) Number of information bits
c) Number of code bits
d) Code distance
Answer: d
Explanation: The ability of a block code to correct errors is a function of the code distance. Block codes can be used to improve the performance of communication systems when other means of improvement are impractical.
5. The weight of code is given by number of _______
a) Non-zero elements in the codeword
b) Zero elements in the codeword
c) Total elements in the codeword
d) Elements in parity bits
Answer: a
Explanation: The weight of a code is given by number of non-zero elements in the codeword. For a binary code, the weight is basically the number of ones in the codeword.
6. Which of the following is not a property of block code?
a) Linearity
b) Systematic
c) Cyclic
d) Non linearity
Answer: d
Explanation: Block codes are linear, systematic and cyclic in nature. Encoding and decoding techniques make use of mathematical constructs known as finite fields.
7. In systematic codes, parity bits are appended at the __________
a) Beginning
b) End
c) End
d) Odd places
Answer: b
Explanation: A systematic code is one in which the parity bits are appended to the end of the information bits. For an code, the first k bits are identical to the information bits, and the remaining bits of each code word are linear combinations of k information bits.
8. Which of the following is not an example of block code?
a) Hamming code
b) Cyclic code
c) Convolution code
d) BCH codes
Answer: c
Explanation: Hamming codes, cyclic codes and BCH codes are the example of block codes. Convolution code does not come in the category of block code. Some other examples of block codes are Reed Solomon codes, Golay codes and Hadamard codes.
9. Which of the following code is a class of non-binary BCH?
a) Hamming code
b) Hadamard code
c) Golay code
d) Reed Solomon codes
Answer: d
Explanation: The most important and most common class of non binary is the family of codes known as Reed Solomon codes. BCH codes are among the most popular block codes that exist for a wide range of rates, achieve significant coding gains.
10. Which of the following linear codes achieve largest possible minimum distance?
a) Hamming code
b) Hadamard code
c) Golay code
d) Reed Solomon codes
Answer: d
Explanation: RS codes achieve the largest possible minimum distance, d min of any linear code. They are non-binary codes which are capable of correcting errors that appears in bursts.
11. CDPD stands for ___________
a) Cellular Digital Packet Data
b) Cellular Decoded Packet Data
c) Cellular Demodulated Packet Data
d) Cellular Decoded Plane Data
Answer: a
Explanation: CDPD is a wide area mobile data service which uses unused bandwidth. It was mostly used in AMPS phones. Reed Solomon code in US CDPD uses m=6 bits per code symbol.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Convolutional Codes”.
1. Block codes can achieve a larger coding gain than convolution coding.
a) True
b) False
Answer: b
Explanation: Convolution code can achieve a larger coding gain that can be achieved using a block coding with the same complexity. Their mapping is highly structured, enabling a decoding method considerably different from block codes.
2. Which of the following indicates the number of input bits that the current output is dependent upon?
a) Constraint length
b) Code length
c) Search window
d) Information rate
Answer: a
Explanation: Constraint length determines the number of input data bits that the current output is dependent upon. The constraint length determines how powerful and complex the code is.
3. Which of the following is not a way to represent convolution code?
a) State diagram
b) Trellis diagram
c) Tree diagram
d) Linear matrix
Answer: d
Explanation: Linear matrix is not a way to represent convolution code. Various ways of representing convolution codes are generator matrix, generator polynomial, logic tables, state diagram, tree diagram and trellis diagram.
4. Which of the following is not an algorithm for decoding convolution codes?
a) Viterbi algorithm
b) Stack algorithm
c) Fano’s sequential coding
d) Ant colony optimization
Answer: d
Explanation: There are a number of techniques for decoding convolution codes. The most important of these methods is Viterbi algorithm. Other decoding algorithms for convolutional codes are Fano’s sequential coding, stack algorithm and feedback coding.
5. Viterbi algorithm performs ____________ decoding of convolutional codes.
a) Maximum likelihood
b) Maximum a posteriori
c) Minimum square
d) Minimum mean square
Answer: a
Explanation: Viterbi algorithm performs maximum likelihood decoding of convolutional codes. The algorithm was first developed by A.J. Viterbi. It is one of the most important algorithm used for decoding convolutional codes.
6. Fano’s algorithm searches all the paths of trellis diagram at same time to find the most probable path.
a) True
b) False
Answer: b
Explanation: Fano’s algorithm searches for the most probable path through the trellis diagram by examining one path at a time. The error rate performance of Fano’s algorithm is comparable to Viterbi’s algorithm.
7. Which of the following is not an advantage of Fano’s algorithm in comparison to Viterbi’s algorithm?
a) Less storage
b) Large constraint length
c) Error rate
d) Small delays
Answer: d
Explanation: In comparison to Viterbi decoding, sequential decoding has a significantly larger delay. In advantage over Viterbi decoding is that it requires less storage, and thus codes with larger constraint lengths can be employed.
8. In comparison to stack algorithm, Fano’s algorithm is simpler.
a) True
b) False
Answer: b
Explanation: In comparison to Fano’s algorithm, the stack algorithm is computationally simpler since there is no retracting over the same path. But stack algorithm requires more storage than Fano’s algorithm.
9. Which of the following is not an error correction and detection code?
a) Block code
b) Convolutional codes
c) Passive codes
d) Turbo codes
Answer: c
Explanation: There are three basic types of error correction and detection codes. They are block codes, convolutional codes and turbo codes. A channel coder operates on digital message data by encoding the source information into a code sequence.
10. Which decoding method involves the evaluation by means of Fano’s algorithm?
a) Maximum Likelihood Decoding
b) Sequential Decoding
c) Maximum a priori
d) Minimum mean square
Answer: b
Explanation: Fano’s algorithm involves sequential decoding. It searches for the most probable path through the trellis by examining one path at a time.
11. In Viterbi’s algorithm, the selected paths are regarded as __________
a) Survivors
b) Defenders
c) Destroyers
d) Carriers
Answer: a
Explanation: In Viterbi’s algorithm, the selected paths are regarded as survivors. The path thus defined is unique and corresponds to the decoded output.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Characteristics of Speech Signals”.
1. The higher the bit rate, the more speech channels can be compressed within a given bandwidth.
a) True
b) False
Answer: b
Explanation: The lower the bit rate at which the coder can deliver toll quality speech, the more speech channels can be compressed within a given bandwidth. Thus, manufacturers are continuously in search of speech coders that provide toll quality speech at lower bit rates.
2. Which of the following are two types of speech coders?
a) Waveform coders and source coders
b) Active coders and passive coders
c) Direst coders and indirect coders
d) Time and frequency coders
Answer: a
Explanation: Speech coders can be categorised into waveform coders and source coders. Waveform coders can further be categorised into time domain and frequency domain. Source coders can be classified into linear predictive coders and vocoders.
3. Waveform coders has _______ complexity and achieves _______ economy in transmission bit rate.
a) Maximum, moderate
b) Maximum, high
c) Minimal, moderate
d) Minimal, high
Answer: c
Explanation: Waveform coders have minimal complexity. This class of coders achieves only moderate economy in transmission bit rate. They are designed to be source independent and hence code equally well a variety of signals.
4. Vocoders has _______ complexity and achieves _______ economy in transmission bit rate.
a) Maximum, moderate
b) Maximum, high
c) Minimal, moderate
d) Minimal, high
Answer: b
Explanation: Vocoders achieve very high economy in transmission bit rate. They are in general more complex. They are based on using a priori knowledge about the signal to be coded, and for this reason, they are signal specific.
5. Which of the following is not a property that is utilized in coder design?
a) Non zero autocorrelation between successive speech signals
b) Non flat nature of speech signal
c) Quasiperiodicity of voiced speech signals
d) Uniform probability distribution of speech amplitude
Answer: d
Explanation: Speech waveforms have a number of useful properties that can be exploited when designing efficient coders. They are non uniform probability distribution of speech amplitude, non-zero autocorrelation between successive speech samples, the nonflat nature of the speech spectra and quasiperiodicity of voiced speech signals.
6. Speech waveforms are _______
a) Bandlimited
b) Bandpass
c) High pass
d) Infinite bandwidth
Answer: a
Explanation: The most basic property of speech waveforms that are exploited by all speech coders is that they are bandlimited. A finite bandwidth means that it can be time-discretized at a finite rate and reconstructed complexity from its samples.
7. Which of the following is not a property of pdf of speech signals?
a) Non uniformity
b) Very high probability of non-zero amplitudes
c) Significant probability of very high amplitudes
d) Increasing function of amplitudes between these extremes
Answer: d
Explanation: There is a non-uniform probability distribution of speech amplitude. The pdf of a speech signal is in general characterized by a very high probability of non-zero amplitudes, a significant probability of very high amplitudes, and a monotonically decreasing function of amplitudes between these extremes.
8. Auto correlation function measures______ between samples of a speech signal as a function of _______
a) Similarity, frequency
b) Dissimilarity, time
c) Similarity, time
d) Dissimilarity, frequency
Answer: c
Explanation: The autocorrelation function gives a quantitative measure of the closeness or similarity between samples of a speech signal as a function of their time separation. In every sample of speech, there is a large component that is easily predicted from the values of the previous samples.
9. Power spectral density of speech is flat.
a) True
b) False
Answer: b
Explanation: There is a nonflat characteristic in power spectral density of speech. It makes it possible to obtain significant compression by coding speech in the frequency domain.
10. Spectral flatness measure is the ratio of ______ and _____
a) Variance, Geometric mean
b) Geometric Mean, Variance
c) Arithmetic mean, geometric mean
d) Geometric mean, arithmetic mean
Answer: c
Explanation: Spectral flatness measure is defined as ratio of arithmetic to geometric mean of the samples of the PSD taken at uniform intervals in frequency. Spectral flatness measure is a qualitative measure of the theoretical maximum coding gain that can be obtained by exploiting the nonflat characteristics of speech spectra.
11. Low frequency signals contribute very little to the total speech signals.
a) True
b) False
Answer: b
Explanation: Lon term averaged PSD’s of speech show that high frequency signals contribute very little to the total speech energy. However, high frequency components are insignificant in energy, they are very important carriers of speech information.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Quantization Techniques”.
1. Quantization is a process of mapping a ________ range of amplitude of a signal into a finite set of __________ amplitudes.
a) Continuous, discrete
b) Discrete, continuous
c) Discrete, discrete
d) Continuous, continuous
Answer: a
Explanation: Quantization is a process of mapping a continuous range of amplitude of a signal into a finite set of discrete amplitudes. Quantizers are the devices that remove the irrelevancies in the signal.
2. For a n bit quantizer, number of levels is equal to __________
a) n
b) 2 n
c) n 2
d) 2n
Answer: b
Explanation: A quantizer that uses n bits can have M = 2n discrete amplitude levels. Amplitude quantization is an important step in any speech coding process, and it determines to a great extent the overall distortion as well as bit rate necessary to represent each waveform.
3. Distortion produced by quantization is directly proportion to ________ and inversely proportional to ______
a) Number of levels, step size
b) Amplitude of the signal, step size
c) Square of step size, number of levels
d) Amplitude of the signal, number of levels
Answer: c
Explanation: The distortion introduced by any quantization operation is directly proportional to the square of the step size. It is inversely proportional to the number of levels for a given amplitude range.
4. Signal to quantization noise ratio of a PCM encoder is given by _________
a) 6.02n + α
b) 3n + 6.02α
c) n + 6.02α
d) 6.02n + α
Answer: d
Explanation: Signal to quantization noise ratio of a PCM encoder is related to the number of bits, n used for encoding. It follows the following relation, SQNR= 6.02n + α. Here, α = 4.77 dB for peak SQNR and α = 0 dB for average SQNR.
5. Companding technique used in the US is called ____________
a) ÎĽ law
b) A law
c) Hybrid companding
d) Direct companding
Answer: a
Explanation: Companding technique known as ÎĽ law is used in US. In Europe, A law companding technique is used.
6. An adaptive quantizer varies its ___________ in accordance to input speech signal power.
a) Level
b) Step size
c) Amplitude
d) Frequency
Answer: b
Explanation: An adaptive quantizer varies its step size in accordance to the input speech signal power. Its characteristics shrink and expand in time like an accordion.
7. Shannon predicted that better performance can be achieved by coding one sample at a time.
a) True
b) False
Answer: b
Explanation: Shannon predicted that better performance can be achieved by coding many samples at a time instead of one sample at a time.
8. ________ is a delayed decision coding technique.
a) Adaptive quantization
b) Uniform quantization
c) Vector quantization
d) Non-uniform quantization
Answer: c
Explanation: Vector quantization is a delayed decision coding technique. It maps a group of input samples called a vector to a code book index. A code book is set up consisting of a finite set of vectors covering the entire anticipated range of values.
9. The characteristics of compressor in ÎĽ-law companding are _________
a) Continuous in nature
b) Logarithmic in nature
c) Linear in nature
d) Discrete in nature
Answer: a
Explanation: In the ÎĽ law companding, the compressor characteristic is continuous. It approximates the linear dependence on x for low input signals and a logarithmic one for high input signals.
10. What is the sequence of operations in PCM?
a) Sampling, quantizing, encoding
b) Quantizing, encoding, sampling
c) Quantizing, sampling, encoding
d) None of the mentioned
Answer: a
Explanation: Sequence of operation in PCM is sampling, quantization and encoding. Sampling and quantizing operations transform an analogue signal to a digital signal. The quantizing and encoding operations are usually performed in the same circuit at the transmitter, which is called an Analogue to Digital Converter . At the receiver end the decoding operation converts the pulse back into an analogue voltage in a Digital to Analogue Converter .
11. In Adaptive Delta Modulation, the slope error reduces and ___________
a) Quantization error decreases
b) Quantization error increases
c) Quantization error remains same
d) None of the mentioned
Answer: b
Explanation: ADM reduces slope error, at the expense of increasing quantizing error. This error can be reduced by using a low-pass filter.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Frequency Domain Coding of Speech”.
1. Frequency domain coders divides the speech signal into __________
a) A set of frequency components
b) A set of different amplitudes
c) A set of time delays
d) A set of phase components
Answer: a
Explanation: In the class of frequency domain coders, the speech signal is divided into a set of frequency components. Each frequency component is quantized and encoded separately.
2. In frequency domain coding of speech, the number of bits used to encode each frequency component is constant.
a) True
b) False
Answer: b
Explanation: Frequency domain coders have an advantage that number of bits used to encode each frequency component can be dynamically varied. They can also be shared among different bands.
3. Quantization is a ________ process.
a) Linear
b) Direct
c) Non-linear
d) Indirect
Answer: c
Explanation: Quantization is a non-linear process. It produces distortion products that are typically broad in spectrum.
4. Sub band coding codes the short time transform of a windowed signal.
a) True
b) False
Answer: b
Explanation: It is function of block transform coding. However, a sub band coder divides the speech signal into many smaller sub bands and encodes each sub band separately according to some perceptual criterion.
5. Which of the following is one of the most frequently used transform in speech coding?
a) Fourier transform
b) Wavelet transform
c) Shearlet transform
d) Discrete cosine transform
Answer: d
Explanation: DCT is one of the most attractive and frequently used transforms for speech coding. Fast algorithms developed for computing the DCT in a computationally efficient manner are used.
6. What does ATC stands for in speech coders?
a) Automatic transform code
b) Air traffic controller
c) Active thermal convection
d) Adaptive transform coding
Answer: d
Explanation: In speech coding, ATC stands for adaptive transform coding. It is form of frequency domain coder that encodes the speech at bit rates in the range of 9.6 kbps and 20 kbps.
7. Waveform coders and Vocoders are the types of ____________
a) Speech coders
b) Modulation technique
c) Frequency translation methods
d) Channel allocation for transmission
Answer: a
Explanation: Speech coders can be classified into waveform coders and Vocoders. Waveform coders convert the analog signal into digital signal. Vocoders exploit the special properties of speech signal to reduce the bit rate.
8. The type of frequency domain coding that divides the speech signal into sub bands is _____
a) Waveform coding
b) Vocoders
c) Block transform coding
d) Sub-band coding
Answer: d
Explanation: Sub band coding is a method where the speech signal is subdivided into several frequency bands and each band is digitally encoded separately. The audible frequency spectrum 20Hz-20 KHz is divided into frequency sub-bands using a bank of finite impulse response filter and output of each filter is sampled and encoded.
9. Speech coders are categorized on the basis of __________
a) Signal compression techniques
b) Frequency of signal
c) Bandwidth of the signal
d) Phase of the signal
Answer: a
Explanation: Speech coders are categorised on the basis of signal compression techniques. Speech coding is an art of compressing and then encoding speech signals.
10. The speech coding technique that is dependent on the prior knowledge of the signal is __________
a) Waveform coders
b) Vocoders
c) Sub band coding
d) Block transform
Answer: b
Explanation: Vocoders are dependent on the prior knowledge of the signals. They capture the characteristic elements of audio signal and then uses this characteristic signal to affect other audio signals.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Vocoders”.
1. Vocoders analyse the speech signals at ______
a) Transmitter
b) Receiver
c) Channel
d) IF Filter
Answer: a
Explanation: Vocoders are a class of speech coding systems. They analyse the speech signal at the transmitter. And then transmit the parameters derived from the analysis.
2. Vocoders __________ the voice at the receiver.
a) Analyse
b) Synthesize
c) Modulate
d) Evaluate
Answer: b
Explanation: Vocoders synthesize the voice at the receiver. All vocoder systems attempt to model the speech generation process as a dynamic system and try to quantify certain physical constraints of the system.
3. Vocoders are simple than the waveform coders.
a) True
b) False
Answer: b
Explanation: Vocoders are much more complex than the waveform coders. They can achieve very high economy in transmission bit rate but are less robust.
4. Which of the following is not a vocoding system?
a) Linear predictive coder
b) Channel vocoder
c) Waveform coder
d) Formant vocoder
Answer: c
Explanation: Waveform coder is not a vocoding system. LPC is the most popular vocoding system. Other vocoding systems are channel vocoder, formant vocoder, cepstrum vocoder etc.
5. Which of the following pronunciations lead to voiced sound?
a) ‘f’
b) ‘s’
c) ‘sh’
d) ‘m’
Answer: d
Explanation: Voiced sounds are ‘m’, ‘n’ and ‘v’ pronounciations. They are a result of quasiperiodic vibrations of the vocal chord.
6. Speech signal can be categorised in _____ and ______
a) Voiced, unvoiced
b) Active, passive
c) Direct, indirect
d) Balanced, unbalanced
Answer: a
Explanation: Speech signal is of two types, voiced and unvoiced. Voiced sound is a result of quasiperiodic vibrations of the vocal chord. Unvoiced signals are fricatives produced by turbulent air flow through a constriction.
7. Channel vocoders are the time domain vocoders.
a) True
b) False
Answer: b
Explanation: Channel vocoders are frequency domain vocoders. They determine the envelope of the speech signal for a number of frequency bands and then sample, encode and multiplex these samples with the encoded outputs of the other filters.
8. ________ is often called the formant of the speech signal.
a) Pitch frequency
b) Voice pitch
c) Pole frequency
d) Central frequency
Answer: c
Explanation: The pole frequencies correspond to the resonant frequencies of the vocal tract. They are often called the formants of the speech signal. For adult speakers, the formants are centered around 500 Hz, 1500 Hz, 2500 Hz and 3500 Hz.
9. Formant vocoders use large number of control signals.
a) True
b) False
Answer: b
Explanation: Formant vocoders use fewer control signals. Therefore, formant vocoders can operate at lower bit rates than the channel vocoder. Instead of transmitting the power spectrum envelope, formant vocoders attempt to transmit the position of peaks of spectral envelope.
10. Cepstrum vocoder uses __________
a) Wavelet transform
b) Inverse wavelet transform
c) Cosine transform
d) Inverse Fourier transform
Answer: d
Explanation: Cepstrum vocoders use inverse Fourier transform. It separates the excitation and vocal tract spectrum by Fourier transforming spectrum to produce the cepstrum of the signal.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Linear Predictive Coders”.
1. Linear predictive coders belong to _______ domain class of vocoders.
a) Time
b) Frequency
c) Direct
d) Indirect
Answer: a
Explanation: Linear predictive vocoders belong to the time domain class of vocoders. This class of vocoders attempts to extract the significant features of the speech from the time waveform.
2. Linear predictive coders are computationally simple.
a) True
b) False
Answer: b
Explanation: Linear predictive coders are computationally intensive. But, they are the most popular among the class of low bit vocoders. With LPC, it is possible to transmit good quality voice at 4.8 kbps and poorer quality voice at even lower rates.
3. Linear predictive coding system models the vocal tract as __________ linear filter.
a) Pole and zero
b) All zero
c) All pole
d) No pole
Answer: c
Explanation: The linear predictive coding system models the vocal tract as an all pole linear filter. The excitation to this filter is either a pulse at the pitch frequency or random white noise depending on whether the speech segment is voiced or unvoiced.
4. Linear predictive vocoders use __________ to estimate present sample.
a) Weighted sum of past samples
b) Multiplication of past samples
c) One past sample
d) Do not use past samples
Answer: a
Explanation: The linear predictive coder uses a weighted sum of p past samples. Using this technique, the current sample can be written as linear sum of the immediately precoding samples.
5. Which of the following LPC uses code book?
a) Multiple excited LPC
b) Residual excited LPC
c) LPC Vocoders
d) Code excited LPC
Answer: d
Explanation: Code excited LPC uses code book. In this method, the coder and decoder have a predetermined code book of stochastic excitation signals.
6. How many past samples are used by linear predictive coders to estimate present sample?
a) 100-150
b) 10-15
c) 1
d) 1000-1100
Answer: b
Explanation: LPCs uses weighted sum of past p samples to estimate the present samples. The number of past samples used by linear predictive coders ranges from 10 to 15.
7. Which of the non-linear transform is generally used to improve the coding of reflection coefficient?
a) Long area ratio transform
b) Mutual information
c) Least square
d) Interpolation
Answer: a
Explanation: Long area ratio transform is generally used to improve the coding of reflection coefficient. This non linear transformation reduces the sensitivity of reflection coefficients to quantization errors. LAR performs an inverse hyperbolic tangent mapping of reflection coefficients.
8. Which of the following LPC uses two sources at the receiver?
a) Multiple excited LPC
b) Residual excited LPC
c) LPC Vocoders
d) Code excited LPC
Answer: c
Explanation: LPC vocoder uses two sources at the receiver, one of white noise and the other with a series of pulses at the current pitch rate. The selection of either of these excitation methods is based on voiced/unvoiced decision made at the transmitter.
9. Which of the following LPC produces a buzzy twang in the synthesized speech?
a) Multiple excited LPC
b) Residual excited LPC
c) LPC Vocoders
d) Code excited LPC
Answer: c
Explanation: LPC vocoder requires that the transmitter extract pitch frequency information which is often very difficult. Moreover, the phase coherence between the harmonic components of the excitation pulse tends to produce a buzzy twang in the synthesized speech.
10. The problem of buzzy twang in synthesized speech is mitigated by multipulse excited LPC or code excited LPC.
a) True
b) False
Answer: a
Explanation: LPC vocoder produces buzzy twang in the synthesized speech due to phase coherence between the harmonic components of the excitation pulses. This problem is mitigated by multipulse excited or code excited LPC.
11. Multipulse excited LPC requires pitch detection.
a) True
b) False
Answer: b
Explanation: Multipulse excited LPC does not require pitch detection and the prediction residual is better approximated by several pulses per pitch period. This is the reason for better speech quality.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Speech Codecs”.
1. The choice of speech coder does not depend on cell size used.
a) True
b) False
Answer: b
Explanation: The choice of speech coder depends on the cell size used. When the cell size is sufficiently small such that high spectral efficiency is achieved through frequency reuse, it may be sufficient to use a simple high rate speech codec.
2. Which of the following is an important factor in determining spectral efficiency of the system?
a) Multiple access technique
b) Cell size
c) Modulation technique
d) Vocoder
Answer: a
Explanation: The type of multiple access technique used is an important factor in determining the spectral efficiency of the system. It strongly influences the choice of speech codec.
3. The type of modulation does not affect the choice of speech codec.
a) True
b) False
Answer: b
Explanation: The type of modulation employed has a considerable impact on the choice of speech codec. Using bandwidth efficient modulation scheme can lower the bit rate reduction requirements on the speech codec and vice versa.
4. Which of the following is the name of original speech coder used in the pan European digital cellular standard GSM?
a) Multipulse excited codec
b) Residual excited codec
c) Regular pulse excited long term prediction
d) Code excited codec
Answer: c
Explanation: The original speech coder used in the pan European digital cellular standard GSM goes by a rather grandiose name of regular pulse excited long term prediction codec. This codec has a bit rate of 13 kbps.
5. Which of the following is true for baseband RELP codec?
a) Good quality of speech, low complexity
b) Good quality of speech, high complexity
c) Bad quality of speech, low complexity
d) Bad quality of speech, high complexity
Answer: a
Explanation: The advantage of baseband RELP codec is that it provides good quality speech at low complexity. The speech quality is sometimes limited due to tonal noise introduced by the process of high frequency generation.
6. Which of the following is true for MPE-LTP codec?
a) Good quality of speech, low complexity
b) Good quality of speech, high complexity
c) Bad quality of speech, low complexity
d) Bad quality of speech, high complexity
Answer: b
Explanation: The MPE-LTP technique produces excellent speech quality at high complexity. It is not much affected by bit errors present in the channel.
7. RPE-LTP codec combines the advantage of RELP codec and CELP codec.
a) True
b) False
Answer: b
Explanation: The RPE-LTP codec combines the advantages of the earlier French proposed RELP codec with those of the multipulse excited long term prediction codec proposed by Germany.
8. Which of the following codec is used by IS-136?
a) Residual Excited Linear Predictive Coders
b) Multipulse Excited LPC
c) LPC Vocoders
d) Vector sum excited LPC
Answer: d
Explanation: The US digital cellular system, IS-136 uses a vector sum excited linear predictive coder . This coder operates at a raw data rate of 7950 bits/s and a total data rate of 13 kbps after channel coding.
9. VSELP speech coder is a variant of ___________
a) CELP
b) MPE_LTP
c) RELP
d) RPE-LTP
Answer: a
Explanation: The VSELP speech coder is a variant of the CELP type vocoders. The code books in the VSELP encoder are organised with a predefined structure such that a brute-force search is avoided.
10. Which of the following is true for VSELP?
a) Low speech quality, modest computational complexity, robust to channel errors
b) Highest speech quality, low computational complexity, channel errors
c) Highest speech quality, high computational complexity, robust to channel errors
d) Highest speech quality, modest computational complexity, robust to channel errors
Answer: d
Explanation: VSELP speech coder is designed to accomplish the three goals of highest speech quality, modest computational complexity and robustness to channel errors. The code books used by VSELP impart high speech quality and increased robustness to channel errors.
11. What is DAM in speech coding system?
a) Diagnostic Acceptability Measure
b) Digital Acceptability Measure
c) Diagnostic Accessibility Measure
d) Digital Accessibility Measure
Answer: a
Explanation: The diagnostic acceptability measure is used in speech coding system. It is used for evaluation of acceptability of speech coding systems.
12. ________ exaggerates the bit errors originally received at the base station.
a) Non linear transformation
b) Tandem signalling
c) Large cell size
d) Complex vocoders
Answer: b
Explanation: Tandem signalling tends to exaggerate the bit errors originally received at the base station. Tandem signalling is difficult to protect against but is an important evaluation criterion in the evaluation of speech coders.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Introduction to Multiple Access”.
1. Multiple access schemes are used to allow ________ mobile users to share simultaneously a finite amount of radio spectrum.
a) Many
b) One
c) Two
d) Ten-Fifteen
Answer: a
Explanation: Multiple access schemes are used to allow many mobile users to share simultaneously a finite amount of radio spectrum. The sharing of spectrum is required to achieve high capacity by simultaneously allocating the available bandwidth to multiple users.
2. The technique that makes possible the task of listening and talking in communication system is called ________
a) Simplexing
b) Duplexing
c) Modulating
d) Multiple access technique
Answer: b
Explanation: In conventional telephone systems, it is possible to talk and listen simultaneously. This effect is called duplexing and is generally required in wireless telephone systems.
3. Frequency division duplexing provides ________ distinct bands of frequencies for _________ user.
a) Two, two
b) One, two
c) Two, one
d) Two, many
Answer: c
Explanation: Frequency division duplexing provides two distinct bands of frequencies for every user. In FDD, any duplex channel actually consists of two simplex channels.
4. The forward band in FDD provides traffic from the mobile to base station.
a) True
b) False
Answer: b
Explanation: The forward band in FDD provides traffic from the base station to the mobile. The reverse band provides traffic from the mobile to the base station.
5. The frequency separation between each forward and reverse channel changes throughout the system.
a) True
b) False
Answer: b
Explanation:The frequency separation between each forward and reverse channel is constant throughout the system. It is regardless of the particular channel being used. A device called a duplexer is used inside each subscriber unit and base station to allow simultaneous bidirectional radio transmission.
6. Time division duplexing uses ________ to provide both a forward and reverse link.
a) Frequency
b) Time
c) Time and frequency
d) Cell spacing
Answer: b
Explanation: Time division duplexing uses time instead of frequency to provide both a forward and reverse link. In TDD, multiple users share a single radio channel by taking turns in the time domain.
7. TDD is effective for _________
a) Fixed wireless access and users are stationary
b) Dynamic wireless access and users are stationary
c) Fixed wireless access and users are moving
d) Dynamic wireless access and users are moving
Answer: a
Explanation: TDD is effective for fixed wireless access when all users are stationary. This makes the propagation delay does not vary in time among the users. Because of rigid timing required for time slotting, TDD generally is limited to cordless phone or short range potable access.
8. In wideband systems, the transmission bandwidth of a single channel _________ coherence bandwidth of the channel.
a) Equal to
b) Not related to
c) Larger than
d) Smaller than
Answer: c
Explanation: In wideband systems, the transmission bandwidth of a single channel is much larger than the coherence bandwidth of the channel. Thus, multipath fading does not greatly vary the received signal power within a wideband channel.
9. In narrowband system, the channels are usually operated using TDD.
a) True
b) False
Answer: b
Explanation: In narrowband system, channels are usually operated using FDD. To minimize interference between forward and reverse links on each channel, the frequency separation is made as great as possible within the frequency spectrum.
10. Narrowband FDMA allows users to share the same radio channel allocating a unique time slot to each user.
a) True
b) False
Answer: b
Explanation: In narrowband FDMA, a user is assigned a particular channel which is not shared by other users in the vicinity. However narrowband TDMA allows the users to share the same radio channel allocating a unique time slot to each user.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Frequency Division Multiple Access ”.
1. Frequency division multiple access assigns ______ channels to _______ users.
a) Individual, individual
b) Many, individual
c) Individual, many
d) Many, many
Answer: a
Explanation: Frequency division multiple access assigns individual channels to individual users. Each user is allocated a unique frequency band or channel. These channels are assigned on demand to users who request service.
2. During the period of call, other users can share the same channel in FDMA.
a) True
b) False
Answer: b
Explanation: In FDMA systems, no other user can share the same channel during the period of call. In FDD systems, the users are assigned a channel as a pair of frequencies; one is used for the forward channel while the other frequency is used for the reverse channel.
3. The FDMA channel carries ____________ phone circuit at a time.
a) Ten
b) Two
c) One
d) Several
Answer: c
Explanation: The FDMA channel carries one phone circuit at a time. Each individual band or channel is wide enough to accommodate the signal spectra of the transmissions to be propagated.
4. If the FDMA channel is not in use, it can be used by other users.
a) True
b) False
Answer: b
Explanation: If an FDMA channel is not in use, it sits idle and cannot be used by other users to increase or share capacity. It is essentially a wasted resource.
5. The bandwidth of FDMA channel is ______
a) Wide
b) Narrow
c) Large
d) Zero
Answer: b
Explanation: The bandwidth of FDMA channels is relatively narrow as each channel supports only one circuit per carrier. That is, FDMA is usually implemented in narrow band systems.
6. The symbol time in FDMA systems is _________ thus intersymbol interference is ______
a) Large, high
b) Small, low
c) Small, high
d) Large, low
Answer: d
Explanation: The symbol time of a narrowband signal is large as compared to the average delay spread. This implies that the amount of intersymbol interference is low and, thus, little or no equalization is required in FDMA narrowband systems.
7. Due to _________ transmission scheme __________ bits are needed for overhead in FDMA systems.
a) Continuous, few
b) Discontinuous, few
c) Continuous, many
d) Discontinuous, many
Answer: a
Explanation: Since FDMA is a continuous transmission scheme, fewer bits are needed for overhead purposes as compared to TDMA.
8. Which of the following is not true for FDMA systems as compared to TDMA systems?
a) Low complexity
b) Lower cell site system cost
c) Tight RF filtering
d) Narrow bandwidth
Answer: b
Explanation: FDMA systems have higher cell site system costs as compared to TDMA systems. It is due to single channel per carrier design, and the need to use costly bandpass filters to eliminate spurious radiation at the base station.
9. __________ is undesired RF radiation.
a) Intermodulation frequency
b) Intermediate frequency
c) Instantaneous frequency
d) Instrumental frequency
Answer: a
Explanation: Intermodulation frequency is undesired RF radiation which can interfere with other channels in the FDMA systems. The nonlinearities cause signal spreading in the frequency domain and generate IM frequency.
10. __________ is based on FDMA/FDD.
a) GSM
b) W-CDMA
c) Cordless telephone
d) AMPS
Answer: d
Explanation: The first US analog cellular system, the Advanced Mobile Phone System is based on FDMA/FDD. A single user occupies a single channel while the call is in progress.
11. In US AMPS, 416 channels are allocated to various operators with 10 kHz guard band and channel between them is 30 kHz. What is the spectrum allocation given to each operator?
a) 12.5 kHz
b) 30 kHz
c) 12.5 MHz
d) 30 MHz
Answer: c
Explanation: Spectrum allocated to each cellular operator is 12.5 MHz. As B t = NB c + 2B guard ; which is equal to 416*30*10 3 +2(10*10 3 ) = 12.5 MHz.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Time Division Multiple Access ”.
1. TDMA systems transmit in a continuous way.
a) True
b) False
Answer: b
Explanation: TDMA systems transmit data in a buffer and burst method. Thus, the transmission for any user is not continuous.
2. Preamble contains __________
a) Address
b) Data
c) Guard bits
d) Trail bits
Answer: a
Explanation: TDMA frame is made up of a preamble, an information message and the trail bits. In a TDMA frame, the preamble contains the address and synchronization information that both the base station and the subscribers use to identify each other.
3. __________ are utilized to allow synchronization of the receivers between different slots and frames.
a) Preamble
b) Data
c) Guard bits
d) Trail bits
Answer: c
Explanation: Guard times are utilized to allow synchronization of the receivers between different slots and frames. TDMA/FDD systems intentionally induce several time slots of delay between the forward and reverse time slots for a particular user.
4. Which of the following is not true for TDMA?
a) Single carrier frequency for single user
b) Discontinuous data transmission
c) No requirement of duplexers
d) High transmission rates
Answer: a
Explanation: TDMA share a single carrier frequency with several users, where each user makes use of non-overlapping time slots. The number of time slots per frame depends on several factors, such as modulation technique, available bandwidth etc.
5. Because of _______ transmissions in TDMA, the handoff process in __________
a) Continuous, complex
b) Continuous, simple
c) Discontinuous, complex
d) Discontinuous, simple
Answer: d
Explanation: Because of discontinuous transmissions in TDMA, the handoff process is much simpler for a subscriber unit, since it is able to listen for other base stations during idle time slots.
6. __________ synchronization overhead is required in TDMA due to _______ transmission.
a) High, burst
b) High, continuous
c) Low, burst
d) No, burst
Answer: a
Explanation: High synchronization overhead is required in TDMA systems because of burst transmissions. TDMA transmissions are slotted, and this requires the receivers to be synchronized for each data burst.
7. TDMA allocates a single time per frame to different users.
a) True
b) False
Answer: b
Explanation: TDMA has an advantage that it can allocate different numbers of time slots per frame to different users. Thus, bandwidth can be supplied on demand to different users by concatenating or reassigning time slots based on priority.
8. ___________ of TDMA system is a measure of the percentage of transmitted data that contains information as opposed to providing overhead for the access scheme.
a) Efficiency
b) Figure of merit
c) Signal to noise ratio
d) Mean
Answer: a
Explanation: Efficiency of TDMA system is a measure of the percentage of transmitted data that contains information as opposed to providing overhead for the access scheme. The frame efficiency is the percentage of bits per frame which contain transmitted data.
9. A TDMA system uses 25 MHz for the forward link, which is broken into radio channels of 200 kHz. If 8 speech channels are supported on a single radio channel, how many simultaneous users can be accommodated?
a) 25
b) 200
c) 1600
d) 1000
Answer: d
Explanation: For a TDMA system that uses 25 MHz for the forward link, which is broken into radio channels of 200 kHz. If 8 speech channels are supported on a single radio channel, 1000 simultaneous users can be accommodated as N = /.
10. What is the time duration of a bit if data is transmitted at 270.833 kbps in the channel?
a) 270.833 s
b) 3 ÎĽs
c) 3.692 ÎĽs
d) 3.692 s
Answer: c
Explanation: If data is transmitted at 270.833 kbps in the channel, the time duration of a bit will be 3.692 ÎĽs, as T b = = 3.692 ÎĽs.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Spread Spectrum Multiple Access”.
1. SSMA uses signals which have a transmission bandwidth that is smaller than the minimum required RF bandwidth.
a) True
b) False
Answer: b
Explanation: Spread spectrum multiple access uses signals which have a transmission bandwidth that is several orders of magnitude greater than the minimum required RF bandwidth.
2. PN sequence converts _______ signal to ______ signal.
a) Narrowband, wideband
b) Wideband, narrowband
c) Unmodulated, modulated
d) Low frequency, high frequency
Answer: a
Explanation: A pseudo-noise sequence converts a narrowband signal to a wideband noise like signal before transmission. SSMA provides immunity to multipath interference and robust multiple access capability.
3. SSMA is bandwidth efficient when used with a single user.
a) True
b) False
Answer: b
Explanation: SSMA is not very bandwidth efficient when used by a single user. However, since many users can share the same spread spectrum bandwidth without interfering with one another, spread spectrum systems become bandwidth efficient in a multiple user environment.
4. ___________ is a digital multiple access system in which carrier frequencies are varied in pseudorandom order.
a) CDMA
b) FCDMA
c) FHMA
d) SDMA
Answer: c
Explanation: Frequency hopped multiple access is a digital multiple access system in which the carrier frequencies of the individual users are varied in a pseudorandom fashion within a wideband channel.
5. If the rate of change of the carrier frequency is greater than the symbol rate, then the system is referred as ___________
a) Fast frequency hopping system
b) Slow frequency hopping system
c) Time division frequency hopping system
d) Code division multiple access system
Answer: a
Explanation: If the rate of change of the carrier frequency is greater than the symbol rate, then the system is referred as a fast frequency hopping system. If the channel changes at a rate less than or equal to the symbol rate, it is called slow frequency hopping.
6. A frequency hopped system does not provide security.
a) True
b) False
Answer: b
Explanation: A frequency hopped system provides a level of security, especially when a large number of channels are used. Since, an unintended receiver that does not know the pseudorandom sequence of frequency slots must retune rapidly to search for the signal it wishes to intercept.
7. All users in CDMA system uses __________ carrier frequency.
a) Different
b) Two
c) Ten
d) Same
Answer: d
Explanation: All users in CDMA system use the same carrier frequency and may transmit simultaneously. Each user has its own pseudorandom codeword which is approximately orthogonal to all other codewords.
8. ___________ problem occurs when many mobile users share the same channel.
a) Near-far
b) Activation
c) Line of sight
d) Windowing
Answer: a
Explanation: The near far problem occurs when many mobile users share the same channel. In general, the strongest received mobile signal will capture the demodulator at a base station.
9. In CDMA, symbol duration is ___________ than channel delay spread.
a) Equal
b) Greater
c) Slightly greater
d) Much less
Answer: d
Explanation: In CDMA, symbol duration is very short and usually much less than the channel delay spread. Thus, channel data rates are very high in CDMA systems.
10. ____________ is used to improve reception by collecting time delayed versions of the required signal.
a) RAKE receiver
b) Equalizer
c) Frequency modulator
d) High pass filter
Answer: a
Explanation: A RAKE receiver can be used to improve reception by collecting time delayed versions of the required signals. Since PN sequences have low autocorrelation, multipath which is delayed by more than a chip appear as noise.
11. ____________ arises from the fact that the spreading sequences of different users are not orthogonal.
a) Near-far problem
b) Line of sight
c) Windowing
d) Self jamming
Answer: d
Explanation: Self jamming arises from the fact that the spreading sequences of different users are not orthogonal. Hence, in the despreading of a particular PN code, non zero contributions to the receiver decision statistic for the desired user arise from the transmissions of other users in the system.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Packet Radio”.
1. In ___________ technique, many subscribers attempt to access a single channel in an uncoordinated manner.
a) Packet radio
b) Multiple access
c) Modulation
d) Spread spectrum
Answer: a
Explanation: In packet radio access techniques, many subscribers attempt to access a single channel in an uncoordinated manner. Transmission is done by using bursts of data.
2. Packet radio uses _________ and __________ signals for perfect feedback.
a) Forward, reverse
b) ACK, NACK
c) Active, passive
d) Linear, non linear
Answer: b
Explanation: Packet radio uses ACK and NACK signals for perfect feedback. The ACK signal indicates an acknowledgment of a received burst and NACK indicates that the previous burst was not received correctly by base station.
3. Packet radio has high spectral efficiency.
a) True
b) False
Answer: b
Explanation: Packet radio multiple access is very easy to implement, but has low spectral efficiency and may induce delays. The subscriber uses a contention technique to transmit on a common channel.
4. ___________ is defined as average number of messages successfully transmitted per unit time in packet radio multiple access technique.
a) Average delay
b) Figure of merit
c) Throughput
d) Efficiency
Answer: c
Explanation: The performance of contention technique can be evaluated by the throughput , which is defined as the average number of messages successfully transmitted per unit time, and the average delay experienced by a typical message burst.
5. What is the time period during which the packets are susceptible to collisions with transmissions from other users?
a) Delay time
b) Latency period
c) Average delay time
d) Vulnerable period
Answer: d
Explanation: Vulnerable period is the time period during which the packets are susceptible to collisions with transmissions from other users. It is used in order to determine the throughput.
6. For a PR multiple access technique, packet transmissions occur with __________ distribution.
a) Poisson
b) Gaussian
c) Pearson
d) Rayleigh
Answer: a
Explanation: For a PR multiple access technique, packet transmissions occur with Poisson distribution. To study packet radio protocols, it is assumed that all packets sent by all users have a constant packet length and fixed channel rate.
7. If λ is mean arrival rate and τ is the packet duration, then traffic occupancy of a packet radio network is equal to _________
a) λτ
b) λ/τ
c) λ-τ
d) λτ 2
Answer: a
Explanation: If λ is mean arrival rate and τ is the packet duration, then traffic occupancy or throughput of a packet radio network is equal to λτ. The unit of R is in Erlangs.
8. Pure ALOHA is __________
a) Modulation technique
b) Multiple access technique
c) Random access technique
d) Spread spectrum technique
Answer: c
Explanation: The pure ALOHA protocol is a random access protocol used for data transfer. A user accesses a channel as soon as a message is ready to be transmitted.
9. For pure ALOHA protocol, the vulnerable period is ___________ the packet duration.
a) Double
b) Equal
c) Ten times
d) Not equal
Answer: a
Explanation: For the pure ALOHA protocol, the vulnerable period is double the packet duration. After a transmission, the user waits for an acknowledgement on either the same channel or a separate feedback channel.
10. The vulnerable period of slotted ALOHA is _________ packet duration.
a) Two
b) Ten
c) One
d) Three
Answer: c
Explanation: The vulnerable period of slotted ALOHA is one packet duration. Since, partial collisions are prevented through synchronization.
11. In slotted ALOHA, time is divided into different length of time slots.
a) True
b) False
Answer: b
Explanation: In slotted ALOHA, time is divided into equal time slots of length greater than the packet duration. The subscribers each have synchronized clocks and transmit a message only at the beginning of a new time slot.
12. CSMA stands for _________
a) Carrier sense multiple access
b) Code sense multiple access
c) Carrier sense modulation access
d) Carry sense multiple access
Answer: a
Explanation: CSMA stands for carrier sense multiple access. CSMA protocol is based on the fact that each terminal on the network is able to monitor the status of the channel before transmitting information.
13. __________ is the time required for a terminal to sense whether or not the channel is idle.
a) Propagation delay
b) Average delay
c) Detection delay
d) Time delay
Answer: c
Explanation: In CSMA protocols, detection delay and propagation delay are two important parameters. Detection delay is a function of the receiver hardware and is the time required for a terminal to sense whether or not the channel is idle.
14. Which of the following is based on time division multiplexing?
a) Slotted ALOHA
b) Pure ALOHA
c) CSMA
d) Reservation ALOHA
Answer: d
Explanation: Reservation ALOHA is a packet access scheme based on time division multiplexing. In this protocol, certain packet slots are assigned with priority, and it is possible for users to reserve slots for the transmission of packets.
15. Discrete packet time technique, PRMA stands for _________
a) Packet reservation multiple access
b) Photo reflector multiple access
c) Proton reflector modulating access
d) Packet reflection multiple access
Answer: a
Explanation: PRMA stands for Packet Reservation Multiple Access. PRMA uses a discrete packet time technique similar to reservation ALOHA and combines the cyclical frame structure of TDMA in a manner that allows each TDMA time slot to carry either voice or data, where voice is the priority.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Capacity of Cellular Systems”.
1. ____________ is the maximum number of channels that can be provided in a fixed frequency band.
a) Channel capacity
b) Radio capacity
c) Spectral capacity
d) Carrier capacity
Answer: a
Explanation: Channel capacity for a radio system can be defined as the maximum number of channels or users that can be provided in a fixed frequency band. It is useful in computer science, in electrical engineering, and in other disciplines evaluating the capacity of a channel or conduit.
2. Which of the following measures spectrum efficiency of a wireless system?
a) Channel capacity
b) Radio capacity
c) Spectral capacity
d) Carrier capacity
Answer: b
Explanation: Radio capacity is a parameter which measures spectrum efficiency of a wireless system. This parameter is determined by the required carrier to interference ratio and the channel bandwidth.
3. If D is the distance between co-channel cells and R be the cell radius, co-channel reuse ratio is given by __________
a) D*R
b) D 2 /R
c) D/R
d) D/R 2
Answer: c
Explanation: Let, D is the distance between co-channel cells and R be the cell radius. Then the minimum ratio of D/R that is required to provide a tolerable level of co-channel interference is called the co-channel reuse ratio.
4. __________ is the interference at a base station receiver that comes from the subscriber units in the surrounding cells.
a) Forward channel interference
b) Carrier interference
c) Receiver interference
d) Reverse channel interference
Answer: d
Explanation: In a cellular system, the interference at a base station receiver will come from the subscriber units in the surrounding cells. This is called reverse channel interference.
5. In practice, TDMA systems improve capacity by ____________ compared to analog cellular systems.
a) Three to six times
b) Equal capacity
c) Two time
d) Ten to twenty times
Answer: a
Explanation: In practice, TDMA systems improve capacity by a factor of three to six times as compared to analog cellular radio systems. Powerful error control and speech coding enable better link performance in high interference environment.
6. __________ allows subscribers to monitor neighbouring base stations.
a) TDMA
b) MAHO
c) FDMA
d) ACA
Answer: b
Explanation: Mobile assisted handoff allows subscribers to monitor the neighbouring base stations, and the best base station choice may be made by each subscriber. MAHO allows deployment of densely packed microcells, thus giving substantial capacity gains in a system.
7. Time division multiple access introduces ACA, which stands for ________
a) Acute carrier allocation
b) Adaptive carrier allocation
c) Adaptive channel allocation
d) Acute channel allocation
Answer: c
Explanation: TDMA makes it possible to introduce adaptive channel allocation . ACA eliminates system planning since it is not required to plan frequencies for cells.
8. Which of the following standard had not adopted digital TDMA for high capacity?
a) USDC
b) GSM
c) PDC
d) IS-95
Answer: d
Explanation: Various proposed standards such as the GSM, US digital cellular , and Pacific Digital Cellular have adopted digital TDMA for high capacity. IS-95 does not use TDMA.
9. Capacity of CDMA is bandwidth limited.
a) True
b) False
Answer: b
Explanation: The capacity of CDMA systems is interference limited. It is bandwidth limited in FDMA and TDMA. Therefore, any reduction in the interference will cause a linear increase in the capacity of CDMA.
10. In a CDMA system, link performance for each user __________ as the number of users _______
a) Increase, decrease
b) Increase, increase
c) Decrease, increase
d) Decrease, decrease
Answer: a
Explanation: In a CDMA system, the link performance for each user increases as the number of users decreases. A straightforward way to reduce interference is to use multisectorized antennas, which results in spatial isolation of users.
11. Capacity of CDMA can be increased by operating in DTX, which stands for _______
a) Discrete transmission mode
b) Discrete transmission modulation
c) Discontinuous transmission mode
d) Digital transmission mode
Answer: c
Explanation: Capacity of CDMA can be increased by operating in DTX, which stands for Discontinuous transmission mode. It takes advantage of the intermittent nature of speech. In DTX, the transmiiter is turned off during the periods of silence in speech.
12. Frequency reuse factor for CDMA system is ___________
a) One
b) Two
c) Zero
d) Ten
Answer: a
Explanation: The amount of out-of-cell interference determines the frequency reuse factor, f, of a CDMA cellular system. Ideally, each cell shares the same frequency and the maximum possible value of f is achieved.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Wireless and Fixed Telephone Networks”.
1. In a wireless communication, base station is connected to central hub called _______
a) PSTN
b) MSC
c) CO
d) PBX
Answer: b
Explanation: To provide wireless communication within a particular geographic region, an integrated network of base stations must be deployed to provide sufficient radio coverage to all mobile users. The base station, in turn must be connected to a central hub called MSC .
2. PSTN stands for ________
a) Public switched telephone network
b) Personal switched telephone network
c) Personal switched telephone node
d) Public switched telephone node
Answer: a
Explanation: PSTN stands for public switched telephone network. The PSTN forms the global telecommunication grid which connects conventional telephone centers with MSCs throughout the world.
3. MSCs provide connectivity between PSTN and the base stations.
a) True
b) False
Answer: a
Explanation: The MSC provide connectivity between the public switched telephone network and the numerous base stations. It ultimately provides connectivity between all of the wireless subscribers in the system.
4. Communication protocol, CAI stands for ___________
a) Common air interchange
b) Control air interchange
c) Common air interface
d) Control air interchange
Answer: c
Explanation: To connect mobile subscribers to the base stations, radio links are established using a carefully defined communication protocol called common air interface which in essence is a precisely defined handshake communication protocol.
5. At the base station, the air interface portion of mobile transmission is passed to MSC.
a) True
b) False
Answer: b
Explanation: At the base station, the air interface portion of the mobile transmission is discarded, and the remaining voice traffic is paased along to the MSC on fixed networks.
6. PSTN is ___________ and wireless networks are ________
a) Highly dynamic, virtually static
b) Static, virtually static
c) Highly dynamic, virtually dynamic
d) Virtually static, highly dynamic
Answer: d
Explanation: The network configurations in the PSTN are virtually static, since the network connections may only be changed when a subscriber changes residence. Wireless networks are highly dynamic with the network configuration being rearranged everytime a new subscriber moves into the coverage region of a different base station.
7. In public switched telephone network, LATA stands for ______
a) Local access and transport area
b) Land area and transport area
c) Local access and telephone access
d) Local area and telephone access
Answer: a
Explanation: In public switched telephone network, LATA stands for Local access and transport area. LATA is a city or a geographic grouping of towns in PSTN.
8. LATAs are connected by a company called _________
a) Land exchange carrier
b) Local exchange carrier
c) Local control exchange
d) Land area exchange
Answer: b
Explanation: Surrounding LATAs are connected by a company called a local exchange carrier . A LEC is a company that provides telephone services and may be a local telephone company.
9. A long distance telephone company that collects toll is called _________
a) LATA
b) LEC
c) PSTN
d) IXC
Answer: d
Explanation: A long distance telephone company collects toll fees to provide connections between different LATAs over its long distance network. These companies are referred to as interexchange carriers .
10. Wireless networks are extremely hostile and random nature of radio channel.
a) True
b) False
Answer: a
Explanation: A problem unique to wireless networks is the extremely hostile and random nature of the radio channel, and since users may request service from any physical location while traveling over a wide range of velocities.
11. The technique for separate but parallel signalling channel is called ________
a) Common channel signalling
b) Forward channel signalling
c) Reverse channel signalling
d) Separate channel signalling
Answer: a
Explanation: During the mid 1980s, the PSTN was transformed into two parallel networks- one dedicated to user traffic, and one dedicated to call signalling traffic. This technique is called common channel signalling.
12. In North America, the cellular telephone signalling network uses _______
a) SS7
b) IXC
c) IS-41
d) PSTN
Answer: a
Explanation: In North America, the cellular telephone signalling network uses No. 7 Signalling System , and each MSC uses the IS-41 protocol to communicate with other MSCs on the continent.
This set of Wireless & Mobile Communications Questions and Answers for Entrance exams focuses on “Development of Wireless Networks”.
1. First generation cellular and cordless telephone networks are based on _________
a) Analog technology
b) Digital technology
c) Active technology
d) Passive technology
Answer: a
Explanation: First generation cellular and cordless telephone networks are based on analog technology. All first generation cellular systems use FM modulation, and cordless telephones use a single base station to communicate with a single portable terminal.
2. Which of the following is true for first generation wireless systems?
a) Efficient
b) Digital technology
c) High data rate
d) Low rate
Answer: d
Explanation: First generation wireless systems provide analog speech and inefficient, low rate, data transmission between the base station and mobile user.
3. Which of the following network protocol is used by US cellular carriers to automatically accommodate subscribers who roam into their coverage regions?
a) IS-36
b) IS-121
c) IS-41
d) IS-14
Answer: c
Explanation: US cellular carriers implemented the network protocol standard IS-41 to allow different cellular systems to automatically accommodate subscribers who roam into their coverage region.
4. IS-41 does not rely on autonomous registration.
a) True
b) False
Answer: b
Explanation: IS-41 relies on a feature of AMPS called autonomous registration. Autonomous registration is a process by which a mobile notifies a serving MSC of its presence and location.
5. MSC distinguishes home users from roaming users based on MIN.
a) True
b) False
Answer: a
Explanation: The MSC is able to distinguish home users from roaming users based on the MIN of each active user, and maintains a real time user list in the home location register and visitor location register .
6. The visited system creates a __________ record for each roamer and notifies home system to update __________
a) HLR, VLR
b) VLR, MIN
c) MIN, ESN
d) VLR, HLR
Answer: d
Explanation: The visited system creates a VLR record for each new roamer and notifies the home system via IS-41 so it can update its own HLR. IS-41 allows the MSCs of neighboring systems to automatically handle the registration and location validation of roamers so that users no longer need to manually register as they travel.
7. Which of the following is not an example of second generation wireless networks?
a) GSM
b) CT2
c) AMPS
d) PACS
Answer: c
Explanation: AMPS is an example of first generation wireless networks. Second generation wireless systems employ digital modulation and advanced call processing capabilities. Example includes GSM, CT2, PACS and DECT.
8. In second generation wireless networks, the handoff process is called __________
a) MAHO
b) Soft handoff
c) Hard handoff
d) Inter system handoff
Answer: a
Explanation: In second generation wireless networks, the handoff process is mobile controlled and is known as mobile assisted handoff . The mobile units in these networks perform several other functions not performed by first generation subscriber units.
9. __________ is an example of second generation cordless telephone standard which allows each cordless phone to communicate with any number of base stations.
a) PACS
b) CT2
c) GSM
d) DECT
Answer: d
Explanation: DECT is an example of second generation cordless telephone standard which allows each cordless phone to communicate with any number of base stations. In DECT, the base stations have greater control in terms of switching, signalling, and controlling handoffs.
10. In PACS/ WACS, the BSC is called a _________
a) Radio port control unit
b) HLR
c) VLR
d) MIN
Answer: a
Explanation: In PACS/ WACS, the BSC is called a radio port control unit. This architectural change has allowed the data interface between the base station controller and the MSC to be standardized.
11. FPLMTS is recently known by the name _________
a) GSM
b) ISDN
c) IMTS-2000
d) DECT
Answer: c
Explanation: FPLMTS is more recently been called International Mobile Telecommunication . It implies emerging third generation wireless systems for hand held devices.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Traffic Routing in Wireless Networks”.
1. The type of traffic carried by a network determines the routing service.
a) True
b) False
Answer: a
Explanation: The type of traffic carried by a network determines the routing services. It also determines protocols, and call handling techniques which must be employed.
2. Connection oriented services are also called __________
a) Datagram services
b) Virtual circuit routing
c) Connectionless services
d) Routing service
Answer: b
Explanation: Connection oriented services are also called virtual circuit routing. In connection oriented routing, the communications path between the message source and destination is fixed for the entire duration of the message.
3. Connectionless services are also called __________
a) Datagram services
b) Virtual circuit routing
c) Connection oriented services
d) Routing service
Answer: a
Explanation: Connectionless services are also called datagram services. In a connectionless service, a call set up procedure is not required at the beginning of a call, and each message burst is treated independently by the network.
4. First generation cellular systems provide connectionless services for each voice user.
a) True
b) False
Answer: b
Explanation: First generation cellular systems provide connection oriented services for each voice user. Voice channels are dedicated for users at the serving base station.
5. Wireless data networks are not well supported by ___________
a) Datagram services
b) Circuit switching
c) Connectionless services
d) Routing service
Answer: b
Explanation: Wireless data networks are not supported by circuit switching. It is due to their short, bursty transmissions which are often followed by periods of inactivity. Circuit switching is best suited for dedicated voice only traffic.
6. Packet switching breaks each message into smaller units.
a) True
b) False
Answer: a
Explanation: Packet switching breaks each message into smaller units for transmission and recovery. When a message is broken into packets, a certain amount of control information is added to each packet to provide source and destination identification, as well as error recovery provisions.
7. Header specifies the ___________ of a new packet.
a) Ending
b) Middle part
c) Beginning
d) Data information
Answer: c
Explanation: The header specifies the beginning of a new packet. It contains the source address, packet sequence number, and other routing and billing information.
8. The control field contains the ________
a) CRC
b) User data
c) Address
d) ARQ
Answer: d
Explanation: The control field defines functions such as transfer of acknowledgments, automatic repeat requests , and packet sequencing. The information field contains user data. And the final field is the frame check sequence field or the CRC .
9. Circuit switching provides greater efficiency in comparison to circuit switching.
a) True
b) False
Answer: b
Explanation: In contrast to circuit switching, packet switching provides excellent channel efficiency for bursty data transmission of short length. An advantage of packet switched data is that the channel is utilized only when sending or receiving bursts of information.
10. __________ was developed by CCITT to provide standard connectionless network access protocols for three layers of OSI model.
a) Packet switching
b) Datagram routing
c) X.25
d) CDPD
Answer: c
Explanation: X.25 was developed by CCITT to provide standard connectionless network access protocols for the three layers of the open systems interconnection model.
11. X.25 protocols specify particular data rates.
a) True
b) False
Answer: b
Explanation: The X.25 protocol does not specify particular data rates or how packet switched networks are implemented. Rather, X.25 provides a series of standard functions and formats which give structure to the design of the software.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Wireless Data Services”.
1. US cellular standard CDPD stands for _________
a) Cellular Digital Packet Data
b) Cellular Digital Packet Data
c) Cellular Digital Pocket Data
d) Cellular Discrete Pocket Data
Answer: a
Explanation: US cellular standard CDPD stands for Cellular Digital Packet Data. In 1993, US cellular industry developed CDPD to coexist with the conventional voice only cellular system.
2. CDPD is data service for third generation US cellular systems.
a) True
b) False
Answer: b
Explanation: CDPD is a data service for first and second generation US cellular systems and uses a full 30 kHz AMPS channel on a shared basis. CDPD provides mobile packet data connectivity to existing data networks.
3. Which of the following is true for CDPD?
a) Expensive to install
b) Uses different infrastructure
c) Complex
d) Inexpensive to install
Answer: d
Explanation: CDPD directly overlays with existing cellular infrastructure and uses existing base station equipment, making it simple and inexpensive to install. CDPD does not use the MSC, but rather has its own traffic routing capabilities.
4. CDPD channel is _________
a) Simplex
b) Complex
c) Duplex
d) Expensive
Answer: c
Explanation: Each CDPD channel is duplex in nature. The forward channel serves as a beacon and transmits data from the PSTN side of the network, while the reverse channel links all mobile users to the CDPD network and serves as the access channel for each subscriber.
5. CDPD transmissions are carried out using ________ blocks.
a) Fixed length
b) Variable length
c) Long
d) Short
Answer: a
Explanation: CDPD transmissions are carried out using fixed length blocks. User data is protected using a Reed-Solomon block code with 6 bit symbols.
6. MDLP is _________
a) Layer protocol used in CDPD
b) Layer protocol used in GSM
c) Layer protocol used in ARDIS
d) Layer protocol used in CDMA
Answer: a
Explanation: MDLP is a lower layer protocol of CDPD. It is used to convey information between data link layer entities across the CDPD air interface.
7. The layer protocol, RRMP stands for _________
a) Radio Resource Management Protocol
b) Radio Resource Mobile Protocol
c) Radio Rate Management Protocol
d) Radio Rate Mobile Protocol
Answer: a
Explanation: The radio resource management protocol is a higher, layer 3 protocol used to manage the radio channel resources of the CDPD system and enables an M-ES to find and utilize a duplex radio channel without interfering with standard voice services.
8. ARDIS is a public network service.
a) True
b) False
Answer: b
Explanation: ARDIS is a private network service provided by Motorola and IBM. It is based on MDC 4800 and RD-LAP protocols developed at Motorola.
9. RMD is a _______ and ________ data service.
a) Private, two way
b) Public, one way
c) Public, two way
d) Private, one way
Answer: c
Explanation: RAM Mobile Data is a public, two way data service based upon the Mobitex protocol developed by Ericsson. RAM provides street level coverage for short and long messages for users moving in an urban environment.
10. The MSC provides subscriber access to the PSTN via the _______
a) Signalling transfer points
b) Service management system
c) Database service management system
d) Switching end points
Answer: d
Explanation: The MSC provides subscriber access to the PSTN via the switching end points . The SEP implements a stored program control switching system known as the service control point .
11. ________ controls the switching of messages between nodes in the CCS network.
a) Signalling transfer points
b) Service management system
c) Database service management system
d) Switching end points
Answer: a
Explanation: The signalling transfer point controls the switching of messages between nodes in the CCS network. For higher reliability of transmission, SEPs are required to be connected to the SS7 network via at least two STPs.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Integrated Services Digital Network ”.
1. ISDN stands for ___________
a) Integrated Services Digital Network
b) Integrated Services Discrete Network
c) Integrated Services Digital Node
d) Integrated Services Discrete Node
Answer: a
Explanation: ISDN stands for integrated services digital network. It defines the dedicated signalling network that has been created to complement the PSTN for more flexible and efficient network access and signalling.
2. ISDN is based on the concept of __________
a) SS7
b) CCS
c) ARDIS
d) CDPD
Answer: b
Explanation: ISDN is a complete network framework designed around the concept of common channel signalling. While telephone users throughout the world rely on the PSTN to carry conventional voice traffic, new end-user data and signalling services can be provided with a parallel, dedicated signalling network.
3. Signalling component that supports traffic between the end-user and network is called ________
a) Network signalling
b) Digital subscriber signalling
c) Access signalling
d) Subscriber system signalling
Answer: c
Explanation: Access signalling supports traffic between the end user and the network. It defines how end users obtain access to the PSTN and the ISDN for communication or services.
4. Access signalling is governed by the SS7 suite of protocols.
a) True
b) False
Answer: b
Explanation: The second signalling component of ISDN, network signalling is governed by the SS7 suite of protocols. For wireless communication systems, the SS7 protocols within ISDN are critical to providing backbone network connectivity between MSCs throughout the world.
5. Information bearing channels in ISDN are called ______
a) D channels
b) Data channels
c) B channels
d) Voice channels
Answer: c
Explanation: The ISDN interface is divided into three different types of channels. Information bearing channels called bearer channels are used exclusively for end user traffic .
6. ISDN provides integrated end user access to only packet switched networks.
a) True
b) False
Answer: b
Explanation: ISDN provides integrated end-user access to both circuit switched and packet switched networks with digital end-to-end connectivity.
7. PRI interface in ISDN serves small capacity terminals.
a) True
b) False
Answer: b
Explanation: ISDN end users may select between two different interfaces. The BRI is intended to serve small capacity terminals while the PRI is intended for large capacity terminals.
8. Which of the following is based on ATM technology?
a) SS7
b) CCS
c) ARDIS
d) B-ISDN
Answer: d
Explanation: Emerging networking technique, broadband ISDN is based on asynchronous transfer mode technology. It allows packet switching rates up to 2.4 Gbps and total switching capacities as high as 100 Gbps.
9. Which of the following is true for ATM?
a) Circuit switched
b) Multiple access technique
c) Multiplexing technique
d) Handle only voice users
Answer: c
Explanation: ATM is a packet switching and multiplexing technique which has been specifically designed to handle both voice users and packet data users in a single physical channel.
10. ATM supports unidirectional transfer of data.
a) True
b) False
Answer: b
Explanation: ATM supports bidirectional transfer of data packets of fixed length between two end points. It preserves the order of transmission.
11. The data unit of ATM is ___________
a) Cell
b) Atom
c) Molecule
d) Packet
Answer: a
Explanation: The data unit of ATM is cell. They are routed based on header information in each unit that identifies the cell as belonging to a specific ATM virtual connection.
12. ATM cells have fixed length of __________
a) 48 bytes
b) 47 bytes
c) 5 bytes
d) 53 bytes
Answer: d
Explanation: ATM cells have a fixed length of 53 bytes. It consists of 48 bytes of data and 5 bytes of header information. Fixed length packets result in simple implementation of fast packet switches.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Signalling System No. 7 ”.
1. Which of the following is not a service offered by SS7?
a) Touchstar
b) 800 services
c) Alternate billing services
d) 400 services
Answer: d
Explanation: There are three main types of services offered by the SS7 network. They are the Touchstar, 800 services, and alternate billing services.
2. Which of the following service is called CLASS?
a) Touchstar
b) 800 services
c) Alternate billing services
d) 400 services
Answer: a
Explanation: Touchstar is also known as CLASS. It is a group of switch controlled services that provide its users with certain call management capabilities.
3. Which of the following service of SS& provides call return, call forwarding?
a) Touchstar
b) 800 services
c) Alternate billing services
d) 400 services
Answer: a
Explanation: Touchstar is one of the three main type of services offered by SS7. It provides services such as call return, call forwarding, repeat dialling, call block, call tracing and caller ID.
4. Alternating billing service uses CCS network.
a) True
b) False
Answer: a
Explanation: Alternate billing service and line information database use the CCS network to enable the calling party to bill a call to a personal number from any number.
5. SS7 was first developed by _______
a) ITU
b) Ericsson
c) CCITT
d) Motorola
Answer: c
Explanation: SS7 is an outgrowth of the out of band signalling first developed by the CCITT under common channel signalling standard, CCS no. 6.
6. Which of the layer of OSI is associated with NSP of SS7?
a) All layers
b) Upper three
c) Lowest three
d) Middle two
Answer: c
Explanation: The lowest three layers of the OSI model are handled in SS7 by the network service part of the protocol. NSP is made up of three message transfer parts and the signalling connection control part of the SS7 protocol.
7. The function of _______ is to transfer and deliver signalling network.
a) MTP
b) CCS
c) ARDIS
d) CDPD
Answer: a
Explanation: The function of the MTP is to ensure that signalling traffic can be transferred and delivered reliably between the end users and the network. MTP is provided at three levels.
8. Signalling data functions are provided by __________
a) MTP level 1
b) MTP level 2
c) MTP level 3
d) TCAP
Answer: a
Explanation: Signalling data link functions provide an interface to the actual physical channel over which communication takes place. Physical channels may include copper wire, twisted pair, fibre, mobile radio and are transparent to the higher layers.
9. Signalling data functions are provided by __________
a) MTP level 1
b) MTP level 2
c) MTP level 3
d) TCAP
Answer: b
Explanation: Signalling data link functions are provided by MTP level 2. They correspond to the second layer in the OSI reference model and provide a reliable link for the transfer of traffic between two directly connected signalling points.
10. _________ provide procedures that transfer messages between signalling nodes.
a) MTP level 1
b) MTP level 2
c) MTP level 3
d) TCAP
Answer: c
Explanation: Signalling network functions provide procedures that ransfer messages between signalling nodes. As in ISDN, there are two types of MTP level 3 functions: signalling message handling and signalling network management.
11. Which of the following does not include in SS7 user part?
a) MTP
b) ISUP
c) TCAP
d) OMAP
Answer: a
Explanation: The SS7 user part includes the ISDN user part , the transaction capabilities application part and operations maintenance and administration part . The telephone user part and the data user part are included in the ISUP.
12. The TCAP part in SS7 refers to ________ layer of OSI.
a) Physical
b) Network
c) Data Link
d) Application
Answer: d
Explanation: The transaction capabilities application part in SS7 refers to the application layer which invokes the services of the SCCP and the MTP in a hierarchical format. TCAP is concerned with remote operations.
This set of Wireless & Mobile Communications online test focuses on “Personal Communication Services/Networks ”.
1. PCS/PCN provides only wired communication.
a) True
b) False
Answer: b
Explanation: The objective of personal communication systems or personal communication networks is to provide ubiquitous wireless communications coverage, enabling users to access telephone network and Internet.
2. Concept of PCS/PCN is based on _________
a) Advanced intelligent network
b) Artificial intelligent network
c) CDPD
d) SS7
Answer: a
Explanation: The concept of PCS/PCN is based on an advanced intelligent network . The mobile and fixed networks will be integrated to provide universal access to the network and its databases.
3. AIN has different telephone numbers for both wireline and wireless services.
a) True
b) False
Answer: b
Explanation: AIN allow its users to have a single telephone number to be used for both wireless and wireline services.
4. Circuit switching has more advantage than packet switching for PCS.
a) True
b) False
Answer: b
Explanation: Packet switching technology will have more advantages for PCS/PCN than circuit switching. PCS is required to serve a wide range of services including voice, data, e-mail and digital video.
5. ______ is used for transmission of packets in the cellular switched architecture.
a) Packet switching techniques
b) Circuit switching techniques
c) Packet and circuit switched technique
d) Datagram technique
Answer: a
Explanation: Packet switching techniques are used for transmission of packets in the cellular switched architecture. Packet switching is attractive for wireless networks because the addresses and other information in packet headers make it possible for dispersed network elements to respond to a mobile user without intervention central controllers.
6. Which of the following has the function to accept information from PSTN?
a) TIU
b) WIU
c) BIU
d) CIU
Answer: a
Explanation: The function of TIU is to accept information from the PSTN. TIU acts as the physical layer and transforms the standard format of the PSTN into the wireless access physical layer.
7. PTI is the address of ____________
a) TIU
b) WIU
c) BIU
d) CIU
Answer: a
Explanation: PTI is the address of TIU. It is the address from where he call is originated.
8. _______ is the information contained in the packet header of TIU.
a) PTI
b) VCI
c) PRMA
d) CDPD
Answer: b
Explanation: VCI is the information contained in the packet header of TIU. It is used to identify the route through which the transmission will take place.
9. Transmission protocol, PRMA stands for _________
a) Pocket reservation multiple access
b) Packet register multiple access
c) Pocket register multiple access
d) Packet reservation multiple access
Answer: d
Explanation: Transmission protocol, PRMA stands for packet reservation multiple access. PRMA is a time division multiplex based multiple access protocol that allows a group of spatially dispersed terminals to transmit packet voice and low bit rate data over a common channel.
10. UMTS stands for ____________
a) Universal mobile telecommunication system
b) Universal mobile telephone system
c) United multiplex telecommunication system
d) Universal mobile telecommunication system
Answer: d
Explanation: UMTS stands for Universal mobile telecommunication system. It is a system that is capable of providing a variety of mobile services to a wide range of global mobile communication standards.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “AMPS and ETACS”.
1. US cellular telephone system, AMPS stands for __________
a) Analog Mobile Phone System
b) Analog Modulation Packet System
c) Analog Machine Precision System
d) Analog Mobile Precision System
Answer: a
Explanation: AMPS stands for Analog Mobile Phone System. It was first deployed in late 1983 in urban and suburban areas of Chicago by Ameritech.
2. Who developed the first US cellular telephone system called AMPS?
a) Motorola
b) Ericsson
c) AT & T Bell Laboratories
d) ETSI
Answer: c
Explanation: In the late 1970s, AT & T Bell Laboratories developed the first US cellular telephone system called the Advanced Mobile Phone System . In 1983, a total of 40 MHz of spectrum in the 800 MHz band was located by the Federal Communication Commission for AMPS.
3. The AMPS system uses a ___________ cell reuse pattern.
a) One
b) Five
c) Three
d) Seven
Answer: d
Explanation: The AMPS system uses a seven cell reuse pattern. It also provides sectoring and cell splitting to increase capacity when needed.
4. Which of the following is virtually identical to AMPS?
a) ETACS
b) GSM
c) CDMA
d) IS-54
Answer: a
Explanation: ETACS is virtually identical to AMPS. It was developed in middle 1980s and stands for European Total Access Communication System.
5. AMPS and ETACS use ___________ and _____ for radio transmission.
a) FM, TDD
b) FM, FDD
c) AM, TDD
d) FM, FDD
Answer: b
Explanation: Like all other first generation, analog, cellular systems, AMPS and ETACS use frequency modulation and frequency division duplex for radio transmission.
6. US AMPS system provides 42 control channel for each of the two service providers.
a) True
b) False
Answer: b
Explanation: In the US AMPS system, there are twenty one control channels for each of the two service providers in each market. However, ETACS supports forty two control channels for a single provider.
7. In each cellular market, the non wireline service provider is assigned an odd SID.
a) True
b) False
Answer: a
Explanation: In each cellular market, the non wireline service provider is assigned an odd SID . The wireline provider is assigned an even SID.
8. ETACS use ___________ instead of SID .
a) Area identification number
b) Analog identification number
c) Digital identification number
d) Dual identification number
Answer: a
Explanation: For ETACS, area identification numbers are used instead of SIDs. ETACS subscriber units are able to access any control or voice channel in the standard.
9. _________ allows base and mobile to distinguish each other from co-channel users located in different cells.
a) SAT
b) VMAC
c) MAC
d) FVC
Answer: a
Explanation: The SAT has one of the three different frequencies which allow the base and mobile to distinguish each other from co channel users located in different cells.
10. AMPS and ETACS use same physical channels for transmission of voice and control channels.
a) True
b) False
Answer: b
Explanation: AMPS and ETACS use different physical rate channels for transmission of voice and control channels. A control channel is used by each base station in the system to simultaneously page subscriber units to alert them of incoming calls.
11. To increase the capacity of AMPS, Motorola developed ____________
a) P-AMPS
b) N-AMPS
c) H-APMS
d) R-AMPS
Answer: b
Explanation: To increase capacity in large AMPS markets, Motorola developed an AMPS like system called N-AMPS in 1991. N-AMPS did not become widespread as 2G digital technologies displaced many of the original FM analog systems.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “United States Digital Cellular ”.
1. The dual mode USDC/AMPS system was standardized as ____________
a) IS-54
b) IS-136
c) PSTN
d) GSM
Answer: a
Explanation: The dual mode USDC/AMPS system was standardized as Interim Standard 54 . It was standardized by Electron Industries Association and Telecommunication Industry Association in 1990. It was later upgraded to IS-136.
2. USDC is also known as ___________
a) SADC
b) NADC
c) PADC
d) N-AMPS
Answer: b
Explanation: USDC is also known as North American Digital Cellular . It has been installed in North America in the countries like Canada and Mexico.
3. USDC was designed to share different frequencies as compared to AMPS.
a) True
b) False
Answer: b
Explanation: The USDC system was designed to share the same frequencies, frequency reuse plan, and base stations as AMPS, so that base stations and subscriber units could be equipped with both AMPS and USDC channels within the same piece of equipment.
4. Because of the compatibility with AMPS, USDC is also known as ___________
a) SADC
b) PADC
c) D-AMPS
d) K-AMPS
Answer: c
Explanation: USDC maintains compatibility with AMPS in a number of ways. Therefore, USDC is also known as D-AMPS .
5. USDC forward and reverse control channels use exactly the same signalling techniques as AMPS.
a) True
b) False
Answer: a
Explanation: To maintain compatibility with AMPS phones, USDC forward and reverse control channels use exactly the same signalling techniques as AMPS. USDC voice channels use 4-ary π/4 DQPSK modulation with a channel rate of 48.6 kbps.
6. USDC has ___________ as many control channels as AMPS.
a) Same
b) Three times
c) Twice
d) Four times
Answer: c
Explanation: USDC has twice as many control channels as AMPS. In addition to the forty two primary AMPS control channels, USDC specifies forty two additional control channels called the secondary control channels.
7. Which of the following data channels of USDC carries the user information?
a) DTC
b) CVDCC
c) SACCH
d) FACCH
Answer: a
Explanation: DTC is the most important data channel as far as the end user is concerned. It carries user information .
8. Which of the following channel is similar in functionality to the SAT used in AMPS?
a) DTC
b) CDVCC
c) SACCH
d) FACCH
Answer: b
Explanation: CDVCC is a 12 bit message sent in every time slot and is similar in functionality to the SAT used in AMPS. The CDVCC is an 8 bit number ranging between 1 and 255, which is protected with four additional channel coding bits from a shortened Hamming code.
9. Which of the following channel in USDC provides a signalling channel?
a) DTC
b) CVDCC
c) SACCH
d) FACCH
Answer: c
Explanation: The SACCH is sent in every time slot and provides a signalling channel in parallel with the digital speech. The SACCH carries various control and supervisory messages between the subscriber unit and the base station.
10. Which of the following channel in USDC is used to send important control or specialized traffic data between the base station and mobile units?
a) DTC
b) CVDCC
c) SACCH
d) FACCH
Answer: d
Explanation: FACCH is used to send important control or specialized traffic data between the base station and mobile units. The FACCH data, when transmitted, takes the place of user information data within a frame.
11. Which of the following speech coder is used in USDC?
a) Vector Sum Excited Linear Predictive coder
b) Formant Vocoder
c) Multipulse Excited Linear Predictive coder
d) Residual Excited Linear Predictive coder
Answer: a
Explanation: The USDC speech coder is called the Vector Sum Excited Linear Predictive coder . This belongs to the class of Code Excited Linear Predictive coders or Stochastically Excited Linear Predictive coders.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Global System for Mobile ”.
1. Which of the following is the world’s first cellular system to specify digital modulation and network level architecture?
a) GSM
b) AMPS
c) CDMA
d) IS-54
Answer: a
Explanation: GSM was the world’s first cellular system to specify digital modulation and level architectures and services. It is the world’s most popular 2G technology. It was developed to solve the fragmentation problems of the first cellular systems in Europe.
2. Previously in 1980s, GSM stands for ____________
a) Global system for mobile
b) Groupe special mobile
c) Global special mobile
d) Groupe system mobile
Answer: b
Explanation: In the mid-1980s GSM was called by the name Groupe special mobile. In 1992, GSM changed its name to Global System for Mobile Communication for marketing reasons.
3. Who sets the standards of GSM?
a) ITU
b) AT & T
c) ETSI
d) USDC
Answer: c
Explanation: The setting of standards for GSM is under the aegis of the European Technical Standards Institute . GSM task was to specify a common mobile communication system for Europe in the 900 MHZ band.
4. Which of the following does not come under the teleservices of GSM?
a) Standard mobile telephony
b) Mobile originated traffic
c) Base originated traffic
d) Packet switched traffic
Answer: d
Explanation: GSM services follow ISDN guidelines and are classified as either teleservices or data services. Teleservices include standard mobile telephony and mobile originated or base originated traffic.
5. Which of the following comes under supplementary ISDN services?
a) Emergency calling
b) Packet switched protocols
c) Call diversion
d) Standard mobile telephony
Answer: c
Explanation: Supplementary ISDN services are digital in nature. They include call diversion, closed user groups, and caller identification, and are not available in analog mobile networks. Supplementary services also include short messaging service .
6. Which of the following memory device stores information such as subscriber’s identification number in GSM?
a) Register
b) Flip flop
c) SIM
d) SMS
Answer: c
Explanation: SIM is a memory device that stores information such as the subscriber’s identification number, the networks and countries where the subscriber is entitled to service, privacy keys, and other user specific information.
7. Which of the following feature makes impossible to eavesdrop on GSM radio transmission?
a) SIM
b) On the air privacy
c) SMS
d) Packet switched traffic
Answer: b
Explanation: The on the air privacy feature of GSM makes impossible to eavesdrop on a GSM radio transmission. The privacy is made possible by encrypting the digital bit stream sent by a GSM transmitter, according to a specific secret cryptographic key that is known only to the cellular carrier.
8. Which of the following does not come under subsystem of GSM architecture?
a) BSS
b) NSS
c) OSS
d) Channel
Answer: d
Explanation: The GSM architecture consists of three major interconnected subsystems that interact between themselves and with the users through certain network interfaces. The subsystems are BSS , NSS and OSS .
9. Which of the following subsystem provides radio transmission between mobile station and MSC?
a) BSS
b) NSS
c) OSS
d) BSC
Answer: a
Explanation: The BSS provides and manages radio transmission paths between the mobile stations and the Mobile Switching Center . It also manages the radio interface between the mobile stations and all other subsystems of GSM.
10. ___________ manages the switching function in GSM.
a) BSS
b) NSS
c) OSS
d) MSC
Answer: b
Explanation: NSS manages the switching functions of the system. It allows the MSCs to communicate with other networks such as PSTN and ISDN.
11. __________ supports the operation and maintenance of GSM.
a) BSS
b) NSS
c) OSS
d) MSC
Answer: c
Explanation: The OSS supports the operation and maintenance of GSM. It allows system engineers to monitor, diagnose, and troubleshoot all aspects of GSM.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “GSM Channel Types”.
1. __________ carries digitally encoded user data.
a) Traffic channels
b) Control channels
c) Signalling channels
d) Forward channels
Answer: a
Explanation: Traffic channels carry digitally encoded user speech or user data. It has identical functions and formats on both the forward and reverse links.
2. ____________ carries signalling and synchronizing commands.
a) Traffic channels
b) Control channels
c) Signalling channels
d) Forward channels
Answer: b
Explanation: Control channels carry signalling and synchronizing commands between the base station and mobile station. Certain types of control channels are defined for just the forward or reverse link.
3. Which of the following is not a control channel of GSM?
a) BCH
b) CCCH
c) DCCH
d) TCH
Answer: d
Explanation: There are three main control channels in the GSM system. These are the broadcast channel , the common control channel and the dedicated control channel . Each control channel consists of several logical channels.
4. Which of the following is the forward control channel that is used to broadcast information?
a) BCCH
b) CCCH
c) DCCH
d) TCH
Answer: a
Explanation: The broadcast control channel is a forward channel that is used to broadcast information such as cell and network identity, and operating characteristics of the cell.
5. Which of the following channel does not come under CCCH?
a) PCH
b) RACH
c) DCCH
d) AGCH
Answer: c
Explanation: CCCH consists of three different channels. They are paging channel , which is a forward link channel, the random access channel which is a reverse link channel, and the access grant channel which is a forward link channel.
6. Which of the following channel provides paging signals from base station to all mobiles in the cell?
a) RACH
b) AGCH
c) DCCH
d) PCH
Answer: d
Explanation: The PCH provides paging signals from the base station to all mobiles in the cell. It notifies a specific mobile of an incoming call which originates from the PSTN.
7. ___________ is a reverse link channel used by a subscriber unit to acknowledge.
a) RACH
b) AGCH
c) DCCH
d) PCH
Answer: a
Explanation: The RACH is a reverse link channel used by a subscriber unit to acknowledge a page from the PCH. It is also used by mobiles to originate a call.
8. Which of the following channel is used by base station to provide forward link communication to mobile?
a) RACH
b) AGCH
c) DCCH
d) PCH
Answer: b
Explanation: The AGCH is used by the base station to provide forward link communication to the mobile. It carries data which instructs the mobile to operate in a particular physical channel with particular dedicated control channel.
9. Which of the following burst is used to broadcast the frequency and time synchronization control messages?
a) FCCH and SCH
b) TCH and DCCH
c) RACH and TCH
d) FCCH and DCCH
Answer: a
Explanation: FCCH and SCH burst are used to broadcast the frequency and time synchronization control messages. They are used in TS0 of specific frames.
10. Which of the following burst is used to access service from any base station?
a) TCH
b) RACH
c) SCH
d) FCCH
Answer: b
Explanation: Each user transmits a burst of data during the time slot assigned to it. The RACH burst is used by all mobiles to access service from any base station, and dummy burst is used as filter information for unused timeslots on forward link.
11. Group of superframes in GSM is called multiframe.
a) True
b) False
Answer: b
Explanation: Each of the normal speech frames are grouped into larger structures called multiframes. These multiframes are grouped into superframes and hyperframes.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “CDMA Digital Cellular Standard ”.
1. US digital cellular system based on CDMA was standardized as ________
a) IS-54
b) IS-136
c) IS-95
d) IS-76
Answer: c
Explanation: A US digital cellular system based on CDMA was standardized as Interim Standard 95 . It was standardized by US Telecommunication Industry Association and promised increased capacity.
2. IS-95 was not compatible with existing AMPS frequency band.
a) True
b) False
Answer: b
Explanation: Like IS-136, IS-95 system was designed to be compatible with the existing US analog cellular system frequency band. Hence, mobile and base stations can be economically produced for dual mode operation.
3. Which of the following is used by IS-95?
a) DSSS
b) FHSS
c) THSS
d) Hybrid
Answer: a
Explanation: IS-95 uses a direct sequence spread spectrum CDMA system. It allows each user within a cell to use the same radio channel, and users in adjacent cell also use the same radio channel.
4. Each IS-95 channel occupies ___________ of spectrum on each one way link.
a) 1.25 MHz
b) 1.25 kHz
c) 200 kHz
d) 125 kHz
Answer: a
Explanation: To facilitate graceful transition from AMPS to CDMA, each IS-95 channel occupies 1.25 MHz of spectrum on each one way link, or 10% of the available cellular spectrum for a US cellular provider.
5. IS-95 uses same modulation technique for forward and reverse channel.
a) True
b) False
Answer: b
Explanation: IS-95 uses different modulation and spreading technique for the forward and reverse links. On the forward link, the base station simultaneously transmits the user data for all mobiles in the cell by using different spreading sequence for each mobile.
6. IS-95 is specified for reverse link operation in _________ band.
a) 869-894 MHz
b) 849-894 MHz
c) 849-869 MHz
d) 824-849 MHz
Answer: d
Explanation: IS-95 is specified for reverse link operation in the 824-849 MHz band and 869-894 MHz for the forward link. The PCS version of IS-95 has also been designed for international use in the 1800-2000 MHz bands.
7. User data in IS-95 is spread to a channel chip rate of ________
a) 1.2288 Mchip/s
b) 9.6 Mchip/s
c) 12.288 Mchip/s
d) 0.96 Mchip/s
Answer: a
Explanation: User data is spread to a channel chip rate of 1.2288 Mchip/s using a combination of techniques. The spreading process is different for the forward and reverse links in the original CDMA specification.
8. __________ are used to resolve and combine multipath components.
a) Equalizer
b) Registers
c) RAKE receiver
d) Frequency divider
Answer: c
Explanation: At both the base station and the subscriber, RAKE receivers are used to resolve and combine multipath components, thereby reducing the degree of fading. A RAKE receiver exploits the multipath time delays in a channel and combines the delayed replicas of transmitted signal.
9. CT2 was the first generation of cordless telephones.
a) True
b) False
Answer: b
Explanation: CT2 was the second generation of cordless telephones introduced in Great Britain in 1989. It is used to provide telepoint services which allow a subscriber to use CT2 handsets at a public telepoint.
10. CT2 is analog version of first generation cordless telephones.
a) True
b) False
Answer: b
Explanation: CT2 is a digital version of the first generation, analog, cordless telephones. When compared with analog cordless phones, CT2 offers good speech quality and is more resistant to interference.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Digital European Cordless Telephone ”.
1. Cordless telephone standard, DECT stands for ____________
a) Discrete European Cordless Telephone
b) Digital European Cellular Telephone
c) Discrete European Cellular Telephone
d) Digital European Cordless Telephone
Answer: d
Explanation: The Digital European Cordless Telephone is a universal cordless telephone standard. It was developed by the European Telecommunications Standard Institute .
2. Which of the following is not true for DECT?
a) High traffic density
b) Long range telecommunication
c) Broad range of application
d) First pan European standard
Answer: b
Explanation: DECT provides a cordless communication framework for high traffic density, short range telecommunications. It covers a broad range of applications and environments.
3. DECT can be used by users in an in-building PBX.
a) True
b) False
Answer: a
Explanation: The main function of DECT is to provide local mobility to portable users in an in-building Private Branch Exchange . It provides excellent quality and services for voice and data applications.
4. The DECT system is based on __________ principles.
a) TCP
b) IP
c) OSI
d) AMPS
Answer: c
Explanation: The DECT system is based on OSI principles in a manner similar to ISDN. A control plane and a user plane use the services provided by the lower layers.
5. Which of the following layer consists of paging channel and control channel?
a) Physical layer
b) Network layer
c) Data link layer
d) MAC layer
Answer: d
Explanation: The MAC layer consists of a paging channel and a control channel for the transfer of signalling information to the C-plane. The U-plane is served with channels for the transfer of user information.
6. Which of the following layer is responsible for providing reliable data links?
a) Physical layer
b) Network layer
c) Data link control layer
d) MAC layer
Answer: c
Explanation: The DLC layer is responsible for providing reliable data links to the network layer. It divides up the logical and physical channels into time slots for each user.
7. __________ is the main signalling layer of DECT.
a) Physical layer
b) Network layer
c) Data link layer
d) MAC layer
Answer: b
Explanation: The network layer is the main signalling layer of DECT. It is based on GSM and ISDN protocols. It provides call control and circuit switched services selected from one of the DLC services.
8. Which of the following supports the physical layer of DECT common air interface?
a) Portable Handset
b) Network specific Interface Unit
c) Radio Fixed Part
d) Cordless controller
Answer: c
Explanation: RFP supports the physical layer of the DECT common air interface. Every RFP covers one cell in the microcellular system. A full duplex operation is achieved using TDD.
9. ________ supports the call completion facility in a multihandset environment.
a) Portable Handset
b) Network specific Interface Unit
c) Radio Fixed Part
d) Cordless controller
Answer: b
Explanation: Network specific Interface Unit supports the call completion facility in a multihandset environment. The interface recommended by the CCITT is the G.732 based on ISDN protocols.
10. __________ is a third generation Personal Communication System.
a) PACS
b) AMPS
c) IS-95
d) GSM
Answer: a
Explanation: PACS is a third generation Personal Communication System. It was originally developed and proposed by Bellcore in 1992.
11. _________ is also known as Japanese Digital Cellular.
a) PACS
b) AMPS
c) PDC
d) GSM
Answer: c
Explanation: PDC is also known as Japanese Digital Cellular. It was developed in 1991 to provide for needed capacity in congested cellular bands in Japan.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Network Protocols”.
1. Several protocols for upper layers in bluetooth use _________
a) UDP
b) HSP
c) ITC
d) L2CAP
Answer: d
Explanation: L2CAP is Logical Link, Control Adaptation Protocol Layer. The logical unit link control adaptation protocol is equivalent to logical link control sub layer of LAN. The ACL link uses L2CAP for data exchange. The various function of L2CAP is segmentation and reassembly, multiplexing and quality of service.
2. Protocols are set of rules to govern _________
a) Communication
b) Standard
c) Metropolitan communication
d) Bandwidth
Answer: a
Explanation: A protocol is a set of rules that governs the communications between computers on a network. These rules include guidelines that regulate the characteristics of a network including access method, allowed physical topologies, types of cabling, and speed of data transfer.
3. An internet is a __________
a) Collection of WANS
b) Network of networks
c) Collection of LANS
d) Collection of identical LANS and WANS
Answer: b
Explanation: Internet is a massive network of networks, a networking infrastructure. It connects millions of computers together globally, forming a network in which any computer can communicate with any other computer as long as they are both connected to the Internet.
4. Checksum is used in Internet by several protocols although not at the _________
a) Session layer
b) Transport layer
c) Network layer
d) Data link layer
Answer: d
Explanation: The checksum is used in the Internet by several protocols although not at the data link layer. Like linear and cyclic codes, the checksum is based on the concept of redundancy. Several protocols still use the checksum for error detection.
5. In version field of IPv4 header, when machine is using some other version of IPv4 then datagram is __________
a) Discarded
b) Accepted
c) Interpreted
d) Interpreted incorrectly
Answer: a
Explanation: A 4 bit field defines the version of IPv4 protocol. This field tells the software running in the processing machine that the datagram has the format of version 4. If the machine is using some other version of IPv4, the datagram is discarded rather than interpreted incorrectly.
6. Network layer at source is responsible for creating a packet from data coming from another ________
a) Station
b) Link
c) Node
d) Protocol
Answer: d
Explanation: ‘The network layer at the source is responsible for creating a packet from the data coming from another’ protocol . The network layer is responsible for checking its routing table to find the routing information.
7. Header of datagram in IPv4 has _________
a) 0 to 20 bytes
b) 20 to 40 bytes
c) 20 to 60 bytes
d) 20 to 80 bytes
Answer: c
Explanation: IP header length is a minimum of 20 bytes and a maximum of 60 bytes. The minimum value for this field is 5, which is a length of 5×32 = 160 bits = 20 bytes. Being a 4-bit value, the maximum length is 15 words or 480 bits = 60 bytes.
8. In IPv4 layer, datagram is of ________
a) Fixed length
b) Variable length
c) Global length
d) Zero length
Answer: b
Explanation: ‘IPv4 is a connectionless protocol used for packet switched networks. It operates on best effort delivery model, in which neither delivery is guaranteed, nor proper sequencing or avoidance of duplicate delivery is assured. The size of the datagram header can be of variable length from 20 bytes to 60 bytes.’
9. In IPv4, service type of service in header field, first 3 bits are called ______
a) Type of service
b) Code bits
c) Sync bits
d) Precedence bits
Answer: d
Explanation: The 8-bit ToS in IPv4 uses 3 bits for IP Precedence, 4 bits for ToS with the last bit not being used. The 4-bit ToS field, although defined, has never been used.
10. Which is a link layer protocol?
a) ARP
b) TCP
c) UDP
d) HTTP
Answer: b
Explanation: ‘In computer networking, the link layer is the lowest layer in the Internet Protocol Suite. It is commonly known as TCP/IP, the networking architecture of the Internet. It is described in RFC 1122 and RFC 1123.’
11. Which protocol is commonly used to retrieve email from a mail server?
a) FTP
b) IMAP
c) HTML
d) TELNET
Answer: b
Explanation: ‘The Internet Message Access Protocol is a mail protocol used for accessing email on a remote web server from a local client. IMAP is the most commonly used Internet mail protocols for retrieving emails. It is supported by all modern email clients and web servers.’
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “TCP/IP Protocol”.
1. What layer in the TCP/IP stack is equivalent to the Transport layer of the OSI model?
a) Application
b) Host to host
c) Internet
d) Network Access
Answer: b
Explanation: The four layers of the TCP/IP stack are Application/Process, Host-to-Host, Internet, and Network Access. The Host-to-Host layer is equivalent to the Transport layer of the OSI model.
2. You want to implement a mechanism that automates the IP configuration, including IP address, subnet mask, default gateway, and DNS information. Which protocol will you use to accomplish this?
a) SMTP
b) SNMP
c) DHCP
d) ARP
Answer: c
Explanation: Dynamic Host Configuration Protocol is used to provide IP information to hosts on your network. DHCP can provide a lot of information, but the most common is the IP address, subnet mask, default gateway, and DNS information.
3. The DoD model has four layers. Which layer of the DoD model is equivalent to the Network layer of the OSI model?
a) Application
b) Host to Host
c) Internet
d) Network Access
Answer: c
Explanation: The four layers of the DoD model are Application/Process, Host-to-Host, Internet, and Network Access. The Internet layer is equivalent to the Network layer of the OSI model.
4. Which of the following protocols uses both TCP and UDP?
a) FTP
b) SMTP
c) Telnet
d) DNS
Answer: d
Explanation: DNS and some other services work on both TCP and the UDP protocols. DNS uses TCP for zone exchanges between servers and UDP when a client is trying to resolve a hostname to an IP address.
5. Length of Port address in TCP/IP is _________
a) 4bit long
b) 16bit long
c) 32bit long
d) 8 bit long
Answer: b
Explanation: TCP and UDP port numbers are 16 bits in length. So, valid port numbers can theoretically take on values from 0 to 65,535. These values are divided into ranges for different purposes, with certain ports reserved for particular uses.
6. TCP/IP layer is equivalent to combined Session, Presentation and _________
a) Network layer
b) Application layer
c) Transport layer
d) Physical layer
Answer: b
Explanation: TCP/IP network model is a hierarchical protocol made up of interactive modules, each of which provides a specific functionality; however, the modules are not necessarily interdependent. It is equivalent to combined session, presentation and application layer.
7. How many levels of addressing is provided in TCP/IP protocol?
a) One
b) Two
c) Three
d) Four
Answer: d
Explanation: Four levels of addresses are used in the internet employing the TCP/IP protocols. They are physical addresses, logical addresses, port addresses, and specific addresses.
8. Virtual terminal protocol is an example of _________
a) Network layer
b) Application layer
c) Transport layer
d) Physical layer
Answer: b
Explanation: In open systems, a virtual terminal is an application service. It allows host terminals on a multi-user network to interact with other hosts regardless of terminal type and characteristics.
9. TCP/IP is related to __________
a) ARPANET
b) OSI
c) DECNET
d) ALOHA
Answer: a
Explanation: In 1983, TCP/IP protocols replaced NCP as the ARPANET’s principal protocol. And ARPANET then became one component of the early Internet. The starting point for host-to-host communication on the ARPANET in 1969 was the 1822 protocol, which defined the transmission of messages to an IMP.
10. A device operating at network layer is called __________
a) Router
b) Equalizer
c) Bridge
d) Repeater
Answer: a
Explanation: A router is a networking device that forwards data packets between computer networks. Routers perform the traffic directing functions on the Internet. It supports different network layer transmission standards. Each network interface is used to enable data packets to be forwarded from one transmission system to another.
11. A device operating at physical layer is called __________
a) Router
b) Equalizer
c) Bridge
d) Repeater
Answer: d
Explanation: A repeater connects two segments of your network cable. It retimes and regenerates the signals to proper amplitudes and sends them to the other segments. Repeaters work only at the physical layer of the OSI network model.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “TCP over Wireless”.
1. A packet in Transmission Control Protocol is called a ____________
a) Transmittable slots
b) Packet
c) Segment
d) Source Slots
Answer: c
Explanation: Applications working at the Application Layer transfers a contiguous stream of bytes to the bottom layers. It is the duty of TCP to pack this byte stream to packets, known as TCP segments, which are passed to the IP layer for transmission to the destination device.
2. Cable TV and DSL are examples of ____________
a) Interconnection of network
b) LAN
c) MAN
d) WAN
Answer: c
Explanation: A MAN often acts as a high speed network to allow sharing of regional resources. It typically covers an area of between 5 and 50 km diameter. Examples of MAN are telephone company network that provides a high speed DSL to customers and cable TV network.
3. Station on a wireless ALOHA network is maximum of ________
a) 400 Km
b) 500 Km
c) 600 Km
d) 700 Km
Answer: c
Explanation: The stations on wireless ALOHA networks are a maximum of 600 km apart. It was designed for a radio LAN, but it can be used on any shared medium. It is obvious that there are potential collisions in this arrangement. The medium is shared between the stations.
4. IEEE 802.11 defines basic service set as building block of a wireless ___________
a) LAN
b) WAN protocol
c) MAN
d) ALOHA
Answer: a
Explanation: The IEEE 802.11 topology consists of components interacting to provide a wireless LAN. It enables station mobility transparent to higher protocol layers, such as the LLC.
5. In wireless LAN, there are many hidden stations so that __________ cannot be detected.
a) Frames
b) Collision
c) Signal
d) Data
Answer: b
Explanation: In wireless networking, the hidden node problem or hidden terminal problem occurs when a node is visible from a wireless access point , but not from other nodes communicating with said AP. This leads to difficulties in media access control and collisions could not be detected.
6. A set that makes stationary or mobile wireless station and also have optional central base station is known as ___________
a) Basic service set
b) Extended service set
c) Network point set
d) Access point
Answer: a
Explanation: A set that makes stationary or mobile wireless station and also have optional central base station is known as a basic service set. BSS is made of stationary or mobile wireless stations and an optional central base station, known as the access point .
7. Wireless communication started in _________
a) 1869
b) 1895
c) 1879
d) 1885
Answer: b
Explanation: In England, Guglielmo Marconi began his wireless experiments in 1895. On 2 June 1896, he filed his provisional specification of a patent for wireless telegraphy. He demonstrated the system to the British Post Office in July.
8. Wireless transmission is divided into ___________
a) 3 broad groups
b) 6 broad groups
c) 9 broad groups
d) 8 broad groups
Answer: a
Explanation: We can divide wireless transmission into three broad groups: radio waves, microwaves, and infrared waves. Radio waves are used for multicast communications, such as AM and FM radio, television, maritime radio, cordless phones and paging systems. Microwave propagation is line-of-sight.
9. Transmission Control Protocol/Internet Networking Protocol have ___________
a) Four Layers
b) Five Layers
c) Six Layers
d) Seven Layers
Answer: a
Explanation: TCP/IP functionality is divided into four layers, each with its own set of agreed-upon protocols: The datalink layer consists of methods and protocols that operate only on a link. The Internet layer connects independent networks to transport the packets. The Transport layer handles communications between. The Application layer standardizes data exchange for applications.
10. Packets of data that is transported by IP is called __________
a) Datagrams
b) Frames
c) Segments
d) Encapsulate message
Answer: a
Explanation: The format of data that can be recognized by IP is called an IP datagram. It consists of two components, the header and data, which need to be transmitted. The fields in the datagram, except the data, have specific roles to perform in the transmission of data.
11. Parameter that is normally achieved through a trailer added to end of frame is ___________
a) Access Control
b) Flow Control
c) Error Control
d) Physical addressing
Answer: c
Explanation: The data link layer adds reliability to the physical layer by adding mechanisms to detect and retransmit damaged or lost frames. It also uses a mechanism to recognize duplicate frames. Error control is normally achieved through a trailer added to the end of the frame.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Paging System”.
1. IPv6 has a larger address space of _________
a) 2 16
b) 2 128
c) 2 32
d) 2 8
Answer: b
Explanation: The main advantage of IPv6 over IPv4 is its larger address space. The length of an IPv6 address is 128 bits, compared with 32 bits in IPv4. The address space therefore has 2 128 or approximately 3.4×10 38 addresses.
2. Three strategies used to handle transition from version 4 to version 6 are dual-stack, tunneling and ________
a) Header Switching
b) Header Translation
c) Header Transfer
d) Header Transmission
Answer: b
Explanation: Three strategies used to handle transition from version 4 to version 6 are dual-stack, tunneling and header translation. Header translation techniques are more complicated than IPv4 NAT because the protocols have different header formats.
3. MTU stands for _________
a) Minimum Transfer Unit
b) Maximum Transfer Unit
c) Maximum Transport Unit
d) Maximum Transmission Unit
Answer: d
Explanation: In computer networking, the maximum transmission unit is the size of the largest network layer protocol data unit that can be communicated in a single network transaction.
4. In IPv6, real-time audio or video, particularly in digital form, requires resources such as __________
a) Fixed Bandwidth
b) Variable Bandwidth
c) High Bandwidth
d) Low Bandwidth
Answer: c
Explanation: In IPv6, real-time audio or video, particularly in digital form, requires resources such as high bandwidth, large buffers, long processing times, and so on. A process can make a reservation for these resources beforehand to guarantee that real time data will not be delayed.
5. In practical IPv6 application, a technology encapsulates IPv6 packets inside IPv4 packets, this technology is called _______
a) Tunneling
b) Hashing
c) Routing
d) NAT
Answer: a
Explanation: IPv6 tunneling enables IPv6 hosts and routers to connect with other IPv6 hosts and routers over the existing IPv4 Internet. The main purpose of IPv6 tunneling is to deploy IPv6 as well as maintain compatibility with large existing base of IPv4 hosts and routers.
6. Which one of the following descriptions about IPv6 is correct?
a) Addresses are not hierarchical and are assigned at random
b) Broadcasts have been eliminated and replaced with multicasts
c) There are 2.7 billion available addresses
d) An interface can only be configured with one IPv6 address
Answer: b
Explanation: In IPv6, there’s no longer any broadcast, sending one packet to a large number of unspecified hosts. There’s only multicast, unicast and anycast. In IPv6 all nodes are required to support multicast.
7. The header length of an IPv6 datagram is ___________
a) 10bytes
b) 25bytes
c) 30bytes
d) 40bytes
Answer: d
Explanation: IPv6 datagram has fixed header length of 40bytes. It results in faster processing of the datagram. Fixed length IPv6 header allows the routers to process the IPv6 datagram packets more efficiently.
8. In the IPv6 header, the traffic class field is similar to which field in the IPv4 header?
a) Fragmentation field
b) Fast switching
c) TOS field
d) Option field
Answer: c
Explanation: This field enables to have different types of IP datagram. In an IPv6 packet, the Traffic Class byte is used in the same way as the ToS byte in an IPv4 packet. A ToS/Traffic Class byte includes a DSCP and precedence bits.
9. Which are the features present in IPv4 but not in IPv6?
a) Fragmentation
b) Header checksum
c) Options
d) All of the mentioned
Answer: d
Explanation: All the features are only present in IPv4 and not IPv6. IPv6 no longer has a header checksum to protect the IP header, meaning that when a packet header is corrupted by transmission errors, the packet is very likely to be delivered incorrectly.
10. IPv6 is designed to allow extension of the _________
a) Protocol
b) Dataset
c) Headers
d) Routes
Answer: a
Explanation: IPv6 is designed to allow the extension of the protocol if required by new technologies or applications. IPv6 uses a new header format in which options are separated from the base header and inserted, when needed, between the base header and upper-layer data.
11. In IPv6, base header can be followed by, up to _________
a) Six Extension Layers
b) Six Extension Headers
c) Eight Extension headers
d) Eight Extension layers
Answer: b
Explanation: The length of the base header is 40 bytes. However, to give more functionality to the IP datagram, the base header can be followed by up to six extension headers.
12. In an IPv6 datagram, M bit is 0, value of HLEN is 5, value of total length is 200 and offset value is ___________
a) 400
b) 350
c) 300
d) 200
Answer: d
Explanation: In an IPv6 datagram, M bit is 0, value of HLEN is 5, the value of total length is 200 and offset value is 200. If the M bit is 0, it means there are no more fragments, the fragment is the last name.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Paging System”.
1. A helical antenna is used for satellite tracking because of _________
a) Circular polarization
b) Maneuverability
c) Beamwidth
d) Gain
Answer: a
Explanation: In helical antenna, the diameter and pitch of the helix are comparable to a wavelength. The antenna functions as a directional antenna radiating a beam off the ends of the helix. It radiates circularly polarized radio waves. These are used for satellite communication.
2. Repeaters inside communications satellites are known as ___________
a) Transceivers
b) Transponders
c) Transducers
d) TWT
Answer: b
Explanation: A communications satellite’s transponder is the series of interconnected units that form a communications channel between the receiving and the transmitting antennas. It is mainly used in satellite communication to transfer the received signals.
3. ___________ is the geographical representation of a satellite antenna radiation pattern.
a) Footprint
b) Spot
c) Earth
d) Region
Answer: a
Explanation: The geographical representation of a satellite’s antenna radiation pattern is called a footprint or footprint map. In essence, a footprint of a satellite is the area on Earth’s surface that the satellite can receive from or transmit to.
4. The smallest beam of a satellite antenna radiation pattern is ________
a) Zone beam
b) Hemispheric beam
c) Spot beam
d) Global beam
Answer: c
Explanation: The size of the antenna that generates these beams on earth is related directly to the peak gain at the center of the spot beams and the smallest spot beam size. The spot beams are typically defined by the contours at 3 or 4 dB down from the peak power at the center of the beam.
5. _________ detects the satellite signal relayed from the feed and converts it to an electric current, amplifies and lowers its frequency.
a) Horn antenna
b) LNA
c) Satellite receiver
d) Satellite dish
Answer: b
Explanation: LNA detects the satellite signal relayed from the feed and converts it to an electric current, amplifies and lower its frequency. The most common device used as an LNA is tunnel diode.
6. A satellite signal transmitted from a satellite transponder to earth’s station is _________
a) Uplink
b) Downlink
c) Terrestrial
d) Earthbound
Answer: b
Explanation: In satellite telecommunication, a downlink is the link from a satellite down to one or more ground stations or receivers, and an uplink is the link from a ground station up to a satellite.
7. __________ is a loss of power of a satellite downlink signal due to earth’s atmosphere.
a) Atmospheric loss
b) Path loss
c) Radiation loss
d) RFI
Answer: b
Explanation: The path loss is the loss in signal strength of a signal as it travels through free space. This value is usually calculated by discounting any obstacles or reflections that might occur in its path.
8. Which of the following is the point on the satellite orbits closest to the Earth?
a) Apogee
b) Perigee
c) Prograde
d) Zenith
Answer: b
Explanation: The point where satellite is closest to the Earth is known as the perigee. Here, the satellite moves at its fastest. The high point of the orbit, when the satellite is moving the slowest is called the apogee.
9. What kind of battery panels are used in some advanced satellites?
a) Germanium based panels
b) Silicon based panel
c) Gallium Phosphate solar panel array
d) Gallium Arsenide solar panel array
Answer: d
Explanation: Gallium Arsenide solar panel arrays are used for battery panels in some advanced satellites. These new types of cells allow smaller solar arrays to be used on future space missions.
10. A satellite battery has more power but lighter _________
a) Lithium
b) Leclanche
c) Hydrogen
d) Magnesium
Answer: a
Explanation: Lithium batteries have more power and are lighter in weight. Any mass that could be saved by the use of lighter batteries would allow a corresponding increase in the amount of useful payload equipment.
11. INTELSAT stands for ___________
a) Intel Satellite
b) International Telephone Satellite
c) International Telecommunications Satellite
d) International Satellite
Answer: c
Explanation: INTELSAT is a communications satellite services provider. INTELSAT operates a fleet of 52 communications satellites, which is one of the world’s largest fleet of commercial satellites.
12. The frequency of Ku band for satellite communications is __________
a) 6/4 GHz
b) 14/11 GHz
c) 12/14 GHz
d) 4/8 GHz
Answer: b
Explanation: The Ku band is a portion of the electromagnetic spectrum in the microwave range of frequencies ranging from 11.7 to 12.7GHz. and 14 to 14.5GHz .
13. The most common device used as an LNA is ________
a) Zener diode
b) Tunnel diode
c) IMPATT
d) Shockley diode
Answer: b
Explanation: The LNA must provide a relatively flat response for the frequency range of interest, preferably with less than 1 dB of gain variation. The most common device used as an LNA is tunnel diode. It is a highly sensitive, low-noise device.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Paging System”.
1. Which of the following technology is mainly designed for indoor coverage?
a) Femtocell network
b) 3GPP
c) LTE
d) AMPS
Answer: a
Explanation: Femtocell Network is a small-size Macro cell network designed for better indoor coverage. It began attracting attention from both industry and academy in late 2007.
2. Coverage of Femtocell Network is much larger than a regular Macro cell Network.
a) True
b) False
Answer: b
Explanation: The “femto” means 10-15. Coverage of Femtocell Network is much smaller than a regular Macro cell Network that is why this name is given. Femtocell Network, installed by end users at home or in an office environment.
3. Which of the following is not a component of Femtocell network?
a) F-BS
b) Internet link
c) FGW
d) BSC
Answer: d
Explanation: Similar to UMTS Terrestrial Radio Access Network architecture, the Femtocell Network consists of three components: Femtocell Base Stations , Internet Link, and Femtocell Gateway .
4. Which of the following is not true for Femtocell base stations?
a) Short range
b) Low power
c) High cost
d) Wireless handsets
Answer: c
Explanation: Femtocell Base Stations are short-range, low-cost, low-power indoor devices to provide service for wireless handsets. F-BS, which looks like WLAN Access Point , is a small device with at least two wireless and internet interfaces.
5. Any existing wireless telecommunication standard cannot be used at F-BS wireless interface.
a) True
b) False
Answer: b
Explanation: Wireless interface provides wireless radio access to Femtocell MSs. Any existing wireless telecommunication standard, such as UMTS/ CDMA200/ WIMAX/ LTE/ EV-DO, can be used at the F-BS wireless interfaces.
6. ________ acts as a gateway between the Internet and communication network.
a) F-BS
b) Internet link
c) FGW
d) BSC
Answer: c
Explanation: Femtocell Gateway is a service provider’s device that acts as a gateway between the Internet and the communication network. One side of FGW connects a large number of F-BSs via broadband Internet, and the other side of FGW is connected to the telephony core network.
7. Femtocell network has improved seamless coverage and enhanced capacity.
a) True
b) False
Answer: a
Explanation: Femtocell Network is a so-called “double-win” strategy that brings benefits to both cellular users and cellular providers. It provides improved seamless coverage and enhanced capacity for cellular users.
8. In _________ all cellular users belong to open subscribers group and can access F-BS unconditionally.
a) Closed access mode
b) Open access mode
c) Hybrid Access mode
d) Zero access mode
Answer: b
Explanation: In open access mode, all cellular users belong to open subscribers group and can access F-BS unconditionally. Several cellular service providers have plans to deploy F-BS for better service quality to cover public hole areas.
9. In _______ a closed subscribers group is set by F-BS owner to allow only small portion of cellular users to be served in the Femtocell Network.
a) Closed access mode
b) Open access mode
c) Hybrid Access mode
d) Zero access mode
Answer: a
Explanation: In closed access mode, a closed subscribers group is set by F-BS owner to allow only small portion of cellular users to be served in the Femtocell Network. For example, people can install F-BS in their house and only household members can access F-BS to attain better service.
10. Femtocell Network has the capability to help Macro cell Network achieve seamless coverage.
a) True
b) False
Answer: a
Explanation: Femtocell Network has the capability to help Macro cell Network achieve seamless coverage. It can attain higher network capacity by transmitting over an Internet link.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Paging System”.
1. _________ also known as impulse or zero-carrier radio technology.
a) Ultra wideband technology
b) Femtocell technology
c) Multicasting
d) Multiplexing
Answer: a
Explanation: Ultra-wideband technology is also known as impulse or zero-carrier radio technology. It appears to be one of the most promising wireless radio communication technologies of recent time.
2. UWB operates across narrow bandwidth.
a) True
b) False
Answer: b
Explanation: Unlike conventional radio systems, which operate within a relatively narrow bandwidth, the UWB radio system operates across a wide range of the frequency spectrum by transmitting a series of extremely narrow and low-power pulses.
3. The basic element in DSC–UWB technology is the monocycle wavelet.
a) True
b) False
Answer: b
Explanation: The basic element in TM–UWB technology is the monocycle wavelet. Typically, wavelet pulse widths are between 0.2 and 1.5 nanoseconds, corresponding to center frequencies between 600 MHz and 5 GHz.
4. In TM–UWB, the system uses a modulation technique called _______
a) Pulse width modulation
b) Pulse code modulation
c) Pulse position modulation
d) Pulse amplitude modulation
Answer: c
Explanation: In TM–UWB, the system uses a modulation technique called pulse position modulation. The TM–UWB transmitter emits ultra-short monocycle wavelets with tightly controlled pulse-to-pulse intervals, which are varied on a pulse-by-pulse basis in accordance with an information signal and a channel code.
5. DSC-UWB uses _________
a) Pulse width modulation
b) Pulse code modulation
c) Pulse position modulation
d) Direct sequence modulation
Answer: d
Explanation: In DSC-UWB, the signal is spread by direct sequence modulating a wavelet pulse trains at duty cycles approaching that of a sine wave carrier. The spectrum spreading, channelization, and modulation are provided by a PN sequence, and the chipping rate is maintained as some fraction of the carrier center frequency.
6. The coherent interaction of signals in UWB arriving by many paths causes ____________
a) Ricean fading
b) Nakagami fading
c) Rayleigh fading
d) Multicast fading
Answer: c
Explanation: The coherent interaction of signals arriving by many paths causes the Rayleigh or multipath fading in RF communications. Inside buildings, when continuous sine waves are transmitted wherein the channels exhibit multipath differential delays in the nanosecond range, the multipath fading occurs naturally.
7. UWB technology supports low bit rate and low speed.
a) True
b) False
Answer: b
Explanation: UWB technology is appropriate for the high-performance wireless home network, which mandates support for large bit rate , high-speed, affordable connectivity between devices, and simultaneous data transmission from multiple devices, and full-motion video capability.
8. Which of the following is not true for UWB?
a) Large spectrum
b) Lower price
c) Pulse data
d) Large interference
Answer: d
Explanation: The combination of a larger spectrum, lower power, and pulsed data means that UWB causes less interference than narrowband radio designs while yielding low probability of detection and excellent multipath immunity.
9. UWB systems are very complex, since they use radio frequency/intermediate frequency conversion stages.
a) True
b) False
Answer: b
Explanation: UWB systems are much less complex, since they do not use any radio frequency/intermediate frequency conversion stages, local oscillators, mixers, and other expensive surface acoustic wave filters common to traditional radio technologies.
10. Which of the following is a drawback of UWB technology?
a) Not appropriate for WAN
b) Power limited
c) Small spectrum
d) Limited jitter requirements
Answer: d
Explanation: UWB devices are power limited because they must coexist on a noninterfering basis with other licensed and unlicensed users across several frequency bands. For UWB systems using PPM as their modulation technique, limited jitter requirements could be an issue.
This set of Wireless & Mobile Communications Question Bank focuses on “Multicast in Wireless Networks”.
1. _______ is responsible for tunneling multicast packets to the MS’s currently subscribed FA.
a) Multicast home agent
b) Mobile multicast
c) Mobile station
d) Base station
Answer: a
Explanation: Multicast home agent is responsible for tunneling multicast packets to the MS’s currently subscribed FA. MHA serves MSs that are roaming around the foreign networks and are within its service range.
2. Every MS can have only one MHA.
a) True
b) False
Answer: a
Explanation: Every MS can have only one MHA, which dynamically changes based on the location of the MS, whereas a HA of an MS never changes. The protocol requires that each MHA must be a multicast group member.
3. _______ provides a fast and efficient handoff for MSs in foreign networks.
a) MHA
b) MMP
c) CBT
d) MS
Answer: b
Explanation: MMP provides a fast and efficient handoff for MSs in foreign networks. It also enables location-independent addressing.
4. MMP combines the concepts of _______ and ________
a) Mobile IPs, GSM
b) Core based trees, GSM
c) Mobile IPs, core based trees
d) Core based trees, LTE
Answer: c
Explanation: MMP combines the concepts of Mobile IP and core-based trees . Here the former controls communication up to the foreign network, and the latter manages movement of the MSs inside them.
5. ________ designed for an Internet work environment with small wireless cells.
a) MMP
b) RMDP
c) RM2
d) Mobicast
Answer: d
Explanation: Mobicast is designed for an Internet work environment with small wireless cells. It assumes that a set of cells are grouped together and are served by a domain foreign agent .
6. ________ serve as multicast forwarding agents and are meant to isolate the mobility of the mobile host from the main multicast delivery tree.
a) DFA
b) MHA
c) FA
d) MMP
Answer: a
Explanation: DFAs serve as multicast forwarding agents and are meant to isolate the mobility of the mobile host from the main multicast delivery tree. This hierarchical mobility management approach tries to isolate the mobility of the FAs from the main multicast delivery tree.
7. Mobicast is based on a method proposed by the IETF to support multicast over Mobile-IP.
a) True
b) False
Answer: a
Explanation: Mobicast is based on a method proposed by the IETF to support multicast over Mobile-IP. To handle the case when an MS is both the source and recipient of a multicast session, one needs to minimize the possibility of rebuilding the complete multicast tree at each foreign domain the MS visits.
8. _______ is meant to be implemented for use on the MBONE.
a) MMP
b) RMDP
c) RM2
d) Mobicast
Answer: b
Explanation: The reliable multicast data distribution protocol is meant to be implemented for use on the MBONE. It relies on the use of FEC and ARQ information to provide reliable multicast service.
9. __________ is a reliable multicast protocol and is used for both wired and wireless environments.
a) MMP
b) RMDP
c) RM2
d) Mobicast
Answer: c
Explanation: Reliable mobile multicast is a reliable multicast protocol and is used for both wired and wireless environments. RM2 guarantees sequential packet delivery to all its multicast members without any packet loss.
10. ____________ relies on IGMP.
a) MMP
b) RMDP
c) RM2
d) Mobicast
Answer: c
Explanation: RM2 relies on the Internet group management protocol to manage multicast group membership. RM2 is a hierarchical protocol that divides a multicast tree into subtrees, whereby subcasting within these smaller regions is applied using a tree of retransmission servers.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Long Term Evolution ”.
1. Which UE category supports 64 QAM on the uplink?
a) Only category 5
b) Only category 4
c) Only category 3
d) Category 3,4 and 5
Answer: a
Explanation: Category information is used to allow the eNB to communicate effectively with all the UEs connected to it. The UE-category defines a combined uplink and downlink capability. Only UE category 5 supports 64 QAM on the uplink.
2. What type of handovers is supported by LTE?
a) Hard handover only
b) Soft handover only
c) Hard and soft handover
d) Hard, soft and softest handover
Answer: a
Explanation: LTE supports only hard handover. It does not receive data from two frequencies at the same time because switching between different carrier frequencies is very fast so soft handover is not required.
3. What is the minimum amount of RF spectrum needed for an FDD LTE radio channel?
a) 1.4 MHz
b) 2.8 MHz
c) 5 MHz
d) 20 MHz
Answer: b
Explanation: In telecommunication, Long-Term Evolution is a standard for high-speed wireless communication for mobile devices and data terminals, based on the GSM/EDGE and UMTS/HSPA technologies. The minimum amount of RF spectrum needed for an FDD LTE radio channel is 2.8 MHz.
4. Which organization is responsible for developing LTE standards?
a) UMTS
b) 3GPP
c) 3GPP2
d) ISO
Answer: b
Explanation: The 3rd Generation Partnership Project is a collaboration between groups of telecommunications standards associations, known as the Organizational Partners. LTE introduced in 3GPP R8, is the access part of the Evolved Packet System .
5. Which channel indicates the number of symbols used by the PDCCH?
a) PHICH
b) PDCCH
c) PBCH
d) PCFICH
Answer: d
Explanation: PCFIH channel indicates the number of symbols used by the PDCCH. The actual number of OFDM symbols occupied in any given subframe is indicated in the PCFICH , which is located in the first OFDM symbol of each subframe.
6. How often can resources be allocated to the UE?
a) Every symbol
b) Every slot
c) Every subframe
d) Every frame
Answer: c
Explanation: Resources can be located to the UE every subframe. CCE Index is the CCE number at which the control channel data is allocated. Normally this index changes for each subframe, i.e. even the same PDCCH data allocated in each subframe changes subframe by subframe.
7. What is the largest channel bandwidth a UE is required to support in LTE?
a) 10 MHz
b) 20 MHz
c) 1.4 MHz
d) 5 MHz
Answer: b
Explanation: The LTE format was first proposed by NTT DoCoMo of Japan and has been adopted as the international standard. LTE-Advanced accommodates the geographically available spectrum for channels above 20 MHz.
8. In LTE, what is the benefit of PAPR reduction in the uplink?
a) Improved uplink coverage
b) Lower UE power consumption
c) Reduced equalizer complexity
d) Improved uplink coverage, lower UE power consumption and reduced equalizer
Answer: d
Explanation: PAPR is the relation between the maximum power of a sample in a given OFDM transmit symbol divided by the average power of that OFDM symbol. PAPR reduction in the uplink leads to improved uplink coverage, lower UE power consumption and reduced equalizer complexity.
9. Which RLC mode adds the least amount of delay to user traffic?
a) Unacknowledged mode
b) Acknowledged mode
c) Low latency mode
d) Transparent mode
Answer: d
Explanation: The transparent mode entity in RLC does not add any overhead to the upper layer SDUs. The entity just transmits the SDUs coming from upper layer to MAC.
10. How much bandwidth is required to transmit the primary and secondary synchronization signals?
a) 1.08 MHz
b) 1.4 MHz
c) 930 kHz
d) 20 MHz
Answer: a
Explanation: Cell synchronization is the very first step when UE wants to camp on any cell. 1.08 MHZ is required to transmit the primary and secondary synchronization signals.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “4G Network Architecture”.
1. Which type of cell provides the best level of service for average subscribers?
a) Acceptance cell
b) Barred cell
c) Reserved cell
d) Suitable cell
Answer: d
Explanation: A suitable cell is a cell on which the UE may camp on to obtain normal service. The UE shall have a valid USIM and such a cell shall fulfil all the following requirements. It provides the best level of service for average subscribers.
2. With the normal cyclic prefix, how many symbols are contained in 1 frame?
a) 7
b) 140
c) 12
d) 40
Answer: b
Explanation: There are two different type of Cyclic Prefix. One is normal Cyclic Prefix and the other is ‘Extended Cyclic Prefix’ which is longer than the Normal Cyclic Prefix. Normal cyclic prefix contains 140 symbols in 1 frame.
3. What is the PBCH scrambled with?
a) Current frame number
b) Physical cell ID
c) UE’s CRNTI
d) Not scrambled
Answer: b
Explanation: The PBCH is scrambled prior to modulation with a cell-specific sequence that depends on the cells’ identity. In contrast to the synchronization signals, the PBCH is transmitted on the 72 reserved subcarriers, which are QPSK-modulated.
4. What is the length of the shortest possible PDCCH in bits?
a) 144
b) 288
c) 72
d) 576
Answer: c
Explanation: PDCCH is a physical channel that carries downlink control information . Shortest possible PDCCH is 72 bits.
5. What is the average uploading speed of 4G LTE network?
a) 1-3 Gbps
b) 2-5 Gbps
c) 1-3 Mbps
d) 2-5 Mbps
Answer: d
Explanation: Verizon 4G LTE wireless broadband is 10 times faster than 3G able to handle download speeds between 5 and 12 Mbps and upload speeds between 2 and 5 Mbps.
6. Which of the following is not a part of the characteristic of 4G network?
a) Multirate management
b) Fully converged services
c) Software dependency
d) Diverse user devices
Answer: a
Explanation: 4G is the fourth generation of broadband cellular network technology, succeeding 3G. Its characteristics include fully converged services, software dependency and diverse user devices.
7. What does SGSN stands for?
a) Serving GPRS Support Node
b) Supporting GGSN Support Node
c) Supporting GPRS Support Node
d) Supporting Gateway Support Node
Answer: a
Explanation: The Serving GPRS Support Node is a main component of the GPRS network, which handles all packet switched data within the network, e.g. the mobility management and authentication of the users. The SGSN performs the same functions as the MSC for voice traffic.
8. What location management feature is supported by 4G?
a) Concatenated Location Registration
b) Concurrent Location Register
c) Concatenated Management
d) Collated Location Registration
Answer: a
Explanation: 4G supports concatenated location registration. Concatenated location registration reports to the network that they are concatenated to a common object.
9. In 2007 ____________ announced its plan to transmit its network to 4G standard LTE with joint efforts of Vodafone group.
a) Verizon Wireless
b) AirTouch
c) Netflix
d) V Cast
Answer: a
Explanation: In 2007, Verizon announced plans to develop and deploy its fourth generation mobile broadband network using LTE, the technology developed within the Third Generation Partnership Project standards organization.
10. Hybrid ARQ is part of the ____________ layer.
a) PDCP
b) RLC
c) MAC
d) PHY
Answer: c
Explanation: Hybrid automatic repeat request is a combination of high-rate forward error-correcting coding and ARQ error-control. It is part of the MAC layer.
This set of Wireless & Mobile Communications online quiz focuses on “Orthogonal Frequency Division Multiple Access ”.
1. OFDMA stands for ________
a) omnidirectional frequency division multiple access
b) orthogonal frequency duplex multiple access
c) orthogonal frequency divider multiple access
d) orthogonal frequency division multiple access
Answer: d
Explanation: Orthogonal frequency-division multiple access is a multi-user version of the popular orthogonal frequency-division multiplexing digital modulation scheme. Multiple access is achieved in OFDMA by assigning subsets of subcarriers to individual users.
2. Why is a cyclic prefix required in an OFDMA?
a) To ensure symbol time is an integer number
b) To help overcome multipath and ISI
c) To maintain orthogonality
d) To make OFDMA scalable
Answer: b
Explanation: Use of cyclic prefix is a key element of enabling the OFDM signal to operate reliably. The cyclic prefix acts as a buffer region or guard interval to protect the OFDM signals from intersymbol interference.
3. What does the DC subcarrier indicate?
a) Identity of the cell
b) Antenna configuration
c) Center of OFDM channel
d) Format of data channel
Answer: c
Explanation: All the subcarriers of an OFDM symbol do not carry useful data. In OFDM and OFDMA PHY layers, the DC subcarrier is the subcarrier whose frequency is equal to the RF centre frequency of the transmitting station.
4. What processing step combines multiple OFDM subcarriers into a single signal for transmission?
a) FFT
b) IFFT
c) RF combining
d) Channel mapping
Answer: b
Explanation: IFFT combines multiple OFDM subcarriers into a single signal for transmission. These transforms are important from the OFDM perspective because they can be viewed as mapping digitally modulated input data onto orthogonal subcarriers.
5. Which property of OFDMA system allows adjacent subcarriers to be used without interference?
a) Orthogonality
b) Orthodoxy
c) Octagonality
d) Originality
Answer: a
Explanation: Orthogonality of sub-carriers simply means their correlation is zero. Orthogonality in OFDMA system allows adjacent subcarriers to be used without interference.
6. In OFDMA, what is the relationship between the subcarrier spacing f and symbol time t?
a) f=t
b) f=1/2t
c) f=1/t
d) no relation
Answer: c
Explanation: In OFDMA, the relationship between the subcarrier spacing f and symbol time t is f=1/t. They are inversely proportional.
7. OFDM is a technique for 3G mobile communication.
a) True
b) False
Answer: b
Explanation: OFDM has developed into a popular scheme for wideband digital communication, used in applications such as digital television and audio broadcasting, DSL internet access, wireless networks, power line networks, and 4G mobile communications.
8. OFDM uses complex equalizers.
a) True
b) False
Answer: b
Explanation: ODMA does not use complex equalizers. In OFDM, the equalizer only has to multiply each detected sub-carrier in each OFDM symbol by a constant complex number, or a rarely changed value.
9. When we divide band of Orthogonal Frequency Division Multiplexing into sub bands, it diminishes effects of __________
a) noise
b) collision
c) interference
d) signals absence
Answer: c
Explanation: OFDM uses the same bandwidth to deliver roughly the same data rate as a single carrier modulation by introducing multiple lower-bandwidth channels. Each of the lower-bandwidth channels has a lower rate, and by combining them together, the original rate is achieved.
10. Common data rates of IEEE 802.11 OFDM are ____________
a) 18 Mbps
b) 200 Mbps
c) 50 Mbps
d) 54 Mbps
Answer: a
Explanation: The IEEE 802.11a standard specifies a modulation that divides a high-speed serial information signal into multiple lower-speed sub signals. Common data rates of IEEE 802.11 OFDM is 18 Mbps.
This set of Wireless & Mobile Communications Multiple Choice Questions & Answers focuses on “Multiple Input Multiple Output ”.
1. MIMO stands for _______
a) Many input many output
b) Multiple input multiple output
c) Major input minor output
d) Minor input minor output
Answer: b
Explanation: MIMO stands for Multiple Input and Multiple Output. It refers to the technology where there are multiple antennas at the base station and multiple antennas at the mobile device.
2. In MIMO, which factor has the greatest influence on data rates?
a) The size of antenna
b) The height of the antenna
c) The number of transmit antennas
d) The area of receive antennas
Answer: c
Explanation: By increasing the number of receiving and transmitting antennas, it is possible to linearly increase the throughput of the channel with every pair of antennas added to the system.
3. MIMO was initially developed in the year __________
a) 1980
b) 1990
c) 1980
d) 1975
Answer: b
Explanation: Since the initial development in the year 1990, MIMO Wireless Communications have become integral part of the most forthcoming commercial and next generation wireless data communication systems.
4. MIMO is a smart antenna technology.
a) True
b) False
Answer: a
Explanation: MIMO is one of several forms of smart antenna technology, the others being MISO and SIMO . It is an antenna technology for wireless communications in which multiple antennas are used at both the source and the destination.
5. MIMO technology makes advantage of a natural radio wave phenomenon called _________
a) Reflection
b) Multipath
c) Refraction
d) Diffraction
Answer: b
Explanation: MIMO technology makes use of multipath phenomenon to maximize transmission by receiving bounced signals from obstructions. Multipath is a phenomenon in wave propagation.
6. Which of the following technology does not use MIMO?
a) 4G
b) Wifi
c) WiMax
d) AMPS
Answer: d
Explanation: MIMO is used in mobile radio telephone standards such as recent 3GPP and 3GPP2. In 3GPP, High-Speed Packet Access plus and Long Term Evolution standards take MIMO into account. Moreover, MIMO is also used in Wifi and WiMax.
7. MIMO means both transmitter and receiver have multiple antennas.
a) True
b) False
Answer: a
Explanation: MIMO provides a way of utilising the multiple signal paths that exist between a transmitter and receiver to significantly improve the data throughput available on a given channel with its defined bandwidth. It uses multiple antennas at the transmitter and receiver along with some complex digital signal processing.
8. _______ is a technique of transmit diversity used in UMTSS third-generation cellular systems.
a) STTD
b) SM
c) Collaborative Uplink MIMO
d) MU-MIMO
Answer: a
Explanation: Space Time Transmit Diversity is a technique of transmit diversity used in UMTSS third-generation cellular systems. Space Time Transmit Diversity is optional in the UTRANN air interface, but compulsory for user equipment.
9. _______ is a transmission method used in MIMO wireless communications to transmit encoded data signals independently.
a) STTD
b) SM
c) Collaborative Uplink MIMO
d) MU-MIMO
Answer: b
Explanation: Spatial multiplexing is a transmission method used in MIMO wireless communications to transmit encoded data signals independently and separately from each of the multiple transmit antennas.
10. _______ is an additional open-loop MIMO technique considered by the WiMAX vendors.
a) STTD
b) SM
c) Collaborative Uplink MIMO
d) MU-MIMO
Answer: c
Explanation: Collaborative MIMO is an open-loop MIMO technique considered by the WiMAX vendors to surge the spectral efficiency and capacity of the uplink communications path. It is compared with the regular spatial multiplexing, wherein data streams are transmitted multiplying from multiple antennas on the same device.
