Digital Communication Pune University MCQs
Digital Communication Pune University MCQs
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Digital Communication Signal Processing”.
1. Which circuit is called as regenerative repeaters?
a) Analog circuits
b) Digital circuits
c) Amplifiers
d) A/D converters
Answer: b
Explanation: The main advantage of digital communication is that the signals can be reproduced easily. Thus digital circuits are called as regenerative repeaters.
2. What are the advantages of digital circuits?
a) Less noise
b) Less interference
c) More flexible
d) All of the mentioned
Answer: d
Explanation: Digital circuits are less subject to noise, distortion and interference as it works on digital pulses and also the pulses can be regenerated.
3. How many different combinations can be made from a n bit value?
a) 2
b) 2
c) 2 +1
d) None of the mentioned
Answer: b
Explanation: 2 different combinations can be made from n bit value. For example, from 2 bit value 2 2 different combinations-00,01,10,11 can be made.
4. How many bytes does a gigabyte have?
a) 1 million bytes
b) 10 million bytes
c) 1 billion bytes
d) 10 billion bytes
Answer: c
Explanation: One gigabyte has 1 billion bytes.
5. What is the ASCII value of space?
a) 32
b) 48
c) 96
d) 65
Answer: a
Explanation: The ASCII value of space is 32 and ASCII value of 0 is 48.
6. Which block or device does the data compression?
a) Channel encoder
b) Source encoder
c) Modulator
d) None of the mentioned
Answer: b
Explanation: Source encoder converts the digital or analog signal to a sequence of binary digits. This process is called as source encoding or compression.
7. What is the code rate?
a) k/n
b) n/k
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: Here n is the total bits of sequence and k bits are mapped. Amount of redundancy introduced is given by n/k and its reciprocal is the code rate.
8. Pulse shaping is done by which block or system?
a) Encoder
b) Baseband modulator
c) Pulse code modulator
d) Demodulator
Answer: c
Explanation: Pulse code modulator does filtering process to build pulses that occupy more than one bit time.
9. Equalizer is used for?
a) Filtering
b) Diminish distortion
c) All of the mentioned
d) None of the mentioned
Answer: c
Explanation: Equalizer is used as a filtering option and also diminishes or reduces the distortion.
10. Source coding block is used for?
a) Compressing
b) Digitizing
c) A/D conversion
d) All of the mentioned
Answer: d
Explanation: Source encoding does all these processes-compression, digitizing the signal and performs analog to digital conversion.
11. Which measurement considers phase as an important parameter?
a) Coherent
b) Non-coherent
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: Coherent measurement considers phase as an important parameter.
12. The size of the alphabet M in symbol is calculated as?
a) 2
b) 2 k
c) 2
d) 1+2 k
Answer: b
Explanation: The size of the alphabet is calculated using 2^k where k is the number of bits in the symbol.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Classification of Signals”.
1. What is the period of a signal x?
a) T
b) 2T
c) T/2
d) None of the mentioned
Answer: a
Explanation: A signal is said to be periodic if the duration of one complete cycle is T and it repeats itself after this duration.
2. Which of the given signals are periodic?
a) x = 4 cos
b) x = u – 1/2
c) x = 4u + 2sin
d) x[n] = 2sin
Answer:a
Explanation: T=2Ď€ / 5Ď€ = 2/5, periodic with period 2/5.
3. Check whether the signal is periodic or not?
x = cos + 2sin
a) Periodic with period π/2
b) Periodic with period 2
c) Periodic with period 2/Ď€
d) Not periodic
Answer: a
Explanation: T1=2Ď€ /4 T2=2Ď€ /8, T1/T2 = 2 T=T1 or 2T2, T=Ď€ /2.
4. Find the periodicity of the following signal. x=cost)sint)
a) 30
b) 7
c) 35
d) 5/3
Answer: c
Explanation: T1=2Ď€/2Ď€/7 = 7, T2=2Ď€/3Ď€/5=5/3, T1/T2=5/21, T=21T1 or 5T2, T=35.
5. Find the fundamental period of 1+sin^2 n.
a) 10/3
b) 5
c) 3Ď€/5
d) None of the mentioned
Answer: b
Explanation: sin 2 n = n)/2 = 1/2-1/2cosn ; period = 6Ď€/ = k/N = 3/5 ; N=5.
6. Which signal is called as energy signal?
a) Finite energy and zero power
b) Finite energy and non-zero power
c) Infinite energy and zero power
d) Infinite energy and non-zero power
Answer: a
Explanation: A signal is said to be an energy signal if it has finite energy and zero average power.
7. Which signal is said to be power signal?
a) Infinite power and zero energy
b) Infinite power and non-zero energy
c) Finite power and infinite energy
d) Finite power and zero energy
Answer: c
Explanation: A signal is said to be power signal if it has finite power and infinite energy.
8. Determine the periodicity and also find whether it is a power or energy signal?
e u
a) Periodic and energy signal
b) Non periodic and energy signal
c) Periodic and power signal
d) Periodic and energy signal
Answer: b
Explanation: This is non periodic as it does not repeat itself after any period. And on calculating we will get energy as 1/2a which is finite. Thus it is an energy signal.
9. Continuous Impulse signal is a power or energy signal?
a) Power signal
b) Energy signal
c) Both power and energy
d) Neither power nor energy signal
Answer: d
Explanation: On calculating using the power and energy calculation formula we will arrive at it is neither power nor energy signal.
10. Discrete impulse signal is a power or energy signal?
a) Power signal
b) Energy signal
c) Both power and energy signal
d) Neither power or energy signal
Answer: b
Explanation: On calculating using energy formula we will get the energy of this signal as 1 which is a finite value. Thus it is a energy signal.
11. A unit impulse function has?
a) Large amplitude
b) Zero pulse width
c) Unity weight
d) All of the mentioned
Answer: d
Explanation: An impulse function has infinite or large amplitude, zero pulse width and unity weight which is concentrated at zero.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Spectral Density and Autocorrelation”.
1. Power spectral density function is a?
a) Real and even function
b) Non negative function
c) Periodic
d) All of the mentioned
Answer: d
Explanation: A power signal is periodic signal and its function is a real even and non negative function as per the definition.
2. Energy spectral density defines
a) Signal energy per unit area
b) Signal energy per unit bandwidth
c) Signal power per unit area
d) Signal power per unit bandwidth
Answer: b
Explanation: Energy spectral density gives the signal energy equal to the area under the waveform energy spectral density versus frequency curve.
3. Power spectrum describes distribution of _________ under frequency domain.
a) Mean
b) Variance
c) Gaussian
d) None of the mentioned
Answer: b
Explanation: Power spectrum gives the distribution of variance of a signal in the frequency domain, sampled into spectral components.
4. How can power spectral density of non periodic signal be calculated?
a) By integrating
b) By truncating
c) By converting to periodic
d) None of the mentioned
Answer: b
Explanation: A power signal usually is a periodic signal. But power spectral density of non periodic signal can be calculated by truncating it and observing it in the range of .
5. What is Wiener-Khinchin theorem?
a) Spectral density and auto-covariance makes a fourier transform pair
b) Spectral density and auto-correlatioon makes a fourier tranform pair
c) Spectral density and variance makes a fourier tranform pair
d) None of the mentioned
Answer: b
Explanation: According to the theorem spectral density of a signal x and auto correlation function makes a fourier tranform pair.
6. According to Parseval’s theorem the energy spectral density curve is equal to?
a) Area under magnitude of the signal
b) Area under square of the magnitude of the signal
c) Area under square root of magnitude of the signal
d) None of the mentioned
Answer: b
Explanation: According to Parseval/s theorem the energy spectral density function can be given as equal to the square of the magnitude of the signal x.
7. Spectogram is the graph plotted against?
a) Frequency domain
b) Time domain
c) Frequency & Time domain
d) None of the mentioned
Answer: b
Explanation: Spectral density function is plotted against frequency domain and spectogram is a graph which is plotted against time domain.
8. Autocorrelation is a function which matches
a) Two same signals
b) Two different signal
c) One signal with its delayed version
d) None of the mentioned
Answer: c
Explanation: Autocorrelation is a function that matches a signal with its delayed version.
9. Autocorrelation is a function of
a) Time
b) Frequency
c) Time difference
d) Frequency difference
Answer: c
Explanation: Autocorrelation is a function of time difference as it matches the signal with its delayed version.
10. Autocorrelation is maximum at _______
a) Unity
b) Origin
c) Infinite point
d) None of the mentioned
Answer: b
Explanation: According to its properties autocorrelation is maximum at origin.
11. Autocorrelation function of periodic signal is equal to _______
a) Energy of the signal
b) Power of the signal
c) Its area in frequency domain
d) None of the mentioned
Answer: b
Explanation: Autocorrelation function of a real valued signal is equal to the energy of the signal and auto-correlation function of the periodic signal is equal to the average power of the signal.
12. Autocorrelation is a _______ function.
a) Real and even
b) Real and odd
c) Complex and even
d) Complex and odd
Answer: a
Explanation: According to properties of autocorrelation function it a even function when the frequency value f is real.
13. Autocorrelation function of white noise will have?
a) Strong peak
b) Infinite peak
c) Weak peak
d) None of the mentioned
Answer: a
Explanation: Autocorrelation function curve of continuous time white noise signal has a strong peak.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Random signals”.
1. Random variables give relationship between _____
a) Two random events
b) Probability of occurence of two random events
c) Random event and a real number
d) Random event and its probability of occurrence
Answer: c
Explanation: A random variable gives a functional relationship between a random event and a real number.
2. The distribution function of random variable is
a) P
b) P
c) P
d) P
Answer: a
Explanation: The distribution function of a random variable is the probability that the value taken by the random variable is less than or equal to the real number x.
3. The distribution function of - and is _____
a) 0 and 1
b) 1 and 0
c) Both 0
d) Both 1
Answer: a
Explanation: F is 0 and F is 1.
4. The value of the probability density function of random variable is
a) Positive function
b) Negative function
c) Zero
d) One
Answer: a
Explanation: The probability density function is always greater than 0. It is a non negative function with the area of 1.
5. Which gives the measure of randomness of the random variable?
a) Mean
b) Variance
c) Standard variance
d) Pdf
Answer: b
Explanation: Variance gives the randomness of the random variable. It is the difference between the mean square value and square of the mean.
6. Random process is a function of ______
a) Random event and time
b) Random event and frequency
c) Random event and real number
d) None of the mentioned
Answer: a
Explanation: Random process is a function of two variables: a random event and its time of occurrence.
7. A random process is called as stationary in strict sense if
a) Its statistics vary with shift in time origin
b) Its statistics does not vary with shift in time origin
c) Its autocorrelation vary with shift in time
d) Its autocorrelation does not vary with shift in time
Answer: a
Explanation: A random process is defined to be stationary in a strict sense if its statistics varies with a shift in time origin.
8. For a stationary process, autocorrelation function depends on
a) Time
b) Time difference
c) Does not depend on time
d) None of the mentioned
Answer: b
Explanation: Autocorrelation function depends on the time difference between t1 and t2.
9. The autocorrelation function is maximum at
a) Origin
b) Infinity
c) Origin & Infinity
d) None of the mentioned
Answer: a
Explanation: On substituting different values in the formula for autocorrelation function it wil be maximum at the origin.
10. Standard deviation is ______
a) Rms value of dc
b) Rms value or ac
c) Either ac or dc
d) Neither dc nor ac
Answer: b
Explanation: The standard deviation of a random variable gives the rms value of an ac component.
11. The average power of white noise is
a) Zero
b) Unity
c) Infinity
d) Between zero and one
Answer: c
Explanation: The average power of white noise is infinity because its bandwidth is infinite.
12. White noise has _____ mean and ______ variance.
a) Zero and zero
b) Finite and zero
c) Zero and finite
d) One and zero
Answer: c
Explanation: White noise is a zero mean function with infinite average power, finite variance and infinite bandwidth.
This set of Digital Communication Puzzles focuses on “Baseband System and Signal Transmission through Linear Systems”.
1. The process of data conversion along with formatting the data is called as ______
a) Formatting
b) Modulation
c) Source coding
d) Amplifying
Answer: c
Explanation: The process of converting source information to a digital signal and also formatting it is called as source coding.
2. Formatting is the process which includes
a) Pulse code modulation
b) Sampling
c) Quantization
d) All of the mentioned
Answer: d
Explanation: Formatting is the process in which the source information is converted to digital signals using methods like sampling, pulse code modulation etc.
3. Analog information is converted to digital data using
a) Sampling
b) Quantization
c) Coding
d) All of the mentioned
Answer: d
Explanation: Textual information, analog information and digital information undergoes different processes before converting to binary or digital data. The method used for converting analog information are the given methods.
4. The process that transforms text into binary digits is called as _______
a) Binary coding
b) Data coding
c) Character coding
d) Sampling
Answer: c
Explanation: According to the definition the process of converting textual data to binary digits is called as character coding.
5. For the number of bits, k=1 and number of symbols, M=2 the system is called as
a) Unary
b) Binary
c) Quarternary
d) None of the mentioned
Answer: b
Explanation: For k number of bits, the number of symbols are M=2 k . For k=1 and M=2 the system is called as binary coding.
6. Perform the bit stream partitioning and find the 8-ary waveform for the word ‘THINK’.
a) 1204443464
b) 4643444021
c) 1240443464
d) none of the mentioned
Answer: a
Explanation: “THINK”
ASCII value: T-001010 ; H-000100; I-100100; N-011100; K-110100
8-ary symbols: 1 2 0 4 4 4 3 4 6 4.
7. Find the 32-ary waveform for the word ‘THINK’.
a) 514172520
b) 202517415
c) 541172520
d) None of the mentioned
Answer: a
Explanation: “THINK”
ASCII value: T-001010; H-000100; I-100100; N-011100; K-110100
32-ary symbols: 5 1 4 17 25 20.
8. Find the 8-ary waveform for the word ‘HAPPY’.
a) 4040020246
b) 0440020246
c) 6420200440
d) None of the mentioned
Answer: b
Explanation: “HAPPY”
ASCII value: H-000100; A-100000; P-000010; P-000010; Y-100110
8-ary symbols: 0 4 4 0 0 2 0 2 4 6.
9. Find the 32-ary waveform for the word ‘HAPPY’.
a) 280856
b) 658082
c) 288056
d) None of the mentioned
Answer: a
Explanation: “HAPPY”
ASCII value: H-000100; A-100000; P-000010; P-000010; Y-100110
32-ary symbols: 2 8 0 8 5 6.
10. What are the characteristics of an ideal transmission line?
a) Different amplitude
b) No distortion
c) Time delay
d) All of the mentioned
Answer: d
Explanation: The output from the ideal transmission line has some time delay different amplitude but it should not have distortion it should have the same shape as the input.
11. The point at which the output signal power has fallen to 0.707 of its peak value is called as _____
a) 3db point
b) Half power point
c) 3db & Half power point
d) None of the mentioned
Answer: c
Explanation: The frequency at which the output signal power falls to half the peak value is called as half power point or 3db point.
12. The ratio of the filter bandwidth at -60db and -6db amplitude response point is called as _____
a) Half point factor
b) Normalized factor
c) Shape factor
d) None of the mentioned
Answer: c
Explanation: Shape factor is a measure of how well realizable filter approximates to ideal filter. It is the ratio of filter bandwidths -60db and -6db amplitude points.
13. As the order of the filter increases, the cost ______ and complexity ______
a) Increases, increases
b) Increases, decreases
c) Decreases, increases
d) Decreases, decreases
Answer: a
Explanation: According to the design of the filter as the order increases the complexity and cost of the filter also increases.
14. Which filter has maximum flatness?
a) Chebyshev filter
b) Butter-worth filter
c) High pass filter
d) Normalized filter
Answer: b
Explanation: Butter-worth filter as the best approximation to ideal filter and has the maximum flatness in the filter pass-band.
15. The measure of the width of the main lobe is called as ________
a) Null to null bandwidth
b) Half power bandwidth
c) Normalized bandwidth
d) Absolute bandwidth
Answer: a
Explanation: The main lobe consists of the maximum signal power. The measure of the width of the main lobe is called as null to null bandwidth.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Noise”.
1. SQNR can be improved by _______ sampling rate.
a) Increasing
b) Decreasing
c) Does not depend
d) None of the mentioned
Answer: a
Explanation: SQNR can be improved by increasing the sampling rate.
2. Which offers better SQNR?
a) Delta modulation
b) PCM
c) Delta modulation & PCM
d) None of the mentioned
Answer: b
Explanation: PCM offers better SQNR than delta modulation.
3. Two functions are called as orthogonal if on integrating the product we get
a) One
b) Zero
c) Infinity
d) None of the mentioned
Answer: b
Explanation: Two functions are called as orthogonal if the integral value of integrating the product is zero.
4. Eigen functions are not orthogonal.
a) True
b) False
Answer: b
Explanation: An important property of eigen function is that they are orthogonal.
5. Shot noise is produced by
a) Electrons
b) Photons
c) Electrons & Photons
d) None of the mentioned
Answer: c
Explanation: Shot noise occurs due to particle like the behaviour of electron and photon.
6. Shot noise is avoidable when current is
a) Zero
b) One
c) Infinity
d) None of the mentioned
Answer: a
Explanation: Shot noise is avoidable only when the current is zero.
7. Shot noise is
a) Stochastic process
b) Poisson process
c) Stochastic & Poisson process
d) None of the mentioned
Answer: b
Explanation: Shot noise is a Poisson process.
8. Which occurs due to equilibrium fluctuations?
a) Thermal noise
b) Johnson noise
c) Thermal & Johnson noise
d) None of the mentioned
Answer: c
Explanation: Thermal noise occurs due to equilibrium fluctuations. This noise is also called as Johnson noise.
9. Flicker noise is called as
a) White noise
b) Pink noise
c) Brown noise
d) None of the mentioned
Answer: b
Explanation: Flicker noise is called as pink noise.
10. Which has the same power spectral density?
a) White noise
b) Brown noise
c) White & Brown noise
d) None of the mentioned
Answer: a
Explanation: White noise has same power spectral density where as it decreases in case of brown noise.
This set of Digital Communication Interview Questions & Answers focuses on “Noise -2”.
1. Which noise is useful for dithering?
a) White noise
b) Pink noise
c) Brown noise
d) None of the mentioned
Answer: c
Explanation: Brown noise is sometimes useful for dithering.
2. Capacity of a channel can be increased by
a) Increasing channel bandwidth
b) Increasing signal power
c) Increasing channel bandwidth &signal power
d) None of the mentioned
Answer: c
Explanation: Capacity of a channel can be increased by increasing channel bandwidth and also by increasing signal power.
3. Capacity of the channel can be increased by reducing in band noise power.
a) True
b) False
Answer: a
Explanation: Capacity of a channel can be increased by reducing the in band noise power.
4. Noise has
a) Infinite energy
b) Infinite power
c) Infinite energy & power
d) None of the mentioned
Answer: a
Explanation: Noise has infinite energy signal.
5. Thermal noise is
a) Energy signal
b) Power signal
c) Energy & Power signal
d) None of the mentioned
Answer: b
Explanation: Thermal noise is considered as power signal as it has infinite energy.
6. Thermal noise is a wide sense stationary process.
a) True
b) False
Answer: a
Explanation: Thermal noise is modeled as a wide sense stationary stochastic process.
7. The maximum likelihood function is
a) Positive
b) Negative
c) Any of the mentioned
d) None of the mentioned
Answer: a
Explanation: The maximum likelihood function is always positive.
8. Matched filtering operation includes
a) Spectral phase matching
b) Spectral amplitude matching
c) Spectral phase & amplitude matching
d) None of the mentioned
Answer: c
Explanation: Matched filtering includes two operations – spectral amplitude matching and spectral phase matching.
9. Matched filter is
a) Linear
b) Non linear
c) Linear & Non linear
d) None of the mentioned
Answer: a
Explanation: Matched filter is a linear filter.
10. Which consists of less ISI?
a) Nyquist filter
b) Raised cosine filter
c) Nyquist & Raised cosine filter
d) None of the mentioned
Answer: b
Explanation: Raised cosine filter results in less ISI than Nyquist filter.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Digital communication”.
1. Digital communication is _______ to environmental changes?
a) Less sensitive
b) More sensitive
c) Does not depend
d) None of the mentioned
Answer: a
Explanation: Digital communication is less sensitive to environmental changes like temperature etc.
2. Advantages of digital communication are
a) Easy multiplexing
b) Easy processing
c) Reliable
d) All of the mentioned
Answer: d
Explanation: Digital communication is a very reliable communication. It is easy for multiplexing, easy for signalling and processing etc.
3. What is necessary for digital communication?
a) Precision timing
b) Frame synchronization
c) Character synchronization
d) All of the mentioned
Answer: d
Explanation: Bit, character, frame synchronization and precision timing is necessary for digital communication. This is considered as a disadvantage of digital communication.
4. What are the disadvantages of digital communication?
a) Needs more bandwidth
b) Is more complex
c) Needs more bandwidth & Is more complex
d) None of the mentioned
Answer: c
Explanation: Digital communication needs more bandwidth, has higher complexity and little performance degradation occurs during analog to digital conversion and vice versa.
5. Examples of digital communication are
a) ISDN
b) Modems
c) Classical telephony
d) All of the mentioned
Answer: d
Explanation: Some of the examples of digital communication systems are classical telephony, ISDN, Modems, LANs, PCM TDM etc.
6. Which system uses digital transmission?
a) ISDN
b) LANs
c) ISDN & LANs
d) None of the mentioned
Answer: c
Explanation: Though the signal type is analog or digital, the transmission takes place in the digital domain in ISDN and LANs.
7. The interval of frequencies outside which the spectrum is zero is called as ________
a) null to null bandwidth
b) normalized bandwidth
c) absolute bandwidth
d) none of the mentioned
Answer:c
Explanation: The measure of frequencies outside which spectrum is zero is called as absolute bandwidth. It is usually infinite.
8. The attenuation level in bounded power spectral density is
a) 35
b) 50
c) 35 & 50
d) none of the mentioned
Answer: c
Explanation: Bounded power spectral density is the bandwidth outside which the spectrum must have fallen to a stated level below that found at the band center.
9. Synchronization available in digital communication are
a) Symbol synchronization
b) Frame synchronization
c) Carrier synchronization
d) All of the mentioned
Answer: d
Explanation: The synchronization techniques available in digital communication are symbol synchronization, frame synchronization and carrier synchronization.
10. Digital system includes
a) Better encryption algorithm
b) Difficult data multiplexing
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: Digital system has an advantage of better encryption algorithm, easier data multiplexing and more reliability.
11. Analog to digital conversion includes
a) Sampling
b) Quantization
c) Sampling & Quantization
d) None of the mentioned
Answer: c
Explanation: Analog to digital conversion is a two step process which includes sampling and quantization.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Digital communication system”.
1. What are the main features of a receiver?
a) Synchronization
b) Multiple parallel receiver chain
c) Synchronization & Multiple parallel receiver chain
d) None of the mentioned
Answer: c
Explanation: The main features of a receiver which increases its complexity are synchronization of carrier, phase, and timing and multiple parallel receiver chain.
2. What conditions must be fulfilled in a good digital communication system?
a) High data rate
b) High fidelity
c) Low transmit power
d) All of the mentioned
Answer: d
Explanation: Some of the conditions which must be satisfied in a digital communication system are high data rate, high fidelity, low bandwidth, low transmit power and low transmitter and receiver complexity.
3. Wired channels are
a) Lossy
b) Lossless
c) Lossy & Lossless
d) None of the mentioned
Answer: a
Explanation: Wired channels are lossy channels.
4. The equivalent temperature in a receiver design must be kept
a) Low
b) High
c) Does not affect the receiver
d) None of the mentioned
Answer: a
Explanation: The equivalent temperature is the function of the receiver design and it must be always kept low.
5. Which corrects the sampling time problem in a digital system?
a) Interpolator
b) Decimator
c) Equalizer
d) Filter
Answer: a
Explanation: Interpolator corrects the sampling time problem using discrete time processing.
6. What are the main features of a transmitter?
a) Higher clock speed
b) Linear power amplifier
c) Directional antennas
d) All of the mentioned
Answer: d
Explanation: Some of the main features which make the transmitter complex are higher clock speed, higher transmit power, directional antennas and need for a linear amplifier.
7. Transmission media used in low frequency band are
a) Air
b) Water
c) Copper cable
d) All of the mentioned
Answer: d
Explanation: Air, water and copper cable can be used as transmission media in low frequency band communication.
8. Transmission media used for medium frequency band are
a) Coaxial cable
b) Copper cable
c) Optical fiber
d) All of the mentioned
Answer: b
Explanation: For medium frequency band communication air and copper cable can only be used as a transmission medium.
9. Matched filter technique is used to
a) Increase SNR
b) Decrease SNR
c) SNR is not affected
d) None of the mentioned
Answer: a
Explanation: Matched filter technique is a demodulation process used to increase SNR.
10. Matched filter can also be used as least squares estimator.
a) True
b) False
Answer: a
Explanation: A matched filter can also be used as a least squares estimator.
11. Digital communication system can handle
a) Analog signals
b) 1D signals
c) 2D signals
d) All of the mentioned
Answer: d
Explanation: Digital communication system can handle signals that are analog or those that are already digital. It can also handle 1D and 2D signals.
12. The information source of a digital communication system can be
a) Packetized
b) Continuous
c) Packetized & Continuous
d) None of the mentioned
Answer: c
Explanation: Information source comes from a higher networking layer. It can be continuous or packetized.
This set of tough Digital Communication Questions & Answers focuses on “Digital Communication System”.
1. Which are the common transmission media used in digital communication system?
a) Coaxial cable
b) Twisted copper cable
c) Radio frequency bands
d) All of the mentioned
Answer: d
Explanation: Some of the commonly used physical transmission media are twisted copper cable, good quality coaxial cable and radio frequency bands.
2. The basic transmission-reception system is a ______ system.
a) Two block system
b) Three block system
c) Four block system
d) None of the mentioned
Answer: b
Explanation: The basic transmission-reception consists of three blocks – transmitter, transmission medium, receiver.
3. Modulation channel consists of
a) Amplifier
b) Signal processing units
c) Amplifier & Signal processing units
d) None of the mentioned
Answer: c
Explanation: When information is transmitted through large distance it must be amplified. Modulation channel consists of an amplifier and other signal processing units.
4. Modulation channel does not accept an analog input signal.
a) True
b) False
Answer: b
Explanation: Modulation channel accepts analog signal as input and delivers another version of modulated signal as analog waveform.
5. If operating frequency bands are higher ______ is available.
a) Smaller bandwidth
b) Larger bandwidth
c) Smaller & Larger bandwidth
d) Cannot be determined
Answer: b
Explanation: Larger bandwidth is available when operating frequency bands are higher.
6. Ground wave communication occurs in
a) Low frequency band
b) Medium frequency band
c) Low & Medium frequency band
d) None of the mentioned
Answer: c
Explanation: Ground wave communication can take place in both low frequency and medium frequency bands.
7. Sampling theorem is used for converting
a) Continuous time signal to discrete
b) Discrete to continuous time signal
c) Continuous time signal to discrete & vice versa
d) None of the mentioned
Answer: c
Explanation: Sampling theorem is used for converting continuous time signal to discrete type signal and vice versa.
8. A signal can be recovered from its sample by using
a) Low pass filter
b) High pass filter
c) Band pass filter
d) Band stop filter
Answer: a
Explanation: A signal can be recovered from its sampled version by using an ideal low pass filter.
9. Which is practically realizable?
a) A train of pulses
b) Impulse train
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: A train of pulses with narrow bandwidth is realizable than the impulse train.
10. In flat top sampling scheme, ______ is kept constant after sampling.
a) Amplitude
b) Phase
c) Frequency
d) Time period
Answer: a
Explanation: In flat top sampling scheme, the amplitude is kept constant after sampling.
11. Loop filter is a ______ used to reduce noise.
a) Low pass filter
b) High pass filter
c) Band pass filter
d) Band reject filter
Answer: a
Explanation: Loop filter is a low pass filter used to reduce noise.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Types of Processes”.
1. A stationary stochastic process has
a) Finite energy signal
b) Infinite zero signal
c) Zero energy signal
d) None of the mentioned
Answer: b
Explanation: A stationary stochastic process is an infinite energy signal and hence its fourier transform does not exist.
2. The power spectral density function of the stochastic process is
a) Real
b) Odd
c) Real & odd
d) None of the mentioned
Answer: a
Explanation: The power spectral density function of a stochastic function is real and even.
3. For a periodic function, the spectral density and auto correlation functions form
a) Fourier transform pair
b) Laplace transform pair
c) Hilbert transform pair
d) Z transform pair
Answer: a
Explanation: For a periodic function, the spectral density and auto correlation function forms fourier transform pair.
4. The capacity of a channel is given by
a) Number of digits used in coding
b) Volume of information it can take
c) Maximum rate of information transmitted
d) Bandwidth requires information
Answer: c
Explanation: The capacity of the channel is given by a maximum rate of information transmitted.
5. In a communication system, a process in which statistical averages and time averages are equal is called as
a) Stationary
b) Ergodic
c) Gaussian
d) Poisson
Answer: b
Explanation: In ergodic process statistical averages and time averages are equal.
6. A rectangular pulse of duration T is applied to a matched filter. The output of the filter is a
a) Rectangular pulse of duration T
b) Rectangular pulse of duration 2T
c) Triangular pulse
d) Sine function
Answer: c
Explanation: The output of the matched filter when a rectangular pulse of duration T is applied is a triangular pulse.
7. The line code has a zero dc component for pulse transmission of random binary data is
a) NRZ
b) RZ
c) Alternate mark inversion
d) None of the mentioned
Answer: c
Explanation: The line code has a zero dc component for pulse transmission of random binary data is alternate mark inversion.
8. The auto-correlation of white noise is
a) A delta function
b) A constant
c) Gaussian
d) None of the mentioned
Answer: a
Explanation: The auto-correlation of white noise is a delta function.
9. Ionospheric communication can take place in
a) High frequency band
b) Very high frequency band
c) Ultra high frequency band
d) Super high frequency band
Answer: a
Explanation: Ionospheric communication takes place in high frequency band.
10. Satellite communication takes place in
a) Ultra high frequency band
b) Super high frequency band
c) Ultra & Super high frequency band
d) None of the mentioned
Answer: c
Explanation: Satellite communication takes place in ultra high and super high frequency bands.
11. The weighing matrix is
a) Positive quantity
b) Inverse of covariance matrix of the input vector
c) Positive quantity & Inverse of a covariance matrix of the input vector
d) None of the mentioned
Answer: c
Explanation: W is a positive definite weighing matrix and it is inverse of covariance matrix of the input vector.
12. Non uniform tree _____ bit rate.
a) Increases
b) Decreases
c) Does not affect
d) None of the mentioned
Answer: a
Explanation: Non uniform tree for binary search increases the bit rate.
13. Two clock with rates T1 and T2 are incommensurate if its ratio is
a) Rational
b) Irrational
c) Equal
d) None of the mentioned
Answer: b
Explanation: Two clock with rates T1 and T2 are incommensurate if its ratio is irrational.
14. Possible timing error detection methods are
a) Early late
b) Zero crossing
c) Early late & Zero crossing
d) None of the mentioned
Answer: c
Explanation: Methods for timing error detections are early late timing error detection and zero crossing timing error detection method.
15. Which has minimum power loss during transmission?
a) Twisted Copper cable at 1KHz
b) Optical fiber
c) Twisted copper cable at 100KHz
d) Wave guide
Answer: a
Explanation: Twisted copper wire at 1KHz have low power loss.
16. Voltage generated by lightning is an example for
a) Energy signal
b) Power signal
c) Energy & Power signal
d) None of the mentioned
Answer: a
Explanation: Voltage generated by lightning is an example for energy signal and it has very high power.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Signals”.
1. In the digital communication system, signals in different frequency bands are
a) Orthogonal
b) Non orthogonal
c) Orthogonal & Non orthogonal
d) None of the mentioned
Answer: a
Explanation: In digital communication system, signals from different frequency bands are orthogonal thus interference won’t occur.
2. Properties of impulse function are
a) Symmetry
b) Time scaling
c) Shifting
d) All of the mentioned
Answer: d
Explanation: Some of the properties of impulse function are symmetry, time scaling and shifting.
3. Properties of Fourier transform are
a) Duality property
b) Time shifting property
c) Modulation property
d) All of the mentioned
Answer: d
Explanation: Some of the properties of Fourier transform are duality property, time scaling property, time shifting property, modulation property and many more.
4. A base-band signal can be up-converted using
a) Sine wave
b) Cosine wave
c) Filtering
d) None of the mentioned
Answer: b
Explanation: A base-band signal can be up-converted to band-pass filter by applying cosine wave.
5. A band-pass signal can be down-converted using
a) Sine wave
b) Cosine wave
c) Time delayed wave
d) None of the mentioned
Answer: b
Explanation: For down-conversion of a band-pass signal also cosine signal is used and multiplied with it.
6. In down-conversion multiplication with cosine wave is followed by
a) Low pass filter
b) High pass filter
c) Bandpass filter
d) Bandstop filter
Answer: a
Explanation: Downconversion of bandpass signal includes multiplication with sine wave followed by low pass filtering.
7. ADSL has _____ information channels.
a) One
b) Three
c) Four
d) None of the mentioned
Answer: b
Explanation: Asymmetric digital subscriber line has three information channels – a high speed downstream channel, ISDN channel and medium speed duplex channel.
8. Fourier transform of a signal gives the
a) Frequency content
b) Bandwidth
c) Frequency content & Bandwidth
d) None of the mentioned
Answer: c
Explanation: Fourier transform of a signal give the frequency content and also determines the bandwidth of the signal.
9. Random things in a communication system are
a) Timing offset
b) Device frequency
c) Attenuation
d) All of the mentioned
Answer: d
Explanation: Some of the random things in the communication system are noise in the channel, attenuation, fading, channel filtering, device frequency, phase and timing offset.
10. Which can be used for periodic and non periodic?
a) Fourier series
b) Fourier transforms
c) Fourier series & transforms
d) None of the mentioned
Answer: b
Explanation: Fourier series is limited to only periodic signals where as Fourier transforms and laplace transforms can be used for both periodic and non periodic signals.
11. A band-pass signal has a Fourier transform equal to
a) One
b) Zero
c) Infinity
d) Cannot be determined
Answer: b
Explanation: A band-pass signal has a Fourier transform equal to zero for all value in both frequency and time domain.
12. A band-pass signal has
a) DC component
b) No DC component
c) No side lobes
d) Cannot be determined
Answer: b
Explanation: A band-pass signal has no DC components and has Fourier transform equal to zero. Outside the band it will not be exactly zero. Thus this results in presence of side lobes.
13. Which are orthonormal signal representation?
a) Sine and cosine at same frequency
b) Wavelets
c) Complex sinusoids at a different frequency
d) All of the mentioned
Answer: d
Explanation: Some of the common orthonormal signal representations are sine and cosine at the same frequency, Fourier serier, sinc functions centered at sampling times, wavelets etc.
14. Constellation diagram is plotted in
a) Constellation space
b) Signal space
c) Orthogonal space
d) Boundary space
Answer: b
Explanation: The constellation diagram is plotted in a space called as signal space.
15. Cumulative distributive function is
a) Non negative
b) Non decreasing
c) Non negative & decreasing
d) None of the mentioned
Answer: c
Explanation: Cumulative distribution function is non negative and non decreasing function.
16. Which are non negative functions?
a) PDF
b) PMF
c) PDF & PMF
d) None of the mentioned
Answer: c
Explanation: PDF, PMF and CDF are non negative functions.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Formatting analog information”.
1. The sampling process includes methods such as
a) Filtering
b) Sample and hold
c) Amplifying
d) None of the mentioned
Answer: b
Explanation: The analog data is converted to digital data through sampling. Sampling is done using sample and hold mechanism which uses transistor, capacitor or shutter etc.
2. The output of sampling process are called as ________
a) Pulse code modulation
b) Pulse amplitude modulation
c) Frequency modulation
d) Amplitude modulation
Answer: b
Explanation: In sampling process, input data is split up into samples whose output samples are called as pulse amplitude modulation as the amplitude of the samples is derived from the input waveform.
3. According to Sampling theorem
a) Ts is greater than 1/2fm
b) Ts is lesser than 1/2fm
c) Ts is equal to 1/2fm
d) Ts is lesser than or equal to 1/2fm
Answer: d
Explanation: By sampling theorem the input can be derived from the output samples if Ts is lesser than or equal to 1/2fm.
4. The fourier tranform of one impulse train is also another impulse train with a period of the output equal to the
a) Period of the input
b) Reciprocal of the period of input signal
c) Half the period of input
d) Twice the period of the input
Answer: b
Explanation: When we perform fourier tranform on one impulse train we will get another impulse train with its period reciprocally related to each other.
5. The process in which the top of each pulse in the output samples retains the shape of the analog segment is called as ________
a) Natural sampling
b) Ideal sampling
c) Aliasing
d) None of the mentioned
Answer: a
Explanation: In the method of natural sampling, the top of each pulse in the sampled sequence retains the same shape of the analog input signal.
6. The effects of aliasing are ________
a) Attenuation of high frequency spectral replicates
b) Non uniform spectral gain applied to desired baseband spectrum
c) Attenuation and non uniform spectral gain
d) None of the mentioned
Answer: c
Explanation: Aliasing is due to undersampling and its effects are attenuation and non uniform spectral gain.
7. Aliasing can be removed using
a) Prefiltering
b) Postfiltering
c) Prefiltering & Postfiltering
d) None of the mentioned
Answer: c
Explanation: Aliasing can be removed using both prefiltering and postfiltering but prefiltering is most effective and preferred.
8. Which process is more economical?
a) Undersampling
b) Oversampling
c) Aliasing
d) None of the mentioned
Answer: b
Explanation: Oversampling is most economic way of sampling or for converting analog information to digital as performing signal processing using digital system is less costlier than doing it with high performace analog system.
9. Flat top sampling or practical sampling has
a) Same frequency
b) Same amplitude
c) Same time difference
d) None of the mentioned
Answer: b
Explanation: In flat top sampling the top of the pulses are flat which in turn means that they have the same amplitude.
10. Multiplication of input signal with pulse train is done in ________ sampling.
a) Impulse sampling
b) Natural sampling
c) Flat top sampling
d) None of the mentioned
Answer:b
Explanation: In impulse sampling the input signal is multiplied with impulse train and in natural sampling it is multiplied with pulse train.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Sources of corruption”.
1. The main sources of corruption are
a) Sampling and quantizing effects
b) Channel effects
c) Sampling, quantizing and channel effects
d) None of the mentioned
Answer: c
Explanation: The analog signal obtained from sampling quantization and transmitted pulses will have corruption from several sources in which the two main sources are quantizing and sampling effect and channel effect.
2. The distortion in quantization is called as
a) Round off error
b) Truncation error
c) Round off & Truncation error
d) None of the mentioned
Answer: c
Explanation: After sampling and quantization of input signals, the ouput sampled sequence consists of some distortion which can be called as round off error or truncation error.
3. In quantization process, the amount of quantization noise is _______________ to number of levels.
a) Directly proportional
b) Inversely proportional
c) Independent
d) None of the mentioned
Answer: b
Explanation: The distortion introduced to approximate the analog signal is called as quantization noise. The amount of this noise is inversely proportional to number of levels employed in quantization process.
4. Saturation noises can be avoided or reduced by
a) Automatic gain control
b) Amplifying
c) Filtering
d) None of the mentioned
Answer: a
Explanation: When the difference between input and output signal increases, we say that analog to digital converter is working in saturation. This introduces saturation noise or error. This can be reduced by using automatic gain control.
5. Timing jitter can be reduced by
a) Good power supply isolation
b) Stable clock reference
c) Good power supply isolation & Stable clock reference
d) None of the mentioned
Answer: c
Explanation: Jitter occurs when there is a slight position change in the sampled signals. This timing jitter can be controlled by power supply isolation and clock reference.
6. The reasons for the threshold effect are
a) Thermal noise
b) Interference from other users
c) Interference from circuit switching transients
d) All of the mentioned
Answer: d
Explanation: The channel noise which is caused by thermal noise interference from other users and circuit switching transients is called as threshold effect.
7. When channel bandwidth is greater than the pulse bandwidth, it causes
a) Intersignal interference
b) Intersymbol interference
c) Bandwidth error
d) None of the mentioned
Answer:b
Explanation: When channel bandwidth is greater than pulse bandwidth, the signal widens and expands exceeding the symbol duration which causes intersymbol interference.
8. The _____________ corresponds to average quantization noise power.
a) Mean
b) Variance
c) Probability density function
d) None of the mentioned
Answer: b
Explanation: The variance corresponds to average quantization noise power. It is calculated assuming the quantization noise and probability distribution function.
9. Signal to noise ratio increases as ___________ increases.
a) Quantization level
b) Square of quantization level
c) Square root of quantization level
d) None of the mentioned
Answer: b
Explanation: On calculating the number of levels, quantization error and power and also signal to noise ration we can find that signal to noise ratio depends directly on square of number of quantization levels.
10. Signal to noise ratio is infinite when
a) Quantization noise is zero
b) Number of levels are infinite
c) Quantization noise is zero & Number of levels are infinite
d) None of the mentioned
Answer: c
Explanation: In the limit L tends to infinity and signal to quantization noise ratio tends to infinity when quantization levels are infinite and quantization noise is zero.
11. The ratio of average signal power and quantization noise is
a) 3L 2
b) L 2/3
c) 2L 3
d) L 3/2
Answer: a
Explanation: On calculating the signal power and the quantization noise, and on taking its ratio it depends on the number of quantization level L and we get as 3L 2 .
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Pulse code modulation”.
1. The signals which are obtained by encoding each quantized signal into a digital word is called as
a) PAM signal
b) PCM signal
c) FM signal
d) Sampling and quantization
Answer: b
Explanation: Pulse code modulation is the name for the class of signals which are obtained by encoding the quantized signals into a digital word.
2. The length of the code-word obtained by encoding quantized sample is equal to
a) l=logL
b) l=logL
c) l=2logL
d) l=logL/2
Answer: a
Explanation: The quantized sample which are digitally encoded into l bit value code-word. The length l can be calculated as l=logL.
3. Quantization noise can be reduced by ________ the number of levels.
a) Decreasing
b) Increasing
c) Doubling
d) Squaring
Answer: b
Explanation: The process of quantization replaces the true signal with the approximation. By increasing the number of quantization level the quantization noise can be reduced.
4. In PCM encoding, quantization level varies as a function of ________
a) Frequency
b) Amplitude
c) Square of frequency
d) Square of amplitude
Answer: b
Explanation: In linear PCM the quantization levels are uniform. But in normal PCM encoding the quantization level vary according to the amplitude, based of A-law of Myu-law.
5. What is bit depth?
a) Number of quantization level
b) Interval between two quantization levels
c) Number of possible digital values to represent each sample
d) None of the mentioned
Answer: c
Explanation: One of the properties of PCM signal which determines its stream fidelity is bit depth which is the number of possible digital values that can be used to represent each sample.
6. Choosing a discrete value that is near but not exactly at the analog signal level leads to
a) PCM error
b) Quantization error
c) PAM error
d) Sampling error
Answer: b
Explanation: One of the limitations of PCM is quantization error which occurs when we choose a discrete value at some near by value and not at the analog signal level.
7. In PCM the samples are dependent on ________
a) Time
b) Frequency
c) Quanization leavel
d) Interval between quantization level
Answer: a
Explanation: The samples depend on time,an accurate clock is required for accurate reproduction.
8. DPCM encodes the PCM values based on
a) Quantization level
b) Difference between the current and predicted value
c) Interval between levels
d) None of the mentioned
Answer: b
Explanation: Differential PCM encodes the PCM value based on the difference between the previous sample and the present sample value.
9. Delta modulation uses _____ bits per sample.
a) One
b) Two
c) Four
d) Eight
Answer: a
Explanation: Delta modulation is used for analog to digital conversion and vice versa. It is a simple form of DPCM. Its uses 1 bit per sample. It also depends on the difference between the current and previous sample values.
10. Sample resolution for LPCM ____ bits per sample.
a) 8
b) 16
c) 24
d) All of the mentioned
Answer: d
Explanation: Common sampling resolution for LPCM are 8, 16, 20, 24 bits per sample.
11. Adaptive DPCM is used to
a) Increase bandwidth
b) Decrease bandwidth
c) Increase SNR
d) None of the mentioned
Answer: b
Explanation: Adaptive DPCM is used to decrease required bandwidth for the given SNR.
This set of Digital Communication Questions & Answers for freshers focuses on “Uniform and Non Uniform Quantization”.
1. The size of the quantile interval is called as
a) Inter level
b) Step size
c) Quantile size
d) Level width
Answer: b
Explanation: The interval between the quantization levels is called as step size.
2. Uniform quantization provides better quantization for
a) Weak signals
b) Strong signals
c) Weak & Strong signals
d) None of the mentioned
Answer: b
Explanation: Signal to noise ratio is worse for weak level signals.so it provides better quantization for high level signals.
3. Non uniform quantization provides better quantization for
a) Weak signals
b) Coarse signals
c) Weak & Coarse signals
d) None of the mentioned
Answer: a
Explanation: According to signal to noise level ratio non uniform quantization provides better quantization for weak signals.
4. In non uniform quantization, the quantization noise is _______ to signal size.
a) Inversely proportional
b) Directly proportional
c) Equal
d) Double
Answer: b
Explanation: In sampling and quantization, the quantization noise is directly dependent on signal size.
5. The output SNR can be made independent of input signal level by using
a) Uniform quantizer
b) Non uniform quantizer
c) Uniform & Non uniform quantizer
d) None of the mentioned
Answer: b
Explanation: The weak signal experiences poorer SNR compared to high level signals. So if non uniform quantizer like logarithmic compressor is used the SNR ratio can be made independent of input signal level.
6. Companding is the process of
a) Compression
b) Expansion
c) Compression & Expansion
d) None of the mentioned
Answer: c
Explanation: The given signal is first compressed using a logarithmic compressor and then it is given as input to the uniform quantizer. Both these steps together is called as companding.
7. Which value of ÎĽ corresponds to linear amplification?
a) ÎĽ=0
b) ÎĽ=1
c) ÎĽ>0
d) ÎĽ<0
Answer: a
Explanation: In ÎĽ-law compression characteristics, we get linear amplification or uniform quantization when ÎĽ=0.
8. What is the standard value of ÎĽ in ÎĽ-law ?
a) 128
b) 255
c) 256
d) 0
Answer: b
Explanation: The standard value of ÎĽ in ÎĽ-law is 255.
9. The standard value of A in A-law is
a) 87
b) 88
c) 86.7
d) 87.6
Answer: d
Explanation: Another famous compression characteristic used is A-law. In this law, the standard value of A is 87.6.
10. Which type of quantization is most preferable for audio signals for a human ear?
a) Uniform quantization
b) Non uniform quantization
c) Uniform & Non uniform quantization
d) None of the mentioned
Answer: b
Explanation: The human ear is sensitive to quantization error in small values so non uniform quantization is more preferable than uniform quantization.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Baseband transmission – 1”.
1. Which waveforms are also called as line codes?
a) PCM
b) PAM
c) FM
d) AM
Answer: a
Explanation: When pulse modulation is applied to binary symbol we obtain pulse code modulated waveforms. These waveforms are also called as line codes.
2. When pulse code modulation is applied to non binary symbols we obtain waveform called as
a) PCM
b) PAM
c) M-ary
d) line codes
Answer: c
Explanation: When pulse code modulation is applied to binary symbols we get PCM waveforms and when it is applied to non binary symbols we obtain M-ary waveforms.
3. Examples of PCM waveforms are
a) Non return to zero
b) Phase encoded
c) Multilevel binary
d) All of the mentioned
Answer: d
Explanation: Some of the examples or classification of pulse code modulated signals are non return to zero, return to zero, phase encoded, multilevel binary etc.
4. Which type is used and preferred in digital logic circuits?
a) NRZ-L
b) NRZ-M
c) NRZ-S
d) None of the mentioned
Answer: a
Explanation: NRZ-L is extensively used in digital logic circuits. In this method, logic 1 is represented by one voltage level and logic 0 is represented by another voltage level.
5. Which method is called as differential encoding?
a) NRZ-L
b) NRZ-M
c) NRZ-S
d) None of the mentioned
Answer: b
Explanation: In NRZ-M, logic 1 is represented by a change in voltage level and logic 0 is represented by no change in level. This is called as differential encoding.
6. Which method is preferred in magnetic tape recording?
a) NRZ-L
b) NRZ-M
c) NRZ-S
d) None of the mentioned
Answer: b
Explanation: NRZ-M is also called as differential encoding and it is most preferred in magentic tape recording.
7. NRZ-S is complement of _______
a) NRZ-L
b) NRZ-M
c) NRZ-L & NRZ-M
d) None of the mentioned
Answer: b
Explanation: NRZ-S is a complement of NRZ-M. Logic 0 is represented by a change in voltage level and logic 1 is represented as no change in voltage level.
8. The return to zero waveform consists of
a) Unipolar RZ
b) Bipolar RZ
c) RZ-AMI
d) All of the mentioned
Answer: d
Explanation: Different types of return to zero waveforms are unipolar RZ, bipolar RZ, RZ-AMI. These are used in baseband transmission and in magnetic recording.
9. Phase encoded group consists of
a) Manchester coding
b) Bi-phase-mark
c) Miller coding
d) All of the mentioned
Answer: d
Explanation: Different types of phase encoded waveform consists of manchester coding, bi-phase-mark, bi-phase-space, delay modulation.
10. In which waveform logic 1 is represented by half bit wide pulse and logic 0 is represented by absence of pulse?
a) Unipolar RZ
b) Bipolar RZ
c) RZ-AMI
d) Manchester coding
Answer: a
Explanation: In unipolar RZ waveform, logic 1 is represented by half bit wide pulse and logic 0 is represented by the absence of a pulse.
11. In which waveform logic 1 and logic 0 are represented by opposite one half bit wide pulses?
a) Unipolar RZ
b) Bipolar RZ
c) RZ-AMI
d) Manchester coding
Answer: b
Explanation: In bipolar return to zero waveform ones and zeroes are represented by opposite level pulses one half bit wide pulses.
12. In which waveform logic 1 is represented by equal amplitude alternating pulses?
a) Unipolar RZ
b) Bipolar RZ
c) RZ-AMI
d) Manchester coding
Answer: c
Explanation: In RZ-AMI logic 1 is represented by equal amplitude alternating pulses and logic 0 is represented by the absence of a pulse.
This set of Digital Communication Interview Questions & Answers for freshers focuses on “Baseband Transmission”.
1. Application of phase encoded binary signals are
a) Optical communication
b) Magnetic recording
c) Satellite telemetry
d) All of the mentioned
Answer: d
Explanation: Some of the fields where phase encoded waveforms is being used are optical communication, magnetic tape recording, satellite telemetry etc.
2. In which waveform one is represented by half bit wide pulse positioned during the first half and zero is represented by half bit wide pulse positioned in the second half?
a) Bi-p-L
b) Bi-p-M
c) Bi-p-S
d) Delay modulation
Answer: a
Explanation: In bi-phase-level one is represented by half bit wide pulse positioned during the first half and zero is represented by half bit wide pulse positioned in the second half.
3. Which binary waveform uses three levels?
a) Bipolar RZ
b) RZ-AMI
c) Bipolar RZ & RZ-AMI
d) None of the mentioned
Answer: c
Explanation: In PCM waveforms signals generally use two levels. But few signals use three levels such as bipolar RZ, RZ-AMI, dicode, duobinary etc.
4. Which waveform type has the feature of clocking?
a) Manchester coding
b) Bbi-p-M
c) Delay modulation
d) NRZ-L
Answer: a
Explanation: In manchester coding transition occurs in the middle of every bit interval. Thus it has a feature of clocking.
5. Which waveform has the feature of error detection?
a) NRZ-L
b) RZ-AMI
c) Manchester coding
d) Duobinary
Answer: d
Explanation: Duobinary scheme does error detection without introducing any additional error bits into the data sequence.
6. Which waveform scheme introduces bandwidth compression?
a) Duobinary
b) Manchester coding
c) Phase encoded waveform
d) Multilevel codes
Answer: d
Explanation: Multilevel codes increases the bandwidth efficiency by reducing the bandwidth utilization for the given data rate.
7. Which waveform type has better noise immunity?
a) NRZ
b) RZ
c) Phase encoded
d) Multilevel codes
Answer: a
Explanation: NRZ waveforms has better error performance than RZ signal waveforms.
8. In pulse modulation, reciprocal of T is
a) Bandwidth
b) Symbol rate
c) Signal voltage
d) None of the mentioned
Answer: b
Explanation: T is the time period of the signal. In pulse modulation, the symbol rate Rs can be given as reciprocal of T.
9. PCM word size can be described by
a) Time period
b) Symbol rate
c) Number of quantization levels
d) None of the mentioned
Answer: c
Explanation: Each analog sample is transformed into PCM word made up of a group of bits. The PCM word size can be described by number of quantization levels allowed for each sample.
10. Some of the M-ary waveforms are
a) PAM
b) PPM
c) PDM
d) All of the mentioned
Answer: d
Explanation: When we pulse modulate non binary signals we get M-ary waveform. Some of the examples of M-ary waveforms are PAM, PPM, PDM.
11. Which method should be implemented for reducing bandwidth?
a) Multilevel codes
b) Multilevel signalling
c) PAM
d) PDM
Answer: b
Explanation: The transmission bandwidth required for binary digital waveforms is large. To reduce the bandwidth, multilevel signalling method can be used.
12. In M-ary PPM waveform, modulation is effected by
a) Delaying
b) Advancing
c) Delaying & Advancing
d) None of the mentioned
Answer: c
Explanation: In M-ary PPM waveform, modulation is effected by delaying or advancing pulse occurrence by an amount that corresponds to information symbol.
13. For both PPM and PDM _______ is kept constant.
a) Amplitude
b) Time period
c) Frequency
d) Number of levels
Answer: a
Explanation: In PPM waveform delaying or advancing of pulse is done. In PDM the pulse width is varied. Thus in both the cases amplitude is maintained constant.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Correlative Coding”.
1. The method in which small amount of controlled ISI is introduced into the data stream rather than trying to eliminate it completely is called as
a) Correlative coding
b) Duobinary signalling
c) Partial response signalling
d) All of the mentioned
Answer: d
Explanation: The interference at the detector can be cancelled out using these methods in which some controlled amount of ISI is introduced into the data stream.
2. From digital filter we will get the output pulse as the _______ of the current and the previous pulse.
a) Summation
b) Difference
c) Product
d) Ratio
Answer: a
Explanation: The digital filter incorporates one digit delay and thus it adds the incoming pulse with the value of the previous pulse.
3. In duobinary signalling method, for M-ary transmission, the number of output obtained is
a) 2M
b) 2M+1
c) 2M-1
d) M2
Answer: c
Explanation: In duobinary coding, the number of output obtained for M-ary transmission is 2M-1.
4. The method using which the error propagation in dubinary signalling can be avoided is
a) Filtering
b) Precoding
c) Postcoding
d) None of the mentioned
Answer: b
Explanation: In duobinary signalling method if one error occurs it repeats everywhere through out the next steps. To avoid this precoding method can be used.
5. In precoding technique, the binary sequence is _____ with the previous precoded bit.
a) And-ed
b) Or-ed
c) EXOR-ed
d) Added
Answer: c
Explanation: To avoid error propogation precoding method is used. In this each bit is encoded individually without having any effect due to its prior bit or decisions.
6. The duobinary filter, He is called as
a) Sine filter
b) Cosine filter
c) Raised cosine filter
d) None of the mentioned
Answer: b
Explanation: The transfer function is 2T cos which is called as cosine filter.
7. The method which has greater bandwidth efficiency is called as
a) Duobinary signalling
b) Polybinary signalling
c) Correlative coding
d) All of the mentioned
Answer:b
Explanation: If more than three levels are introduced in duobinary signalling technique the bandwidth efficiency increases This method is called as polybinary signalling.
8. In polybinary signalling method the present bit of binary sequence is algebraically added with ______ number of previous bits.
a) j
b) 2j
c) j+2
d) j-2
Answer: d
Explanation: In polybinary signalling method the present binary digit of the sequence is formed from the modulo-2 addition of the j-2 preceding digits of the sequence and the present digit.
9. The primary advantage of this method is
a) redistribution of spectral density
b) to favor low frequencies
c) redistribution of spectral density & to favor low frequencies
d) none of the mentioned
Answer: c
Explanation: Each bit can be independently detected in-spite of strong correlation and this provides redistribution of spectral density and also favors low frequencies.
10. Source encoding procedure does
a) Sampling
b) Quantization
c) Compression
d) All of the mentioned
Answer: d
Explanation: Source encoding includes a sampling of continuous time signals, quantization of continuous valued signals and compression of those sources.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Sampling and quantization”.
1. Which signals are function of time?
a) Random signal
b) Deterministic signal
c) Random & Deterministic signal
d) None of the mentioned
Answer: b
Explanation: Deterministic signals are function of time.
2. Auto-correlation function is a
a) Even function
b) Odd function
c) Even & Odd function
d) None of the mentioned
Answer: a
Explanation: Auto-correlation function is an even function of time.
3. Shot noise occurs in
a) Transistors
b) Valves
c) Transistors & Valves
d) None of the mentioned
Answer: c
Explanation: Shot noise occurs in both valves and transistors.
4. Source coding reduces
a) Redundancy
b) Average bit rate
c) Redundancy & Average bit rate
d) None of the mentioned
Answer: c
Explanation: Source coding reduces both average bit rate and reduces redundancy.
5. Delay element in delta modulation acts as
a) First order predictor
b) Second order predictor
c) Third order predictor
d) Fourth order predictor
Answer: a
Explanation: Delay element act as first order predictor.
6. Non uniform quantization includes
a) Compression
b) Expansion
c) Compression & Expansion
d) None of the mentioned
Answer: c
Explanation: Compression and expansion give the feature of non uniform quantization.
7. The quantization will be finer when
a) Smaller the number of discrete amplitudes
b) Larger the number of discrete amplitudes
c) Does not depend on amplitudes
d) None of the mentioned
Answer: b
Explanation: Larger the number of discrete amplitudes, finer will be the quantization.
8. Different cases of sampling include
a) Ideal impulse sampling
b) Flat-topped sampling
c) Sampling with rectangular pulses
d) All of the mentioned
Answer: d
Explanation: The three cases of sampling are ideal impulse sampling, sampling with rectangular pulses and flat topped sampling.
9. Transmitted pulse becomes distorted due to
a) Ideal transmission characteristic
b) Non ideal transmission characteristic
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: A transmitted pulse gradually becomes distorted due to non ideal transmission characteristic of the channel.
10. In which mixing is easier?
a) Analog signal
b) Digital signal
c) Analog & Digital signal
d) None of the mentioned
Answer: b
Explanation: Mixing of digital signals are easier than that of mixing analog signals.
This set of Digital Communication Questions & Answers for experienced focuses on “Sampling and Quantization”.
1. Which filter does not have sharp output?
a) Linear phase filter
b) Delayed symmetric filter
c) Linear phase & Delayed symmetric filter
d) None of the mentioned
Answer: c
Explanation: Linear phase filter or delayed symmetric filter does not have sharp output but sampling can be done in real time.
2. Using ARMA filter
a) Sampling can be done in real time
b) Gives sharp output
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: ARMA filter gives sharp output but sampling cannot be done in real time.
3. To avoid aliasing
a) Reduce the bandwidth
b) Cut out high frequency
c) Reduce the bandwidth & Cut out high frequency
d) None of the mentioned
Answer: c
Explanation: To avoid aliasing bandwidth should be reduced and high frequency should be cut out.
4. Which requires interpolation filtering?
a) Up-sampler
b) D to A converter
c) Up-sampler & D to A converter
d) None of the mentioned
Answer: c
Explanation: Up-sampler and D to A conversion need interpolation filtering.
5. A to D conversion process uses
a) Sampler
b) Quantizer
c) Sampler & Quantizer
d) None of the mentioned
Answer: c
Explanation: A to D conversion process requires both sampler and quantizer.
6. Reconstruction filter is difficult to implement in hardware.
a) True
b) False
Answer: b
Explanation: Reconstruction filter is simpler and easy to implement in hardware.
7. Which process requires low pass filter?
a) Up-sampling
b) Down-sampling
c) Up-sampling & Down-sampling
d) None of the mentioned
Answer: c
Explanation: Up-sampling requires low pass filter after increasing the data rate and down-sampling requires low pass filter before decimation.
8. Decreasing the data rate is called as
a) Aliasing
b) Down sampling
c) Up sampling
d) None of the mentioned
Answer: b
Explanation: Decreasing the data rate is called as down sampling and increasing the data rate is called as up sampling.
9. Original signal came to be retraced from sampled version using
a) Low pass filtering
b) High pass filtering
c) Low & High pass filtering
d) None of the mentioned
Answer: a
Explanation: Original signal can be obtained from its sampled version by using low pass filtering.
10. The signal can be reconstructed
a) At Nyquist rate
b) Above Nyquist rate
c) At & above the Nyquist rate
d) None of the mentioned
Answer: c
Explanation: The signal can be reconstructed from the sampled version at or above Nyquist rate using simple low pass filtering.
11. Which device is needed for the reconstruction of signal?
a) Low pass filter
b) Equalizer
c) Low pass filter & Equalizer
d) None of the mentioned
Answer: c
Explanation: Equalizer followed by a low pass filter is necessary for reconstruction of a signal from its sampled version.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Signals and Noises”.
1. The term heterodyning refers to
a) Frequency conversion
b) Frequency mixing
c) Frequency conversion & mixing
d) None of the mentioned
Answer: c
Explanation: The method heterodyning means frequency conversion and mixing and this results in a spectral shift.
2. The causes for error performance degradation in communication systems are
a) Interference
b) Electrical noise
c) Effect of filtering
d) All of the mentioned
Answer: d
Explanation: The main causes of error performance degradation are interference electrical noise effect of filtering and also due to the surroundings.
3. Thermal noise in the communication system due to thermal electrons
a) Can be eliminated
b) Cannot be eliminated
c) Can be avoided upto some extent
d) None of the mentioned
Answer: b
Explanation: Thermal noise which cannot be eliminated is caused by the motion of thermal electrons causes degradation in system.
4. White noise has _______ power spectral density.
a) Constant
b) Variable
c) Constant & Variable
d) None of the mentioned
Answer: a
Explanation: The AWGN has constant power spectral density.
5. Which are called as hard decisions?
a) Estimates of message symbol with error correcting codes
b) Estimates of message symbol without error correcting codes
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: If error correcting codes are not present, the detector output consists of estimates of the message symbol which is also called as hard decisions.
6. The filter which is used to recover the pulse with less ISI is called as
a) Matched filter
b) Correlator
c) Matched filter & Correlator
d) None of the mentioned
Answer: b
Explanation: The optimum filter used to recover the pulse with best possible signal to noise ratio and less or no ISI is called as correlator or matched filter.
7. The composite equalizing filter is the combination of
a) Receiving and equalizing filter
b) Transmitting and equalizing filter
c) Amplifier and equalizing filter
d) None of the mentioned
Answer: a
Explanation: The functions of both receiving and equalizing filter can be performed by only the equalizing filter alone. Thus equalizing filter is the combination of equalizing and receiving filter.
8. The sample from the demodulation process consists of sample which is _______ to energy of the received symbol and _____ to noise.
a) Directly and inversely proportional
b) Inversely and directly proportional
c) Both directly proportional
d) Both inversely proportional
Answer: a
Explanation: The output symbol of the sampler consists of sample which is directly proportional to the energy of the received signal and inversely proportional to the noise.
9. The average noise power of white noise is
a) 0
b) Infinity
c) 1
d) None of the mentioned
Answer: b
Explanation: White noise is a idealized process with two sided spectral density equal to constant N0/2 and frequencies varying from minus infinity to plus infinity. Thus the average noise power is infinity.
10. The channel may be affected by
a) Thermal noise
b) Interference from other signals
c) Thermal noise & Interference from other signals
d) None of the mentioned
Answer: c
Explanation: A channel can be modelled as a linear filter with additional noise. The noise comes from thermal noise source and also from interference from other signals.
11. Channels display multi-path due to
a) Scattering
b) Time delayed reflections
c) Diffraction
d) All of the mentioned
Answer: d
Explanation: Wireless wide-band channels display multi-path due to time delayed reflections, diffraction and also scattering.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Intersymbol interference 1”.
1. The method in which the tail of one pulse smears into adjacent symbol interval is called as
a) Intersymbol interference
b) Interbit interference
c) Interchannel interference
d) None of the mentioned
Answer: a
Explanation: Due to the effect of system filtering the received pulse can overlap on one and another. The tail of one pulse smears into the adjacent symbol interval thereby interfering the detection process. This process is called as intersymbol interference.
2. If each pulse of the sequence to be detected is in _____ shape, the pulse can be detected without ISI.
a) Sine
b) Cosine
c) Sinc
d) None of the mentioned
Answer: c
Explanation: The sinc shaped pulse is the ideal nyquist pulse. If each pulse in the sequence to be detected is in sinc shape the pulses can be detected without ISI.
3. What is symbol rate packing?
a) Maximum possible symbol transmission rate
b) Maximum possible symbol receiving rate
c) Maximum bandwidth
d) Maximum ISI value allowed
Answer: a
Explanation: A system with bandwidth Rs/2 can support a maximum transmission rate of Rs without ISI. Thus for ideal Nyquist filtering the maximum possible symbol transmission rate is called as symbol rate packing and it is equal to 2 symbols/s/Hz.
4. A nyquist pulse is the one which can be represented by _____ shaped pulse multiplied by another time function.
a) Sine
b) Cosine
c) Sinc
d) None of the mentioned
Answer: c
Explanation: A nyquist filter is one whose frequency transfer function can be represented by a rectangular function convolved with any real even symmetric frequency function and a nyquist pulse is one whose shape can be represented by sinc function multiplied by another time function.
5. Examples of nyquist filters are
a) Root raised cosine filter
b) Raised cosine filter
c) Root raised & Raised cosine filter
d) None of the mentioned
Answer: c
Explanation: The most popular among the class of nyquist filters are raised cosine and root raised cosine filter.
6. The minimum nyquist bandwidth for the rectangular spectrum in raised cosine filter is
a) 2T
b) 1/2T
c) T 2
d) 2/T
Answer: b
Explanation: For raised cosine spectrum the minimum nyquist bandwidth is equal to 1/2T.
7. Roll off factor is the fraction of
a) Excess bandwidth and absolute bandwidth
b) Excess bandwidth and minimum nyquist bandwidth
c) Absolute bandwidth and minimum nyquist bandwidth
d) None of the mentioned
Answer: b
Explanation: The roll off factor is defined by a fraction of excess bandwidth and the minimum nyquist bandwith. It ranges from 0 to 1.
8. Which value of r is considered as Nyquist minimum bandwidth case?
a) 0
b) 1
c) Infinity
d) None of the mentioned
Answer: a
Explanation: For the roll off factor of 0 an ideal rectangular nyquist pulse is obtained. This is called as nyquist minimum bandwidth case.
9. A pulse shaping filter should satisfy two requirements. They are
a) Should be realizable
b) Should have proper roll off factor
c) Should be realizable & have proper roll off factor
d) None of the mentioned
Answer: c
Explanation: A pulse shaping filter should provide the desired roll off and should be realizable, that is the impulse response needs to be truncated to a finite length.
10. Examples of double side band signals are
a) ASK
b) PSK
c) ASK & PSK
d) None of the mentioned
Answer: c
Explanation: ASK and PSK needs twice the transmission bandwidth of equivalent baseband signals. Thus these are called as double side band signals.
This set of Digital Communication Interview Questions & Answers for experienced focuses on “Intersymbol Interference”.
1. The likelihood ratio test is done between
a) Likelihood of s1 by likelihood of s2
b) Likelihood of s2 by likelihood of s1
c) Likelihood of s1 by likelihood of s1
d) None of the mentioned
Answer: a
Explanation: The likelihood ratio is given with the help of conditional probabilities.
2. According to the rule of minimizing the error probabilities, the hypothesis should be like if the priori probabilities are ________ than the ratio of likelihoods.
a) Lesser
b) Greater
c) Equal
d) None of the mentioned
Answer: a
Explanation: The rule of minimizing the error probabilities show that the hypothesis should be like if the ratio of likelihoods should be greater than the priori probabilities.
3. The detector that minimizes the error probability is called as
a) Maximum likelihood detector
b) Minimum likelihood detector
c) Maximum & Minimum likelihood detector
d) None of the mentioned
Answer: a
Explanation: For the signals that are equally likely, the detector used to minimize the error probability is called as maximum likelihood detector.
4. For a M-ary signal or symbol the number of likelihood functions are
a) M
b) M+1
c) M-1
d) 2M
Answer: a
Explanation: For a M-ary signal there will be M likelihood functions representing M signal classes to which a received signal might belong.
5. An error in binary decision making occurs when the channel noise is
a) Greater than the optimum threshold level
b) Lesser than the optimum threshold level
c) Greater or Lesser than the optimum threshold level
d) None of the mentioned
Answer: c
Explanation: An error will occur when s1 or s2 is sent and if the channel noise is greater or lesser than the optimum threshold level.
6. Optimum threshold value is given by
a) a0+a1/2
b) a0-a1/2
c) a0/2
d) a1/2
Answer: a
Explanation: The optimum threshold level value for error probability reduction is given by a0+a1/2.
7. The symbol of the probability under the tail of Gaussian pdf is called as
a) Complementary error function
b) Co error function
c) Complementary error & Co error function
d) None of the mentioned
Answer: c
Explanation: The co error function or complementary error function is a commonly used symbol for probability under the tail of Gaussian pdf.
8. Matched filter provides _____ signal to noise ratio.
a) Maximum
b) Minimum
c) Zero
d) Infinity
Answer: a
Explanation: A matched filter is a linear filter designed to give a maximum signal to noise ratio power at the output.
9. The impulse response of the filter is the ________ of the mirror image of the signal waveform.
a) Delayed version
b) Same version
c) Delayed & Same version
d) None of the mentioned
Answer: a
Explanation: The matched filter’s basic property is that the impulse response of the filter is the delayed version of the mirror image of the signal waveform.
10. Example for antipodal bandpass signaling is
a) BPSK
b) ASK
c) FSK
d) MSK
Answer: a
Explanation: One type of antipodal bandpass signaling is binary phase shift keying.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Equalization “.
1. Channel’s phase response must be a linear function of
a) Time
b) Frequency
c) Time & Frequency
d) None of the mentioned
Answer: b
Explanation: To achieve ideal transmission characteristics, the signal’s bandwidth must be constant and the channel’s phase response must be a linear function of frequency.
2. Amplitude distortion occurs when
a) Impulse response is not constant
b) Impulse response is constant
c) Frequency transfer function is constant
d) Frequency transfer function is not constant
Answer: d
Explanation: When modulus of channel’s transfer function is not constant within W then amplitude distortion occurs.
3. Phase distortion occurs when
a) Phase response is function of frequency
b) Phase response is not a function of frequency
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: When channels phase response is not a linear function of frequency within W then phase distortion occurs.
4. Equalization process includes
a) Maximum likelihood sequence estimation
b) Equalization with filters
c) Maximum likelihood sequence estimation & Equalization with filters
d) None of the mentioned
Answer: c
Explanation: Equalization process can be divided into two major categories called as maximum likelihood sequence estimator and equalizer with a filter.
5. The maximum likelihood sequence estimator adjusts _______ according to _____ environment.
a) Receiver, transmitter
b) Transmitter, receiver
c) Receiver, receiver
d) None of the mentioned
Answer: a
Explanation: In maximum likelihood sequence estimator it measures the impulse response value and adjusts the receiver environment according to that of the transmitter.
6. The filters used with the equalizer is of ______ types.
a) Feed forward
b) Feed backward
c) Feed forward and feedback
d) None of the mentioned
Answer: c
Explanation: The filters used with equalizers can be linear devices that contain only feed forward elements or non linear devices with both feed forward and feed back elements.
7. Transversal equalizers are ________ and decision feedback equalizers are ______
a) Feed forward, feed back
b) Feed back, feed forward
c) Feed forward, feed forward
d) Feedback, feedback
Answer: a
Explanation: Linear devices with only feed forward elements are called as transversal equalizers and non linear devices with both feed forward and feed back elements are called as decision feedback equalizers.
8. Symbol spaced has ____ sample per symbol and fractionally spaced has ___ samples per symbol.
a) One, many
b) Many, one
c) One, one
d) Many, many
Answer: a
Explanation: Predetection samples are provided only on symbol boundaries. A condition in which only one sample per symbol are provided is called as symbol spaced and the condition in which multiple samples are provided per symbol is called as fractionally spaced.
9. The _______ of the opening of eye pattern indicates the time over which the sampling for detection might be performed.
a) Length
b) Width
c) X-axis value
d) Y-axis value
Answer: b
Explanation: An eye pattern is a display that results from measuring a system’s response to baseband signals in a prescribed way. The width of the opening gives the time over which the sampling for detection might be performed.
10. Range of time difference of the zero crossing gives the value of
a) Width
b) Distortion
c) Timing jitter
d) Noise margin
Answer: c
Explanation: The time difference of the zero crossing in the eye pattern gives the value of timing jitter.
This set of Digital Communication test focuses on “Equalization and Non Coherent Detection”.
1. The range of amplitude difference gives the value of
a) Width
b) Distortion
c) Timing jitter
d) Noise margin
Answer: b
Explanation: In the eye pattern, the amplitude difference gives the value of distortion caused by ISI.
2. As the eye opens, ISI _______
a) Increases
b) Decreases
c) Remains the same
d) None of the mentioned
Answer: b
Explanation: As the eye closes, ISI increases and as the eye opens ISI decreases.
3. Pseudo noise signal has _______ and _______ SNR for the same peak transmitted power.
a) Larger, smaller
b) Smaller, larger
c) Larger, larger
d) Smaller, smaller
Answer: c
Explanation: A training pulse is applied to the equalizer and corresponding impulse response is observed. Pseudo noise is preferred as the training pulse as it has larger SNR value and larger average power value.
4. The index value n, in transversal filter can be used as.
a) Time offset
b) Filter coefficient identifier
c) Time offset & Filter coefficient identifier
d) None of the mentioned
Answer: c
Explanation: The index n can be used as both time offset and the filter coefficient identifier, which is the address in the filter.
5. The over-determined set of equations can be solved using
a) Zero forcing
b) Minimum mean square error
c) Zero forcing & Minimum mean square error
d) None of the mentioned
Answer: c
Explanation: The matrix x in transversal equalizer if non square with dimensions 4N+1 and 2N+1. Such equations are called as over-determined set. This can be solved by two methods called as zero forcing method and minimum mean square error method.
6. If the filter’s tap weight remains fixed during transmission of data, then the equalization is called as
a) Preset equalization
b) Adaptive equalization
c) Fixed equalization
d) None of the mentioned
Answer: a
Explanation: If the weight remains fixed during transmission of data then the equalization is called as preset equalization. It is a simple method which consists of setting the tap weight according to some average knowledge of the channel.
7. Equalization method which is done by tracking a slowly time varying channel response is
a) Preset equalization
b) Adaptive equalization
c) Variable equalization
d) None of the mentioned
Answer: b
Explanation: This method is implemented to perform tap weight adjustment periodically or continually. Equalization is done by tracking a slowly varying channel response.
8. Preamble is used for
a) Detect start of transmission
b) To set automatic gain control
c) To align internal clocks
d) All of the mentioned
Answer: d
Explanation: The receiver uses preamble for detecting the start of transmission, to set automatic gain control, and to align internal clocks and local oscillator with the received signal.
9. The disadvantage of preset equalizer is that
a) It doesnot requires initial training pulse
b) Time varying channel degrades the performance of the system
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: The disadvantage of preset equalization is that it requires an initial training period that must be invoked at the start of any new transmission. Also time varying channel can degrade system performance due to ISI, since the tap weights are fixed.
10. For AWGN, the noise variance is
a) N0
b) N0/2
c) 2N0
d) N0/4
Answer: b
Explanation: The noise variance out of the correlator for AWGN is N0/2.
11. Performance of BFSK signal is ________ than BPSK.
a) 3db worse
b) 3db better
c) 6db worse
d) 6db better
Answer: a
Explanation: The performance of BFSK is 3db worse than BPSK signalling, since for a given signal power, the distance squared between orthogonal vectors is a factor of two less than the distance squared between orthopodal signals.
12. A Gaussian distribution into the non linear envelope detector yields
a) Rayleigh distribution
b) Normal distribution
c) Poisson distribution
d) Binary distribution
Answer: a
Explanation: The two output signals of Gaussian distribution yields Rayleigh and Rician distribution.
13. The non coherent FSK needs ________ Eb/N0 than coherent FSK.
a) 1db more
b) 1db less
c) 3db more
d) 3db less
Answer: a
Explanation: The non coherent receiver is easier to implement. The non coherent FSK needs 1db more Eb/N0 than coherent FSK.
14. The DPSK needs ________ Eb/N0 than BPSK.
a) 1db more
b) 1db less
c) 3db more
d) 3db less
Answer: a
Explanation: The DPSK system is easier to implement than PSK and it needs 1db more Eb/N0 than BPSK.
15. Coherent PSK and non coherent orthogonal FSK have a difference of ______ in PB.
a) 1db
b) 3db
c) 4db
d) 6db
Answer: c
Explanation: The difference of PB is approximately 4db for the best and the worst .
16. Which is easier to implement and is preferred?
a) Coherent system
b) Non coherent system
c) Coherent & Non coherent system
d) None of the mentioned
Answer: b
Explanation: A non coherent system is desirable because there may be difficulty is establishing and maintaining a coherent reference.
17. Which is the main system consideration?
a) Probability of error
b) System complexity
c) Random fading channel
d) All of the mentioned
Answer: d
Explanation: The major system considerations are error probability, complexity and random fading channel. Considering all this non coherent system is more desirable than coherent.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Digital bandpass modulation techniques”.
1. Wavelength and antenna size are related as
a) λ/2
b) λ/4
c) 2λ
d) 4λ
Answer: b
Explanation: The transmission of Em field in space is done with the help of antennas. Antenna size depends of the wavelength. The length of the antenna is equal to λ/4.
2. The detection method where carrier’s phase is given importance is called as
a) Coherent detection
b) Non coherent detection
c) Coherent detection & Non coherent detection
d) None of the mentioned
Answer: a
Explanation: When the receiver uses carrier’s phase as major factor for detection then it is called as coherent detection and when carrier is not given importance it is called as non coherent detection.
3. The coherent modulation techniques are
a) PSK
b) FSK
c) ASK
d) All of the mentioned
Answer: d
Explanation: Some of the examples of coherent modulation techniques are phase shift keying, amplitude shift keying, frequency shift keying and continuous phase modulation.
4. The real part of a sinusoid carrier wave is called as
a) Inphase
b) Quadrature
c) Inphase & Quadrature
d) None of the mentioned
Answer: a
Explanation: The two main parts of sinusoid carrier wave – real part is called as inphase and the imaginary part is called as quadrature.
5. Antipodal signal sets are those vectors that can be illustrated as
a) Two 180 opposing vector
b) Two 90 opposing vector
c) Two 360 opposing vector
d) None of the mentioned
Answer: a
Explanation: In BPSK the two vectors, the signal set can be illustrated as two 180 opposing vector which is called as antipodal signal sets.
6. The FSK signal which has a gentle shift from one frequency level to another is called as
a) Differential PSK
b) Continuous PSK
c) Differential & Continuous PSK
d) None of the mentioned
Answer: b
Explanation: In general FSK the signal change from one frequency to another will be abrupt but in continuous FSK the signal change from one frequency to another will be gentle and gradual.
7. Which modulation scheme is also called as on-off keying method?
a) ASK
b) FSK
c) PSK
d) GMSK
Answer: a
Explanation: In ASK modulation scheme the signal attains either maximum amplitude or zero point. Thus it is also called as on-off keying.
8. In amplitude phase keying each phase vector is separated by
a) 90
b) 0
c) 45
d) 180
Answer: c
Explanation: The combination of ASK and PSK is called as APK. In APK each vector is separated by 45.
9. The term heterodyning refers to
a) Frequency conversion
b) Frequency mixing
c) Frequency conversion & mixing
d) None of the mentioned
Answer: c
Explanation: The term heterodyning means frequency conversion and frequency mixing that yields a spectral shift in the signal.
10. The transformation of the waveform into a single point in signal space is called as
a) Vector point
b) Predetection point
c) Preamplification point
d) Transformation point
Answer: b
Explanation: The first step of the detection process is to reduce waveform into a single or group of random variable. This first step in the transformation of waveform into a point in signal space. This point is called as predetection point.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Coherent detection”.
1. The correlating detector is also known as
a) Maximum likelihood detector
b) Minimum likelihood detector
c) Maximum & Minimum likelihood detector
d) None of the mentioned
Answer: a
Explanation: Coherent detector considers phase as the most important parameter. This coherent detector is also known as maximum likelihood detector.
2. The minimum nyquist sampling rate is given as, fs =
a) 1/T
b) T
c) 2/T
d) 2T
Answer: a
Explanation: The input signal comprises of a prototype signal plus noise and the bandwidth is 1/2T where T is the symbol time and minimum nyquist sampling rate can be given as 2W or 1/T.
3. Phase-locked loop circuitry is used for
a) Carrier wave recovery
b) Phase estimation
c) Carrier wave recovery & Phase estimation
d) None of the mentioned
Answer: c
Explanation: The phase locked loop circuitry locks on the arriving carrier wave and estimates its phase.
4. In differential PSK the date is
a) Encoded differentially
b) Decoded differentially
c) Encoded & Decoded differentially
d) None of the mentioned
Answer: a
Explanation: In this method, the data is encode deferentially that is the presence of binary zero or one is manifested by symbol’s similarity or difference when compared to that of the previous symbol.
5. Envelope detector consists of
a) Rectifier and high pass filter
b) Rectifier and low pass filter
c) Amplifier and low pass filter
d) Amplifier and high pass filter
Answer: b
Explanation: An envelope detector consists of a rectifier and a low pass filter. The detectors are matched to the envelopes and not to the signal themselves.
6. The minimum required spacing is the difference between the
a) Center of the spectral main lobe and first zero crossing
b) First and second zero crossing
c) First and last zero crossing
d) None of the mentioned
Answer: a
Explanation: The frequency difference between the center of the spectral main lobe and the first zero crossing is called as the minimum required spacing in non coherent detection.
7. The minimum tone separation corresponds to
a) T
b) 1/T
c) 2T
d) T/2
Answer: b
Explanation: In non coherent detection the minimum tone separation corresponds to 1/T.
8. Matched filter is used for
a) Coherent detection
b) Non coherent detection
c) Coherent & Non coherent detection
d) None of the mentioned
Answer: a
Explanation: Matched filter is used for coherent detection. It cannot be used for non coherent detection because matched filter output is a function of unknown angle α.
9. In differential encoding the _________ different between two wave forms is measured.
a) Magnitude
b) Frequency
c) Phase
d) Time period
Answer: c
Explanation: In differential encoding, the phase of the present signal waveform is compared with the phase of the previous signal.
10. The error probability of DPSK is ________ worse than PSK.
a) Twice
b) 3 db
c) Twice
d) None of the mentioned
Answer: c
Explanation: DPSK has twice as much as noise than in PSK. The error probability in DPSK is twice worse than PSK.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “M-ary signalling and performance”.
1. The limit which represents the threshold Eb/N0 value below which reliable communication cannot be maintained is called as
a) Probability limit
b) Error limit
c) Shannon limit
d) Communication limit
Answer: c
Explanation: Eb/N0 curve has waterfall shape. Shannon limit gives the threshold value below which reliable communication cannot be maintained.
2. M-ary signalling produces _______ error performance with orthogonal signalling and _______ error performance with multiple phase signalling.
a) Degraded, improved
b) Improved, degraded
c) Improved, improved
d) Degraded, degraded
Answer: b
Explanation: In M-ary signalling as k increases, the curve moves towards the degraded error performance. It produces improved error performance in case of orthogonal signalling and degraded error performance in case of multiple phase signalling.
3. Which is more vulnerable to noise?
a) 2-ary system
b) 4-ary system
c) Binary system
d) None of the mentioned
Answer: b
Explanation: The minimum energy noise vector for 4-ary system is smaller than 2-ary system. So 4-ary system is more vulnerable to noise.
4. In which system, bit stream is portioned into even and odd stream?
a) BPSK
b) MSK
c) QPSK
d) FSK
Answer: c
Explanation: In QPSK bit stream is portioned into even and odd stream, I and Q bit streams. Each new stream modulates as orthogonal component at half bit rate.
5. The error performance of MPSK ______ as M or k increases.
a) Increases
b) Decreases
c) Stays constant
d) None of the mentioned
Answer: b
Explanation: The error performance of MPSK degrades as M or k increases.
6. In MPSK adding new signals _______ make it vulnerable to noise and in MFSK _______ make it vulnerable.
a) Does, does not
b) Does not, does
c) Does, does
d) Does not, does not
Answer: a
Explanation: In MPSK adding new signals that is on crowding the signals it makes it vulnerable to noise where as in MFSK it does not.
7. In orthogonal signalling with symbols containing more number of bits we need ____ power.
a) More
b) Less
c) Double
d) None of the mentioned
Answer: a
Explanation: In orthogonal signalling with symbols having more number of bits need more power but requirement per bit is reduced.
8. For FSK signalling, WT is equal to
a) 0
b) 1
c) 0.737
d) Infinity
Answer: b
Explanation: For FSK signalling the detection bandwidth is typically equal to symbol rate 1/T that is WT is nearly equal to 1.
9. Energy per symbol Es is given as
a) Es=Eb
b) Es=Eb/
c) Es=2Eb
d) Es=Eb/2
Answer: a
Explanation: In M-ary PSK signalling, the energy per symbol is given as Es=Eb.
10. The relation between the probability of bit error and probability of symbol error in M-ary orthogonal signalling is
a) M/M-1
b) 2M/M-1
c) /M-1
d) M/M+1
Answer: c
Explanation: The relationship between the probability of bit error and probability of symbol error is /M-1 in M-ary orthogonal signalling.
11. As limit of k increases, the ratio of PB/PE becomes
a) 1:2
b) 2:1
c) 1:3
d) 3:1
Answer: a
Explanation: As the limit of k increases, the ratio of PB/PE becomes 1:2.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Link budget analysis”.
1. Link budget consists of calculation of
a) Useful signal power
b) Interfering noise power
c) Useful signal & Interfering noise power
d) None of the mentioned
Answer: c
Explanation: The link analysis and its output, the link budget consists of calculations and tabulations of useful signal power and interfering noise power at the receiver.
2. Link budget can help in predicting
a) Equipment weight and size
b) Technical risk
c) Prime power requirements
d) All of the mentioned
Answer: d
Explanation: Link budget can help to predict equipment weight, size, prime power requirements, technical risk and cost. Link budget is one of the system manager’s useful document.
3. Which is the primary cost for degradation of error performance?
a) Loss in signal to noise ratio
b) Signal distortion
c) Signal distortion & Loss in signal to noise ratio
d) None of the mentioned
Answer: c
Explanation: There are two primary causes for the degradation of error performance. They are loss in signal to noise ratio and the second is signal distortion caused by intersymbol interference.
4. Which factor adds phase noise to the signal?
a) Jitter
b) Phase fluctuations
c) Jitter & Phase fluctuations
d) None of the mentioned
Answer: c
Explanation: When a local oscillator is used in signal mixing, phase fluctuations and jitter adds phase noise to the signal.
5. Antennas are used
a) As transducer
b) To focus
c) As transducer & To focus
d) None of the mentioned
Answer: c
Explanation: Antennas are used as transducer that converts electronic signals to electromagnetic fields and vice versa. They are also used to focus the electromagnetic energy in the desired direction.
6. Mechanism contributing to a reduction in efficiency is called as
a) Amplitude tapering
b) Blockage
c) Edge diffraction
d) All of the mentioned
Answer: d
Explanation: Mechanism contributing to a reduction in efficiency is called as amplitude tapering, spillover, edge diffraction, blockage, scattering, re-radiation and dissipative loss.
7. Space loss occurs due to a decrease in
a) Electric field strength
b) Efficiency
c) Phase
d) Signal power
Answer: a
Explanation: Due to the decrease in electric field strength there will be a decrease in signal strength as a function of distance. This is called as space loss.
8. Antenna’s efficiency is given by the ratio of
a) Effective aperture to physical aperture
b) Physical aperture to effective aperture
c) Signal power to noise power
d) Losses
Answer: a
Explanation: The larger the antenna aperture the larger is the resulting signal power density in the desired direction. The ratio of effective aperture to physical aperture is the antenna’s efficiency.
9. Effective radiated power of an isotropic radiator can be given as a product of
a) Radiated power and received power
b) Effective area and physical area
c) Transmitted power and transmitting gain
d) Receiving power and receiving gain
Answer: c
Explanation: An effective radiated power with respect to an isotropic radiator EIRP can be defined as the product of transmitted power and the gain of the transmitting antenna.
10. According to reciprocity theorem, _____ and _____ are identical.
a) Transmitting power and receiving power
b) Transmitting gain and receiving gain
c) Effective area and physical area
d) None of the mentioned
Answer: b
Explanation: The reciprocity theorem states that for a given antenna and carrier wavelength the transmitting and receiving gain are identical.
This set of Digital Communication quiz focuses on “Link Budget Analysis and Error Control”.
1. Field of view is ______ to antenna gain.
a) Proportional
b) Inversely proportional
c) Half the
d) Double
Answer: b
Explanation: The antenna field of view is the measure of a solid angle in which most of the field power is concentrated. It is inversely related to antenna gain.
2. Antenna gain ______ as effective area increases.
a) Increases
b) Decreases
c) Remains same
d) None of the mentioned
Answer: a
Explanation: Antenna gain increases with a decrease in wavelength increase in frequency and increase in effective area.
3. The beam-width becomes narrower on
a) Increasing frequency
b) Increasing antenna size
c) Increasing frequency & antenna size
d) None of the mentioned
Answer: c
Explanation: Increasing either the signal frequency or antenna size results in narrower beam-width.
4. Path loss Ls is dependent on
a) Signal power
b) Effective area
c) Wavelength
d) Antenna size
Answer: c
Explanation: The path loss is wavelength dependent.
5. Thermal noise is generated due to
a) Lossy coupling
b) Lossless coupling
c) Lossy & Lossless coupling
d) None of the mentioned
Answer: a
Explanation: Thermal noise occurs due to the thermal motion of electrons in all conductors. It occurs due to lossy coupling between an antenna and the receiver.
6. In analog receivers, noise bandwidth is _________ signal bandwidth.
a) Lesser than
b) Greater than
c) Equal to
d) Not related
Answer: b
Explanation: With analog receivers, noise bandwidth seen by the demodulator is usually greater than the signal bandwidth.
7. The link availability measures the percentage of time the link can be ______
a) Open
b) Closed
c) Used
d) None of the mentioned
Answer: b
Explanation: Link availability is the measure of long term link utility stated on an average annual basis for a given geographical location, the link availability measures the percentage of time the link can be closed.
8. Visibility is ________ to required availability.
a) Proportional
b) Inversely proportional
c) Equal
d) Not related
Answer: b
Explanation: For a fixed link margin, visibility is inversely proportional to the required availability, and for a fixed availability visibility increases monotonically with margin.
9. The stop and wait ARQ needs _______ connection.
a) Half duplex
b) Full duplex
c) Simplex
d) None of the mentioned
Answer: a
Explanation: The stop and wait ARQ needs half duplex system since the transmitter needs ACK for each transmission.
10. In continuous ARQ will pullback uses _____ system.
a) Half duplex
b) Full duplex
c) Simplex
d) None of the mentioned
Answer: b
Explanation: The ARQ procedure called continuous ARQ with pullback requires full duplex connection.
11. Which method requires less redundancy?
a) Error detection
b) Error correction
c) Error detection & correction
d) None of the mentioned
Answer: a
Explanation: Error detection requires much simpler decoding equipment and much less redundancy than error correction.
12. Hybrid automatic repeat request is a combination of
a) ARQ and error correction code
b) ARQ and error detection code
c) Error detection and correction codes
d) One of the mentioned
Answer: a
Explanation: An alternate approach for error control is a hybrid automatic repeat request method which is a combination of ARQ and error correction code.
13. Automatic repeat request is also called as
a) Forward error correction
b) Backward error correction
c) Forward & Backward error correction
d) None of the mentioned
Answer: b
Explanation: Automatic repeat request is also referred to as backward error correction method.
14. Error detection is realized using
a) Hash function
b) Check sum
c) Hash function & Check sum
d) None of the mentioned
Answer: b
Explanation: Error detection is commonly realized using a suitable hash function or check-sum algorithm.
15. Check-sum scheme has
a) Check bits
b) Parity bits
c) Longitudinal redundancy bits
d) All of the mentioned
Answer: d
Explanation: Check-sum scheme includes parity bits, longitudinal redundancy bits and check bits.
16. Example of block code is
a) Hamming code
b) Reed-solomon code
c) Repetition code
d) All of the mentioned
Answer: d
Explanation: Some examples of block codes are repetition codes, hamming codes, parity check bit codes, reed-solomon codes, turbo codes etc.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Noise parameters”.
1. Noise figure measures the
a) Power degradation
b) Noise degradation
c) SNR degradation
d) None of the mentioned
Answer: c
Explanation: Noise figure relates the SNR of the input to SNR of the output. It measures the SNR degradation caused by the network.
2. Noise figure is a parameter that represents a ______ of the system.
a) Noisiness
b) Efficiency
c) Maximum output
d) Maximum power handling capacity
Answer: a
Explanation: Noise figure is a parameter that represents the noisiness of a two port network or device such as an amplifier, compared with a reference noise.
3. When two networks are connected in series, its composite noise figure can be given as
a) F1+/G1
b) F1-/G1
c) F2+/G1
d) F1G1+
Answer: a
Explanation: When two networks are connected in series, the composite noise figure can be written as F1+/G1.
4. The space window region ranges from
a) 1GHz to 10 GHz
b) 10GHz to 100GHZ
c) 5GHz to 50GHz
d) 1MHz to 10MHz
Answer: a
Explanation: There is a region between 1GHz and 10GHz where the temperature will be lowest. This region is called as microwave window or space window.
5. Atmospheric window region is used for
a) Satellite communication
b) Deep-space communication
c) Satellite & Deep-space communication
d) None of the mentioned
Answer: c
Explanation: The region between 1GHz and 10GHz is called as space window or atmospheric window. Here the temperature will be the lowest. This region is used for satellite communication and deep-space communication.
6. Which parameter is referred to as receiver sensitivity?
a) S/N ratio
b) G/T ratio
c) EIRP
d) PR/N0
Answer: b
Explanation: The figure of parameter G/T0 is called as receiver sensitivity.
7. What are the functions of satellite repeaters?
a) Re transmits
b) Regenerates
c) Reconstitutes
d) All of the mentioned
Answer: d
Explanation: Satellite repeaters re transmit the message received. It regenerates, demodulates and reconstitutes the digital information embedded in the received waveform.
8. What are the functions of non-regenerative repeaters?
a) Amplifies
b) Re transmits
c) Amplifies & Re transmits
d) None of the mentioned
Answer: c
Explanation: A non regenerative repeaters can be used with many different modulation formats. It only amplifies and re transmits the message received.
9. In which repeaters uplink is decoupled from the downlink?
a) Regenerative
b) Non regenerative
c) Regenerative & Non regenerative
d) None of the mentioned
Answer: a
Explanation: The principal advantage of regenerative repeaters over non regenerative repeaters is that the uplink is decoupled from the downlink so that the uplink noise is not transmitted on the downlink.
10. AM-to-AM conversion brings about
a) Amplitude variation
b) Phase variation
c) Frequency variation
d) None of the mentioned
Answer: a
Explanation: AM-to-AM conversion is a phenomenon common to non linear devices and it undergoes a non linear transformation and results in amplitude variation.
11. AM-to-PM conversion brings about
a) Amplitude variation
b) Phase variation
c) Frequency variation
d) None of the mentioned
Answer: b
Explanation: AM-to-PM conversion is another common phenomenon in non linear devices. It brings about phase variation that can effect the error performance.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Waveform Coding”.
1. Channel coding relates to area such as
a) Waveform coding
b) Structured sequence
c) Waveform coding & Structured sequence
d) None of the mentioned
Answer: c
Explanation: Channel coding can be partitioned into two area of studies, one is waveform coding or signal design and the other is a structured sequence or structured redundancy.
2. Examples of structured sequences are
a) Block codes
b) Convolutional codes
c) Turbo codes
d) All of the mentioned
Answer: d
Explanation: Some examples of structured sequence or structured redundancy codes are block codes, convolutional codes and turbo codes.
3. Antipodal signals are
a) Mirror images
b) 1800 apart
c) One is negative of the other
d) All of the mentioned
Answer: d
Explanation: Antipodal signals are mirror images, one is negative of the other and they are 1800 apart.
4. Orthogonal vectors are
a) Perpendicular to each other
b) Their dot product must be zero
c) One signal cannot interfere with the other
d) All of the mentioned
Answer: d
Explanation: Orthogonal signals are perpendicular to each other, the inner or dot product between those two signals must equal zero, one vector has zero projection over the other because they do not share same signal space.
5. In orthogonal signalling as k increase there is
a) Improved error performance
b) Degraded error performance
c) Improved bandwidth
d) None of the mentioned
Answer: a
Explanation: In orthogonal signalling as k increases there is improved error performance or a reduction in required Eb/N0 at the expense of bandwidth.
6. In non orthogonal signalling as k increases there is
a) Degraded error performance
b) Improved bandwidth efficiency
c) Increase in required Eb/N0
d) All of the mentioned
Answer: d
Explanation: Non orthogonal signalling manifests improved bandwidth efficiency at the expense of degraded error performance and increases in required Eb/N0.
7. In orthogonal signal, all cross correlation coefficients are
a) One
b) Zero
c) Negative
d) None of the mentioned
Answer: b
Explanation: In orthogonal set all cross correlation coefficients can be made zero.
8. The smaller the cross correlation, the ______ is the distance between the signal vectors.
a) More
b) Less
c) Double
d) Half
Answer: a
Explanation: The cross correlation is the measure of distance between two vectors. The smaller the cross correlation value the greater the distance between the two vectors.
9. Hadamard matrix is given as
a)
H2 = H1 H1
H1 H1
b)
H2 = H1 H1
H1'H1
c)
H2 = H1 H1
H1 H1'
d)
H2 = H1'H1
H1 H1
Answer: c
Explanation: We can construct a code-word set Hk of dimension 2k*2k called a Hadamard matrix.
10. Biorthogonal set consists of
a) Orthogonal signals
b) Antipodal signals
c) Combination of orthogonal and antipodal
d) None of the mentioned
Answer: c
Explanation: The biorthogonal set consists of a combination of both orthogonal and antipodal signals.
11. Which codes perform better?
a) Orthogonal
b) Biorthogonal
c) Orthogonal & Biorthogonal
d) None of the mentioned
Answer: b
Explanation: Since the antipodal vector has a better distance property than the orthogonal signals, biorthogonal codes perform better than the orthogonal codes.
12. Biorthogonal codes needs ________ bandwidth as orthogonal codes.
a) Equal
b) Double
c) Half
d) Triple
Answer: c
Explanation: Biorthogonal codes perform better than the orthogonal codes and it needs only half the bandwidth as orthogonal codes.
13. Transorthogonal code is obtained by
a) Deleting last digit of each orthogonal codeword
b) Deleting first digit of each orthogonal codeword
c) Doubling each digit of orthogonal codeword
d) Taking negative of each digit of orthogonal codeword
Answer: b
Explanation: A code generated from an orthogonal set by deleting the first digit of each codeword is called as transorthogonal code or simplex code.
14. Which code requires minimum Eb/N0?
a) Orthogonal
b) Biorthogonal
c) Simplex
d) All of the mentioned
Answer: c
Explanation: A simplex code represents the minimum energy equivalent and it requires minimum Eb/N0 for a specified symbol error rate.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Structured sequence”.
1. Structured sequences are
a) Block codes
b) Convolutional codes
c) Turbo codes
d) All of the mentioned
Answer: d
Explanation: Structured sequences are those which add structured redundancy to the source code. Examples are turbo codes, block codes, and convolutional codes.
2. In binary symmetric channel, in the demodulator _____ decision is made.
a) Hard
b) Soft
c) Hard & Soft
d) None of the mentioned
Answer: c
Explanation: In binary symmetric channel the demodulator consists of only two discrete elements. Thus it makes hard or firm decision.
3. In Gaussian channel demodulator ______ decision is made.
a) Hard
b) Soft
c) Hard & Soft
d) None of the mentioned
Answer: b
Explanation: In Gaussian channel demodulator consists of continuous alphabet or quantized approximation thus soft decision is made.
4. Which decoders are less complex?
a) Soft decision decoder
b) Hard decision decoder
c) Hard & Soft decision decoder
d) None of the mentioned
Answer: b
Explanation: The design of a soft decision decoder is more complex than the design of hard decision decoder.
5. The ratio of redundant bits to data bits is called as
a) Code rate
b) Redundancy rate
c) Symbol rate
d) Transmission rate
Answer: b
Explanation: The ratio of redundant bits to data bits is called as redundancy of the code and the ratio of data bits to total bits is called as code rate.
6. If the parity bit takes value one then the summation of code-word gives
a) Even result
b) Odd result
c) Even & Odd result
d) None of the mentioned
Answer: a
Explanation: The parity bit takes on the value of one or zero to ensure that the summation of all the bits in the code-word yields even or odd result.
7. The summation operation is performed using _____ logical operation.
a) EX-OR
b) AND
c) OR
d) EX-NOR
Answer: a
Explanation: The summation operation is performed using modulo 2 arithmetic or exclusive OR operation.
8. The probability of message error is written as
a) Block error
b) Word error
c) Block & Word error
d) None of the mentioned
Answer: c
Explanation: We can write the probability of message error also called as block error or word error.
9. In a real time communication system addition of redundant bits leads to
a) More bandwidth requirement
b) Faster rate of transmission
c) More bandwidth requirement & Faster rate of transmission
d) None of the mentioned
Answer: c
Explanation: In a real time communication system the addition of redundant bits dictates a faster transmission rate and which means more bandwidth.
10. Coding gain is defined as
a) Reduction in Eb/N0
b) Increase in Eb/N0
c) Same as Eb/N0
d) None of the mentioned
Answer: a
Explanation: For a given bit error probability coding gain can be defined as reduction in Eb/N0 that can be realized through the use of code.
11. The capacity per cell is ______ Eb/N0.
a) Directly proportional
b) Inversely proportional
c) Double the
d) Half the
Answer: b
Explanation: The capacity that is the number of users per cell is inversely proportional to Eb/N0.
This set of Digital Communication MCQs focuses on “Error Detecting and Correcting Capability and Sampling”.
1. Hamming distance can be given by the number of elements in which
a) They are same
b) They differ
c) Which are non zero
d) None of the mentioned
Answer: b
Explanation: Hamming distance between two code words can be given by the number of elements in which they differ.
2. Code strength is characterized by its
a) Minimum distance
b) Maximum distance
c) Code weight
d) Code size
Answer: a
Explanation: The smallest number of the set given the minimum distance of the code. This minimum distance characterizes the strength of the code.
3. The distance between two code-words is equal to the _____ of the third code-word which is the sum of the first two code-words.
a) Size
b) Weight
c) Minimum distance
d) None of the mentioned
Answer: b
Explanation: The sum of two code words gives a third code which is also a linear code. The distance between two code words is equal to the weight of the third code word.
4. Error detecting capability is given as
a) Dmin + 1
b) Dmin -1
c) Dmin
d) Dmin/2
Answer: b
Explanation: The error detecting capability is given as e = Dmin – 1, where Dmin is the minimum distance of the code word.
5. The minimum distance Dmin can also be given as
a) Dmin >= α + β + 1
b) Dmin <= α + β + 1
c) Dmin >= α + β – 1
d) Dmin <= α + β + 1
Answer: a
Explanation: If a code can correct α errors and detect β errors, then the minimum distance can be given by the relation Dmin >= α + β + 1.
6. The number of errors that can be corrected without erasure information is
a) Dmin+1
b) Dmin – 1
c) /2
d) /2
Answer: d
Explanation: If a code has a minimum distance dmin then dmin-1 erasures can be reconstituted. The number of errors that can be corrected without erasure information is /2.
7. For better efficiency and simplicity, n should be
a) Maximum
b) Minimum
c) Zero
d) Infinity
Answer: b
Explanation: For real world codes, n should be minimum for better efficiency and simplicity.
8. Nyquist frequency is given by
a) fs
b) 2fs
c) fs/2
d) none of the mentioned
Answer: c
Explanation: Nyquist frequency is given as fs/2 where fs is the sampling frequency.
9. Some various types of distortion are
a) Jitter
b) Noise
c) Aperture error
d) All of the mentioned
Answer: d
Explanation: Various types of distortion are aliasing, jitters, aperture error, noise, error due to non linear effects, error due to quantization etc.
10. Noise which can affect sampling are
a) Thermal sensor noise
b) Analog circuit noise
c) Thermal sensor & Analog circuit noise
d) None of the mentioned
Answer: c
Explanation: Some of the noises that can cause distortion in sampling are thermal sensor noise, analog circuit noise etc.
11. Oversampling can completely eliminate
a) Aperture error
b) Non linearity
c) Quantization error
d) All of the mentioned
Answer: a
Explanation: Oversampling can completely eliminate aperture error and aliasing but can only reduce quantization error and non linearity to some extent.
12. Sampling can be used in
a) Audio
b) Speech
c) Video
d) All of the mentioned
Answer: d
Explanation: Sampling can be used in audio sampling, speech sampling and also in video sampling.
13. What is the bit depth used for audio recording?
a) 8 bit
b) 16 bit
c) 24 bit
d) All of the mentioned
Answer: d
Explanation: Audio is recorded at 8 bit, 16 bit, and also 24 bit depth which yield theoretically maximum SNR.
14. Which factors are measured using the units of lines per picture height?
a) Resolution
b) Sampling rate
c) Resolution & Sampling rate
d) None of the mentioned
Answer: c
Explanation: Sampling rate and resolution in spatial directions can be measured in units of lines per picture height.
15. Sampling of simultaneously two different but related wave-forms is called as
a) Over sampling
b) Complex sampling
c) Inter sampling
d) None of the mentioned
Answer: b
Explanation: Complex sampling is the process of simultaneously sampling two different but related signals.
16. Sampling can be done for functions varying in
a) Space
b) Time
c) Space & Time
d) None of the mentioned
Answer: c
Explanation: Sampling can be done for functions varying in time, frequency, space or any other dimension.
17. Reconstruction of continuous signals is done using
a) Decimation algorithm
b) Interpolation algorithm
c) Decimation & Interpolation algorithm
d) None of the mentioned
Answer: b
Explanation: Reconstruction of signals from the samples is done using interpolation algorithm.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Cyclic codes and linear block codes”.
1. The cyclic codes are designed using
a) Shift registers with feedback
b) Shift registers without feedback
c) Flipflops
d) None of the mentioned
Answer: a
Explanation: The cyclic codes are a subclass of linear codes. It is designed using feedback shift registers.
2. A cyclic code can be generated using
a) Generator polynomial
b) Generator matrix
c) Generator polynomial & matrix
d) None of the mentioned
Answer: a
Explanation: A cyclic code can be generated using generator polynomial and block codes can be generated using generator matrix.
3. The feedback shift register circuit is called as
a) Multiplying circuit
b) Dividing circuit
c) Feedback circuit
d) Shifting circuit
Answer: b
Explanation: The cyclic shift of a code-word polynomial and encoding involves division of one polynomial by another. Thus this feedback shift register is also called as dividing circuit.
4. In the dividing circuit, the parity polynomial is obtained by the
a) Quotient
b) Remainder
c) Dividend
d) Divisor
Answer: b
Explanation: The parity polynomial is the remainder after diving by the generator polynomial it is available in the register after n shifts through the n-k stage feedback registers.
5. The received code contains an error if the syndrome vector is
a) Zero
b) Non zero
c) Infinity
d) None of the mentioned
Answer: b
Explanation: If the syndrome is an all zero vector then the received code-word is a valid code. If the syndrome is a non zero vector then the received code has errors.
6. Block codes are generated using
a) Generator polynomial
b) Generator matrix
c) Generator polynomial & matrix
d) None of the mentioned
Answer: b
Explanation: Block codes are generated using generator matrix and cyclic codes are generated using generator polynomial.
7. Extended go-lay code is formed by
a) Adding overall parity bit to perfect go-lay code
b) Ex-oaring overall parity bit with perfect go-lay code
c) Ex-oaring each bit of go-lay code
d) Dividing the overall parity bit with perfect go-lay code
Answer: a
Explanation: Extended go-lay code is formed by adding overall parity bit with the perfect bit known as the golay code.
8. Block length is the _____________ in the code word.
a) Number of elements
b) Distance between elements
c) Number of parity bits
d) None of the mentioned
Answer: a
Explanation: The block length n is the number of elements in the code word.
9. The rate of a block code is the ration of
a) Block length to message length
b) Message length to block length
c) Message weight to block length
d) None of the mentioned
Answer: b
Explanation: The rate of a block code is the ratio between its message length and the block length, R=k/n.
10. Linear codes are used for
a) Forward error correction
b) Backward error correction
c) Forward error detection
d) Backward error detection
Answer: a
Explanation: Linear codes are used in forward error correction. It allows for more efficient encoding and decoding procedures.
11. The k-bit message forms ____ distinct messages which is referred to as k-tuples.
a) 2k
b) K2
c) 2k
d) 21/k
Answer: c
Explanation: The k bit messages for 2k distinct message sequences which are referred to as k-tuples or sequence of k digits.
12. The sum of any two vectors in subset S is also in S. This is called as
a) Addition property
b) Subset property
c) Closure property
d) Similarity property
Answer: c
Explanation: The closure property states that the sum of any two vectors in subset S is also in S.
13. To avoid corruption during transmission, the code-word should be
a) Near
b) Far apart
c) Far
d) None of the mentioned
Answer: b
Explanation: The code-words should be far apart from one and another as possible so even when the vectors experience some corruption they may still be correctly decoded.
14. In a standard matrix set code-word there are _______ cosset.
a) 2k
b) 2n+k
c) 2n-k
d) 2n
Answer: c
Explanation: Each n-tuple appears in only one location none are missing and none are replicated. There are 2n-k cosets.
15. Syndrome is calculated by
a) HT/r
b) rHT
c) rH
d) None of the mentioned
Answer: b
Explanation: The syndrome is calculated using S=rHT.
16. The _____ of the code-word is the number of non zero elements.
a) Size
b) Weight
c) Distance
d) Subspace
Answer: b
Explanation: The size of the code-word is the number of code words. The weight of the code word can be given as the number of non zero elements and the distance between two code words is the hamming distance between them.
17. Some examples of linear codes
a) Hamming code
b) Reed-Solomon code
c) Parity code
d) All of the mentioned
Answer: d
Explanation: Some examples of linear codes are block codes, parity codes, reed-Solomon codes, hamming code, cyclic codes, polynomial codes, go-lay codes etc.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Convolution encoding and decoding”.
1. The measure of the amount of redundancy is given by
a) Code size
b) Code weight
c) Code rate
d) Minimum distance
Answer: c
Explanation: The rate of the code gives the measure of the amount of redundancy. The rate is given by the ratio of number of data bits that form an input to a block encoder to the total number of bits.
2. The number of k bit shift over which a single information bit influences the encoder output is given by
a) Code rate
b) Constraint length
c) Code length
d) Code weight
Answer: b
Explanation: The constraint length represents the number of k bit shifts over which a single information bit influences the encoder output.
3. The method used for representing convolution encoder are
a) Connection pictorial
b) State diagram
c) Tree diagram
d) All of the mentioned
Answer: d
Explanation: Several methods are used for representing convolution encoder such as connection pictorial, connection vector or polynomials, state diagram, tree diagram and trellis diagram.
4. Periodic truncation is done by
a) Appending zero to end of the input data sequence
b) Appending zero to the beginning of the input data sequence
c) Appending one’s to end of the data sequence
d) Appending one’s to beginning of the data sequence
Answer: a
Explanation: Convolution code are forced into a block structure by periodic truncation which is done by appending zero to the end of the input data sequence, for the purpose of clearing or flushing the encoding shift register of the data bits.
5. Finite state machines have
a) Past memories also
b) Only present and future memories
c) Only future memories
d) Only present memories
Answer: a
Explanation: A convolution encoder belongs to a class of device called as finite state machines which are those machines that have a memory of past signals.
6. How many number of transitions can be made at each bit time?
a) One
b) Two
c) Three
d) Half
Answer: b
Explanation: An a consequence of shifting-in-one bit at a time there are only two possible transitions that the register can make at each bit time.
7. In trellis diagram, the number of nodes ______ at successive branching.
a) Increases by 1
b) Doubles
c) Triples
d) None of the mentioned
Answer: b
Explanation: As per the algorithm, in the trellis diagram at successive branching the number of nodes doubles.
8. Example for convolution encoder state diagram is
a) Tree diagram
b) Trellis diagram
c) Tree & Trellis diagram
d) None of the mentioned
Answer: c
Explanation: Tree diagram and trellis diagram are both examples for convolution encoding state diagram.
9. In maximum likelihood decoding technique, the likelihood function compares the
a) Joint probabilities
b) Individual probabilities
c) Conditional probabilities
d) None of the mentioned
Answer: c
Explanation: If all input messages are equally likely a decoder that achieves the minimum probability of error is that which compares the conditional probabilities also called the likelihood function.
10. In maximum likelihood detector the error probability is
a) Maximum
b) Minimum
c) Zero
d) None of the mentioned
Answer: b
Explanation: The maximum likelihood detector is an optimal detector which minimizes the error probability.
11. For a binary code, the maximum number of possible sequence made up of L branch words is
a) L2
b) 2L
c) 2L
d) L/2
Answer: b
Explanation: In a binary code, all possible different code word sequences that can be made up of L branch words are 2L.
12. If the quantization level of a demodulator output is ________ then it is called as soft decision decoding.
a) Equal to 2
b) More than 2
c) Less than 2
d) None of the mentioned
Answer: b
Explanation: If the quantization level of a demodulator output is 2 it is called as hard decision decoding and if it is greater than 2 it is called as soft decision decoding.
13. Soft decision decoding requires
a) Less memory
b) More memory
c) More speed
d) None of the mentioned
Answer: b
Explanation: The price paid for soft decision decoding is an increase in required memory size at the decoder.
14. A binary symmetric channel is a
a) Discrete memoryless channel
b) Continuous memoryless channel
c) Discrete memory channel
d) Continuous memory channel
Answer: a
Explanation: A binary symmetric channel is a discrete memoryless channel that has binary input and output alphabets and symmetric transition probabilities.
15. Binary symmetric channel is hard decision channel.
a) True
b) False
Answer: a
Explanation: A binary symmetric channel is an example for hard decision channel which means that even though continuous values may be received by the demodulator the BSC only allows firm decisions.
16. Gaussian channel is a hard decision channel.
a) True
b) False
Answer: b
Explanation: The quantized Gaussian channel is referred to as the soft decision channel.
17. Branch word synchronization is the process of determining the
a) Beginning of received sequence
b) Beginning of transmitted sequence
c) End of received sequence
d) End of transmitted sequence
Answer: a
Explanation: Branch word synchronization is the process of determining the beginning of a branch word in the received sequence. Such synchronization can take place without adding new information to the transmitted symbols.
This set of Digital Communication Multiple Choice Questions & Answers focuses on “Properties of Convolution Codes and Reed-Solomon Codes”.
1. Which distance is related to the error correcting capability of the code?
a) Maximum distance
b) Minimum distance
c) Maximum & Minimum distance
d) None of the mentioned
Answer: b
Explanation: Minimum distance between all pair of code words is related to the error correcting capability of the code.
2. A catastrophic error is an event whereby ______ number of code symbol error causes _______ number of decoded data error bits.
a) Finite, finite
b) Finite, infinite
c) Infinite, finite
d) Infinite, infinite
Answer: b
Explanation: A catastrophic error is defined as an event whereby a finite number of code symbols error causes an infinite number of decoded data bit errors.
3. A code should be selected such that it should
a) Not have catastrophic error propagation
b) Have minimum free distance
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: Generally code’s are selected based on the free distance property. The criteria for selecting code is that it should not have catastrophic error propagation and it should have maximum free distance.
4. The error correcting capability of a code scheme increases as the
a) Number of channel symbols per information bit increases
b) Bandwidth increases
c) Information per bit increases
d) All of the mentioned
Answer: d
Explanation: The error correcting capability of a code scheme increases as the number of channel symbols n per information bit k increases or the rate k/n decreases. As n increases bandwidth and complexity also increases.
5. Which has a limited resolution?
a) Hard decision decoder
b) Soft decision decoder
c) Hard & Soft decision decoder
d) None of the mentioned
Answer: b
Explanation: The soft decision algorithm has limited resolution than hard decision algorithm and thus it cannot use hamming distance metric.
6. Which parameter is used in soft decision algorithm?
a) Euclidean distance
b) Euclidean distance squared
c) Euclidean distance & distance squared
d) None of the mentioned
Answer: c
Explanation: As soft decision decoding algorithm has limited resolution it cannot use hamming distance metric. Instead it uses Euclidean and monotonic metric such as Euclidean distance squared.
7. Which factor is desirable?
a) Minimum correlation
b) Maximum correlation
c) Maximum distance
d) None of the mentioned
Answer: b
Explanation: While using Euclidean distance squared metric, maximum correlation is desirable rather than minimum distance.
8. Soft decision results in
a) Increase in complexity
b) Decrease in storage
c) Increase in complexity & Decrease in storage
d) None of the mentioned
Answer: a
Explanation: Soft decisions require more storage and it also increases the complexity. For this reason soft decisions are preferred less when compared to hard decisions.
9. A feedback decoder makes hard decision.
a) True
b) False
Answer: a
Explanation: The feedback decoder makes hard decision on data bits.
10. The look ahead length L is given by
a) m+1
b) m-1
c) /2
d) 1-m
Answer: a
Explanation: Look ahead length L is given by m+1 where m is the preselected positive integer.
11. On increasing L
a) Coding gain increases
b) Complexity increases
c) Coding gain & Complexity increases
d) None of the mentioned
Answer: c
Explanation: On increasing the look ahead length the coding gain increases and also increases the decoder implementation complexity.
12. Reed Solomon codes are
a) Non binary
b) Cyclic
c) Non binary & Cyclic
d) None of the mentioned
Answer: c
Explanation: Reed Solomon codes are non binary cyclic codes with symbols made up of m bit sequences where m is any positive integer with value greater than 2.
13. The minimum distance for Reed Solomon code is given by
a) n+k+1
b) n-k+1
c) 1-n-k
d) 1+n-k
Answer: b
Explanation: Reed Solomon code achieves the largest possible code minimum distance for any linear code and it is given by n-k+1.
14. R-S codes have
a) High code rate
b) Low redundancy
c) High code rate & Low redundancy
d) None of the mentioned
Answer: c
Explanation: Reed Solomon codes have high code rate and low redundancy. The complexity of a high speed implementation increases with redundancy.
15. Greater the redundancy greater will be the error correcting capability.
a) True
b) False
Answer: a
Explanation: In coding, the improving mechanism, the greater the redundancy the greater will be the error correcting capability of the code.
16. A primitive element are those which yield ______ when raised to higher order exponents.
a) All zero elements
b) Non zero elements
c) Unity elements
d) None of the mentioned
Answer: b
Explanation: In a primitive polynomial at least one of its roots will be primitive element. A primitive elements are those which yield non zero elements when raised to a higher order exponent.
17. If the inversion was performed correctly, then the multiplication of the original matrix with the inverted matrix should yield identity matrix. This is known as
a) Identity theorem
b) Safety theorem
c) Multiplication theorem
d) Inversion theorem
Answer: b
Explanation: According to safety theorem, if the inversion is performed correctly, the multiplication of original matrix by inverted matrix will yield identity matrix.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Concatenated and interleaving codes and its application in CD”.
1. Compact disc is used for
a) Digital storage
b) Reproduction of audio signals
c) Digital storage & Reproduction of audio signals
d) None of the mentioned
Answer: c
Explanation: Compact disc digital audio system is used for both digital storage and reproduction of audio signals.
2. Sources of channel errors are
a) Finger prints
b) Air bubbles
c) Unwanted particles
d) All of the mentioned
Answer: d
Explanation: Some of the sources of channel errors are some unwanted particles, air bubbles, finger prints, scratches etc.
3. The CIRC error control scheme includes
a) Correction
b) Concealment
c) Correction & Concealment
d) None of the mentioned
Answer: c
Explanation: The cross interleave reed Solomon error control scheme in the CD system involves both correction and concealment of errors.
4. Decoding step consists of
a) De-interleaving
b) Decoding
c) De-interleaving & Decoding
d) None of the mentioned
Answer: c
Explanation: The decoder step consists of both De-interleaving and decoding which are performed in the reverse order of encoding steps.
5. The ______ the error correcting capability used, the _____ will be the erasure correcting capability.
a) Larger, smaller
b) Smaller, larger
c) Smaller, smaller
d) Larger, larger
Answer: a
Explanation: There is a trade off between error correction and erasure correction, the larger the error correcting capability used the smaller will be the erasure correcting capability.
6. In interpolation
a) New samples are added
b) Unreliable samples are removed
c) New samples are added & Unreliable samples are removed
d) None of the mentioned
Answer: c
Explanation: The function of the interpolation function is to add new samples estimated from the reliable ones in place of unreliable ones.
7. The channel that exhibits multi-path fading
a) Has memory
b) Exhibits mutually dependent signal transmission impairments
c) Received signal will be distorted
d) All of the mentioned
Answer: d
Explanation: A channel that exhibits multi-path fading where the signal arrives at the receiver over two or more different paths. This channel has memory, exhibits mutually dependent signal transmission impairment.
8. Separating the symbols makes the channel memory-less.
a) True
b) False
Answer: a
Explanation: Separating the symbols in time effectively changes a channel with memory into a memory-less one.
9. The minimum end to end delay in block interleaving is
a) 2MN+2M+2
b) 2MN-2M+2
c) 2MN-2M-2
d) 2MN+2M-2
Answer: b
Explanation: The inter leaver or De-inter leaver delay is 2MN symbol times. M+1 memory cells needs to be filled before transmission can begin. Thus the minimum end to end delay is 2MN-2M+2 symbol times, not including any channel propagation delay.
10. A concatenated code uses
a) One level of coding
b) Two levels of coding
c) Three levels of coding
d) None of the mentioned
Answer: b
Explanation: A concatenated code is the one that uses two levels of coding, inner code and outer code to achieve desired error performance.
11. A concatenated code has
a) Low error rate
b) High complexity
c) Low error rate & High complexity
d) None of the mentioned
Answer: a
Explanation: A concatenated code has low error rate with an overall implementation complexity which is less than that would be required for single level coding.
12. The performance of non binary concatenated code depends on
a) Bit errors
b) Symbols errors
c) Bit & Symbols errors
d) None of the mentioned
Answer: b
Explanation: The performance of non binary code like reed-Solomon code which is also an example for concatenated code depends only on the number of symbol errors in the block.
13. Concatenated code is an example of
a) Error detecting code
b) Error correcting code
c) Error detecting & correcting code
d) None of the mentioned
Answer: b
Explanation: Concatenated codes belong to the class of error correcting codes.
14. Turbo code is a
a) Serial concatenation code
b) Parallel concatenation code
c) Block code
d) None of the mentioned
Answer: b
Explanation: Turbo codes implemented a parallel concatenation of two convolution codes with an inter leaver and an iterative decoder.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Shannon hartley theorem and turbo codes”.
1. The minimum nyquist bandwidth needed for baseband transmission of Rs symbols per second is
a) Rs
b) 2Rs
c) Rs/2
d) Rs2
Answer: c
Explanation: Theoretical minimum nyquist bandwidth needed for the baseband transmission of Rs symbols per second without ISI is Rs/2.
2. The capacity relationship is given by
a) C = W log2
b) C = 2W log2
c) C = W log2
d) C = W log10
Answer: a
Explanation: The capacity relationship from Shannon-hartley capacity theorem is given by C = W log2 .
3. Which parameter is called as Shannon limit?
a) PB/N0
b) EB/N0
c) EBN0
d) None of the mentioned
Answer: b
Explanation: There exists a limiting value for EB/N0 below which they can be no error free communication at any information rate. This EB/N0 is called as Shannon limit.
4. Entropy is the measure of
a) Amount of information at the output
b) Amount of information that can be transmitted
c) Number of error bits from total number of bits
d) None of the mentioned
Answer: a
Explanation: Entropy is defined as the average amount of information per source output.
5. Equivocation is the
a) Conditional entropy
b) Joint entropy
c) Individual entropy
d) None of the mentioned
Answer: a
Explanation: Shannon uses a correction factor called equivocation to account for uncertainty in the received signal. It is defined as the conditional entropy of the message X given Y.
6. For a error free channel, conditional probability should be
a) Zero
b) One
c) Equal to joint probability
d) Equal to individual probability
Answer: a
Explanation: For a error free channel conditional probability should be zero, because having received Y there is complete certainty about the message X.
7. Average effective information is obtained by
a) Subtracting equivocation from entropy
b) Adding equivocation with entropy
c) Ratio of number of error bits by total number of bits
d) None of the mentioned
Answer: a
Explanation: According to Shannon the average effective information is obtained by subtracting the equivocation from the entropy of the source.
8. Turbo codes are
a) Forward error correction codes
b) Backward error correction codes
c) Error detection codes
d) None of the mentioned
Answer: a
Explanation: Turbo codes are a class of high performance forward error correction codes.
9. Components used for generation of turbo codes are
a) Inter leavers
b) Punching pattern
c) Inter leavers & Punching pattern
d) None of the mentioned
Answer: c
Explanation: There are many instances of turbo codes, using different component encoders, input/output ratios, inter leavers, punching patterns.
10. Decoders are connected in series.
a) True
b) False
Answer: a
Explanation: Two elementary decoders are connected in serial connection for decoding the turbo codes.
11. The inter leaver connected between the two decoders is used to
a) Remove error bursts
b) Scatter error bursts
c) Add error bursts
d) None of the mentioned
Answer: b
Explanation: An inter leaver installed between the two decoders connected in series is used to scatter error bursts.
12. In soft decision approach what does -127 mean?
a) Certainly one
b) Certainly zero
c) Very likely zero
d) Very likely one
Answer: b
Explanation: The decoder front end produces an integer for each bit in the data stream. This integer is the measure of how likely it is that the bit 0 or 1 and is called as soft bit. It ranges from -127 to 127. Here -127 represents certainly zero.
13. In soft decision approach 100 means?
a) Certainly one
b) Certainly zero
c) Very likely zero
d) Very likely one
Answer: d
Explanation: The decoder front end produces an integer for each bit in the data stream. This integer is the measure of how likely it is that the bit 0 or 1 and is called as soft bit. It ranges from -127 to 127. Here 100 represents very likely one.
14. In soft decision approach 0 represents
a) Certainly one
b) Certainly zero
c) Very likely zero
d) Could be either zero or one
Answer: d
Explanation: The decoder front end produces an integer for each bit in the data stream. This integer is the measure of how likely it is that the bit 0 or 1 and is called as soft bit. It ranges from -127 to 127. Here 0 represents ‘could be either zero or one’.
This set of Digital Communication online test focuses on “Bandwidth Efficiency Plane and Bandwidth Efficient Modulation”.
1. QPSK is a composite of
a) Two BPSK
b) Three BPSK
c) Two FSK
d) Two M-ary PSK
Answer: a
Explanation: QPSK is effectively a composite of two BPSK signals transmitted on orthogonal components of the carrier.
2. The IF transmission bandwidth of MFSK is
a) M/T
b) T/M
c) 2M/T
d) M/2
Answer: a
Explanation: In non coherent orthogonal MFSK modulation, the IF transmission bandwidth is WIF = M/T.
3. The bandwidth efficiency of MFSK is
a) M/log2M
b) log2M/M
c) M/log10M
d) log10M/M
Answer: b
Explanation: For non coherent orthogonal MFSK modulation the bandwidth efficiency is R/W=log2M/M.
4. The bandwidth efficiency increases as M increases.
a) True
b) False
Answer: b
Explanation: The bandwidth efficiency decreases as M increases.
5. The bandwidth efficiency of QFSK is _______ that of BFSK.
a) Greater than
b) Less than
c) Equal to
d) None of the mentioned
Answer: c
Explanation: The bandwidth efficiency of QPSK is equal to that of the BFSK.
6. Which modulation is the most efficient one?
a) BPSK
b) BFSK
c) QPSK
d) QAM
Answer: d
Explanation: Of all the modulation showed, QAM is the most bandwidth efficient one.
7. The primary communication resource is
a) Transmitted power
b) Received power
c) Efficiency
d) None of the mentioned
Answer: a
Explanation: The primary communication resources are transmitted power and channel bandwidth.
8. Greater the redundancy lesser is the bandwidth expansion.
a) True
b) False
Answer: b
Explanation: Within the same block size, the code with greater redundancy requires more bandwidth expansion.
9. Which modulation scheme uses nonlinear transponders?
a) MSK
b) Offset QPSK
c) MSK & Offset QPSK
d) None of the mentioned
Answer: c
Explanation: Offset QPSK and MSK are two examples of constant envelope modulation schemes that are attractive for systems using non linear transponders.
10. Which modulation requires more bandwidth?
a) QPSK
b) OQPSK
c) BPSK
d) BFSK
Answer: c
Explanation: BPSK requires more bandwidth than the other modulation techniques for the given level of spectral density.
11. Which modulation has lower side lobe levels?
a) QPSK
b) OQPSK
c) BPSK
d) MSK
Answer: d
Explanation: MSK spectrum has wider main lobe level and lower side lobe levels than the other modulation spectrum.
12. Which modulation is spectrally more efficient?
a) BPSK
b) MSK
c) QPSK
d) OQPSK
Answer: b
Explanation: MSK is spectrally more efficient than the other modulation schemes. It has wider main lobe level and lower side lobe levels.
13. Which modulation spectrum has narrow main lobe?
a) QPSK
b) OQPSK
c) BPSK
d) MSK
Answer: a
Explanation: QPSK has narrower main lobe than MSK because for a given rate the QPSK symbol rate is half the MSK symbol rate.
14. Which technique can be used for bandwidth reduction?
a) BPSK
b) QPSK
c) MPSK
d) MFSK
Answer: c
Explanation: Coherent M-ary PSK modulation is a well known technique for achieving bandwidth reduction.
15. QPSK amplitude modulates
a) Sine function
b) Cosine function
c) Sine & Cosine function
d) None of the mentioned
Answer: c
Explanation: QPSK modulation consists of two independent streams. One stream amplitude modulates cosine wave and the other amplitude modulates sine wave.
16. QAM is a combination of
a) ASK and FSK
b) ASK and PSK
c) PSK and FSK
d) None of the mentioned
Answer: b
Explanation: QAM is a combination of both ASK and PSK and is also called as amplitude phase keying.
This set of Digital Communication online quiz focuses on “Evaluating Digital Communication System and Allocation of Communication Resources”.
1. Which refers to the presence of error correction coding scheme?
a) Coded
b) Uncoded
c) Coded & Uncoded
d) None of the mentioned
Answer: a
Explanation: The term coded refers to the presence of error correction coding scheme involving the use of redundant bits and expanded bandwidth.
2. The system will be more bandwidth efficient as WTb decreases.
a) True
b) False
Answer: a
Explanation: Any digital communication system will become more bandwidth efficient as its WTb product decreases.
3. In MPSK, R/W
a) Increases with increase in M
b) Increases with decrease in M
c) Decreases with increase in M
d) Is not dependent on M
Answer: a
Explanation: The bandwidth efficiency R/W increases as M increases in case of MPSK modulation.
4. In MFSK, R/W
a) Increases with increase in M
b) Increases with decrease in M
c) Decreases with increase in M
d) Is not dependent on M
Answer: c
Explanation: In MFSK, the denominator increases faster as the numerator when M increases. Thus R/W decreases with increase in M.
5. Which modulation is selected for power limited system?
a) MPSK
b) MFSK
c) MPSK & MFSK
d) None of the mentioned
Answer: b
Explanation: For an uncoded system, if the channel is power limited MFSK is selected and if it is bandwidth limited MPSK is selected.
6. The transmission rate is given as
a) R/log2M
b) log2M/R
c) R/log10M
d) log10M/R
Answer: a
Explanation: The relationship between symbol transmission rate and data rate R is given by Rs= R/log2M.
7. Code’s redundancy is given as
a) n/k
b) k/n
c) nk
d) None of the mentioned
Answer: a
Explanation: The rate of code is given as k/n and its inverse n/k gives the redundancy of the code.
8. For more than two quantization level, _______ is used.
a) Hard decision
b) Soft decision
c) Hard & Soft decision
d) None of the mentioned
Answer: b
Explanation: When the output is quantized for more than two levels, soft decision is made.
9. How can the throughput of the system be increased?
a) Increasing EIRP
b) Increasing channel bandwidth
c) Decreasing system losses
d) All of the mentioned
Answer: d
Explanation: The basic ways to increase the throughput of the system is to increase the EIRP, by reducing the system losses, providing channel bandwidth, and efficient allocation of CR.
10. In which technique spot beam antennas are used to point the radio signals in different directions?
a) Code division
b) Space division
c) Frequency division
d) None of the mentioned
Answer: b
Explanation: In space division multiplexing, spot beam antennas are used to separate radio signals by pointing in different directions.
11. In polarization division technique ______ polarization is used.
a) Orthogonal
b) Non orthogonal
c) Orthogonal & Non orthogonal
d) None of the mentioned
Answer: a
Explanation: In polarization division technique orthogonal polarization is used to separate signals for reuse.
12. Channelization characterized by orthogonal spectra is called as
a) Time division multiplexing
b) Frequency division multiplexing
c) Time division & Frequency division multiplexing
d) None of the mentioned
Answer: b
Explanation: Channelization characterized by orthogonal wave-forms is called as time division multiplexing and those characterized by orthogonal spectra is called as frequency division multiplexing.
13. In double side band spectrum which side-band is called as inverted side-band?
a) LSB
b) USB
c) LSB & USB
d) None of the mentioned
Answer: a
Explanation: The double side band spectrum consists of two side bands. LSB is called as inverted side-band and the USB is called as erect side-band.
14. The mixing and filtering of DSB removes
a) LSB
b) USB
c) LSB & USB
d) None of the mentioned
Answer: b
Explanation: The mixing and filtering yields frequency shifted channels and it removes the USB.
15. The super group is made up of
a) Five groups
b) Sixty channels
c) Five groups & Sixty channels
d) None of the mentioned
Answer: c
Explanation: In FDM multiplex hierarchy, the first level is composed of five channels. The second level consists of five groups or 60 channels.
16. The orbital period of communication satellites is _______ as that of earth’s rotational period.
a) Same
b) Greater
c) Lesser
d) None of the mentioned
Answer: a
Explanation: The communication satellite is in circular orbit and has orbital period same as that of the earth’s rotational period.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Trellis coded modulation”.
1. Adding improvement to the signal subset choice by starting with the densest possible lattice for the space is called as
a) Optimum signal constellation boundaries
b) Higher density lattice structures
c) Trellis coded modulation
d) None of the mentioned
Answer: b
Explanation: Higher density lattice structures are formed by adding improvement to the signal subset choice by starting with the densest possible lattice for the space.
2. Choosing a closely packed signal subset from any regular lattice or array of candidate points is called as
a) Optimum signal constellation boundaries
b) Higher density lattice structures
c) Trellis coded modulation
d) None of the mentioned
Answer: a
Explanation: Optimum signal constellation boundaries are obtained by choosing closely packed signal subset from any regular lattice or array of candidate points.
3. Trellis coded modulation uses
a) Non binary method
b) Uses redundant bits
c) No expansion of bandwidth
d) All of the mentioned
Answer: d
Explanation: Trellis coded modulation is a achieves error performance improvements without expansion of bandwidth. It uses redundant non binary modulation in combination with finite state machines.
4. In finite state machines, output is predicted using
a) Past output
b) Present input
c) Past output & Present input
d) None of the mentioned
Answer: c
Explanation: In finite state machines, the set of possible future outputs are predicted using present input and past output.
5. In TCM the signals are decoded using
a) Soft decision decoders
b) Hard decision decoders
c) Soft & Hard decision decoders
d) None of the mentioned
Answer: a
Explanation: In trellis coded modulation the noisy received signals are decoded using soft decision decoders.
6. Trellis coded modulation scheme
a) Has memory
b) Is memory-less
c) Needs more bandwidth
d) None of the mentioned
Answer: a
Explanation: Trellis coded modulation scheme has memory such as the convolutional code.
7. In TCM channel capacity can be increased.
a) True
b) False
Answer: a
Explanation: In trellis coded modulation increase in channel capacity is achieved by signal set expansion.
8. Transmission bandwidth depends on
a) Rate of signalling
b) Density of signal points
c) Reduced distance
d) None of the mentioned
Answer: a
Explanation: Transmission bandwidth of non orthogonal signalling depends on rate of signalling and not on density of signal points in the constellation.
9. Which determines the error performance in TCM?
a) Reduced distance
b) Free distance
c) Bandwidth
d) None of the mentioned
Answer: b
Explanation: Free distance which is the minimum distance between members of set of allowed code sequences determines the error performance.
10. The TCM decoder uses
a) Minimum likelihood detector
b) Maximum likelihood detector
c) Does not use redundant bits
d) None of the mentioned
Answer: b
Explanation: The task of convolutional decoder is to estimate the path that the message had traversed through the encoding trellis using maximum likelihood detector.
This set of Digital Communication Question Bank focuses on “PAM, PWM and TDM”
1. Flat top sampling of low pass signals
a) Gives rise to aperture effect
b) Implies over sampling
c) Leads to aliasing
d) Introduces delay distortion
Answer: a
Explanation: Flat top sampling of low pass signals gives rise to aperture effect.
2. In a delta modulation system, granular noise occurs when the
a) Modulating signal increases rapidly
b) Pulse rate decreases
c) Pulse amplitude decreases
d) Modulating signal remains constant
Answer: d
Explanation: In a delta modulation system, granular noise occurs when the modulating signal remains constant.
3. A PAM signal can be detected using
a) Low pass filter
b) High pass filter
c) Band pass filter
d) All pass filter
Answer: a
Explanation: A PAM signal can be detected by using low pass filter.
4. Coherent demodulation of FSK signal can be performed using
a) Matched filter
b) BPF and envelope detectors
c) Discriminator
d) None of the mentioned
Answer: a
Explanation: Coherent demodulation of FSK signal can be performed using matched filter.
5. The use of non uniform quantization leads to
a) Reduction in transmission bandwidth
b) Increase in maximum SNR
c) Increase in SNR for low level signals
d) Simplification of quantization process
Answer: c
Explanation: The use of non uniform quantization leads to increase in SNR for low level signals.
6. Which of the following requires a synchronizing signal?
a) Single channel PPM system
b) PAM
c) DM
d) All of the mentioned
Answer: b
Explanation: PAM requires a synchronizing signal.
7. A PWM signal can be generated by
a) An astable multi vibrator
b) A monostable multi vibrator
c) Integrating a PPM signal
d) Differentiating a PPM signal
Answer: b
Explanation: A PWM signal can be generated by a mono stable multi vibrator.
8. TDM is less immune to cross-talk in channel than FDM.
a) True
b) False
Answer: b
Explanation: False because different message signals are not applied to the channel simultaneously.
9. In an ideal TDM system, the cross correlation between two users of the system is
a) 1
b) 0
c) Infinity
d) -1
Answer: b
Explanation: In an ideal TDM system, the cross correlation between two users of the system is 0.
10. TDM requires
a) Constant data transmission
b) Transmission of data samples
c) Transmission of data at random
d) Transmission of data of only one measured
Answer: b
Explanation: TDM requires transmission of data samples.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “M-ary modulation and amplifiers”.
1. In M-ary FSK, as M increases error
a) Increases
b) Decreases
c) Does not get effected
d) Cannot be determined
Answer: b
Explanation: In M-ary FSK as M increases error decreases.
2. In M-ary FSK as M tends to infinity, probability of error tends to
a) Infinity
b) Unity
c) Zero
d) None of the mentioned
Answer: c
Explanation: In M-ary FSK as M tends to infinity, probability of error becomes zero.
3. For non coherent reception of PSK _____ is used.
a) Differential encoding
b) Decoding
c) Differential encoding & Decoding
d) None of the mentioned
Answer: c
Explanation: For non coherent reception of PSK, differential encoding is used at the transmitter and decoding is used at the receiver.
4. Which modulation technique have the same bit and symbol error probability?
a) BPSK
b) DPSK
c) OOK
d) All of the mentioned
Answer: d
Explanation: BPSK, DPSK, OOK and non coherent FSK have same bit and symbol error probability.
5. An amplifier uses ______ to take input signal.
a) DC power
b) AC power
c) DC & AC power
d) None of the mentioned
Answer: a
Explanation: An amplifier uses DC power to take an input signal and increase its amplitude at the output.
6. Which has 50% maximum power efficiency?
a) Class A
b) Class B
c) Class AB
d) None of the mentioned
Answer: a
Explanation: Class A amplifiers have 50% maximum power efficiency.
7. Which generates high distortion?
a) Class A
b) Class B
c) Class C
d) Class AB
Answer: c
Explanation: Class C amplifiers generate high distortion and it is closer to switch than an amplifier.
8. Class B linear amplifiers have maximum power efficiency of
a) 50%
b) 75%
c) 78.5%
d) None of the mentioned
Answer: c
Explanation: Class B linear amplifiers have maximum power efficiency of 78.5%.
9. Which has the maximum power efficiency?
a) Class A
b) Class B
c) Class C
d) Class AB
Answer: c
Explanation: Class C has the maximum power efficiency when compared to the other linear amplifiers.
Answer: c
Explanation: Free space is an idealization that consists of only transmitter and receiver.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Modulation and Multiplexing”.
1. Properties used to determine stream’s fidelity
a) Sampling rate
b) Bit depth
c) Sampling rate & Bit depth
d) None of the mentioned
Answer: c
Explanation: Two basic properties to determine stream’s fidelity are bit depth and sampling rate, number of times per second that samples are taken.
2. In bipolar codes, pulses can be
a) Positive
b) Negative
c) Absent
d) All of the mentioned
Answer: d
Explanation: In bipolar codes, the pulses can be positive, negative or absent.
3. Delta modulation is ______ conversion.
a) Analog to digital
b) Digital to analog
c) Analog to digital and digital to analog
d) None of the mentioned
Answer: c
Explanation: Delta modulation is the process of analog to digital and digital to analog conversion technique used for transmission of voice signals.
4. To achieve high signal to noise ratio, delta modulation must use
a) Under sampling
b) Over sampling
c) Aliasing
d) None of the mentioned
Answer: b
Explanation: To achieve high signal to noise ratio, delta modulation must use over sampling techniques.
5. The demodulator in delta modulation technique is
a) Differentiator
b) Integrator
c) Quantizer
d) None of the mentioned
Answer: b
Explanation: The demodulator used in delta modulation is a simple form of integrator.
6. Source of noise in delta modulation is
a) Granularity
b) Slope overload
c) Granularity & Slope overload
d) None of the mentioned
Answer: c
Explanation: Sources of noise in delta modulation are granularity and slope overload.
7. When probability of receiving a symbol is 1 then how much information will be obtained?
a) Little information
b) Much information
c) No information
d) None of the mentioned
Answer: c
Explanation: When the probability of receiving a symbol is 1 then the information obtained is zero.
8. In channel encoding procedure
a) Redundancy bits are added
b) Errors are corrected
c) Redundancy bits are added & Errors are corrected
d) None of the mentioned
Answer: c
Explanation: In channel encoding includes addition of redundancy to the signal such that any bit errors can be corrected.
9. Modulation process includes
a) Analog to digital conversion
b) Digital to analog conversion
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: Modulation procedure includes digital to analog conversion which produces a continuous time signal that can be sent through the channel.
10. Switching exists in
a) Point to point communication
b) Broadcast communication
c) Point to point & Broadcast communication
d) None of the mentioned
Answer: a
Explanation: Circuit connection in point to point communication is called as switching. Switching does not exist in broadcast communication or network.
11. Space division has dedicated
a) Paths
b) Time slots
c) Paths & Time slots
d) None of the mentioned
Answer: a
Explanation: Space division has dedicated paths and time division has dedicated time slots.
12. In time division system, the actual switch is called as
a) Speech memory
b) Cross point
c) Connecting point
d) None of the mentioned
Answer: a
Explanation: In time division, actual switch is called as speech memory and in space division actual switch is called as cross point.
13. In time division, connection is established using
a) Data exchange
b) Galvanic connection
c) Data exchange & Galvanic connection
d) None of the mentioned
Answer: a
Explanation: In time division, connection is established using data exchange and in space division it is established using galvanic connections.
14. Analogue switches provides
a) Good bandwidth
b) Low distortion
c) Low cost
d) All of the mentioned
Answer: d
Explanation: Some of the features of analogue switches are good bandwidth, low cost, low distortion, lower reliability.
15. Operations performed by switching system
a) Path establishment
b) Information exchange
c) Tariff computation
d) All of the mentioned
Answer: d
Explanation: Operations performed by switching network are path establishment, information exchange, tariff computation, maintenance, billing etc.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Modulation and Entropy”.
1. PCM includes the process of
a) Amplitude discretization
b) Time discretization
c) Amplitude & Time discretization
d) None of the mentioned
Answer: c
Explanation: Time discretization and amplitude discretization are the two processes done in PCM system.
2. For which quantization process is used?
a) Amplitude discretization
b) Time discretization
c) Amplitude & Time discretization
d) None of the mentioned
Answer: a
Explanation: Sampling process is used for time discretization and quantization process is used for amplitude discretization.
3. Modulation process corresponds to ______ the amplitude, frequency or phase.
a) Switching
b) Keying
c) Switching or keying
d) None of the mentioned
Answer: c
Explanation: Modulation process corresponds to switching or keying the amplitude, frequency or phase of CW carrier.
4. Matched filter
a) Is a non linear filter
b) Produces maximum output SNR
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: Matched filter is a linear filter which produces maximum output SNR for a given transmitted signal.
5. Which has same probability of error?
a) ASK and FSK
b) ASK and PSK
c) PSK and FSK
d) None of the mentioned
Answer: c
Explanation: Uni-polar base-band signalling, PSK and FSK has same probability of error.
6. Which has higher error probability performance?
a) Uni-polar base-band signalling
b) Bipolar base-band signalling
c) ASK
d) FSK
Answer: b
Explanation: Bipolar base-band signalling has high error probability performance than the others. The probability of error value is A2T.
7. Time division multiplexing uses
a) High pass filter
b) Commutator
c) High pass filter & Commutator
d) None of the mentioned
Answer: b
Explanation: Time division multiplexing uses low pass filter and commutator.
8. In TDM, at the receiver end, ____ filter is used.
a) Low pass
b) High pass
c) Band pass
d) Band stop
Answer: a
Explanation: In TDM, at the receiver end low pass filtering is done to obtain individual signals.
9. Which provides more secure communication?
a) CDMA
b) FDMA
c) TDMA
d) None of the mentioned
Answer: a
Explanation: CDMA provides more secure communication than TDMA and FDMA.
10. Entropy is the measure of
a) Randomness
b) Information
c) Randomness & Information
d) None of the mentioned
Answer: c
Explanation: Entropy can be defined as the measure of randomness or information.
11. Entropy calculation returns
a) Random variable
b) Deterministic number
c) Random number
d) None of the mentioned
Answer: b
Explanation: Entropy calculation returns a deterministic number and not a random variable.
12. Entropy of N random variables is the _____ of the entropy of individual random variable.
a) Sum
b) Product
c) Sum of squares
d) Average
Answer: a
Explanation: Entropy of N random variables is the sum of the entropy of individual random variable.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Modulation Techniques”.
1. Average energy per bit is given by
a) average energy symbol/log2 M
b) average energy symbol * log2 M
c) log2 M/ Average energy symbol
d) none of the mentioned
Answer: a
Explanation: Average energy per bit is given by average energy symbol divided by log2 M.
2. Which FSK has no phase discontinuity?
a) Continuous FSK
b) Discrete FSK
c) Uniform FSK
d) None of the mentioned
Answer: a
Explanation: Continuous frequency shift keying has no phase discontinuity between symbols.
3. FSK reception is
a) Phase Coherent
b) Phase non coherent
c) Phase Coherent & non coherent
d) None of the mentioned
Answer: c
Explanation: Reception of FSK can be either phase coherent or phase non coherent.
4. FSK reception uses
a) Correlation receiver
b) PLL
c) Correlation receiver & PLL
d) None of the mentioned
Answer: c
Explanation: Frequency shift keying uses correlation receiver and phase locked loop.
5. In non coherent reception _____ is measured.
a) Phase
b) Energy
c) Power
d) None of the mentioned
Answer: b
Explanation: In non coherent reception of FSK, energy in each frequency is measured.
6. Every frequency has ____ orthogonal functions.
a) One
b) Two
c) Four
d) Six
Answer: b
Explanation: Every frequency has two orthogonal functions – sine and cosine.
7. If we correlate the received signal with any one of the two orthogonal function, the obtained inner product will be
a) In phase
b) Quadrature
c) Zero
d) Cannot be determined
Answer: c
Explanation: If we correlate the received signal with only one of the orthogonal function for example sine, the inner product obtained will be zero.
8. If we correlate the received signal with both orthogonal function the inner product will be
a) In phase
b) Quadrature
c) In phase and quadrature
d) Unity
Answer: c
Explanation: If we correlate the received signal with both the orthogonal function, the two inner products obtained will be in phase and quadrature.
9. Simulation is used to determine
a) Bit error rate
b) Symbol error rate
c) Bit error
d) Symbol error
Answer: a
Explanation: A simulation of digital communication system is used to estimate bit error rate.
10. Matched filter is a _____ technique.
a) Modulation
b) Demodulation
c) Modulation & Demodulation
d) None of the mentioned
Answer: b
Explanation: Matched filter is a demodulation technique with LTI.
11. Which is called as on-off keying?
a) Amplitude shift keying
b) Uni-polar PAM
c) Amplitude shift keying & Uni-polar PAM
d) None of the mentioned
Answer: c
Explanation: Amplitude shift keying and uni-polar PAM both schemes are called as on off keying.
12. QAM uses ______ as the dimensions.
a) In phase
b) Quadrature
c) In phase & Quadrature
d) None of the mentioned
Answer: c
Explanation: QAM uses in phase and quadrature which is cosine and sine respectively as the dimensions.
13. Which has same probability of error?
a) BPSK and QPSK
b) BPSK and ASK
c) BPSK and PAM
d) BPSK and QAM
Answer: c
Explanation: BPSK is similar to bipolar PAM and both have same probability of error.
14. Which system uses QAM?
a) Digital microwave relay
b) Dial up modem
c) Digital microwave relay & Dial up modem
d) None of the mentioned
Answer: c
Explanation: Digital microwave relay, dial up modem and etc uses QAM modulation technique.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Synchronization and network synchronization”.
1. In phase lock which parameter is synchronized
a) Frequency
b) Phase
c) Frequency & Phase
d) None of the mentioned
Answer: c
Explanation: Being in phase lock means that the receiver’s local oscillator is synchronized in both frequency and phase with the received signal.
2. Coherent modulation requires ____ level of synchronization.
a) One
b) Two
c) Three
d) None of the mentioned
Answer: c
Explanation: Coherent modulation requires three levels of synchronization – phase symbol and frame.
3. Non coherent system requires
a) Phase synchronization
b) Frequency synchronization
c) Phase & Frequency synchronization
d) None of the mentioned
Answer: c
Explanation: Non coherent system requires frequency synchronization.
4. For block codes _____ is necessary.
a) Phase synchronization
b) Frequency synchronization
c) Frame synchronization
d) None of the mentioned
Answer: c
Explanation: Frame synchronization is necessary when the information is organized in blocks or messages of some uniform number of symbols. Thus block code needs frame synchronization.
5. The synchronization necessary for BPSK is
a) Frequency tracking
b) Bit timing
c) Phase tracking
d) All of the mentioned
Answer: d
Explanation: Non coherently detected BPSK is the simplest digital receiver and it requires frequency tracking, bit timing synchronizations and also phase synchronization.
6. Transmitter synchronization implies
a) One way communication
b) Two way communication
c) One & Two way communication
d) None of the mentioned
Answer: b
Explanation: Transmitter plays a role in synchronization by varying the frequency and timing of its transmissions to correspond to expectation of receiver. This transmitter synchronization implies two way communication.
7. As the synchronization levels increases, cost
a) Decreases
b) Increases
c) Remains the same
d) None of the mentioned
Answer: b
Explanation: For each level increase in synchronization levels, cost increases.
8. Synchronization is used in system which has
a) Non coherent modulation techniques
b) Many users accessing
c) Non coherent modulation techniques & Many users accessing
d) None of the mentioned
Answer: c
Explanation: For communication system using non coherent modulation techniques and many users accessing the central communication node, synchronization is necessary.
9. Which method has fast acquisition and can work without return link?
a) Open loop method
b) Closed loop method
c) Open & Closed loop method
d) None of the mentioned
Answer: a
Explanation: The main advantage of open loop method is that acquisition is fast, the procedure can work without return link and the amount of real time computation that is required is small.
10. Which method has two way link?
a) Open loop method
b) Closed loop method
c) Open & Closed loop method
d) None of the mentioned
Answer: b
Explanation: Closed loop method requires return link, large amount of real time processing and has two way link.
11. Which method requires external authority and is inflexible?
a) Open loop method
b) Closed loop method
c) Open & Closed loop method
d) None of the mentioned
Answer: a
Explanation: Open loop method is inflexible, does not adjust quickly to unplanned changes and needs external authority that provides prior knowledge, where as closed loop method does not need all these.
12. Which method needs precorrect time?
a) Open loop method
b) Closed loop method
c) Open & Closed loop method
d) None of the mentioned
Answer: a
Explanation: In open loop method the frequency and time are precorrected.
13. In open loop method, the time reference error __________ with respect to time.
a) Increases exponentially
b) Decreases exponentially
c) Increases quadratically
d) Decreases quadratically
Answer: a
Explanation: In open loop transmitter synchronization system, the time reference error increases quadratically with time.
14. When a terminal is able to utilize the measurements made on the return link it is called as
a) Open loop method
b) Closed loop method
c) Quasi close loop method
d) None of the mentioned
Answer: c
Explanation: The case where a terminal is able to utilize the measurements made on a return link signal is called as quasi closed loop transmitter synchronization.
15. Which reduces the effective distance between signals in signal space?
a) Time error
b) Frequency offset
c) Time error & Frequency offset
d) None of the mentioned
Answer: c
Explanation: Time error or frequency offset or combination of both will reduce the effective distance between signals in signal space and degrade error performance.
16. The system which has ______ signal to noise ratio can tolerate larger timing error.
a) Large
b) Small
c) Large & Small
d) None of the mentioned
Answer: a
Explanation: The system with increased signal to noise ratio will allow it to tolerate larger timing error so the improvement is error performance is rapid.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Receiver synchronization”.
1. Phase locked loops consists of
a) Phase detectors
b) Loop filter
c) Voltage controlled oscillator
d) All of the mentioned
Answer: d
Explanation: Phase locked loops have three basic components – phase detector, loop filter and voltage controlled oscillator. It measures the phase difference between the incoming signal and local replica.
2. In VCO the output frequency is a linear function of its input
a) Frequency
b) Voltage
c) Time period
d) None of the mentioned
Answer: b
Explanation: A VCO is an oscillator whose output frequency is a linear function of its input voltage over some range of input and output.
3. PLL designs have
a) First order loop
b) Second order loop
c) Higher order loop
d) None of the mentioned
Answer: b
Explanation: Most of the PLL designs have second order loop as they can be made unconditionally stable.
4. A cycle slip occurs when the magnitude of the original phase error exceeds
a) Î radians
b) 2Ď€ radians
c) π/2 radians
d) π2 radians
Answer: b
Explanation: A cyclic slip occurs when the magnitude of the original phase error exceeds 2Ď€ radians.
5. When phase noise and phase jitter is doubled, the phase variance increases by
a) Two times
b) Four times
c) 0.5 times
d) 8 times
Answer: b
Explanation: In PLL, when the phase jitter and phase noise increases by 2, the phase variance increases by 4 than that of the original signal.
6. Acquisition is a ______ operation.
a) Linear
b) Non linear
c) Linear & Non linear
d) None of the mentioned
Answer: b
Explanation: Acquisition is accomplished by external signal or circuits or sometimes it might be unaided . This acquisition process is an inherently non linear operation.
7. To avoid the loop to be driven to lock, the sweep rate should be
a) Smaller
b) Larger
c) It does not depend on it
d) None of the mentioned
Answer: a
Explanation: The sweep rate must not be too large, or the loop will be driven through the lock point so fast that it will fail to acquire.
8. Which method is more accurate?
a) Closed loop synchronizer
b) Open loop synchronizer
c) Closed & Open loop synchronizer
d) None of the mentioned
Answer: a
Explanation: Closed loop synchronizers are more accurate, but they are much more complex and costly.
9. What is the function of the edge detector?
a) Differentiation
b) Rectification
c) Differentiation & Rectification
d) None of the mentioned
Answer: c
Explanation: The main operation of the edge detector is differentiation and rectification by the use of square law detector.
10. Continuous phase modulation has
a) More bandwidth efficiency
b) More smoothness
c) More bandwidth efficiency & smoothness
d) None of the mentioned
Answer: c
Explanation: Continuous phase modulation is an important signalling technique because of their bandwidth efficiency. Bandwidth efficiency is obtained by increasing the smoothness of the signal.
11. CPM has
a) Narrower bandwidth
b) Wider bandwidth
c) More bandwidth requirement
d) None of the mentioned
Answer: a
Explanation: The increase in smoothness in the signals concentrate the signal energy in a narrower bandwidth, reducing the amount of bandwidth required.
12. The number of filters in CPM receiver structure is
a) LM
b) ML
c) M/L
d) None of the mentioned
Answer: b
Explanation: The number of filters in CPM receiver structure is given by ML.
This set of Digital Communication Questions for entrance exams focuses on “Spread Spectrum, TDMA and FDMA”.
1. In which modulation schemes, the spreading of spectrum occurs?
a) Frequency modulation
b) Amplitude modulation
c) Frequency & Amplitude modulation
d) None of the mentioned
Answer: a
Explanation: In modulation schemes such as frequency modulation and pulse code modulation spreading of spectrum occurs but it cannot be considered as types of spread spectrum technique.
2. Gaussian noise has _____ power spread _____ over all frequencies.
a) Zero, uniformly
b) Zero, non uniformly
c) Infinite, uniformly
d) Infinite, non uniformly
Answer: c
Explanation: Gaussian noise is a mathematical model which has infinite power spread uniformly over all frequencies.
3. Jamming is caused by
a) Multipath
b) Natural phenomena
c) Multipath & Natural phenomena
d) None of the mentioned
Answer: c
Explanation: Jamming can be caused by natural phenomenon and also is caused by multipath.
4. Which system makes the detection difficult for all users other than the intended users?
a) Low probability of intercept
b) Low probability of detection
c) Low probability of intercept & detection
d) None of the mentioned
Answer: c
Explanation: Low probability of detection or low probability of intercept is a system which makes the detection difficult for anyone but the intended users.
5. The radio meter consists of
a) Low pass filter
b) High pass filter
c) Band pass filter
d) Band reject filter
Answer: c
Explanation: The radio meter consists of band pass filter, squaring circuit and integrating circuit.
6. Spread spectrum signals are used for
a) Ranging
b) Determination of position
c) Ranging & Determination of position
d) None of the mentioned
Answer: c
Explanation: Spread spectrum signals are used for ranging or determination of position location.
7. Uncertainty in the delay measurement is _________ to the bandwidth of signal pulse.
a) Directly proportional
b) Inversely proportional
c) Not related
d) None of the mentioned
Answer: b
Explanation: Uncertainty in the delay measurement is inversely proportional to the bandwidth of the signal pulse.
8. Uncertainty in the delay measurement is _______ to rise time of the pulse.
a) Directly proportional
b) Inversely proportional
c) Not related
d) None of the mentioned
Answer: a
Explanation: Uncertainty in the delay measurement is directly proportional to the rise time of the pulse and inversely proportional to the bandwidth of the signal pulse.
9. In spread spectrum technique, the desired signal is multiplied _____ and interference signal is multiplied _____
a) Once, twice
b) Twice, once
c) Twice, thrice
d) Thrice, twice
Answer: b
Explanation: In spread spectrum technique, the desired signal is multiplied twice and interference signal is multiplied once.
10. Each satellite has ______ transponders.
a) 10
b) 12
c) 15
d) 18
Answer: b
Explanation: Each satellite has 12 transponders with a bandwidth of 36 Mhz each.
11. The preamble consists of
a) Synchronization
b) Addressing
c) Error control sequence
d) All of the mentioned
Answer: d
Explanation: The preamble portion consists of synchronization, addressing and error control sequences.
12. Which method is efficient?
a) Fixed assignment
b) Dynamic allocation
c) Fixed assignment & Dynamic allocation
d) None of the mentioned
Answer: b
Explanation: Dynamic allocation method is more efficient than fixed assignment as any slot won’t be wasted unnecessarily.
13. Which has a lesser average packet delay?
a) FDMA
b) TDMA
c) Dynamic FDMA
d) None of the mentioned
Answer: b
Explanation: Both TDMA and FDMA has equivalent performance but when it comes to average packet delay TDMA has lesser delay than FDMA.
14. The maximum waiting time before transmission of a packet is
a) T/M
b) M/T
c) T/M
d) M/T
Answer: a
Explanation: The maximum waiting time before transmission of a packet is given by T/M.
15. CDMA is
a) Similar to FDMA
b) Similar to TDMA
c) Combination of both
d) None of the mentioned
Answer: c
Explanation: Code division multiple access is the technique which is the combination of both TDMA and FDMA.
16. Frequency modulation technique involves
a) Data modulation
b) Frequency hopping modulation
c) Data & Frequency hopping modulation
d) None of the mentioned
Answer: c
Explanation: The FH modulation is considered as a two step process. It includes data modulation and frequency hopping modulation.
17. Which needs precise time coordination?
a) CDMA
b) TDMA
c) CDMA & TDMA
d) None of the mentioned
Answer: b
Explanation: CDMA when compared to TDMA does not need precise time coordination among various simultaneous transmitters.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Multiple access techniques for LAN and access algorithms”.
1. A local area network can be used to interconnect
a) Computers
b) Terminals
c) Printers
d) All of the mentioned
Answer: d
Explanation: A local area network can be used to interconnect computers, terminals, printers and so on located within a building or a small set of buildings.
2. LAN uses _________ cables.
a) Low bandwidth
b) High bandwidth
c) Low & High bandwidth
d) None of the mentioned
Answer: b
Explanation: Local area network uses high bandwidth cables for designing.
3. The maximum packet size of Ethernet is
a) 1426 bytes
b) 1526 bytes
c) 1256 bytes
d) 1626 bytes
Answer: b
Explanation: The maximum packet size of Ethernet is 1526 bytes where a byte is 8 bits.
4. The minimum packet size of Ethernet is
a) 81 bytes
b) 75 bytes
c) 72 bytes
d) 64 bytes
Answer: c
Explanation: The minimum packet size is 72 bytes, consisting of 8 byte preamble, 14 byte header, 46 byte data, 4 byte parity.
5. The interface between the ring and the stations are passive.
a) True
b) False
Answer: b
Explanation: The interface between the rings and the stations are active rather than passive.
6. Bit stuffing is done by inserting _____, after ____ consecutive ones.
a) Zero, seven
b) Zero, five
c) One, seven
d) One, five
Answer: a
Explanation: Bit stuffing is done by inserting a zero in the data stream after seven consecutive ones.
7. As channel capacity increases, channel throughput
a) Increases
b) Decreases
c) Remains the same
d) None of the mentioned
Answer: a
Explanation: As channel capacity R increases, the channel throughput should also increase.
8. In Aloha technique, the re-transmission takes place immediately.
a) True
b) False
Answer: b
Explanation: In Aloha technique if the negative acknowledgment is received, the re-transmission is done after a random delay so that it doesn’t collide again.
9. In which technique, packet size are constant?
a) Pure Aloha
b) Slotted Aloha
c) Pure & Slotted Aloha
d) None of the mentioned
Answer: a
Explanation: The Pure Aloha technique has constant packet lengths.
10. For a throughput of value less than 0.20, which has less average delay?
a) S-ALOHA
b) Pure ALOHA
c) R-ALOHA
d) None of the mentioned
Answer: a
Explanation: For a throughput value of less than 0.20, S-ALOHA has less average delay than R-ALOHA.
11. For a throughput value of 0.20 to 0.67, which has less average delay?
a) S-ALOHA
b) Pure ALOHA
c) R-ALOHA
d) None of the mentioned
Answer: c
Explanation: For throughput value of 0.20 to 0.67, it is clear that R-ALOHA is superior and has less average delay than S-ALOHA.
12. Binary tree search is used for
a) Multiple users
b) Single users
c) Single & Multiple users
d) None of the mentioned
Answer: a
Explanation: Binary tree search is used for rapid polling of user population and it is used for multiple users.
13. Which method is better suited for large population?
a) Straight polling
b) Binary tree search
c) Straight polling & Binary tree search
d) None of the mentioned
Answer: b
Explanation: Binary tree search is more efficient method than straight polling in case of a large population.
14. For a higher throughput value which method works efficiently?
a) S-ALOHA
b) Pure ALOHA
c) R-ALOHA
d) None of the mentioned
Answer: c
Explanation: At lower throughput values, S-ALOHA works better and for higher throughput values R-ALOHA works better.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “INTELSAT”.
1. INTELSAT II and III operates their TWTA in
a) Saturation region
b) Linear region
c) Hard limiting region
d) None of the mentioned
Answer: b
Explanation: INTELSAT II and III operate its traveling wave tube amplifiers in linear region.
2. As the number of carriers, the capacity of the transponder
a) Increases
b) Decreases
c) Remains the same
d) None of the mentioned
Answer: b
Explanation: The capacity of the transponder drops as the number of carriers increases.
3. As the number of carriers increases, the number of guard bands
a) Increases
b) Decreases
c) Remains the same
d) None of the mentioned
Answer: a
Explanation: Guard bands are needed between the carriers. Thus as carriers increases, guard bands are also increases.
4. TWTA which operates in the linear region
a) Reduces interference
b) Provide less overall power
c) Is power limited
d) All of the mentioned
Answer: d
Explanation: TWTA operating in the linear region reduces interference, and can provide less overall power. The channel becomes power limited and can service fewer carriers.
5. Which method is better when traffic is heavy?
a) SPADE
b) Binary tree search
c) SPADE & Binary tree search
d) None of the mentioned
Answer: c
Explanation: SPADE operation and binary tree search can be used when the traffic is heavy.
6. SPADE uses
a) QPSK
b) BPSK
c) FSK
d) MSK
Answer: a
Explanation: SPADE uses digital voice transmission – QPSK.
7. SPADE is a
a) MAA system
b) DAMA system
c) MAA & DAMA system
d) None of the mentioned
Answer: b
Explanation: SPADE is a DAMA system where all channels are shared.
8. In SPADE system, the carrier is done based on
a) Fixed assignment
b) Demand assignment
c) Fixed & Demand assignment
d) None of the mentioned
Answer: b
Explanation: In SPADE system, the carrier is assigned dynamically, that is on demand.
9. Which is more costlier?
a) TDMA
b) FDMA
c) ALOHA
d) None of the mentioned
Answer: a
Explanation: TDMA earth station equipment is more costlier than the FDMA equipments.
10. Which technique needs precise synchronization?
a) TDMA
b) FDMA
c) TDMA & FDMA
d) None of the mentioned
Answer: a
Explanation: TDMA operation needs very precise synchronization needed to assure orthogonality of time slots.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “TDM and FDM”.
1. Which is based on orthogonality?
a) TDM
b) FDM
c) TDM & FDM
d) None of the mentioned
Answer: b
Explanation: FDM technique is based on the orthogonality of sinusoids.
2. Which provides constant delay?
a) Synchronous TDM
b) Non synchronous TDM
c) Synchronous & Non synchronous TDM
d) None of the mentioned
Answer: a
Explanation: Synchronous time division multiplexing provides constant bandwidth and constant delay.
3. Random access is
a) Distributed
b) Fault tolerant
c) Distributed & Fault tolerant
d) None of the mentioned
Answer: c
Explanation: Random access is simple, distributed and very fault tolerant.
4. Example of reservation system is
a) Synchronous TDM
b) Non synchronous TDM
c) Synchronous & Non synchronous TDM
d) None of the mentioned
Answer: a
Explanation: Synchronous TDM is an example for the reservation system.
5. Reservation systems are useful when
a) Delay is small
b) Delay is large
c) Delay is small or large
d) None of the mentioned
Answer: b
Explanation: Reservation systems are useful when delays are large.
6. In slotted system, access is
a) Synchronous
b) Asynchronous
c) Synchronous & Asynchronous
d) None of the mentioned
Answer: b
Explanation: In slotted system, access is asynchronous and requires addressing information inside.
7. In CSMA, collision window is _____ to cable length.
a) Equal
b) Greater
c) Lesser
d) None of the mentioned
Answer: b
Explanation: The collision window is twice the cable length.
8. Which isolates collision?
a) Bridges
b) Routers
c) Bridges & Routers
d) None of the mentioned
Answer: c
Explanation: Bridges and routers isolates collision.
9. Minimum packet size increases as
a) Speed increases
b) Distance increases
c) Speed & Distance increases
d) None of the mentioned
Answer: c
Explanation: As speed and distance goes up, minimum packet size also goes up.
10. Which are non orthogonal multiplexing?
a) TDM
b) FDM
c) TDM & FDM
d) None of the mentioned
Answer: d
Explanation: TDM and FDM are both orthogonal multiplexing.
This set of Digital Communication Questions for campus interviews focuses on “TDM vs FDM”.
1. Companding is used to
a) Increase the information transmission rate
b) Use only one carrier frequency to handle different signals
c) To use different frequency bands for different signals
d) To protect all small signals in PCM from quantizing noise
Answer: d
Explanation: Companding is used to protect all small signals in PCM from quantizing noise.
2. TDM is used to
a) Increase the information transmission rate
b) Use only one carrier frequency to handle different signals
c) To use different frequency bands for different signals
d) To protect all small signals in PCM from quantizing noise
Answer: b
Explanation: TDM is used to use only one carrier frequency to handle different signals.
3. Which has greater bandwidth?
a) TDM
b) FDM
c) TDM & FDM
d) None of the mentioned
Answer: b
Explanation: FDM has greater bandwidth.
4. Which has lower noise immunity?
a) TDM
b) FDM
c) TDM & FDM
d) None of the mentioned
Answer: a
Explanation: TDM has lower noise immunity.
5. TDM can be employed to transmit channels having unequal bandwidth,
a) True
b) False
Answer: a
Explanation: True. TDM can be employed to transmit channels having unequal bandwidth.
6. Using sampling theorem, analog signal cannot be reconstructed.
a) True
b) False
Answer: b
Explanation: Using sampling theorem analog signal can be reconstructed.
7. Which maintains better fidelity?
a) Analog communication
b) Digital communication
c) Analog & Digital communication
d) None of the mentioned
Answer: b
Explanation: Digital communication provides better fidelity control.
8. In which reusing of components can be made possible?
a) Analog communication
b) Digital communication
c) Analog & Digital communication
d) None of the mentioned
Answer: b
Explanation: In digital communication reusing of components is possible.
9. A mechanism for breaking the problem down is
a) Layering
b) Dividing
c) Abstraction
d) Entity sharing
Answer: c
Explanation: Abstraction is the mechanism of breaking the problem down.
10. In digital communication, interaction between ______ is possible.
a) Any layer to layer
b) One layer to next layer
c) layers
d) None of the mentioned
Answer: b
Explanation: In digital communication, in layers only one layer and its next layer can communicate.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Spread spectrum and CDMA”.
1. Some advantages of spread spectrum are
a) Low susceptibility
b) Immunity to jamming
c) Reduced interference
d) All of the mentioned
Answer: d
Explanation: Advantages of spread spectrum are low susceptibility, reduced interference, immunity to jamming and co existence of multi path system.
2. Processing gain is the ratio of message bandwidth to signal bandwidth.
a) True
b) False
Answer: b
Explanation: Processing gain is the ratio of signal bandwidth to message bandwidth.
3. Which is better for avoiding jamming?
a) Direct sequence spread spectrum
b) Frequency hopping spread spectrum
c) Time hopping spread spectrum
d) None of the mentioned
Answer: b
Explanation: Frequency hopping spread spectrum is better for avoiding jamming.
4. Which is more bandwidth efficient?
a) Direct sequence spread spectrum
b) Frequency hopping spread spectrum
c) Time hopping spread spectrum
d) None of the mentioned
Answer: c
Explanation: Time hopping spread spectrum is more bandwidth efficient.
5. Which is more simpler to implement?
a) Direct sequence spread spectrum
b) Frequency hopping spread spectrum
c) Time hopping spread spectrum
d) None of the mentioned
Answer: a
Explanation: Direct sequence spread spectrum is more simpler to implement.
6. Which uses orthogonal codes?
a) Synchronous CDMA
b) Asynchronous CDMA
c) Synchronous & Asynchronous CDMA
d) None of the mentioned
Answer: a
Explanation: Synchronous CDMA uses orthogonal codes and asynchronous CDMA uses pseudorandom codes.
7. Which is more suitable when large number of transmitters are used?
a) Synchronous CDMA
b) Asynchronous CDMA
c) Synchronous & Asynchronous CDMA
d) None of the mentioned
Answer: b
Explanation: Asynchronous CDMA is more suitable for large number of transmitters.
8. CDMA rejects
a) Narrow band interference
b) Wide band interference
c) Narrow & Wide band interference
d) None of the mentioned
Answer: a
Explanation: CDMA effectively rejects narrow band interference.
9. Frequency planning is very essential in
a) FDMA
b) TDMA
c) FDMA & TDMA
d) None of the mentioned
Answer: c
Explanation: Frequency planning is necessary for both TDMA and FDMA.
10. CDMA uses
a) Hard hand off
b) Soft hand off
c) Hard & Soft hand off
d) None of the mentioned
Answer: b
Explanation: CDMA uses soft hand off which provides more reliable communication.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Pseudo-noise sequence and jamming”.
1. Pseudorandom signal ________ predicted.
a) Can be
b) Cannot be
c) maybe
d) None of the mentioned
Answer: a
Explanation: Random signals cannot be predicted whereas pseudorandom sequence can be predicted.
2. The properties used for pseudorandom sequence are
a) Balance
b) Run
c) Correlation
d) All of the mentioned
Answer: d
Explanation: The three basic properties that can be applied for pseudorandom sequence are balance, run and correlation properties.
3. The shift register needs to be controlled by clock pulses.
a) True
b) False
Answer: a
Explanation: The shift register operation is controlled by clock pulses.
4. A linear feedback shift register consists of
a) Feedback path
b) Modulo 2 adder
c) Four stage register
d) All of the mentioned
Answer: d
Explanation: A linear feedback shift register consists of four stage register for storage and shifting, modulo 2 adder and feedback path.
5. If the initial pulse of 1000 is fed to shift register, after how many clock pulses does the sequence repeat?
a) 15
b) 16
c) 14
d) 17
Answer: a
Explanation: If the initial pulse 1000 is given to shift register, the foregoing sequence repeats after 15 clock pulses.
6. The sequences produced by shift register depends on
a) Number of stages
b) Feedback tap connections
c) Initial conditions
d) All of the mentioned
Answer: d
Explanation: The sequences produced by shift register depends on the number of stages, the feedback tap connections and initial conditions.
7. For maximal length sequence, the sequence repetition clock pulses p is given by
a) 2n + 1
b) 2n -1
c) 2n
d) None of the mentioned
Answer: b
Explanation: For maximal length sequence, produced by n stage linear feedback shift register the sequence repetition clock pulses p is given by 2n -1 .
8. For any cyclic shift, the auto-correlation function is equal to
a) 1/p
b) -1/p
c) –p
d) p
Answer: b
Explanation: For any cyclic shift the auto-correlation function is equal to -1/p.
9. Which method is better?
a) To share same bandwidth
b) To share different bandwidth
c) To share same & different bandwidth
d) None of the mentioned
Answer: b
Explanation: If the jammer noise shares the same bandwidth, the result could be destructive.
10. Pulse jammer consists of
a) Pulse modulated excess band noise
b) Pulse modulated band-limited noise
c) Pulse width modulated excess band noise
d) Pulse width modulated band-limited noise
Answer: b
Explanation: Pulse jammer consists of pulse modulated band-limited noise.
11. Which are the design options for anti jam communicator?
a) Time diversity
b) Frequency diversity
c) Special discrimination
d) All of the mentioned
Answer: d
Explanation: The design options for anti-jam communicator are time diversity, frequency diversity and special discrimination.
12. The ratio reqd gives the measure of
a) Vulnerability to interference
b) Invulnerability to interference
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: The ratio reqd gives the measure of how invulnerable the system is to interference.
13. The system should have
a) Larger reqd
b) Greater system’s noise rejection capability
c) Larger reqd & Greater system’s noise rejection capability
d) None of the mentioned
Answer: c
Explanation: The system will be efficient if it has greater reqd and larger system’s noise rejection capability.
14. The broadband jammer jams the entire
a) W
b) Wss
c) W & Wss
d) None of the mentioned
Answer: b
Explanation: The broadband jammer or wide-band jammer is the one which jams the entire Wss with its fixed power.
15. To increase error probability, the processing gain should be
a) Increased
b) Decreased
c) Exponentially increased
d) Exponentially decreased
Answer: a
Explanation: In a system, to increase the error probability the processing gain should be increased.
16. Which jamming method produces greater degradation?
a) Broadband jamming
b) Partial jamming
c) Broadband & Partial jamming
d) None of the mentioned
Answer: b
Explanation: Greater degradation is possible more with partial jamming than broadband jamming.
17. The jammer which monitors a communicator’s signal is known as
a) Frequency follower jammers
b) Repeat back jammers
c) Frequency follower & Repeat back jammers
d) None of the mentioned
Answer: c
Explanation: The smart jammers which monitor a communicator’s signals is known as frequency follower or repeat back jammers.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Direct sequence and frequency hopping”.
1. DS/BPSK includes
a) Despreading
b) Demodulation
c) Despreading & Demodulation
d) None of the mentioned
Answer: c
Explanation: DS/BPSK is a two step precess which includes despreading and demodulation.
2. In direct sequence process which step is performed first?
a) De-spreading
b) Demodulation
c) Despreading & Demodulation
d) None of the mentioned
Answer: a
Explanation: In direct sequence process, De-spreading correlator is followed by a modulator.
3. The processing gain is given as
a) Wss/R
b) R/Wss
c) Wss/2R
d) R/2Wss
Answer: a
Explanation: The processing gain is given by the ratio of the minimum bandwidth of the data to data rate.
4. Chip is defined as
a) Shortest uninterrupted waveform
b) Largest uninterrupted waveform
c) Shortest diversion
d) None of the mentioned
Answer: a
Explanation: A chip is defined as the shortest uninterrupted waveform in the system.
5. Processing gain is given as
a) Wss/R
b) Rch/R
c) Wss/R & Rch/R
d) None of the mentioned
Answer: c
Explanation: Processing gain is given as both as the ratio of the minimum bandwidth of the data to data rate and also the by the ratio of code chip rate and data rate as minimum bandwidth is approximately equal to code chip rate.
6. Which modulation scheme is preferred for direct sequence spread spectrum process?
a) BPSK
b) QPSK
c) BPSK & QPSK
d) None of the mentioned
Answer: c
Explanation: Both the modulation scheme BPSK and QPSK can be used for direct sequence spread spectrum process.
7. The frequency hopping system uses ______ modulation scheme.
a) FSK
b) BPSK
c) MFSK
d) MPSK
Answer: c
Explanation: The frequency hopping spread spectrum system uses M-ary frequency shift keying modulation scheme.
8. The minimum spacing between consecutive hop positions gives the
a) Minimum number of chips necessary
b) Maximum number of chips necessary
c) Chip rate
d) None of the mentioned
Answer: a
Explanation: The minimum spacing between consecutive hop positions given the minimum number of chips necessary in the frequency word.
9. Which system allows larger processing gain?
a) Direct sequence
b) Frequency hopping
c) Direct sequence & Frequency hopping
d) None of the mentioned
Answer: b
Explanation: Frequency hopping spread spectrum system allows greater processing gain than direct sequence spread spectrum technique.
10. In which technique is phase coherence hard to maintain?
a) Direct sequence
b) Frequency hopping
c) Direct sequence & Frequency hopping
d) None of the mentioned
Answer: b
Explanation: In frequency hopping spread spectrum phase coherence is hard to maintain from hop to hop.
11. Which type of demodulator is used in the frequency hopping technique?
a) Coherent
b) Non coherent
c) Coherent & Non coherent
d) None of the mentioned
Answer: b
Explanation: As it is difficult to maintain phase coherence, non coherent demodulator is used.
12. Robustness gives the inability of a signal to withstand the impairments.
a) True
b) False
Answer: b
Explanation: Robustness gives the ability of a signal to withstand the impairments such as noise, jamming etc.
13. Chips are the
a) Repeated symbols
b) Non repeated symbols
c) Smallest length symbols
d) None of the mentioned
Answer: a
Explanation: The repeated symbols are called as chips.
14. Slow frequency hopping is
a) Several hops per modulation
b) Several modulations per hop
c) Several symbols per modulation
d) None of the mentioned
Answer: b
Explanation: Slow frequency hopping is several modulation per frequency hop.
15. Fast frequency hopping is
a) Several modulations per hop
b) Several modulations per symbol
c) Several symbols per modulation
d) None of the mentioned
Answer: d
Explanation: Fast frequency hopping is several frequency hops per modulation.
16. Which duration is shorter?
a) Hop duration
b) Symbol duration
c) Chip duration
d) None of the mentioned
Answer: a
Explanation: In frequency hopping technique hop duration is shorter than the symbol duration.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Acquisition and Tracking”.
1. Acquisition is a process of bringing two spreading signals into _______ alignment.
a) Fine
b) Coarse
c) Fine & Coarse
d) None of the mentioned
Answer: b
Explanation: Acquisition is a process of bringing two spreading signals into coarse alignment with one another.
2. Tracking maintains the possible waveform ______ alignment using a feedback loop.
a) Fine
b) Coarse
c) Fine & Coarse
d) None of the mentioned
Answer: a
Explanation: Tracking continuously maintains the best possible waveform fine alignment by means of a feedback loop.
3. Acquisition can be of ______ type.
a) Coherent
b) Non coherent
c) Coherent & Non coherent
d) None of the mentioned
Answer: c
Explanation: Acquisition can be categorized as two types coherent and non coherent.
4. Uncertainty in the distance between transmitter and receiver translates into uncertainty in
a) Frequency
b) Propagation delay
c) Efficiency
d) None of the mentioned
Answer: b
Explanation: uncertainty in the distance between the transmitter and the receiver translates into uncertainty in the propagation delay.
5. Acquisition uses ______ likelihood algorithm for acquiring the code.
a) Minimum
b) Maximum
c) Minimum & Maximum
d) None of the mentioned
Answer: b
Explanation: It considers all possible code positions in parallel and maximum likelihood algorithm is used to acquire the code.
6. Acquisition can be accomplished ________ as all possible codes are examined ______
a) Slowly, one by one
b) Rapidly, simultaneously
c) Slowly, simultaneously
d) Rapidly, one by one
Answer: b
Explanation: Acquisition can be accomplished rapidly as all possible codes are examined simultaneously.
7. Matched filter does ______ search.
a) Serial
b) Parallel
c) Serial & Parallel
d) None of the mentioned
Answer: a
Explanation: A single correlator or matched filter does serial search.
8. How many correlators are used for despreading and code tracking function?
a) One
b) Two
c) Three
d) None of the mentioned
Answer: a
Explanation: The design of code tracking function or despreading function needs only one correlator.
9. Which has worse signal to noise performance?
a) Delay locked loop
b) Tau-dither loop
c) Delay locked & Tau-dither loop
d) None of the mentioned
Answer: b
Explanation: The signal to noise performance of TDL is a bit worse than DDL.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Cellular systems”.
1. In the case of direct CDMA, interference will occur if the PN codes are
a) Orthogonal to each other
b) Non orthogonal to each other
c) Orthogonal & Non orthogonal to each other
d) None of the mentioned
Answer: b
Explanation: If the PN codes are not purely orthogonal in direct CDMA interference will occur.
2. If the active users using a cell doubles, the required Eb/I0
a) Doubles
b) Halves
c) Remains the same
d) None of the mentioned
Answer: b
Explanation: If the active users were to double, then the received Eb/I0 would essentially be halved.
3. In CDMA ______ frequency reuse is possible.
a) No
b) 50%
c) 90%
d) 100%
Answer: d
Explanation: In CDMA, 100% frequency reuse can be employed.
4. Bandwidth efficiency improvement involves
a) Dividing the geographical region into cells
b) Allowing frequency allocation
c) Dividing the geographical region into cells & Allowing frequency allocation
d) None of the mentioned
Answer: c
Explanation: The idea of dividing the geographical region into cells and allowing frequency allocation of one cell can be reused represents bandwidth efficiency improvement methods.
5. Which has more user capacity?
a) CDMA
b) AMPS
c) TDMA
d) None of the mentioned
Answer: a
Explanation: CDMA has more user capacity compared to AMPS and TDMA.
6. Which technique is interference limited?
a) TDMA
b) CDMA
c) AMPS
d) None of the mentioned
Answer: b
Explanation: CDMA technique capacity is interference limited.
7. Which technique is dimension limited?
a) TDMA
b) FDMA
c) TDMA & FDMA
d) None of the mentioned
Answer: c
Explanation: TDMA and FDMA capacity is dimension limited.
8. Which type of channel does the forward link contain?
a) Paging
b) Pilot
c) Traffic
d) All of the mentioned
Answer: d
Explanation: The forward link contains different types of channels such as paging, pilot, synchronization and traffic.
9. The types of channels that reverse channel contains are
a) Traffic
b) Access
c) Traffic & Access
d) None of the mentioned
Answer: c
Explanation: The reverse channel contains two types of channels – traffic and access.
10. The steps followed in forward channel CDMA includes
a) Modulo 2 addition
b) Channelization
c) Modulo 2 addition & Channelization
d) None of the mentioned
Answer: c
Explanation: Forward channel CDMA includes steps like modulo 2 addition, channelization, and base station identification.
11. Walsh cover technique is used for
a) Channelization
b) Spreading
c) Channelization & Spreading
d) None of the mentioned
Answer: c
Explanation: Walsh cover is used for channelization plus spreading.
12. Short code is configured using ______ shift register.
a) 4-stage
b) 2-stage
c) 12-stage
d) 15-stage
Answer: d
Explanation: The code in the forward direction, short code is designed using 15 stage shift register.
13. How many channels are transmitted in the forward and reverse direction?
a) 1,64
b) 64,1
c) 2,32
d) 32,2
Answer: b
Explanation: In forward direction 64 channels are transmitter and in reverse direction, only one channel is transmitted.
14. Which channel is more robust?
a) Forward channel
b) Reverse channel
c) Forward & Reverse channel
d) None of the mentioned
Answer: a
Explanation: Forward channel is more robust than the reverse channel.
15. Which serves the hard hand-off process?
a) Base station
b) Mobile switching center
c) Mobile
d) None of the mentioned
Answer: b
Explanation: Mobile unit continuously scans for stronger pilot and provides and serves hand off process.
16. Which serves the soft hand-off process?
a) Base station
b) Mobile switching center
c) Mobile
d) None of the mentioned
Answer: c
Explanation: Soft hand off process is served by the mobile. It scans for stronger pilot and requests handoff to base station.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Cipher Systems and secrecy of cipher systems”.
1. Cryptosystems are used for
a) Privacy
b) Authentication
c) Privacy & Authentication
d) None of the mentioned
Answer: c
Explanation: Cryptosystems which includes the process of encryption and decryption is used for both privacy and authentication.
2. The key also contains a
a) Plaintext
b) Ciphertext
c) Plaintext & Ciphertext
d) None of the mentioned
Answer: c
Explanation: The key is supplied along with a plaintext message for encryption and cipher text message for decryption.
3. The system which has insufficient information available to crypt-analyst is
a) One-time pad
b) Unconditionally secure
c) One-time pad & Unconditionally secure
d) None of the mentioned
Answer: c
Explanation: Unconditionally secure system is the one which has insufficient information available to crypt-analyst. One such system is one-time pad.
4. Caesar cipher was a mono alphabetic cipher.
a) True
b) False
Answer: a
Explanation: An earliest example of a mono alphabetic cipher was the Caesar cipher.
5. In Caesar cipher, each plain text is replaced by
a) Alphabet shift
b) Numerals
c) Symbols
d) None of the mentioned
Answer: a
Explanation: In Caesar cipher each plain text is replaced with a new letter obtained by an alphabetic shift.
6. The trithemius cipher is a mono alphabetical cipher.
a) True
b) False
Answer: b
Explanation: The Trithemius cipher is a poly alphabetical cipher.
7. In auto key method
a) Feedback is necessary
b) Feedback is not necessary
c) Can be either necessary or not
d) None of the mentioned
Answer: a
Explanation: In auto key method feedback is necessary for the encryption process.
8. In Vigenere method ______ is provided.
a) Priming key
b) Feedback
c) Priming key & Feedback
d) None of the mentioned
Answer: c
Explanation: Vigenere method which is also called as vigenere auto method priming key and feedback are used.
9. A cipher system is said to have perfect secrecy if
a) Posteriori probability is equal to priori probability
b) Posteriori probability is greater than priori probability
c) Posteriori probability is less than priori probability
d) None of the mentioned
Answer: a
Explanation: A cipher system is said to have perfect secrecy if for every message and cipher text the posteriori probability is equal to priori probability.
10. The conditions for perfect secrecy are
a) Only one key transforming message should exist
b) All keys should be equally likely
c) One key transforming message to each cipher text & all keys should be equally likely
d) None of the mentioned
Answer: c
Explanation: The conditions for perfect secrecy are, there is only one key transforming message to each cipher text and all keys should be equally likely.
11. The _____ the uncertainty, the ______ is the information content.
a) Lesser, greater
b) Greater, lesser
c) Lesser, lesser
d) Greater, greater
Answer: d
Explanation: The more uncertainty there is in predicting the occurrence of a message, the greater the information content.
12. Equivocation is the
a) Joint probability of X and Y
b) Conditional probability of X given Y
c) Conditional probability of Y given X
d) None of the mentioned
Answer: b
Explanation: Equivocation is defined as the conditional probability of X given Y.
13. The maximum number of message bits contained in each character is called as
a) True rate
b) Absolute rate
c) Optimum rate
d) None of the mentioned
Answer: b
Explanation: Absolute rate is the maximum number of message bits contained in each character.
14. When the key size is finite, equivocation approaches
a) Unity
b) Zero
c) Infinity
d) None of the mentioned
Answer: b
Explanation: With a finite key size, equivocation approaches zero.
15. When a system is called as unbreakable?
a) When unique solution is available
b) When unique solution is not available
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: A system is said to be unbreakable when unique solution is not available because of the number of equations being smaller than the number of unknown key bits.
This set of Digital Communication aptitude tests focuses on “Stream Encryption and Commercial Applications”.
1. Which are called as substitution encryption techniques?
a) Caesar cipher
b) Trithemius cipher
c) Caesar & Trithemius cipher
d) None of the mentioned
Answer: c
Explanation: Examples of substitution encryption techniques are Caesar cipher and Trithemius cipher.
2. In substitution technique encryption protection is more.
a) True
b) False
Answer: b
Explanation: The substitution techniques have less encryption protection.
3. The standard building block uses _______ of a key to transform 64-bit input into 64-bit output.
a) 32 bit
b) 48 bit
c) 64 bit
d) 128 bit
Answer: b
Explanation: The standard building block uses 48 bit keys to transform 64 bit input data into 64 bit output.
4. The 64 bit block input key has _____ number of parity bits.
a) 4
b) 6
c) 8
d) 16
Answer: c
Explanation: The 64 bit input key has 8 parity bits placed in 8,16…..64 bit positions.
5. One-time pad has
a) Random key
b) Unconditional security
c) Can be used only once
d) All of the mentioned
Answer: d
Explanation: One time pad is an encryption system with random key, used one time only, that exhibits unconditional security.
6. Perfect secrecy can be achieved for
a) Finite number of messages
b) Infinite number of messages
c) Finite & Infinite number of messages
d) None of the mentioned
Answer: b
Explanation: Perfect secrecy can be achieved for infinite number of messages since each message would be encrypted with different portion of random key.
7. The segment of the key-stream of plaintext is obtained by _____ of two sequences together.
a) ANDing
b) ORing
c) Addition
d) Modulo 2 addition
Answer: d
Explanation: The crypt-analyst performs modulo 2 addition of two sequences together to get the sequence of key-stream of plaintext.
8. In which method the key-stream is generated independently of the message?
a) Synchronous encryption method
b) Self synchronous encryption method
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: In Synchronous encryption method, the key-stream is generated independently of message.
9. In which method cipher feedback is used?
a) Synchronous encryption method
b) Self synchronous encryption method
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: In self synchronous encryption technique, cipher feedback is used and the key-stream is related to the messages.
10. Spread spectrum multiple access techniques allows the multiple signal to be transmitted
a) One after the other
b) Simultaneously
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: Spread spectrum multiple access technique allows multiple signals occupying the same RF bandwidth to be transmitted simultaneously without interfering with one and another.
11. In CDMA, the users share the bandwidth
a) Synchronously
b) Asynchronously
c) Synchronously & Asynchronously
d) None of the mentioned
Answer: b
Explanation: In CDMA system, all participants can share the entire bandwidth asynchronously.
12. Multi paths may be because of
a) Reflections from buildings
b) Refraction
c) Atmospheric reflections
d) All of the mentioned
Answer: d
Explanation: Multi paths may be caused by atmospheric reflections, refraction, reflections from buildings or other objects and may result in fluctuation in the received signal level.
13. For direct sequence systems what is the minimum required processing gain?
a) 3db
b) 10db
c) 12db
d) 20db
Answer: b
Explanation: For a direct sequence systems, the minimum required processing gain is 10db.
14. For hybrid systems the minimum required processing gain is
a) 5db
b) 10db
c) 17db
d) 25db
Answer: c
Explanation: For hybrid systems employing both direct sequence and frequency hopping, the minimum required processing gain is 17db.
15. Frequency hopping system can provide reliable mitigation only if
a) Hopping rate is greater than the symbol rate
b) Hopping bandwidth is large
c) Hopping rate is greater than the symbol rate & its bandwidth is large
d) None of the mentioned
Answer: c
Explanation: The frequency hopping system can provide reliable mitigation only if its hopping rate is faster than the symbol rate and the hopping bandwidth is large.
16. Direct system allows detection of signals whose psd level is
a) Below noise level
b) Above noise level
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: Direct sequence spread spectrum techniques allow for the detection of signals whose psd is below noise level.
This set of Digital Communication Problems focuses on “Spread Spectrum “.
1. Spread spectrum is a ______ technique.
a) Encoding
b) Decoding
c) Encoding & Decoding
d) None of the mentioned
Answer: a
Explanation: Spread spectrum is an important encoding technique for a wireless communication system.
2. Spread spectrum can be used to transmit
a) Analog data
b) Digital data
c) Analog & Digital data
d) None of the mentioned
Answer: c
Explanation: Spread spectrum is used to transmit analog data and digital data using analog signal.
3. Spread spectrum makes ______ difficult.
a) Jamming
b) Interception
c) Jamming & Interception
d) None of the mentioned
Answer: c
Explanation: Spread spectrum makes jamming and interception difficult and provides improved reception.
4. Spread spectrum has immunity from
a) Noise
b) Multi-path distortion
c) Noise & Multi-path distortion
d) None of the mentioned
Answer: c
Explanation: Spread spectrum has immunity to noise and multi-path distortion.
5. The width of each channel in frequency hopping spread spectrum corresponds to
a) Bandwidth of input signal
b) Pseudorandom sequence used
c) Carrier frequency
d) None of the mentioned
Answer: a
Explanation: The width of the channel in frequency hopping spread spectrum corresponds to bandwidth of the input signal.
6. The transmitter of frequency hopping system is fed with encoding scheme such as
a) Frequency shift keying
b) Binary phase shift keying
c) Frequency & Binary phase shift keying
d) None of the mentioned
Answer: c
Explanation: For transmission, the binary data is fed into modulator using modulation scheme such as frequency shift keying or binary phase shift keying.
7. Which filter is used to get the final FHSS signal?
a) Low pass filter
b) High pass filter
c) Band pass filter
d) Band stop filter
Answer: c
Explanation: Band pass filter is used to block difference frequency and allow the sum frequency to yield final FHSS signal.
8. In CDMA the incoming signal is _____ with the spreading code.
a) Added
b) Multiplied
c) XOR-ed
d) None of the mentioned
Answer: b
Explanation: In CDMA, the incoming signal is multiplied with the spreading code.
9. The original bandwidth of the unspread signal in spread spectrum is ____ to data rate.
a) Proportional
b) Inversely proportional
c) Not related
d) None of the mentioned
Answer: a
Explanation: In spread spectrum, the original bandwidth of the unspread signal is proportional to the data rate.
10. In DSSS the signal is recovered using
a) Low pass filter
b) High pass filter
c) Band pass filter
d) Band stop filter
Answer: c
Explanation: In direct sequence spread spectrum, band pass signal at the demodulator can be used to recover the original signal.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Public key cryptosystem – 1”.
1. In public key cryptosystem _____ keys are used for encryption and decryption.
a) Same
b) Different
c) Encryption Keys
d) None of the mentioned
Answer: b
Explanation: In conventional cryptosystem, same keys are used for encryption and decryption where as in public key cryptosystem different keys are used.
2. In public key cryptosystem which is kept as public?
a) Encryption keys
b) Decryption keys
c) Encryption & Decryption keys
d) None of the mentioned
Answer: a
Explanation: In public key cryptosystem, the encryption keys are kept as public where as decryption keys are kept as secret.
3. In a trapdoor function, the functions are easy to go in
a) One direction
b) Two directions
c) All directions
d) None of the mentioned
Answer: a
Explanation: A trapdoor is a one way function. Such functions can go easily in only one direction.
4. Pretty good privacy program is used for
a) Electronic mails
b) File encryption
c) Electronic mails & File encryption
d) None of the mentioned
Answer: c
Explanation: Pretty good privacy privacy program is used for both electronic mails and file encryption.
5. PGP system uses
a) Private key system
b) Public key system
c) Private & Public key system
d) None of the mentioned
Answer: c
Explanation: Pretty good privacy system uses both private key and public key based systems.
6. Private key algorithm is used for _____ encryption and public key algorithm is used for _____ encryption.
a) Messages, session key
b) Session key, messages
c) Can be used for both
d) None of the mentioned
Answer: a
Explanation: Private key algorithm is used for encryption of messages and public key algorithm is used for encryption of session keys.
7. Which are called the block ciphers?
a) IDEA
b) CAST
c) Triple-DES
d) All of the mentioned
Answer: d
Explanation: Some of the examples of block codes are triple-DES, CAST and IDEA.
8. Which has a key length of 128 bits?
a) IDEA
b) Triple-DES
c) IDEA & Triple-DES
d) None of the mentioned
Answer: a
Explanation: Triple-DES uses 168 bit key size and IDEA and CAST uses 128 bits key length.
9. Which algorithm can be used to sign a message?
a) Public key algorithm
b) Private key algorithm
c) Public & Private key algorithm
d) None of the mentioned
Answer: a
Explanation: Public key algorithms are used to sign a message and private key algorithms are used to encrypt the messages.
10. Examples of hash functions are
a) MD5
b) SHA-1
c) MD5 & SHA-1
d) None of the mentioned
Answer: c
Explanation: Some examples of hash functions are MD5 and SHA-1.
11. A cryptographic hash function has variable output length.
a) True
b) False
Answer: b
Explanation: A cryptographic hash function has fixed output length.
12. A set of numbers is called as super increasing if
a) Each element is greater than previous element
b) Each element greater than sum of all the previous elements
c) Each element greater than the average of all elements
d) Each element lesser than the sum of all the elements
Answer: b
Explanation: A set of numbers is called as super increasing if each element is greater than the sum of all the numbers lesser than it.
13. Knapsack problem can be solved by
a) Public key cryptosystem
b) Private key cryptosystem
c) Public & Private key cryptosystem
d) None of the mentioned
Answer: a
Explanation: The knapsack problem can be solved by public key cryptosystem system using knapsack vector and data vector.
14. Merkle hellman is a symmetric cryptosystem.
a) True
b) False
Answer: b
Explanation: Merkell hellman cryptosystem is a public key asymmetric cryptosystem method.
15. In this Merkle Hellman method same key is used for encryption and decryption.
a) True
b) False
Answer: b
Explanation: In Merkle hellman cryptosystem method, two separate private and public keys are used for encryption and decryption.
This set of Basic Digital Communication Questions & Answers focuses on “Public Key Crptosystems – 2”.
1. Cipher system can be solved effectively by statistically using
a) Time of occurrence
b) Frequency of occurrence
c) Length of the message
d) None of the mentioned
Answer: b
Explanation: By statistically using the frequency of occurrence of individual characters and character combinations we can solve many cipher systems.
2. Encryption transformations are known as
a) Diffusion
b) Confusion
c) Diffusion & Confusion
d) None of the mentioned
Answer: c
Explanation: The encryption transformations were termed confusion and diffusion by Shannon.
3. For n input bits the number of substitution patterns are
a) 2n
b) !
c) 1/2n!
d) 2n!
Answer: b
Explanation: For n input bits the total number of possible substitution or connection patterns are !.
4. In transposition, the plaintext letters are
a) Substituted
b) Rearranged
c) Removed
d) None of the mentioned
Answer: b
Explanation: In permutation or transposition, the positions of the plaintext letters in the message are simply rearranged.
5. The substitution of the letters are done
a) Serially
b) Parallel way
c) Serially & Parallel way
d) None of the mentioned
Answer: b
Explanation: The substitution part of the encryption scheme is performed in parallel on small segment of the blocks.
6. In cipher feedback method, encryption is done on
a) Single individual bits
b) Characters
c) Single individual bits & Characters
d) None of the mentioned
Answer: a
Explanation: In cipher feedback method, single individual bits are encrypted rather than characters and this leads to stream encryption.
7. In stream encryption, the key sequence repeats itself.
a) True
b) False
Answer: b
Explanation: In stream encryption, random key-stream is used. The key sequence never repeats.
8. In which method, key-stream is generated independently?
a) Synchronous
b) Self synchronous
c) Synchronous & Self synchronous
d) None of the mentioned
Answer: a
Explanation: In synchronous encryption system, the key-stream is generated independently of the message.
9. Synchronous ciphers uses
a) Confusion
b) Diffusion
c) Confusion & Diffusion
d) None of the mentioned
Answer: a
Explanation: Synchronous cipher system is designed to use confusion and not diffusion.
10. Public key cryptosystem uses same key for both encryption and decryption.
a) True
b) False
Answer: b
Explanation: Public key cryptosystem uses different keys for encryption and decryption.
11. Which should be kept as a secret in public key cryptosystem?
a) Encryption key
b) Decryption key
c) Encryption & Decryption key
d) None of the mentioned
Answer: b
Explanation: In public key cryptosystem, decryption key needs to be kept as a secret.
12. Public key cryptosystem is also known as
a) One way function
b) Two way function
c) Feedback function
d) None of the mentioned
Answer: a
Explanation: Public key cryptosystem is called as a trap door one way function.
13. Euler’s totient function is determined by
a) pq
b)
c)
d) p/q
Answer: b
Explanation: The Euler’s totient function is determined by , where p and q are kept hidden.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Pretty good privacy”.
1. Pretty good privacy security system uses
a) Public key cryptosystem
b) Private key cryptosystem
c) Public & Private key cryptosystem
d) None of the mentioned
Answer: c
Explanation: PGP uses many encryption techniques such as private key cryptosystem and also public key cryptosystem.
2. Public key cryptosystem is used for the encryption of
a) Messages
b) Session key
c) Session key & Messages
d) None of the mentioned
Answer: b
Explanation: Public key cryptosystem is used for encryption of messages and private key cryptosystem is used for encryption of messages.
3. For digital signatures private key cryptosystem is used.
a) True
b) False
Answer: b
Explanation: Public key cryptosystem is used for the creation of digital signatures.
4. Data compression includes
a) Removal of redundant character
b) Uniform distribution of characters
c) Removal of redundant character & Uniform distribution of characters
d) None of the mentioned
Answer: c
Explanation: Data compression removes redundant character strings in a file and produces a more uniform distribution of characters.
5. PGP offers _____ block ciphers for message encryption.
a) Triple-DES
b) CAST
c) IDEA
d) All of the mentioned
Answer: d
Explanation: Pretty good privacy security system offers three block ciphers for message encryption – Triple-DES, IDEA and CAST.
6. Which block cipher has key length of 128 bits?
a) IDEA
b) CAST
c) IDEA & CAST
d) None of the mentioned
Answer: c
Explanation: The block ciphers IDEA and CAST has key length of 128 bits where as Triple-DES has key size of 168 bits.
7. These three ciphers can operate of ______ of plaintext and cipher text.
a) 128 bit blocks
b) 64 bit blocks
c) 256 bit blocks
d) 156 bit blocks
Answer: b
Explanation: The three ciphers – Triple-DES, IDEA and CAST can operate on 64 bit blocks of plain text and cipher text.
8. The key size of DES is
a) 56 bits
b) 64 bits
c) 128 bits
d) 168 bits
Answer: a
Explanation: The key size of DES algorithm is 56 bits and with Triple-DES, the message to be encrypted is run on DES algorithm 3 times.
9. Which operation is used in encryption using IDEA?
a) Addition modulo 216
b) Bit wise XOR
c) Addition modulo 216 & Bit wise XOR
d) None of the mentioned
Answer: c
Explanation: The arithmetic operations used in IDEA are addition modulo 216, multiplication modulo 216+1, and bit wise XOR.
10. What is the key size allowed in PGP?
a) 1024-1056
b) 1024-4056
c) 1024-4096
d) 1024-2048
Answer: c
Explanation: Pretty good privacy security system allows 1024 to 4096 bits of key size.
11. Which algorithm is used for public key encryption?
a) RSA
b) Diffie-Hellman
c) RSA & Diffie-Hellman
d) None of the mentioned
Answer: c
Explanation: The two algorithms used for public key encryption in PGP are RSA and Diffie-Hellman algorithms.
12. Which should be used first to decrypt?
a) Public key
b) Private key
c) Public & Private key
d) None of the mentioned
Answer: b
Explanation: The recipient should first decrypt the message using private key and then with the public key.
13. The digital signature provides authentication to the
a) Sender
b) Message
c) Sender & Message
d) None of the mentioned
Answer: c
Explanation: The digital signature provides authentication to both the sender and the message.
14. In hash function output length is fixed.
a) True
b) False
Answer: a
Explanation: In hash function the output length is fixed and it is easy to compute.
15. The hash function
a) Is collision free
b) Has manageable collision
c) Has high unmanageable level of collision
d) None of the mentioned
Answer: a
Explanation: Hash function is collision free and it is unfeasible that two different messages will create the same hash value.
16. DSA protocol is based on
a) Discrete logarithm problem
b) Continuous logarithm problem
c) Discrete & Continuous logarithm problem
d) None of the mentioned
Answer: a
Explanation: The DSA algorithm uses and is based on discrete logarithm problem.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Mobile radio propagation”.
1. What is the source for performance degradation?
a) Thermal noise
b) Man made noise
c) Natural noise
d) All of the mentioned
Answer: d
Explanation: The sources for performance degradation are thermal noise, natural noise and man made noise.
2. The received power is attenuated by a factor called
a) Path loss
b) Free space loss
c) Path & Free space loss
d) None of the mentioned
Answer: c
Explanation: The received power expressed in terms of transmitted power is attenuated by a factor called path loss or free space loss.
3. Scintillation describes the fading caused by
a) Atmospheric changes
b) Physical changes
c) Thermal noise
d) Propagating effects
Answer: b
Explanation: Scintillation is used to describe fading caused by physical changes in the propagating medium.
4. Large scale fading refers the attenuation in
a) Amplitude
b) Phase
c) Signal power
d) None of the mentioned
Answer: c
Explanation: large scale fading refers to signal power attenuation or path loss due to motion in large area.
5. Small scale refers to changes in
a) Amplitude
b) Phase
c) Amplitude & Phase
d) None of the mentioned
Answer: c
Explanation: Small scale fading refers to changes in amplitude and phase as a result of small changes in spatial positioning between transmitter and receiver.
6. Small scale fading manifests due to
a) Signal dispersion
b) Time variant behavior
c) Signal dispersion & Time variant behavior
d) None of the mentioned
Answer: c
Explanation: Small scale fading manifests itself in two mechanisms- time spreading of the signal and time variant behavior of the channel.
7. The parameters used to describe large scale fading are
a) Reference distance
b) Path loss exponent
c) Standard deviation of random variable
d) All of the mentioned
Answer: d
Explanation: The parameters used to statistically describe path loss due to large scale fading are reference distance, path loss exponent and standard deviation of random variable.
8. Small scale fading occurs due to
a) Doppler shift
b) Time delay
c) Doppler shift & Time delay
d) None of the mentioned
Answer: c
Explanation: Small scale fading occurs due to time delay and frequency or Doppler shift.
9. What are the types of small scale fading that occurs due to Doppler shift?
a) Slow fading
b) Fast fading
c) Slow & Fast fading
d) None of the mentioned
Answer: c
Explanation: Slow fading and fast fading occurs due to Doppler or frequency shift.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Signal time spreading”.
1. Multiple isolated peaks in multipath components is called as
a) Fingers
b) Returns
c) Fingers & Returns
d) None of the mentioned
Answer: c
Explanation: The received signal usually consists of several discrete multipath components causing multipath intensity profile to exhibit multiple isolated peaks also called as fingers or returns.
2. If delay time is greater than symbol time, _______ fading occurs.
a) Frequency selective
b) Time selective
c) Frequency non selective
d) None of the mentioned
Answer: a
Explanation: When delay time is greater than symbol time, frequency selective fading occurs.
3. If delay time is lesser than symbol time, ______ fading occurs.
a) Frequency non selective
b) Flat
c) Frequency non selective & Flat
d) None of the mentioned
Answer: c
Explanation: If delay time is lesser than the symbol time, frequency non selective or flat fading occurs.
4. When channel coherence bandwidth is greater than the transmitted signal bandwidth ______ occurs.
a) Time selective
b) Frequency selective
c) Frequency non selective
d) None of the mentioned
Answer: c
Explanation: Frequency non selective or flat fading occurs if channel coherence bandwidth is greater than the transmitted signal bandwidth.
5. Channel’s time spreading properties are
a) Coherence bandwidth
b) Signal dispersion
c) Coherence bandwidth & Signal dispersion
d) None of the mentioned
Answer: c
Explanation: Signal dispersion and coherence bandwidth characterizes the channel’s time spreading properties in a local area.
6. When channel coherence time is less than the time duration of a transmission symbol, then ______ fading occurs.
a) Fast
b) Slow
c) Time selective
d) None of the mentioned
Answer: a
Explanation: Fast fading occurs when the channel coherence time is less than the time duration of a transmission symbol.
7. Minimum time required for magnitude change or phase change is called as
a) Shift time
b) Coherence time
c) Delay time
d) None of the mentioned
Answer: b
Explanation: Coherence time is the minimum time required for magnitude or phase change of the channel to become uncorrelated from the previous value.
8. When a user’s moving, the user’s velocity causes shift in frequency. This is called as
a) Doppler shift
b) Frequency shift
c) Velocity shift
d) None of the mentioned
Answer: a
Explanation: When a user’s moving, the user’s velocity causes shift in frequency of the transmitted signal along each signal path. This is called as Doppler shift.
9. Coherence time is _______ to Doppler spread.
a) Directly proportional
b) Inversely proportional
c) Not dependent
d) None of the mentioned
Answer: b
Explanation: Coherence time is inversely proportional to Doppler spread.
10. Block fading occurs in
a) Frequency domain
b) Time domain
c) Frequency & Time domain
d) None of the mentioned
Answer: c
Explanation: Block fading can occur in both frequency domain and time domain. This is called as doubly block fading.
11. If coherence bandwidth is smaller than the bandwidth of the signal, _____ fading occurs.
a) Flat
b) Frequency selective
c) Fast fading
d) Time selective
Answer: b
Explanation: Frequency selective fading occurs when coherence bandwidth is smaller than the bandwidth of the transmitted signal.
12. Flat fading can be avoided or stopped by
a) Error coding
b) Equalization
c) Adaptive bit loading
d) All of the mentioned
Answer: d
Explanation: Flat fading can be combated by means of error coding, equalization and adaptive bit loading.
13. The effect of Doppler shift can be counterattacked by
a) OFDM
b) By using two receivers with separate antennas
c) By using diversity receiver
d) All of the mentioned
Answer: d
Explanation: The effect of Doppler shift can be removed by using diversity scheme such as orthogonal frequency division multiplexing, by using two receivers with separate antennas or by using specially designed diversity receiver.
14. When transmitter and receiver moves towards each other, the magnitude of frequency shift is
a) Positive
b) Negative
c) Zero
d) Infinity
Answer: a
Explanation: When transmitter and receiver moves towards each other, the magnitude of frequency shift is positive, where as if it moves away from each other then it is negative.
15. If signalling rate is less than fading rate, _____ fading occurs.
a) Slow
b) Fast
c) Time selective
d) None of the mentioned
Answer: b
Explanation: A channel is said to be fast fading if symbol rate or signalling rate is less than fading rate.
16. Frequency tracking loop can minimize irreducible error rate by using
a) GMSK
b) DMSK
c) MSK
d) QAM
Answer: b
Explanation: Frequency tracking loop can minimize the irreducible error rate in a mobile system by using differential minimum shift keying modulation.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Diversity technique and rake receivers”.
1. The technique for combining diversity signals are
a) Feedback
b) Maximal ratio
c) Equal gain
d) All of the mentioned
Answer: d
Explanation: The most common techniques used for combining diversity signals are selection, feedback, maximal ratio and equal gain.
2. Diversity technique is used for combating
a) Fading
b) Error bursts
c) Co-channel interference
d) All of the mentioned
Answer: d
Explanation: Diversity technique is a common method used for combating fading, co-channel interference and error bursts.
3. Diversity technique is applied at
a) Base station
b) Mobile receiver
c) Base station & Mobile receiver
d) None of the mentioned
Answer: c
Explanation: Diversity technique can be applied at both base station and at mobile receivers.
4. Which is more effective and commonly preferred technique?
a) Time diversity
b) Spatial diversity
c) Frequency diversity
d) None of the mentioned
Answer: b
Explanation: Most commonly used and more effective diversity technique is spatial diversity technique.
5. Diversity technique
a) Provides significant link improvement
b) Needs training overhead
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: Diversity technique does not require training overhead at the transmitter and also provides significant link performance.
6. What are the modes of adaptive equalizer?
a) Training mode
b) Tracking mode
c) Training & Tracking mode
d) None of the mentioned
Answer: c
Explanation: Adaptive equalizer operates on two modes – training mode and tracking mode.
7. Which signalling scheme is preferred for fading channel?
a) Frequency based modulation
b) Phase based modulation
c) Frequency & Phase based modulation
d) None of the mentioned
Answer: c
Explanation: Frequency based modulation and phase based modulation is preferable for fading channel than amplitude based modulation.
8. Uncorrelated fading occurs when
a) Time span is small
b) Time span is large
c) Does not depend on time span
d) None of the mentioned
Answer: b
Explanation: As the time span increases, the fading is more uncorrelated and effectiveness also increases.
9. Interleaving does
a) Forward error correction
b) Backward error correction
c) Forward & Backward error correction
d) None of the mentioned
Answer: a
Explanation: Inter leaver does forward error correction.
10. Rake receiver does
a) Counter effects multi-path fading
b) Resembles equalizer
c) Resembles equalizer & Counter effects multi-path fading
d) None of the mentioned
Answer: c
Explanation: Rake receiver resembles equalizers and it is used to counter the effects of multipath fading and is also used to exploit the path diversity.
11. Information reliability depends on
a) Magnitude of each component
b) Time of arrival of each component
c) Magnitude & Time of arrival of each component
d) None of the mentioned
Answer: c
Explanation: The information reliability can be improved by computing the amplitude and time of arrival of each component.
12. A rake receiver uses multiple
a) Delay circuits
b) Correlators
c) Detectors
d) None of the mentioned
Answer: b
Explanation: A rake receiver uses multiple correlators to separately detect multiple strongest components.
13. Rake receivers are used in
a) Radio astronomy
b) CDMA
c) W-CDMA
d) All of the mentioned
Answer: d
Explanation: Rake receivers are used in radio astronomy, CDMA, W-CDMA, wireless LAN networks etc.
14. The rake receivers functions are similar to
a) Equalizer
b) Matched filter
c) Equalizer & Matched filter
d) None of the mentioned
Answer: c
Explanation: The functionality of rake receiver resembles that of the equalizer and matched filter.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Effects of fading and decision theory”.
1. Fading channel has memory.
a) True
b) False
Answer: a
Explanation: Fading channel has memory and the received samples are correlated with each other in time.
2. How can frequency selective distortion be minimized?
a) By using pilot signal
b) By adaptive equalization
c) By spread spectrum
d) All of the mentioned
Answer: d
Explanation: Frequency selective distortion can be minimized by using adaptive equalization, by using spread spectrum, by using pilot signal and also by using orthogonal FDM modulation.
3. How can slow fading be minimized?
a) By diversity technique
b) Error correcting codes
c) By diversity technique & Error correcting codes
d) None of the mentioned
Answer: c
Explanation: Slow fading can be minimized by using error correcting codes and also by using diversity technique to get additional uncorrelated estimates of a signal.
4. Fast fading can be minimized by
a) Robust modulation
b) Coding and interleaving
c) Robust modulation, Coding and interleaving
d) None of the mentioned
Answer: c
Explanation: Fast fading can be minimized by using coding and interleaving, by using robust modulation and by introducing signal redundancy to increase signalling rate.
5. The decision feedback equalizer has a linear traversal filter which is
a) Feed forward section
b) Feedback section
c) Feed forward section & Feedback section
d) None of the mentioned
Answer: a
Explanation: The decision feedback filter has a feed forward section which is a linear traversal filter.
6. The ISI and adjacent channel interference is removed by
a) Cancelling filter
b) Port processing equalizer
c) Cancelling filter & Port processing equalizer
d) None of the mentioned
Answer: c
Explanation: The known ISI which is introduced deliberately and the adjacent channel interference is removed by cancelling filter and post processing equalizer.
7. The inter-leaver is more effective if the vehicle is
a) Fast
b) Slow
c) Fast & Slow
d) None of the mentioned
Answer: a
Explanation: The inter-leaver is more effective in fast running vehicles.
8. Channel noise is
a) Additive
b) White and stationary
c) Has infinite bandwidth
d) All of the mentioned
Answer: d
Explanation: Channel noise is stationary, additive and white with infinite bandwidth.
9. Which noise component plays a role in decision making?
a) Relevant noise
b) Non relevant noise
c) Relevant & Non relevant noise
d) None of the mentioned
Answer: a
Explanation: Only the relevant noise components play an important role in decision making. It does not depend on non relevant noise components.
10. Decision making needs
a) Priors
b) Likelihoods
c) Priors & Likelihoods
d) None of the mentioned
Answer: c
Explanation: Decision making needs both priors and likelihoods and Bayes decision rule combines them to achieve minimum probability of error.
11. In matched filter _______ is performed.
a) Convolution
b) Correlation
c) Convolution & Correlation
d) None of the mentioned
Answer: c
Explanation: In matched filter a known signal is correlated with an unknown signal. It is similar to convolving a unknown signal with time reversed version of it.
12. Which needs more signal power?
a) BPSK
b) 16-QAM
c) BPSK & 16-QAM
d) None of the mentioned
Answer: b
Explanation: 16-QAM needs ten times more signal power than BPSK to attain the same probability of error.
13. Which has higher transmission rate?
a) BPSK
b) 16-QAM
c) BPSK & 16-QAM
d) None of the mentioned
Answer: b
Explanation: The rate of transmission of bits in 16-QAM is four times that of BPSK.
14. Symbol error probability can also be determined using upper bound.
a) True
b) False
Answer: a
Explanation: Certain signals lacks symmetry in representation. For those signals symbol error probability can be determined by using upper bound.
15. Which signal sets are called as equivalent signal sets?
a) Simplex
b) Bi-orthogonal
c) Simplex & Bi-orthogonal
d) None of the mentioned
Answer: c
Explanation: Simplex signals and bi-orthogonal signals are examples of equivalent signal sets.
16. For neyman pearson decision criterion, which are important?
a) Probability of false alarm
b) Probability of miss
c) Probability of false alarm & miss
d) None of the mentioned
Answer: c
Explanation: For neyman pearson decision criterion two probabilities are important – probability of false alarm and probability of miss.
17. What are the parameters calculated for an estimator?
a) Error
b) Mean square error
c) Variance
d) All of the mentioned
Answer: d
Explanation: Some of the parameters calculated for an estimator are error, mean square error, variance, sampling deviation etc.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Abstraction and layering”.
1. Top layer is
a) Physical media
b) Application
c) Design
d) None of the mentioned
Answer: b
Explanation: Top layer is application where as bottom layer is physical media.
2. Entities also does
a) Error correction
b) A/D
c) D/A
d) All of the mentioned
Answer: d
Explanation: Entities also does A/D, D/A, encryption and error correction.
3. In channel, delay is
a) Fixed
b) Variable
c) Fixed & Variable
d) None of the mentioned
Answer: c
Explanation: In channel, delay can be either fixed or variable.
4. Channel can provide connectivity to
a) Point to point
b) 1 to many
c) Many to many
d) All of the mentioned
Answer: d
Explanation: Channel can provide connectivity between point to point, 1 to many and also many to many.
5. Embedding is a form of
a) Abstraction
b) Layering
c) Entity sharing
d) None of the mentioned
Answer: b
Explanation: Embedding is a form of layering.
6. Transmission media is
a) Acoustic
b) Guided
c) Optical
d) All of the mentioned
Answer: d
Explanation: Transmission media can be acoustic, electronic or optical form.
7. In transmission media, the symbol type is generally
a) Analog
b) Digital
c) Analog & Digital
d) None of the mentioned
Answer: a
Explanation: In transmission media, the symbol type is generally analog signals.
8. Noise is _____ in nature.
a) Random
b) Systematic
c) Random or Systematic
d) None of the mentioned
Answer: c
Explanation: Noise can be random or systematic.
9. The radiation loss increases as
a) Frequency increases
b) Frequency decreases
c) Does not depend on frequency
d) None of the mentioned
Answer: a
Explanation: The radiation loss increases as frequency increases.
10. Attenuation is a function of
a) Channel length
b) Transmission media
c) Signal frequencies
d) All of the mentioned
Answer: d
Explanation: Attenuation is a function of channel length, transmission media and signal frequencies.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Protocols”.
1. Protocol is a set of
a) Formats
b) Procedures
c) Formats & Procedures
d) None of the mentioned
Answer: c
Explanation: Protocol is a set of procedures and formats that entities use to communicate information.
2. The time required to transmit frame depends on
a) Frame size
b) Channel capacity
c) Frame size & Channel capacity
d) None of the mentioned
Answer: c
Explanation: The time to transmit frame is given by the ratio of frame size and channel capacity.
3. Window given the number of
a) Bytes
b) Frames
c) Bytes & Frames
d) None of the mentioned
Answer: c
Explanation: Window gives the number of frames or bytes that can be in transit unacknowledged.
4. Routing is
a) Static
b) Dynamic
c) Static & Dynamic
d) None of the mentioned
Answer: c
Explanation: Routing can be static and dynamic.
5. Routing is performed only centrally.
a) True
b) False
Answer: a
Explanation: Routing is performed centrally and also in fully distributed way.
6. Which is connection oriented and which is connection-less?
a) Datagrams, virtual circuits
b) Virtual circuits, datagrams
c) Datagrams
d) None of the mentioned
Answer: b
Explanation: Datagram is connection-less and virtual circuits are connection oriented.
7. Which uses UDP?
a) Echo
b) Time
c) Domain name server
d) All of the mentioned
Answer: d
Explanation: The protocols using UDP are echo, time, domain name server, network file system etc.
8. Which uses TCP?
a) Simple mail transfer protocol
b) Simple network management protocol
c) Simple mail transfer & network management protocol
d) None of the mentioned
Answer: a
Explanation: Some of the protocols using TCP are http, telnet, file transfer protocol, simple mail transfer protocol etc.
9. Which tells about low level transmission and framing?
a) V series
b) X series
c) G series
d) Q series
Answer: c
Explanation: G series tells about level transmission, modulation and framing.
10. Which tells about signalling?
a) V series
b) X series
c) G series
d) Q series
Answer: d
Explanation: Q series tells about signalling.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Random processes”.
1. Which method is much better and efficient?
a) Vector quantization
b) Scalar quantization
c) Vector & Scalar quantization
d) None of the mentioned
Answer: a
Explanation: Vector quantization will always equal or exceed the performance of scalar quantization.
2. Analog source coding method are
a) Temporal waveform coding
b) Spectral waveform coding
c) Model based coding
d) All of the mentioned
Answer: d
Explanation: Three types of analog source coding methods are temporal waveform coding, spectral waveform coding and model based coding.
3. The source output in PCM and DPCM is quasi stationary.
a) True
b) False
Answer: b
Explanation: The source output of PCM and DPCM and stationary in nature.
4. Which reduces the dynamic range of quantization noise in PCM?
a) Non uniform quantizer
b) Uniform quantizer
c) Adaptive quantizer
d) None of the mentioned
Answer: c
Explanation: Adaptive quantizer reduces the dynamic range of quantization noise in PCM and DPCM.
5. The type of distortion which occurs in delta modulation is
a) Slope overload distortion
b) Granular noise
c) Slope overload distortion & Granular noise
d) None of the mentioned
Answer: c
Explanation: The two types of distortion that occurs in delta modulation are slope overload distortion and granular noise.
6. Binomial distribution deals with
a) Continuous random variable
b) Discrete random variable
c) Continuous & Discrete random variable
d) None of the mentioned
Answer: b
Explanation: Binomial distribution deals with discrete random variable.
7. In poisson distribution mean is ______ variance.
a) Greater than
b) Lesser than
c) Equal to
d) Does not depend on
Answer: c
Explanation: In poisson distribution, variance is nearly same or equal to mean of the distribution.
8. Stochastic process are
a) Random in nature
b) Are function of time
c) Random in nature and are a function of time
d) None of the mentioned
Answer: c
Explanation: Stochastic process are random in nature and are a function of time.
9. Stochastic processes are
a) Strict sense stationary process
b) Wide sense stationary process
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: Stochastic process is a wide sense stationary process.
10. Gaussian process is a
a) Wide sense stationary process
b) Strict sense stationary process
c) All of the mentioned
d) None of the mentioned
Answer: c
Explanation: If Gaussian process is a wide sense stationary process then it will also be strict sense stationary process.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “ISDN and interface”.
1. ISDN integrates speech and data on
a) Different lines
b) Same lines
c) Different & Same lines
d) None of the mentioned
Answer: b
Explanation: Integrated services for digital domain integrates speech and data signals on the same lines.
2. ISDN is a
a) Packet switched network
b) Circuit switched telephone network
c) Packet switched & Circuit switched telephone network
d) None of the mentioned
Answer: c
Explanation: ISDN is a circuit switched telephone network which also provides access to packet switched network.
3. Circuit switched connection is provided for
a) Voice
b) Data
c) Voice & Data
d) None of the mentioned
Answer: c
Explanation: Circuit switched connection can be provided for either voice or data.
4. Packet switched connection is provided for
a) Voice
b) Data
c) Voice & Data
d) None of the mentioned
Answer: b
Explanation: Packet switched connection can be provided only for data.
5. Channel bonding provides
a) Higher data rate
b) Lower data rate
c) Does not affect data rate
d) None of the mentioned
Answer: a
Explanation: Channel bonding can achieve higher data rate.
6. U-interface is the two wire interface between
a) Computing device and terminal adapter
b) Terminal adapter and exchange unit
c) Exchange unit and network terminating unit
d) Computing device and exchange unit
Answer: c
Explanation: U-interface is the two wire connection or interface between exchange unit and network terminating unit.
7. T interface is the serial interface between
a) Computing device and terminal adapter
b) Terminal adapter and exchange unit
c) Exchange unit and network terminating unit
d) Computing device and exchange unit
Answer: a
Explanation: T interface is the serial interface between computing device and terminal adapter.
8. S-interface is a _______ bus.
a) Two wire bus
b) Four wire bus
c) Single wired bus
d) None of the mentioned
Answer: b
Explanation: The S interface is a four wire bus that ISDN consumer devices plug into.
9. The services provided by ISDN
a) Bearer service
b) Teleservice
c) Secondary service
d) All of the mentioned
Answer: d
Explanation: The services provided by ISDN and classified by attributes are bearer service, teleservice and secondary service.
10. Attributes of ISDN have
a) Definition
b) Allowable values
c) Definition & Allowable values
d) None of the mentioned
Answer: c
Explanation: ISDN services are characterized by attributes and attributes have definition and allowable values.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Error detection and correction 1”.
1. In layering, n layers provide service to
a) n layer
b) n-1 layer
c) n+1 layer
d) none of the mentioned
Answer: c
Explanation: In layering n layer provides service to n+1 layer and use the service provided by n-1 layer.
2. Which can be used as an intermediate device in between transmitter entity and receiver entity?
a) IP router
b) Microwave router
c) Telephone switch
d) All of the mentioned
Answer: d
Explanation: IP router, microwave router and telephone switch can be used as an intermediate device between communication of two entities.
3. Which has comparatively high frequency component?
a) Sine wave
b) Cosine wave
c) Square wave
d) None of the mentioned
Answer: c
Explanation: Square wave has comparatively high frequency component in them.
4. Which has continuous transmission?
a) Asynchronous
b) Synchronous
c) Asynchronous & Synchronous
d) None of the mentioned
Answer: b
Explanation: Synchronous has continuous transmission where as asynchronous have sporadic transmission.
5. Which requires bit transitions?
a) Asynchronous
b) Synchronous
c) Asynchronous & Synchronous
d) None of the mentioned
Answer: b
Explanation: Synchronous transmission needs bit transition.
6. In synchronous transmission, receiver must stay synchronous for
a) 4 bits
b) 8 bits
c) 9 bits
d) 16 bits
Answer: c
Explanation: In synchronous transmission, receiver must stay synchronous for 9 bits.
7. How error detection and correction is done?
a) By passing it through equalizer
b) By passing it through filter
c) By amplifying it
d) By adding redundancy bits
Answer: d
Explanation: Error can be detected and corrected by adding additional information that is by adding redundancy bits.
8. Which is more efficient?
a) Parity check
b) Cyclic redundancy check
c) Parity & Cyclic redundancy check
d) None of the mentioned
Answer: b
Explanation: Cyclic redundancy check is more efficient than parity check.
9. Which can detect two bit errors?
a) Parity check
b) Cyclic redundancy check
c) Parity & Cyclic redundancy check
d) None of the mentioned
Answer: b
Explanation: CRC is more powerful and it can detect various kind of errors like 2 bit errors.
10. CRC uses
a) Multiplication
b) Binary division
c) Multiplication & Binary division
d) None of the mentioned
Answer: c
Explanation: CRC uses more math like multiplication and binary division.
This set of Advanced Digital Communication Questions & Answers focuses on “Error Detection and Correction”.
1. Which needs re-sending of signal?
a) Error correction
b) Error detection
c) Error correction & detection
d) None of the mentioned
Answer: b
Explanation: Error detection needs re-sending of data.
2. Which needs more check bits?
a) Error correction
b) Error detection
c) Error correction & detection
d) None of the mentioned
Answer: a
Explanation: Error correction needs more check bits where as error detection needs less check bits.
3. Which gets less affected by noise?
a) Error correction
b) Error detection
c) Error correction & detection
d) None of the mentioned
Answer: b
Explanation: Error detection is less affected by noise.
4. Which is used to protect privacy of the information?
a) Compression coding
b) Source coding
c) Cipher coding
d) None of the mentioned
Answer: c
Explanation: Cipher coding is used to provide privacy for the information.
5. Digital signals are easy for
a) Storage
b) Handling
c) Time dilation
d) All of the mentioned
Answer: d
Explanation: Digital signals have ease of handing, storage and time dilation.
6. Which are forward error correcting codes?
a) Block codes
b) Convolutional codes
c) Block & Convolutional codes
d) None of the mentioned
Answer: c
Explanation: Block codes and convolutional codes are examples of forward error correcting codes.
7. Which operates on continuous stream of data?
a) Block codes
b) Convolutional codes
c) Block & Convolutional codes
d) None of the mentioned
Answer: b
Explanation: Convolutional codes operates on a continuous stream of data.
8. Which is more complex?
a) Encoding
b) Decoding
c) Encoding & Decoding
d) None of the mentioned
Answer: b
Explanation: In the case of block codes, decoding is a complex method.
9. Which has better minimum distance?
a) Check sum
b) Cyclic redundancy check
c) Check sum & Cyclic redundancy check
d) None of the mentioned
Answer: b
Explanation: Cyclic redundancy check has better distances.
10. In symmetric key cryptosystem, who knows the key?
a) Sender
b) Receiver
c) Sender & Receiver
d) None of the mentioned
Answer: c
Explanation: In symmetric key cryptosystem, both sender and receiver knows the same key.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Codes and coding technique”.
1. While recovering signal, which gets attenuated more?
a) Low frequency component
b) High frequency component
c) Low & High frequency component
d) None of the mentioned
Answer: b
Explanation: High frequency components are attenuated more when compared to low frequency components while recovering the signals.
2. Mutual information should be
a) Positive
b) Negative
c) Positive & Negative
d) None of the mentioned
Answer: c
Explanation: Mutual information can also be negative.
3. ASCII code is a
a) Fixed length code
b) Variable length code
c) Fixed & Variable length code
d) None of the mentioned
Answer: a
Explanation: ASCII code is a fixed length code. It has a fixed length of 7 bits.
4. Which reduces the size of the data?
a) Source coding
b) Channel coding
c) Source & Channel coding
d) None of the mentioned
Answer: a
Explanation: Source coding reduces the size of the data and channel coding increases the size of the data.
5. In digital image coding which image must be smaller in size?
a) Input image
b) Output image
c) Input & Output image
d) None of the mentioned
Answer: b
Explanation: In digital image coding, output image must be smaller than the input image.
6. Which coding method uses entropy coding?
a) Lossless coding
b) Lossy coding
c) Lossless & Lossy coding
d) None of the mentioned
Answer: b
Explanation: Lossy source coding uses entropy coding.
7. Which achieves greater compression?
a) Lossless coding
b) Lossy coding
c) Lossless & Lossy coding
d) None of the mentioned
Answer: b
Explanation: Lossy coding achieves greater compression where as lossless coding achieves only moderate compression.
8. A code is a mapping from
a) Binary sequence to dicrete set of symbols
b) Discrete set of symbols to binary sequence
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: A code is a mapping from discrete set of symbols to finite binary sequence.
9. Which are uniquely decodable codes?
a) Fixed length codes
b) Variable length codes
c) Fixed & Variable length codes
d) None of the mentioned
Answer: a
Explanation: Fixed length codes are uniquely decodable codes where as variable length codes may or may not be uniquely decodable.
10. A rate distortion function is a
a) Concave function
b) Convex function
c) Increasing function
d) None of the mentioned
Answer: b
Explanation: A rate distortion function is a monotone decreasing function and also a convex function.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Interpolation”.
1. Interpolation is done by
a) Curve fitting
b) Regression analysis
c) Curve fitting & Regression analysis
d) None of the mentioned
Answer: c
Explanation: Interpolating the value requires or is done by curve fitting and regression analysis.
2. Interpolation provides a mean for estimating functions
a) At the beginning points
b) At the ending points
c) At the intermediate points
d) None of the mentioned
Answer: c
Explanation: Interpolation provides a mean for estimating the function at the intermediate points.
3. Interpolation methods are
a) Linear interpolation
b) Piecewise constant interpolation
c) Polynomial interpolation
d) All of the mentioned
Answer: d
Explanation: Some of the interpolation techniques are linear interpolation, piecewise constant interpolation, polynomial interpolation, spline interpolation etc.
4. Linear interpolation is
a) Easy
b) Precise
c) Easy & Precise
d) None of the mentioned
Answer: a
Explanation: Linear interpolation is quick and easy but not precise.
5. Error is equal to
a) Distance between the data points
b) Square of the distance between the data points
c) Half the distance between the data points
d) None of the mentioned
Answer: b
Explanation: Error is equal to square of the distance between the data points.
6. Which produces smoother interpolants?
a) Polynomial interpolation
b) Spline interpolation
c) Polynomial & Spline interpolation
d) None of the mentioned
Answer: c
Explanation: Polynomial interpolation and spline interpolation produces smoother interpolants.
7. Which is more expensive?
a) Polynomial interpolation
b) Linear interpolation
c) Polynomial & Linear interpolation
d) None of the mentioned
Answer: a
Explanation: Polynomial interpolation is more expensive than linear interpolation.
8. Gaussian process is a _____ interpolation process.
a) Linear
b) Non linear
c) Not an interpolation process
d) None of the mentioned
Answer: a
Explanation: Gaussian process is a non linear interpolation process.
9. Interpolation means
a) Adding new data points
b) Only aligning old data points
c) Only removing old data points
d) None of the mentioned
Answer: a
Explanation: Interpolation is a method of adding new data points within the range of a discrete set of known data points.
10. Interpolation is a method of
a) Interrelating
b) Estimating
c) Integrating
d) Combining
Answer: b
Explanation: Interpolation is a method of estimating the value of the function.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Information and Coding”.
1. Self information should be
a) Positive
b) Negative
c) Positive & Negative
d) None of the mentioned
Answer: a
Explanation: Self information is always non negative.
2. The unit of average mutual information is
a) Bits
b) Bytes
c) Bits per symbol
d) Bytes per symbol
Answer: a
Explanation: The unit of average mutual information is bits.
3. When probability of error during transmission is 0.5, it indicates that
a) Channel is very noisy
b) No information is received
c) Channel is very noisy & No information is received
d) None of the mentioned
Answer: c
Explanation: When probability of error during transmission is 0.5 then the channel is very noisy and thus no information is received.
4. Binary Huffman coding is a
a) Prefix condition code
b) Suffix condition code
c) Prefix & Suffix condition code
d) None of the mentioned
Answer: a
Explanation: Binary Huffman coding is a prefix condition code.
5. The event with minimum probability has least number of bits.
a) True
b) False
Answer: b
Explanation: In binary Huffman coding the event with maximum probability has least number of bits.
6. The method of converting a word to stream of bits is called as
a) Binary coding
b) Source coding
c) Bit coding
d) Cipher coding
Answer: b
Explanation: Source coding is the method of converting a word to stream of bits that is 0’s and 1’s.
7. When the base of the logarithm is 2, then the unit of measure of information is
a) Bits
b) Bytes
c) Nats
d) None of the mentioned
Answer: a
Explanation: When the base of the logarithm is 2 then the unit of measure of information is bits.
8. When X and Y are statistically independent, then I is
a) 1
b) 0
c) Ln 2
d) Cannot be determined
Answer: b
Explanation: When X and Y are statistically independent the measure of information I is 0.
9. The self information of random variable is
a) 0
b) 1
c) Infinite
d) Cannot be determined
Answer: c
Explanation: The self information of a random variable is infinity.
10. Entropy of a random variable is
a) 0
b) 1
c) Infinite
d) Cannot be determined
Answer: c
Explanation: Entropy of a random variable is also infinity.
11. Which is more efficient method?
a) Encoding each symbol of a block
b) Encoding block of symbols
c) Encoding each symbol of a block & Encoding block of symbols
d) None of the mentioned
Answer: b
Explanation: Encoding block of symbols is more efficient than encoding each symbol of a block.
12. Lempel-Ziv algorithm is
a) Variable to fixed length algorithm
b) Fixed to variable length algorithm
c) Fixed to fixed length algorithm
d) Variable to variable length algorithm
Answer: a
Explanation: Lempel-Ziv algorithm is a variable to fixed length algorithm.
13. Coded system are inherently capable of better transmission efficiency than the uncoded system.
a) True
b) False
Answer: a
Explanation: Yes, the coded systems are capable of better transmission efficiency than the uncoded system.
This set of Digital Communications Multiple Choice Questions & Answers focuses on “Quantization”.
1. Spread spectrum is used for
a) Encrypting signal
b) Hiding signal
c) Encrypting & Hiding signal
d) None of the mentioned
Answer: c
Explanation: Spread spectrum is used for hiding and encrypting signals.
2. Which is a quantization process?
a) Rounding
b) Truncation
c) Rounding & Truncation
d) None of the mentioned
Answer: c
Explanation: Rounding and truncation are examples of quantization process.
3. Quantization is a _____ process.
a) Few to few mapping
b) Few to many mapping
c) Many to few mapping
d) Many to many mapping
Answer: c
Explanation: Quantization is a many to few mapping process.
4. Quantization is a _____ process.
a) Non linear
b) Reversible
c) Non linear & Reversible
d) None of the mentioned
Answer: a
Explanation: Quantization is a non linear and irreversible process.
5. Which conveys more information?
a) High probability event
b) Low probability event
c) High & Low probability event
d) None of the mentioned
Answer: b
Explanation: High probability event conveys less information than a low probability event.
6. If the channel is noiseless information conveyed is ___ and if it is useless channel information conveyed is ___
a) 0,0
b) 1,1
c) 0,1
d) 1,0
Answer: d
Explanation: If the channel is noiseless information conveyed is 1 and if it is useless channel information conveyed is 0.
7. The mutual information between a pair of events is
a) Positive
b) Negative
c) Zero
d) All of the mentioned
Answer: d
Explanation: The mutual information between a pair of events can be positive negative or zero.
8. The output of the source encoder is an analog signal.
a) True
b) False
Answer: b
Explanation: The output of the source encoder is a sequence of binary digits. The conversion of source output to digital form is done here in source encoder.
9. The output of an information source is
a) Random
b) Deterministic
c) Random & Deterministic
d) None of the mentioned
Answer: a
Explanation: The output of any information source is random.
10. When the base of the logarithm is e, the unit of measure of information is
a) Bits
b) Bytes
c) Nats
d) None of the mentioned
Answer: c
Explanation: The unit of measure of information is determined based on the base of logarithm. If the base is e then the unit is nats.
This set of tricky Digital Communication Questions & Answers focuses on “Quantization”.
1. Uniform quantizer is also known as
a) Low rise type
b) Mid rise type
c) High rise type
d) None of the mentioned
Answer: b
Explanation: Uniform quantizer is also known as mid rise type quantizer.
2. The SNR value can be increased by _____ the number of levels.
a) Increasing
b) Decreasing
c) Does not depend on
d) None of the mentioned
Answer: a
Explanation: The signal to noise ratio can be increased by increasing the number of levels.
3. Prediction gain _____ for better prediction.
a) Increases
b) Decreases
c) Remains same
d) None of the mentioned
Answer: a
Explanation: Prediction gain increases for better prediction.
4. Delta modulation is
a) 1 bit DPCM
b) 2 bit DPCM
c) 4 bit DPCM
d) None of the mentioned
Answer: a
Explanation: Delta modulation is also considered as 1 bit DPCM.
5. 1 bit quantizer is a
a) Hard limiter
b) Two level comparator
c) Hard limiter & Two level comparator
d) None of the mentioned
Answer: c
Explanation: 1 bit quantizer is also called as two level comparator and also as hard limiter.
6. If step size is increased _____ occurs.
a) Slope overload distortion
b) Granular noise
c) Slope overload distortion & Granular noise
d) None of the mentioned
Answer: b
Explanation: When step size is increased to prevent slope overload distortion, granular noise occurs.
7. Which helps in maintaining the step size?
a) Delta modulation
b) PCM
c) DPCM
d) Adaptive delta modulation
Answer: d
Explanation: Step size if effectively maintained using adaptive delta modulation.
8. The low pass filter at the output end of delta modulator depends on
a) Step size
b) Quantization noise
c) Bandwidth
d) None of the mentioned
Answer: c
Explanation: The design of low pass filter at the output end of delta modulator depends on bandwidth.
9. In early late timing error detection method if the bit is constant, then the slope will be
a) Close to zero
b) Close to infinity
c) Close to origin
d) None of the mentioned
Answer: a
Explanation: In early timing error detection method if the bit is constant and doesn’t change, then the slope will be close to zero.
10. The theoretical gain in zero crossing TED is greater than early late TED.
a) True
b) False
Answer: a
Explanation: The theoretical gain in zero crossing timing error detection is twice more than that in the early late timing error detection method.
11. Non uniform quantizer ______ distortion.
a) Increases
b) Decreases
c) Does not effect
d) None of the mentioned
Answer: b
Explanation: Distortion can be reduced by using non uniform quantizer.
12. Vector quantization is used in
a) Audio coding
b) Video coding
c) Speech coding
d) All of the mentioned
Answer: c
Explanation: Vector quantization is widely used in speech coding for digital cellular systems.
13. The spectral density of white noise is
a) Exponential
b) Uniform
c) Poisson
d) Gaussian
Answer: b
Explanation: The spectral density of white noise is uniform.
14. The probability density function of the envelope of narrow band noise is
a) Uniform
b) Gaussian
c) Rayleigh
d) Rician
Answer: b
Explanation: The probability density function of the envelope of narrow band noise is Gaussian function.
15. The type of noise that interferes much with high frequency transmission is
a) White
b) Flicker
c) Transit time
d) Shot
Answer: c
Explanation: The type of noise that interferes much with high frequency transmission is transit time.
16. Thermal noise power of a resistor depends upon
a) Its resistance value
b) Noise temperature
c) Bandwidth
d) Ambient temperature
Answer: b
Explanation: Thermal noise power of a resistor depends upon noise temperature.
