Antennas Pune University MCQs

This set of Antennas Multiple Choice Questions & Answers  focuses on “Basics”.

1. The relation between vector magnetic potential and current density is given by ______

a) ∇. A = J

b) ∇× A = H

c) ∇ 2 A =-μ J

d) ∇ 2 A =∇× H

Answer: c

Explanation: Magnetic Flux density B is expressed as B =∇× A

Taking curl on both sides, we get ∇× B =∇×(∇× A ) = ( ∇. A )∇- A

From Maxwell’s equation, ∇. B =0 =>∇. A =0

⇨ ∇× B = -∇ 2 A and B = μ H , ∇× H = J

⇨ ∇×μ H = -∇ 2 A

⇨ ∇ 2 A =-μ J .

2. The induction and radiation fields are equal at a distance of _______

a) λ/4

b) λ/6

c) λ/8

d) λ/2

Answer: b

Explanation: For an Hertzian dipole, equating the magnitudes of maximum induction and radiation fields we get,

=\frac{I_mdl}{4\pi\epsilon} 

 

\)

\(r=\frac{v}{ω}=\frac{\lambda f}{2\pi f}=\frac{\lambda}{6}.\)

3. The ratio of radiation intensity in a given direction from antenna to the radiation intensity over all directions is called as ________

a) Directivity

b) Radiation power density

c) Gain of antenna

d) Array Factor

Answer: a

Explanation: Directivity of antenna is defined as the ratio of radiation intensity in a given direction from antenna to the radiation intensity over all directions. \(D = \frac{U_{max}}{U_0}.\)

Radiation Intensity is power radiated from an antenna for unit solid angle. \(U_0= w_r.r^2\frac{watts}{steradians}.\)

Gain of antenna is ratio of radiation intensity in given direction to the radiation intensity of isotropic radiation. Array factor is a function of geometry of array and the excitation phase.

4. What is the overall efficiency of a lossless antenna with reflection coefficient 0.15?

a) 0.997

b) 0.779

c) 0.669

d) 0.977

Answer: d

Explanation: For a lossless antenna, the radiation efficiency e cd =1.

Overall efficiency of antenna is given by e o =e cd 

\)=1×(1-(0.15 2 ))=0.977.

5. The equivalent area when multiplied by the instant power density which leads to free radiation of power at antenna is called as _______

a) Loss area

b) Scattering area

c) Captured area

d) Effective area

Answer: b

Explanation: Scattering area is the equivalent area when multiplied by the instant power density which leads to free radiation of power. Loss area leads to power dissipation and captured area leads to total power collection by the antenna. The relation among them is given by,

Captured area= effective area + loss area + scattering area.

6. Equivalent circuit representation of an antenna is ______

a) Series R, L, C

b) Parallel R, L, C

c) Series R, L parallel to C

d) Parallel R, C series to L

Answer: a

Explanation: Antenna is represented by a series R, L, C equivalent circuit. Antenna is used for impedance matching and acts like a transducer.

7. Radiation resistance of a Hertzian dipole of length λ/8 is ________

a) 12.33Ω

b) 8.54Ω

c) 10.56Ω

d) 13.22Ω

Answer: a

Explanation: Radiation resistance of a Hertzian dipole of length l is

R=80π 2 \

 

^2=80\pi^2 

 

^2=12.33\Omega\)

8. Relation between directivity and effective area of transmitting and receiving antenna is ________

a) D t A t =D r A r

b) D t A r =D r A t

c) A t D t =∈D r A r

d) D t A t =∈D r A r

Answer: b

Explanation: The power collected by the receiving antenna is

\(P_r = \frac{P_tD_tA_r}{4\pi R^2} => D_tA_r = \frac{P_r}{P_t}4\pi R^2\) = D r A t

∴ D t A r =D r A t

9. The axis of back lobe makes an angle of 180° with respect to the beam of an antenna.

a) True

b) False

Answer: a

Explanation: The axis of back lobe is opposite to the main lobe. So it makes 180° with beam of antenna. It is also a side lobe which is at 180 to main lobe.

10.Radiation resistance of a half-wave dipole is ______

a) 36.56Ω

b) 18.28Ω

c) 73.12Ω

d) 40.24Ω

Answer: c

Explanation: Since radiation resistance of quarter-wave monopole  is 36.56Ω, then for a half-wave dipole  it is given by 36.56×2 = 73.12Ω. Hertzian dipole is an ideal dipole of infinitesimal dipole.

11. The radiation efficiency for antenna having radiation resistance 36.15Ω and loss resistance 0.85Ω is given by ________

a) 0.977

b) 0.799

c) 0.997

d) 0.779

Answer: a

Explanation: The radiation efficiency \(e_{cd}=\frac{R_r}{R_l+R_r}=\frac{36.15}{36.15+0.85}=0.977.\)

Radiation efficiency is also known as conductor-dielectric efficiency. It is the ratio of power delivered to the radiation resistance to the power delivered to it when conductor-dielectric losses are present.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Antenna Characteristics”.

1. A linear antenna having length less than λ/8 is called as _______

a) Short monopole

b) Short dipole

c) Half-wave dipole

d) Quarter-wave monopole

Answer: a

Explanation: Short monopoles have length less than λ/8 and the current distribution is triangular. Short dipole has length less than λ/2. Half-wave dipoles have length equal to λ/2. Quarter-wave monopoles have length equal to λ/4.

2. Find the power radiated by an antenna whose radiation resistance is 100Ω and operating with 3A of current at 2GHz frequency?

a) 900W

b) 1800W

c) 450W

d) 700W

Answer: a

Explanation: Power radiated P r =I 2 R r =100×3 2 =900Watts.

3. Front-to-Back ratio is defined as ratio of power radiated in desired direction to the power radiated in back lobe.

a) True

b) False

Answer: a

Explanation: Front-to-Back ratio  =\(\frac{Power\, radiated\, desired\, direction}{Power\, radiated\, in\, backward\, direction}.\) If more power is diverted backside, then the gain of the antenna decreases.

4. Relation between beam solid angle Ω, horizontal half-power beam width ∅ A , vertical half-power beam width ∅ E is __________

a) Ω≈∅ A .∅ E

b) Ω≈∅ A +∅ E

c) Ω≈∅ A /∅ E

d) Ω≈∅ A -∅ E

Answer: a

Explanation: Half-power beam width is the angular range of antenna pattern in which half power is radiated. The relation between Ω, ∅ A , ∅ E is given by Ω≈∅ A .∅ E .

5. Which of the following field varies inversely with r 2 ?

a) Far field

b) Near field

c) Radiation field

d) Electrostatic field

Answer: b

Explanation: Induction field is also known as ‘Near field’, varies inversely with r 2 . Electrostatic field varies inversely with r 3 . Far field is also known as Radiation field, varies inversely with r.

6. Find the effective area of a half-wave dipole operating at frequency 100MHz and directive gain 1.8?

a) 1.28m 2

b) 2.18m 2

c) 0.128m 2

d) 12.8m 2

Answer: a

Explanation: The effective area \(A_e=\frac{\lambda^2}{4π}D \)

\(\lambda = \frac{c}{f}=3×\frac{10^8}{100×10^6}=3m\)

\(A_e=\frac{3^2}{4π}×1.8=1.28m^2\)

7. Which of the following option is false?

a) Omni-directional antenna is a special case of directional antenna

b) Directional antenna radiates power effectively in particular directions compared to other directions

c) Isotropic antenna radiates power in all directions

d) End-fire array antenna has its main beam normal to the axis containing antenna

Answer: d

Explanation: End-fire array has its main beam parallel to the axis of antenna . For broadside antenna it is normal to the axis of antenna. Omni-directional antenna radiates power in only one direction and is non-radiating in other directions. So it is a special case of directional antenna.

8. The angular distance between two successive nulls of main lobe is called as ______

a) FNBW

b) HPBW

c) Beam width

d) FBR

Answer: a

Explanation: The angular distance between two successive nulls of main lobe is called First Null Beam width. Half-power beam width is the angular distance when 50% of power is radiated. FBR is the Front-to-Back ratio defined as ratio of power radiated at 0° to power radiated at 180°.

9. If beam width of the antenna increases, then directivity ________

a) Decreases

b) Increases

c) Remains unchanged

d) Depends on type of antenna

Answer: a

Explanation: As beam width of antenna increases its area coverage broadens, thereby directivity decreases. Beam area and directivity are inversely proportional. \(D=\frac{4\pi}{Beam \,Area}.\)

10. The receiving antenna is designed to have ____ side-lobe-ratio and ____ SNR.

a) Low, high

b) High, high

c) Low, low

d) High, low

Answer: a

Explanation: Side lobe ratio is ratio of power density in side lobes to main lobe. A receiving antenna is said to be efficient if side lobes are minimized and receives most of the transmitted signal. So it should have low SLR and high SNR.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Radiation Pattern”.

1. Units of radiation intensity is _______

a) Watts/unit Solid angle

b) Watts/m 2

c) Watts- m 2

d) Watts

Answer: a

Explanation: Radiation intensity is defined as the power radiated form an antenna per unit solid angle. So its units are Watts/Steradian. Steradian is the unit for solid angle.

2. The graphical representation of the radiation properties of the antenna as a function of space coordinates is called Radiation pattern.

a) True

b) False

Answer: a

Explanation: The graphical representation of the radiation properties of the antenna as a function of space coordinates is called Radiation pattern. It is also known as antenna pattern. A radiation property includes power flux density, directivity, and radiation intensity.

3. What is the total power radiated in Watts for the power density \

 

 4π 2

b) 8π 2 /3

c) 4π 2 /3

d) 2π 2 /3

Answer: c

Explanation: Total power radiated \

∯

 

 

 

 

d\theta d\emptyset\)

\

 

 

=\frac{4}{3}\) π 2 .

4. Which of the following pattern varies with square of magnitude of field?

a) Power Pattern

b) Electric Field Pattern

c) Current distribution

d) Array Factor

Answer: a

Explanation: Power pattern varies with square of magnitude of field. The average power =\(\frac{\mid \overline{E}\mid^2}{2\eta}\). It is defined as the trace of received power at a constant radius.

5. The radiation lobe containing the direction of maximum radiation is called as _____

a) Major lobe

b) Minor lobe

c) Side lobe

d) Back lobe

Answer: a

Explanation: Major lobe contains the maximum radiated power. Sometimes depending on our requirement there can be more than 2 major lobes.

6. Fresnel zone is also called as ____

a) Near Field

b) Far Field

c) Electrostatic Field

d) Reactive Field

Answer: a

Explanation: Near Field is called Fresnel Field. Far field is called Fraunhofer zone. Near field varies inversely with r 2 . Electrostatic field varies inversely with r 3 .

7. If FNBW is 6°, then resolution is ____

a) 12°

b) 3°

c) 2°

d) 6°

Answer: b

Explanation: Resolution is half of the first null beam width. \(R=\frac{FNBW}{2}=\frac{6°}{2}=3°.\)

8. For a center fed short antenna, current distribution is _____ at center and ____ at ends.

a) Low, high

b) High, high

c) Low, low

d) High, low

Answer: d

Explanation: The current distribution follows a triangular pattern. At the center it is maximum and nearly zero at ends. Amount of power radiates will depend on the wavelength λ. For vertical antennas, with wavelength n\(\frac{\lambda}{2}\) the current distribution at center is high.

9. Which pattern represents a plot with magnitude of field strength Vs θ at a constant φ?

a) E-plane pattern

b) H-plane pattern

c) Horizontal pattern

d) Power pattern

Answer: a

Explanation: Varying θ at a constant φ represents a vertical plot against the field strength. So this pattern is called E-plane pattern. H-plane or horizontal pattern are plotted with magnitude of field strength Vs φ at a constant θ. Power pattern is plotted with square of the magnitude of field strength.

10. The portion of the near field immediate to the surrounding the antenna is called as _____

a) Reactive near-field

b) Radiating near-field

c) Fraunhofer zone

d) Far field

Answer: a

Explanation: The portion of the near field immediate to the surrounding the antenna is called Reactive near-field. The boundary of this region exists at a distance of R < 0.62√(D 3 /λ) where D is the largest dimension of the antenna. Radiating near field lies between reactive near field and far field. Far field is also known as Fraunhofer region.

11. The region of the field that angular field distribution is independent of the distance from the antenna is called as _______

a) Reactive near-field

b) Radiating near-field

c) Fresnel zone

d) Far field

Answer: d

Explanation: The region of the field that angular distribution is independent of the distance from the antenna is called far field. This region is present at d distance greater than 2D 2 /λ. Ideally the outer boundary is taken at infinity.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Isotropic Radiators”.

1. An ideal source in which the power is radiated equally in all directions is called as ________ radiator.

a) Isotropic

b) Omni-directional

c) Directional

d) Transducer

Answer: a

Explanation: Isotropic radiators radiate power in all directions uniformly. Omni-directional antenna radiates only in one direction. Directional antenna radiates maximum power in particular direction. Transducer converts one form of energy to other.

2. In Isotropic radiation, which of the following vector component is absent in pointing vector?

a) \

 \

 \

 Both \(\widehat{a_\theta} \,and\, \widehat{a_\emptyset} \)

Answer: d

Explanation: Isotropic radiators will radiate power equally in all directions. Due to this symmetrical distribution, components of θ, Φ get cancelled. So it will have only radial component.

3. What is the amount of Electric field present at a distance of 10km for an isotropic radiator with radiating power 3kW?

a) 30mV/m

b) 60mV/m

c) 15mV/m

d) 10mV/m

Answer: a

Explanation: For isotropic radiator, power per unit area = \

 

From pointing theorem, \

 

Equating  &  and E rms =E max /√2, we get

\(E_{rms}=\frac{\sqrt{30P_t}}{r}=\frac{\sqrt{30×3000}}{10000}=30mV/m.\)

4. What is the radiation intensity for isotropic antenna having radiation power density \

 

 3sinθ a r W/Steradian

b) 3cosθa r W/Steradian

c) 6πsinθ a r W/Steradian

d) 6πcosθ a r W/steradian

Answer: a

Explanation: Radiation intensity is defined as the power radiated per unit solid angle.

U= Wr.r 2 = 3sinθ a r W/Steradian.

5. For an isotropic source, Radiation intensity will be _____ on θ and ______ on Φ.

a) Dependent, independent

b) Independent, independent

c) Independent, dependent

d) Dependent, dependent

Answer: b

Explanation: Due to symmetric distribution in isotropic source, there will be no θ, ф components. Only radial component will be present. \

∯.\)

∴ Radiation intensity \(U=\frac{P_{total \,rad}}{4π}\)

6. Find the effective length of a receiving antenna with open circuit voltage 1V and incident electric field 200mV/m?

a) 0.2m

b) 50m

c) 5m

d) 5cm

Answer: c

Explanation: Effective length of the receiving antenna \(L_{eff}=\frac{V_{oc}}{E_i}=\frac{1}{0.2}=5m.\)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Directive Gain”.

1. For an isotropic antenna, the average power P av can be expressed in terms of radiated power P r as ____

a) P av =P r /4π

b) P av =P r /2πr 2

c) P av =P r /2π

d) P av =P r /4πr 2

Answer: d

Explanation: Average power is the total power radiated in the unit area. Here for isotropic radiation, area is spherical  and the area is 4πr 2 .

∴ P av =P r /4πr 2

2. Directive gain is defined as a measure of concentration of power in a particular direction.

a) True

b) False

Answer: a

Explanation: Directive gain is the ratio of power density to the average power radiated.

\(G_d = \frac{P_{d}}{P_{avg}}\)

3. What is the directive gain when the magnitude of radiation intensity equals to average radiation intensity?

a) 4π

b) ∞

c) 1

d) 0

Answer: c

Explanation: Directive gain \(G_d = \frac{P_{d}}{P_{avg}} = \frac{P_{d}}{P_r/4\pi r^2} = \frac{P_{d^{r^2}}}{P_r/4\pi} = \frac{U_{d}}{U_{avg}} \)

∴ \(G_d = \frac{U_{}}{U_{avg}}\)=1.

4. Directive gain of antenna when radiation intensity is 5W/Steradian and radiated power 5W is ____

a) 4π

b) 1/4π

c) 25

d) 1

Answer: a

Explanation: Given U d = 5W/steradian , P r =5W

Directive gain \(G_d=\frac{P_{d^{r^2}}}{P_r/4\pi} = \frac{U_{d}}{P_r/4\pi}=4\pi\)

5. The Directive gain is ______ on input power to antenna and _____ on power due to ohmic losses.

a) Independent, independent

b) Dependent, independent

c) Independent, dependent

d) Dependent, dependent

Answer: a

Explanation: Directive gain is the ratio of power density to the average power radiated. \(G_d = \frac{P_{d}}{P_{avg}}\) So, the Directive gain is independent on both input power to antenna and power due to ohmic losses.

Power gain is dependent on input power and ohmic losses to antenna.

6. What is the maximum directive gain of antenna with radiation efficiency 98% and maximum power gain 1?

a) 0.98

b) 1.02

c) 1.98

d) 1

Answer: b

Explanation: G pmax =η r G dmax where η r is radiation efficiency

Therefore G dmax =1/0.98=1.02

7. Which of the following expression is correct for radiation efficiency?

a) \

 

 \

 

 \

 

 \(\eta_r=\frac{R_l}{R_r+R_l}\)

Answer: c

Explanation: Radiation efficiency is defined as the ratio of power radiated to the total input power to the antenna. Total input power is the sum of the radiated power P r and the ohmic losses P l .

\(\eta_r = \frac{P_r}{P_{in}} = \frac{P_r}{P_r+P_l} = \frac{R_r I_{rms}^2}{I_{rms}^2 R_r+I_{rms}^2 R_l} = \frac{R_r}{R_r+R_l} \)

8. For a lossless antenna, maximum Power gain equals to the maximum directive gain.

a) True

b) False

Answer: a

Explanation: For a lossless antenna, ohmic losses will be zero. So, radiation efficiency will be 100%. Hence, maximum power gain will be equal to the maximum directive gain of antenna.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Power Gain”.

1. The ratio of power radiated in a particular direction to the total input power of antenna is called as _____

a) Directive gain

b) Power gain

c) Directivity

d) Partial directivity

Answer: b

Explanation: The ratio of power radiated in a particular direction to the actual power input to antenna is called Power gain. \(G_p=\frac{P_{d}}{P_{in}}\). Directive gain is the ratio of power radiated in desired direction to the average power radiated from the antenna. Partial directivity is the part of radiation intensity in a particular polarization to radiation intensity in all directions.

2. What is the maximum power gain of antenna with radiation efficiency 98% and directive gain 1?

a) 0.98

b) 1.02

c) 1.98

d) 1

Answer: a

Explanation: G pmax =η r G dmax where η r is radiation efficiency

∴ G pmax =0.98×1=0.98.

3. Which of the following expression is correct for radiation efficiency?

a) \

 

 \

 

 \

 

 \(\eta_r = \frac{P_l}{P_r+P_l}\)

Answer: c

Explanation: Radiation efficiency is defined as the ratio of power radiated to the total input power to the antenna. Total input power is the sum of the radiated power P r and the ohmic losses P l .

\(\eta_r = \frac{P_r}{P_{in}} = \frac{P_r}{P_r+P_l}\)

4. Which of the following represents the relation between maximum power gain and maximum directivity gain of the antenna?

a) G pmax = η r G dmax

b) G pmax = η r /G dmax

c) η r = \

 η r = \(\frac{G_{dmax}+G_{pmax}}{G_{dmax}-G_{pmax}}\)

Answer: a

Explanation: Maximum power gain is obtained when there are no ohmic losses. G pmax =\(\frac{U_{max}}{P_{in}/4π}\)

Maximum directive gain G dmax =\(\frac{U_{max}}{P_r/4π}\, and\, \eta_r=\frac{P_r}{P_{in}}\)

∴G pmax =η r G dmax

5. What is the maximum power gain when the radiation resistance is 72Ω, loss resistance is 8Ω and the maximum directive gain is 1.5?

a) 1.15

b) 1.35

c) 1.25

d) 1.53

Answer: b

Explanation: Maximum power gain G pmax =η r G dmax

Radiation efficiency η r =\(\frac{R_r}{R_r+R_l}=\frac{72}{72+8}=\frac{72}{80}=0.9\)

Now, G pmax =η r G dmax =0.9×1.5=1.35

6. The radiation efficiency value is ______

a) 0

b) 1<η<∞

c) 0≤η≤1

d) ∞

Answer: c

Explanation: Radiation efficiency η r =\(\frac{R_r}{R_r+R_l}\)

R r +R l >R r So \(\frac{R_r}{R_r+R_l}\) < 1 and If R l =0 then η r =1

Therefore, the value of efficiency lies in the range 0 to 1.

7. The value of maximum power gain is always greater than or equal to the maximum directive gain.

a) True

b) False

Answer: b

Explanation: Since G pmax =η r G dmax and the value of radiation efficiency lies in range 0 to 1, The maximum power gain will be always less than or equal to the maximum directive gain of antenna.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Directivity”.

1. The ratio of maximum power density in the desired direction to the average power radiated from the antenna is called as _______

a) directivity

b) directive gain

c) power gain

d) partial directivity

Answer: a

Explanation: The ratio of maximum power density in the desired direction to the average power radiated from the antenna is called Directivity. The ratio of power radiated in a particular direction to the actual power input to antenna is called Power gain. Directive gain is the ratio of power radiated in desired direction to the average power radiated from the antenna. Maximum Directive gain is called as Directivity. Partial directivity is the part of radiation intensity in a particular polarization to radiation intensity in all directions.

2. What is the Beam area for Directivity to be 1 in Steradian?

a) 4π

b) 1/2π

c) 2π

d) 1/4π

Answer: a

Explanation: The Directivity in terms of the Beam area Ω A is given by D=\(\frac{4\pi}{\Omega_A}\)

⇨ \(\frac{4\pi}{D}\)=4π steradians.

3. If directivity of antenna increases, then the coverage area _____

a) decreases

b) increases

c) increases and then decreases

d) remains unchanged

Answer: a

Explanation: As the directivity increases, the beam area decreases. So, the coverage area of beam decreases and takes more time to scan a target for target detection.

4. If half power beam width in one plane and other plane orthogonal to it are equal to π then the directivity is ____

a) π

b) 4π

c) 4/π

d) 2π

Answer: c

Explanation: Beam area Ω A ≈ θ 1r θ 2r where θ 1r , θ 2r are half-power beam widths in radians.

Directivity \(D=\frac{4\pi}{\Omega_A}=\frac{4\pi}{\theta_{1r}\theta_{2r}}=\frac{4\pi}{\pi.\pi} = 4/\pi\)

5. Directive gain with maximum radiation intensity is called as Directivity.

a) True

b) False

Answer: a

Explanation: Directivity is defined as ratio of maximum power density in desired direction to the average power radiated in all directions. This is simply, maximum Directive gain. Maximum Directive gain is obtained if maximum radiation intensity is present in desired direction.

6. How the directivity and effective aperture related to each other?

a) Inversely proportional

b) Directly proportional

c) Independent

d) Proportionality depends on input power

Answer: b

Explanation: The directivity D is directly proportional to the effective aperture of antenna. \(A_e=D\frac{\lambda^2}{4\pi}\)

7. What is the directivity of half-wave dipole?

a) 1.64

b) 1.5

c) 1.43

d) 1.44

Answer: a

Explanation: Directivity \(D=\frac{U_{max}}{U_{av}} = \frac{4\pi U_{max}}{P_{rad}} \)

P rad =36.54I 0 2 Since the radiation resistance for Half-wave dipole is 36.54Ω.

Maximum radiation intensity is given by \(U_{max}=P_{max}.r^2=\frac{\mid E_\theta\mid_{max}^2}{2\eta}.r^2\)

Where \(E_\theta=\frac{j\eta I_0 e^{-jkr}cos⁡

 

}{2\pi rsin\theta}\)

⇨ \(D=\frac{4\pi}{1}\frac{\eta I_0^2}{8\pi^2} \frac{1}{36.54I_0^2}=1.64.\)

8. What is the directivity of antenna having effective aperture 1 m 2 ?

a) \

 

 \

 

 1

d) 4π

Answer: a

Explanation: Directivity D=\(\frac{4\pi}{\lambda^2}A_e\)

Given effective aperture A e =1m 2

⇨ D=\(\frac{4\pi}{\lambda^2}A_e = \frac{4\pi}{\lambda^2} × 1 =\frac{4\pi}{\lambda^2}\)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Effective Aperture”.

1. Effective aperture is the ability of antenna to extract energy from the electromagnetic wave.

a) True

b) False

Answer: a

Explanation: Effective aperture is defined as the ratio of power received from load to the average power density produced at that point. So it is the ability of antenna to extract energy from EM wave.

2. Which of the following best describes the condition for Maximum effective aperture?

a) Load impedance must be equal to the antenna impedance

b) Load impedance must be equal complex conjugate to the antenna impedance

c) Receiver power should be minimum

d) Transmitter power should be minimum

Answer: b

Explanation: Foe effective aperture to be maximum, the receiver power should be maximum. Maximum power transfer states that the maximum received power is obtained when the impedance of network matches with the complex conjugate of the load impedance. Hence, Load impedance must be equal to the complex conjugate of the antenna impedance.

3. What is the effective aperture of Hertzian dipole antenna operating at frequency 100 MHz?

a) 1.07m 2

b) 0.17m 2

c) 1.7m 2

d) 1.2m 2

Answer: a

Explanation: Effective aperture for a Hertzian dipole is given by \(A_e=1.5\frac{\lambda^2}{4\pi} \)

Gain of Hertzian dipole is 1.5

\(\lambda=\frac{c}{f}=\frac{3×10^8}{100×10^6}=3m\)

\(A_e=1.5 \frac{\lambda^2}{4\pi} =1.5\frac{3^2}{4\pi} =1.07m^2\)

4. If physical aperture of antenna is 0.02m 2 and aperture efficiency is 0.5, then what is the value of effective aperture?

a) 0.0004m 2

b) 0.001m 2

c) 0.01m 2

d) 25m 2

Answer: c

Explanation: Effective aperture A em =A e η=0.02×0.5=0.01 m 2

5. Expression for aperture efficiency in terms of physical aperture A e and effective aperture A em is ____

a) \

 

 \

 

 \

 

 \(\frac{A_e-A_{em}}{A_e+A_{em}}\)

Answer: b

Explanation: The aperture efficiency is defined as the ratio of effective aperture to physical aperture of antenna. So aperture efficiency \(\eta=\frac{A_{em}}{A_e}.\)

6. What is the effective aperture of a Half-wave dipole operating at 100MHz?

a) 1.07m 2

b) 1.17m 2

c) 1.27m 2

d) 1.77m 2

Answer: b

Explanation: The directivity of half-wave dipole is 1.64

The effective aperture of half-wave dipole is \(A_e=1.64\frac{\lambda^2}{4\pi}\)

\(\lambda=\frac{c}{f}=\frac{3×10^8}{100×10^6}=3m\)

\(A_e=1.64 \frac{\lambda^2}{4\pi} = 1.64 \frac{3^2}{4\pi}\)=1.17m 2 .

7. What is the relation between effective length and Effective aperture of antenna?

a) \

 

 \

 

 \

 

 \(A_e = \frac{dL^2 \eta ^2}{4R_{rad}}\)

Answer: a

Explanation: Maximum effective aperture is ratio of maximum power received to the average power density. \

 

⇨

 

⇨

 

^2 R_{rad}=\frac{V_{oc}^2}{8R_{rad}}=\frac{\mid E_\theta\mid^2 dL^2}{8R_{rad}} and P_{avg}=\frac{\mid E_\theta\mid^2}{2\eta} \)

\(A_e=\frac{P_{Rmax}}{P_{avg}}=\frac{dL^2\eta}{4R_{rad}}\)

8. The physical aperture of an isotropic radiator is _______

a) \

 

 \

 

 \

 

 \(\frac{\lambda^2\eta}{4\pi}\)

Answer: c

Explanation: For isotropic radiator, directivity is 1. So the effective aperture is given by \(A_{em}=D \frac{\lambda^2}{4\pi} = \frac{\lambda^2}{4\pi}\)

Then physical aperture =\(\frac{Effective\, aperture}{Aperture\, efficiency}=\frac{\lambda^2}{4\pi \eta}\)

This set of Antenna Parameters Interview Questions and Answers for freshers focuses on “Radiation Resistance”.

1. What is the radiation resistance of an antenna if it radiates 1kW and current in it is I rms =10A?

a) 0.1Ω

b) 1Ω

c) 10Ω

d) 100Ω

Answer: c

Explanation: Power radiated P rad =I rms 2 R rad

⇨ R rad =\(\frac{P_{rad}}{I_{rms}^2} = \frac{1000}{10×10}=10\Omega\)

2. What is the radiation resistance of an antenna if input power to it is 1KW and current in it is 10A having a power loss of 200W?

a) 10Ω

b) 2Ω

c) 12Ω

d) 8Ω

Answer: d

Explanation: Input power P in =P rad +P loss

⇨ P rad =P in -P loss =1000-200=800W

Now, Power radiated P rad =I rms 2 R rad

⇨ R rad =\(\frac{P_{rad}}{I_{rms}^2} = \frac{800}{10×10}\)=8Ω

3. What is the radiation resistance of a short dipole of length L?

a) 20π 2 \

 

^2\)

b) 80π 2 \

 

^2\)

c) 40π 2 \

 

^2\)

d) 160π 2 \

 

^2\)

Answer: a

Explanation: Radiation resistance of a Hertzian dipole of length l is given by R=80π 2 \

 

^2\)

Then short dipole is of length l/2, so the radiation resistance is given by R=80π 2 \

 

^2=20\pi^2 

 

^2.\)

4. If the length of the dipole decreases, then the radiation resistance will________

a) increase

b) decrease

c) depends on current distribution

d) not change

Answer: b

Explanation: Since the radiation resistance of a Hertzian dipole is R=80π 2 \

 

^2\), the radiation resistance will decrease as the length of the dipole decreases. It is directly proportional to the square of the length of the dipole.

5. For a half-wave dipole with length λ/12, what is the antenna efficiency if the Radiation resistance is 2Ω?

a) 0.73

b) 0.073

c) 0.37

d) 0.78

Answer: a

Explanation: Radiation resistance \

 

^2=80π^2 

 

^2\)=5.48Ω

Antenna efficiency \(\eta=\frac{R_{rad}}{R_{rad}+R_{loss}}=\frac{5.48}{5.48+2}=0.73\)

6. Find the radiation resistance of a Hertzian dipole of length 1m and operating at a frequency 1MHz?

a) 0.08Ω

b) 8.8mΩ

c) 8.8Ω

d) 0.88Ω

Answer: b

Explanation: \

 

 

 

^2=80π^2 

 

^2\)=0.0088Ω

7. Radiation resistance doesn’t depend on direction of power radiated but depends on the frequency.

a) True

b) False

Answer: a

Explanation: Radiation resistance doesn’t depend on the direction in which the power is radiated. It depends on frequency through which we will find the wavelength and thereby the radiation resistance.

8. What is the radiation resistance of the antenna radiating at 5kW and having maximum current 2A?

a) 25kΩ

b) 2.5kΩ

c) 0.25kΩ

d) 2.5Ω

Answer: b

Explanation: \(I_{rms}=\frac{I_m}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)

Power radiated \(P_{rad}=I_{rms}^2 R_{rad}\)

⇨ \(R_{rad}=\frac{P_{rad}}{I_{rms}^2} = \frac{5K}{2}=2.5k\Omega\)

9. Power radiated by half-wave dipole with maximum current amplitude 10A is ______

a) 3.65kΩ

b) 3.650Ω

c) 0.365kΩ

d) 36.50Ω

Answer: a

Explanation: For a half-wave dipole the radiation resistance is 73Ω.

\(I_{rms}=\frac{I_m}{\sqrt{2}}=\frac{10}{\sqrt{2}}=5\sqrt{2}\)

Power radiated P rad =I rms 2 R rad = 2 ×73=3650Ω=3.65kΩ

10. The radiation resistance dissipates same amount of power as it radiated by the antenna.

a) True

b) False

Answer: a

Explanation: The radiation resistance is defined as the equivalent resistance that dissipates equal amount of power that is radiated by the antenna. The power radiated by the radiation resistance is given by P rad =I rms 2 R rad . Radiation resistance doesn’t depend on direction of power radiated but it depends on the frequency.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Antenna Bandwidth”.

1. Relation between Quality factor, Bandwidth, and resonant frequency is _________

a) \

 

 \

 

 Q = BW×f 0

d) \(Q=\frac{BW+f_0}{BW-f_0}\)

Answer: b

Explanation: Quality factor is defined as the ratio of resonant or center frequency to the bandwidth.\(Q=\frac{f_0}{BW}\)

2. What is the Bandwidth of the antenna operating at resonant frequency 200MHz with Quality factor 20?

a) 10Hz

b) 5MHz

c) 10MHz

d) 0.1MHz

Answer: c

Explanation: Bandwidth \(Q=\frac{f_0}{Q}=\frac{200MHz}{20}=10MHz\)

3. What is the length of the half-wave dipole with bandwidth 20MHz and Quality factor 30?

a) 5m

b) 0.25m

c) 0.50m

d) 2.5m

Answer: b

Explanation: Operating frequency f 0 =BW×Q=20MHz×30=600MHz

Now, \(λ=\frac{c}{f}=\frac{3×10^8 m/s}{600MHz}=0.5m\)

∴ Length of half-wave dipole is \(l=\frac{λ}{2}=\frac{0.5}{2}=0.25m\)

4. Quality factor is defined as ________

a) \

 

 \

 

 \

 

 \(2π×\frac{Total \,energy\, stored\, by \,antenna}{energy\, radiated \,per \,cycle}\)

Answer: d

Explanation: Quality factor Q=\(2π×\frac{Total \,energy\, stored\, by \,antenna}{energy\, radiated \,per \,cycle}\). Higher Q antennas will have low bandwidth.

5. What is the quality factor of the antenna operating at 650MHz and having a bandwidth of 10MHZ?

a) 65

b) 0.65

c) 15

d) 55

Answer: a

Explanation: Quality factor \(Q=\frac{f_0}{BW}=\frac{650MHz}{10MHz}=65\)

6. In an antenna, the lower frequency limit is determined by pattern, gain or impedance.

a) True

b) False

Answer: a

Explanation: In an antenna, the lower frequency limit is determined by pattern, gain or impedance as the requirements changes. Higher frequency limit is determined by other parameters like the SWR, FBR etc.

7. In the impedance v/s frequency graph of antenna, the antenna impedance at frequencies less than resonant frequency is ____

a) inductive

b) capacitive

c) resistive

d) both inductive and capacitive

Answer: b

Explanation: Antenna is a series R,L,C equivalent circuit.For frequencies less than resonant frequency, the impedance is capacitive.

8. High the Fractional Bandwidth ___________ is the quality factor.

a) low

b) high

c) constant

d) infinity

Answer: a

Explanation: Fractional Bandwidth is the ratio of the frequency range of antenna to the center frequency. If Fractional bandwidth is high, then Bandwidth of the antenna is also more. So the quality factor decreases as it is inversely proportional to the bandwidth.

9. For lower Quality factor antennas, the bandwidth is very high.

a) True

b) False

Answer: a

Explanation: The bandwidth of antenna is inversely proportional to the Quality factor of the antenna.

\(Q=\frac{f_0}{BW}\)

This set of Antenna Parameters Questions and Answers for Experienced people focuses on “Friis Transmission Equation”.

1. Friss transmission is applicable when same antenna is used for both transmission and reception.

a) True

b) False

Answer: b

Explanation: Friss transmission is used to find the receiver power in antenna when power is transmitted from another antenna. These are separated by a far zone distance.

2. What is the distance between antennas to apply the Friss transmission equation in terms of antennas largest dimension?

a) R » 2D 2 /λ

b) R « 2D 2 /λ

c) R » 2λ 2 /D

d) R « 2λ 2 /D

Answer: a

Explanation: The transmitting and receiving antennas are in a far zone to each other. So the separation distance between them is R » 2D 2 /λ.

3. Free space loss factor is given by _____

a) \

 

 \

 

^2\)

c) \

 

 \

 

^2\)

Answer: b

Explanation: The free space loss factor is given by \

 

^2\). It is used to know the amount of losses occurred due to the spreading of energy by an antenna.

4. Which of the following is the Friss transmission equation for the matched polarization of antennas?

a) \

 

 

 \

 

 

 \

 

 

 \(\frac{P_t}{P_r} = \frac{G_t G_r\lambda^2}{4πR^2}\)

Answer: a

Explanation: Friss transmission equation is used to calculate the power received by the receiving antenna when transmitted from other antenna separated by a distance R. the equation is given by \(\frac{P_r}{P_t} = \frac{G_t G_r \lambda^2}{^2}.\)

5. If the operating frequency increases, powers received by the receiving antenna ______

a) will decrease

b) will Increase

c) is Independent of frequency

d) is not predictable

Answer: a

Explanation: From the Friss transmission equation, the received power depends on the wavelength which is inversely proportional to the frequency. So the power decreases as the frequency increases.

\(\frac{P_r}{P_t} = \frac{G_t G_r λ^2}{^2} = \frac{G_t G_r c^2}{^2} \)

6. Power received by the antenna when one antenna is horizontally polarized and the other is vertically polarized is _______

a) 1

b) 0

c) \

 

 

 \(\frac{P_r}{P_t} = \frac{G_t G_r λ^2}{2^2}\)

Answer: b

Explanation: When the receiving and transmitting antennas polarization is not matched, the Friss transmission equation includes a polarization loss factor given by cos 2 θ. Since one is vertically polarized and other is horizontally polarized, the angle difference is 900. PLF=cos 2 θ=0

∴ \(\frac{P_r}{P_t} = PLF\frac{G_t G_r \lambda^2}{^2} =0\)

So, no power is received.

7. Find the power received by the receiving antenna if it is placed at a distance of 20m from the transmitting antenna which is radiating 50W power at a frequency 900MHz and are made-up of half-wave dipoles.

a) 23.65μW

b) 2.365μW

c) 236.5μW

d) 4.73μW

Answer: c

Explanation: given d=20m, P t =50W and f=900MHz

Gain of half-wave dipoles is 1.64

\(λ = \frac{c}{f} = \frac{3×10^8}{900Mhz} = \frac{1}{3} m \)

\(\frac{P_r}{P_t} = \frac{G_t G_r \lambda^2}{^2} = \frac{1.64×1.64×1/3^2}{^2}\)

P r =236.5μW

8. Let’s assume a transmitting antenna having gain 10dB is placed at a distance of 100m from the receiving antenna and radiates a power of 5W. Find the gain of the receiving antenna in dB when the received power is 150μW and transmitter frequency 500MHz?

a) 1.31dB

b) 1.19dB

c) 11.19dB

d) 13.16dB

Answer: c

Explanation:

Given P t =5W, P r =150μW, f=500MHz, R=100m and G t in dB=10dB

G t in dB=10log 10 G t =10dB

G t =10

⇨ \(\lambda = \frac{c}{f} = \frac{3×10^8}{500Mhz} = 0.6m\)

From Friss transmission equation, \(\frac{P_r}{P_t} = \frac{G_t G_r \lambda^2}{^2} \)

⇨ \(G_r=\frac{P_r ^2}{P_t G_t \lambda^2} = \frac{150\mu^2}{5×10×0.6^2}=13.16\)

⇨ G r in dB=10log 10 G r =10log 10 13.16=11.19dB

9. If the distance between the transmitting and receiving antenna is decreased by a factor 2 while other factors remain same, then the new power received by the antenna _______

a) increases by factor 2

b) decreases by factor 2

c) increases by factor 4

d) decreases by factor 4

Answer: c

Explanation: From Friss transmission equation, \(P_r=P_t\frac{G_t G_r \lambda^2}{^2}\)

\(\frac{P_{r1}}{P_{r2}} = \frac{R_2^2}{R_1^2} = \frac{^2}{R^2} = \frac{1}{4}\)

P r2 =4P r1 .

10. Assume two similar antennas for transmitting and receiving. If the operating frequency gets reduced by 3 times then the received power gets _______

a) increases by factor 3

b) decreases by factor 3

c) increases by factor 9

d) decreases by factor 9

Answer: c

Explanation: From Friss transmission equation,

\(\frac{P_r}{P_t} = \frac{G_t G_r λ^2}{^2} = \frac{G_t G_r c^2}{^2}\)

\(\frac{P_{r1}}{P_{r2}} = \frac{f_2^2}{f_1^2} = \frac{^2}{f^2} = \frac{1}{9}\)

P r2 =9P r1

This set of Antennas Multiple Choice Questions & Answers  focuses on “Antenna Noise Temperature”.

1. Relation between brightness temperature T B and physical body temperatureT p is ____

a) T B =\

 T_p\)

b) T B =\

\)

c) T B =\

T_p\)

d) T B =\

^2 T_p\)

Answer: a

Explanation: The relation between brightness temperature and the physical body temperature is given by T B =\

 T_p\)

Here Γ s is the reflection coefficient for a given polarization and emissivity =\(1-\mid\Gamma_s \mid^2.\)

2. If the reflection co-efficient is ½ then emissivity is ___

a) 3/4

b) 1/4

c) 1/2

d) 3/2

Answer: a

Explanation: Emissivity in terms of reflection coefficient is given by \(\epsilon=1-\mid\Gamma_s\mid^2=1-\frac{1}{4}=\frac{3}{4}.\)

3. Overall receiver noise temperature expression if T 1 , T 2 … are amplifier 1, 2, and so on noise Temperature and G 1 , G 2 , and so on are their gain respectively is_____

a) T = \

 

 

 T = T 1 +T 2 (1-G 1 )+T 3 (1-G 1 G 2 )+⋯

c) T = \

 

 

 T = T 1 +T 2 (G 1 )+T 3 (G 1 G 2 )+⋯

Answer: a

Explanation: Overall receiver noise temperature expression is given by T = \(T_1+\frac{T_2}{G_1}+\frac{T_3}{G_1 G_2}+⋯ \) System Temperature is one of the important factors to determine the antenna sensitivity and SNR.

4. Total noise power of the system is P=_____

a) k(T A +T R )B

b) k(T A +T R )/B

c) k(T R )B

d) kB/T sys

Answer: a

Explanation: The overall noise temperature of the system is the sum of noise temperature of antenna T A and the receiver surrounding T R .

⇨ Total noise power of the system is P= k(T A +T R )B

⇨ K is Boltzmann’s constant and B is the bandwidth

5. What is the relation between noise temperature introduced by beam T B and the antenna temperature T A when the solid angle obtained by the noise source is greater than antenna solid angle?

a) T A = T B

b) T A > T B

c) T A < T B

d) T A « T B

Answer: a

Explanation: When the solid angle obtained by the noise source Ω B is greater than antenna solid angle Ω A , then relation between noise temperature introduced by beam T B and the antenna temperatureT A is given by

T A = T B .

For radio astronomy, Ω B <Ω A and T A ≠ T B ; ΔT A =\(\frac{\Omega_B}{\Omega_A}T_B\)

6. Which expression suits best when the solid angle obtained by the noise source is less than antenna solid angle?

a) P A Ω A =P B Ω B and ΔT A =\

 

 P A Ω B =P B Ω A and ΔT A =\

 

 ΔT A =\(\frac{\Omega_A}{\Omega_B} T_B\) and P A Ω B =P B Ω A

d) ΔT A =\(\frac{\Omega_A}{\Omega_B} T_B\) and P A Ω A =P B Ω B

Answer: a

Explanation: When the solid angle obtained by the noise sourceΩ B is greater than antenna solid angle Ω A , then relation between noise temperature introduced by beam T B and the antenna temperatureT A is given by

T A = T B .

For radio astronomy, Ω B <Ω A and T A ≠ T B ; ΔT A =\(\frac{\Omega_B}{\Omega_A} T_B\) and P A Ω A =P B Ω B

7. Expression for noise figure F related to the effective noise temperature T e is ____

a) \

 

 \

 

 \

 

 \(F=1-\frac{T_0}{T_e}\)

Answer: a

Explanation: The noise introduced by antenna is known as the effective noise temperature. The relation between noise figure and effective noise temperature is given by

\(F=1+\frac{T_e}{T_o}, T_o\) is the room temperature.

8. Effective noise temperature T e in terms of noise figure is ____

a) T e =T o 

b) T e =T o /

c) T e =T o /

d) T e =T o 

Answer: a

Explanation: The relation between noise figure and effective noise temperature is given by F=1+\(\frac{T_e}{T_o}\)

⇨ F-1=\(\frac{T_e}{T_o}\)

⇨ T e =T o 

9. Which of the following statement is false?

a) Noise power of antenna depends on the antenna temperature as well as the noise due to the receiver surroundings

b) Noise figure value lies between 0 and 1

c) Any object with physical temperature greater than 0K radiates energy

d) Noise power per unit bandwidth is kT A W/Hz

Answer: b

Explanation: The relation between noise figure and effective noise temperature is given by F=1+\(\frac{T_e}{T_o}\)

And object with physical temperature greater than 0K radiates energy. So \(\frac{T_e}{T_o}\) > 0 and F > 1

Noise power per unit bandwidth of antenna is kT A W/Hz while noise power of antenna is kT A B W

10. Find the effective noise temperature if noise figure is 3 at room temperature ?

a) 290K

b) 580K

c) 289K

d) 195K

Answer: b

Explanation: Room temperature T o =290K

Noise figure F=1+\(\frac{T_e}{T_o}\)

T e =T o =290=580K.

11. What should be the noise figure value at which the effective noise temperature equals to room temperature?

a) 2

b) 1

c) 0

d) 1/T_

Answer: a

Explanation: Noise figure F=1+\(\frac{T_e}{T_o}\)

T e =T o 

F-1=1

F=2.

This set of Antenna Parameters Interview Questions and Answers for Experienced people focuses on “Power Radiation from Half Wave Dipole”.

1. If the current input to the antenna is 100mA, then find the average power radiated from the half-wave dipole antenna?

a) 365mW

b) 0.356mW

c) 0.365mW

d) 356mW

Answer: a

Explanation: Average Power radiated from the half-wave dipole P avg =\

 

^2 R \)

Radiation resistance of a half-wave dipole is 73Ω.

Given I m =100mA => P avg =\

 

^2×73=365mW.\)

2. The average radiated power of half-wave dipole is given by ______

a) \

 \

 \

 \(146I_{rms}^2\)

Answer: a

Explanation: Radiation resistance of a half-wave dipole is 73Ω.

Average Power radiated from the half-wave dipole \(P_{avg}=I_{rms}^2 R=73I_{rms}^2\)

Radiation resistance of a quarter-wave monopole is 36.5Ω.

3. If the power radiated by a quarter-wave monopole is 100mW then power radiated by a half wave dipole under same current input is _____

a) 100W

b) 100mW

c) 200W

d) 200mW

Answer: d

Explanation: Average Power radiated from the half-wave dipole \

⇨

 

 

 

⇨ P avg-hlf = 2P avg mono = 2*100mW=200mW

4. Power radiated by a half wave dipole is how many times the power radiated by a quarter wave monopole under same input current to antennas?

a) 2

b) 3

c) 4

d) 1

Answer: a

Explanation: Average Power radiated from the half-wave dipole \

⇨

 

 

 

5. Find the magnetic field if the electric field radiated by the half-wave dipole is 60mV/m?

a) 159μA/m

b) 195μA/m

c) 159mA/m

d) 195mA/m

Answer: a

Explanation: η=\(\frac{E}{H}\)

⇨ 120π=60m/H

⇨ H = 159μA/m

6. In which of the following the power is radiated through a complete spherical surface?

a) Half-wave dipole

b) Quarter-wave Monopole

c) Both Half-wave dipole & Quarter-wave Monopole

d) Neither Half-wave dipole nor Quarter-wave Monopole

Answer: a

Explanation: In a half-wave dipole the power is radiated in the entire spherical surface and in quarter wave monopole the power is radiated only through a hemispherical surface. Hence its radiation resistance is also twice that of the quarter wave monopole.

7. Find the power radiated from the half wave dipole at 2km away with magnetic field at point \

 

 0.576mW

b) 0.576W

c) 0.756W

d) 0.675W

Answer: b

Explanation: Magnetic field strength \

 

 

 

\)

⇨ 10×10 -6 =\

 

 

 

 

 

\)

⇨ I m =0.125A

Now Average power radiated \

 

^2 R=

 

^2 ×73×=0.576W\)

8. For the same current, the power radiated by half-wave dipole is four times that of the radiation by quarter wave monopole.

a) True

b) False

Answer: b

Explanation: The radiation resistance of a half wave dipole is 73Ω and that of a quarter wave monopole is 36.5Ω. So the power radiated by half-wave dipole is two times that of the radiation by quarter wave monopole.

9. If the power radiated by a half wave dipole is 100mW then power radiated by a quarter wave monopole under same current input is _____

a) 50mW

b) 200mW

c) 100mW

d) 50W

Answer: a

Explanation: Average Power radiated from the half-wave dipole \

⇨

 

 

 

\(P_{avg \,mono}=\frac{P_{avg-hlf}}{2}=\frac{100mW}{2}=50mW.\)

10. If the power radiated by a quarter wave monopole is 100mW, then power radiated by a half wave dipole with doubled current is ______

a) 800mW

b) 400mW

c) 200mW

d) 100mW

Answer: a

Explanation: Average Power radiated from the half-wave dipole \

⇨

 

 

 

P avg-hlf =8(P avg mono )=800mW

This set of Antennas Quiz focuses on “Half Power Beam Width”.

1. Beam area and directivity are related as _____

a) \

 

 \

 

 \

 

 \(D=\frac{B}{4\pi}\)

Answer: b

Explanation: Beam width/ beam area is the measure of directivity of an antenna.

\(D=\frac{4\pi}{B}\)

2. What is the Beam area in radians when the HPBW in perpendicular planes are given by 60°, 30°?

a) \

 

 \

 

 \

 

 2

Answer: c

Explanation: Beam area B=θ E θ H =60°×30°=\(\frac{\pi}{3}×\frac{\pi}{6}=\frac{\pi^2}{18}\)

3. Beam width is the measure of directivity of antenna in terms of angles either radians or degrees.

a) True

b) False

Answer: a

Explanation: Beam width is defined as the measure of angular width at two points on the major lobe where the power decreases to half of the maximum radiated power. So beam width is the measure of directivity of antenna.

\(D=\frac{4\pi}{B}\)

4. Find the directivity when the half power beam widths are 45° and 60° in perpendicular planes.

a) 12.58

b) 17.98

c) 22.91

d) 15.28

Answer: d

Explanation: Directivity in terms of HPBW  is given by \(D=\frac{41257}{θ_E θ_H} =\frac{41257}{45° × 60°}=15.28\)

5. What happens to the beam-width of antenna if the frequency of operation is increased?

a) Beam-width will decrease

b) Beam-width will increase

c) Beam-width is independent of frequency

d) Beam width may increase or decrease

Answer: a

Explanation: If frequency is high, power radiated is low . That means directivity is high as \(D=\frac{4\pi U}{P_{rad}}.\) Beam width and directivity are inversely proportional to each other. So beam width will be low.

6. Half-power beam width is also called as 6-db beam width.

a) True

b) False

Answer: b

Explanation: Half-power beam width is also called as 3-db beam width. It the angular width calculated between the points when the power drops to half of the maximum power radiated.

This set of Antennas Multiple Choice Questions & Answers  focuses on “First Null Beam Width”.

1. Angular width between the first nulls or first side lobes is called as _______

a) half power beam width

b) full null beam width

c) beam area

d) directivity

Answer: b

Explanation: Angular width between the first nulls or first side lobes is called full null beam width. Half power beam width is the angular width measured between the 3dB power points of the major lobe. Beam area is the product of HPBW in perpendicular directions. Directivity is the maximum directive gain.

2. If the HPBW is 30° then FNBW is approximately _____

a) 60°

b) 30°

c) 15°

d) 20°

Answer: a

Explanation: FNBW ≈ 2HPBW=2×30°=60°

3. Total beam area is sum of major and minor lobe areas.

a) True

b) False

Answer: a

Explanation: Total beam area is sum of major lobe area Ω M and minor lobe area Ω m .

Ω A = Ω M +Ω m

4. If beam efficiency is 0.87 then the stray factor is ____

a) 1.87

b) 0.13

c) 1.30

d) 0.87

Answer: b

Explanation: Stray factor ∈ s is the ratio of minor beam area to total beam area.

Given beam efficiency ∈ B =0.87

∈ B +∈ s = 1

∈ s = 1-0.87 = 0.13

5. If the antennas revolution time is 10 sec and 3 dB beam width duration is 150ms then the antenna beam width is _______

a) 5.4°

b) 54°

c) 15°

d) 1.5°

Answer: a

Explanation: Antenna beam width = \(\frac{Beam \,duration}{rotating \,period}×360° = \frac{150m}{10} ×360°=5.4°\)

6. Relation between beam efficiency and stray factor is given by ____

a) ∈ B +∈ s =1

b) ∈ B -∈ s =1

c) \

 

 ∈ B +∈ s =2

Answer: a

Explanation: Stray factor ∈ s is the ratio of minor beam area to total beam area. Beam efficiency ∈ B is the ratio of major beam area to total beam area. Total beam area is sum of major lobe area Ω M and minor lobe area Ω m .

\(\in_B=\frac{\Omega_M}{\Omega_A}, \in_s = \frac{\Omega_m}{\Omega_A}\)

∈ B +∈ s =1

This set of Antennas Multiple Choice Questions & Answers focuses on “Radiation – Basic Maxwell Equations”.

1. The Maxwell equation ∇×E=\

 

 Amperes law

b) Faradays Law

c) Lens law

d) Gauss law

Answer: b

Explanation: Faradays law states that emf generated around a loop of wire in magnetic field is proportional to the rate of change of time-varying magnetic field through the loop.

Amperes law gives ∇×H=J

Lens law gives only the reason for the negative sign in the Faradays law of induction.

Gauss’s law states that the net flux of an electricfield in a closed surface is directly proportional to the enclosed electric charge.

2. The minus sign in the Faradays law of induction is given by ______

a) Lens Law

b) Gauss law

c) Amperes Law

d) Gauss law

Answer: a

Explanation:Lens law gives only the reason for the negative sign in the Faradays law of induction

emf=\(-\frac{\partial \phi}{\partial t}\)

The minus sign indicates the direction of induced current.

3. Which of the following Maxwell equation is obtained from Amperes law?

a) ∇×H=J

b) emf=\

 

 ∇×E=\

 

 ∇×D=ρ v

Answer: a

Explanation:

Faradays Law                : ∇×E=\(-\frac{\partial B}{\partial t}\)

Amperes Law                : ∇×H=J+\(\frac{\partial D}{\partial t}\)

Gauss Law for electric field     : D=ρ v

Gauss law for magnetic field  : ∇.B=0

4. Gauss for the Magnetic Field is given by ______

a) ∇.B=0

b) ∇×B=ρ v

c) ∇×B=0

d) ∇.B=ρ v

Answer: a

Explanation: Gauss law for magnetic field states that the net flux out of any closed surface is zero.

∇.B=0

This Maxwell equation is one of the equation used to determine the boundary conditions.

5. Gauss for the Electric Field is given by ______

a) ∇.D=0

b) ∇×D=ρ v

c) ∇×D=0

d) ∇.D=ρ v

Answer: d

Explanation:Gauss’s law for electric field states that the net flux of an electricfield in a closed surface is directly proportional to the enclosed electric charge. ∇.D=ρ v

This Maxwell equation is one of the equation used to determine the boundary conditions.

6. Which of the Following Maxwell equation is for nonexistence of isolated magnetic charge?

a) ∇×E=-\

 

 ∇×H=J

c) ∇.D=ρ v

d) ∇.B=0

Answer: d

Explanation: Gauss law for magnetic field states that the net flux out of any closed surface is zero.

∇.B=0

This is satisfied only when two different poles of magnet exist. So this Maxwell equation proves for the nonexistence of the isolated magnetic charge.

Faradays Law : ∇×E=\(-\frac{\partial B}{\partial t}\)

Amperes Law : ∇×H=J

Gauss Law for electric field : ∇.D=ρ v

7. In which of the following Integral form of Maxwell equations, the surface is closed?

a) Amperes law

b) Gauss Law

c) Faradays Law

d) Both Amperes and Faraday law

Answer: b

Explanation: The surface integral is closed for the Gauss laws of magnetic and electric fields. It is open for the amperes and Faradays law.

Maxwell Equations:

Gauss law electric field : \(\oint_sD.ds =\int_v\rho_v dv \)

Gauss law magnetic field : \(\oint_sB.ds =0\)

Faradays law : \(\int_cE.dl =-\int_s\frac{\partial B}{\partial t}.dS\)

Amperes law : \

 

.dS\)

8. Divergence of Magnetic field is ______

a) volume charge density ρ v

b) zero

c) infinite

d) dependent on magnetic field vector

Answer: b

Explanation: The Divergence of Magnetic is always zero.It is obtained from the Maxwell equation ∇.B=0 which is derived from the Gauss law of magnetic field.Gauss law for magnetic field states that the net flux out of any closed surface is zero. ∇.D=ρ v .

9. Which of the following Maxwell equation is correct for a non-conducting and lossless medium?

a) ∇.D=ρ v

b) ∇.D=0

c) ∇×D=ρ v

d) ∇×E=0

Answer: b

Explanation: Since it is given non-conducting medium, the charge density ρ v =0 and current density J=0. The Maxwell equations are:

Faradays Law                : ∇×E=\(-\frac{\partial B}{\partial t}\)

Amperes Law                : ∇×H=\(\frac{\partial D}{\partial t}\)

Gauss Law for electric field     : ∇.D=0

Gauss law for magnetic field  : ∇.B=0

10. Find skin depth of 5GHz for silver with a conductivity 6.1×10 7 s/m and relative permittivity 1.

a) 0.00091m

b) 0.9113μm

c) 0.319μm

d) 0.1913μm

Answer: b

Explanation: The skin depth is given by \(δ = \sqrt{\frac{1}{\pi fμσ}}\)

Given f=5GHz

Conductivity σ= 6.1×10 7 s/m

And μ r = 1 =>μ=4π ×10 -7

⇨ \(δ = \sqrt{\frac{1}{\pi fμσ}}=0.9113\mu m.\)

This set of Antennas MCQs focuses on “Radiation – Hertzian Dipole”.

1. Hertzian dipole carries which type of current throughout its length while radiating?

a) Varying

b) Constant

c) Depends on type of polarization

d) Depends on radiation resistance

Answer: b

Explanation: Hertzian dipole is a short linear antenna which carries a constant current throughout its length while radiating. It consists of two equal and opposite charges separated by a very short distance. It is infinitesimal current element.

2. Power radiated by a Hertzian dipole of length λ/30 and carrying a current 2A?

a) 0.87W

b) 3.51W

c) 2.51W

d) 8.77W

Answer: b

Explanation: Power radiated by a Hertzian dipole P rad =R rad I 2

R rad = \

 

^2=80\pi^2

 

 

^2=0.877\Omega\)

P rad = R rad I 2 =0.877×2×2=3.51W

3. A Hertzian dipole consists of two _____ and ______ charges separated by a very short distance.

a) unequal, opposite

b) equal, same

c) equal, opposite

d) unequal, same

Answer: c

Explanation: A Hertzian dipole consists of two equal and opposite charges separated by a very short distance. It is infinitesimal current element. It is a short linear antenna which carries a constant current throughout its length while radiating.

4. When Hertzian dipole is connected to a practical antenna, which of the following fields is observed to be absent when a uniform current flow is observed?

a) Radiation field

b) Induction field

c) Electrostatic field

d) Both radiation and Induction Field

Answer: c

Explanation: Since a constant current flow and there is no any charge accumulation at the ends of the dipole, the term 1/r 3 disappears. Therefore, electrostatic field is absent.

5.Which of the following is the radiation resistance of the Hertzian dipole?

a) \

 

 \

 

 \

 

 \(\frac{\eta_0 w^3 dl^2}{3\pi c^3}\)

Answer: a

Explanation: Radiation resistance of a Hertzian dipole is \

 

^2\)

By simplifying the options given above,

\

 

 

 

^2\)

6. If the radiation resistance of a Hertzian dipole is 100Ω, then the radiation resistance of short dipole is ____Ω.

a) 25

b) 50

c) 73

d) 35.6

Answer: a

Explanation: The radiation resistance of the short dipole is ¼ times the radiation resistance of a current element. So 100/4= 25Ω.

7. The radiation resistance of a monopole of height 1cm and operating at frequency 100MHz is ____ Ω.

a) 4.83m

b) 4.38k

c) 4.38m

d) 4.83k

Answer: c

Explanation: The radiation resistance of a monopole is 1/8 times the current element.

\

^2\)

For a monopole height h= l/2 => l= 2h

\

 

^2=40\pi^2 

 

^2=40\pi^2 

 

^2=4.38m\Omega\)

8. The radiation resistance of a monopole is _____ times the current element.

a) 1/8

b) 1/4

c) 1/2

d) 1/16

Answer: a

Explanation: The radiation resistance of monopole is ½ times the short dipole. But the radiation resistance of short dipole is ¼ time the current element.

\

 

 

 

^2=\frac{1}{8}×80\pi^2

 

^2=10\pi^2 

 

^2=\frac{R_{rad \,Herztian}}{8}\)

The radiation resistance of monopole is 1/8 times the current element.

9. Practically we don’t use Hertzian dipole.

a) True

b) False

Answer: a

Explanation: Since the current distribution at the center is maximum and minimum at ends, there is no uniform distribution of current along length. But Hertzian dipole is derived by assuming a uniform current distribution along length and having infinitesimal length. So that is reason why we don’t use Hertzian dipole practically.

10. If the radiation resistance of a monopole is 18Ω, then the radiation resistance of a Hertzian dipole is _____________

a) 124Ω

b) 144Ω

c) 164Ω

d) 154Ω

Answer: b

Explanation: The radiation resistance of monopole is 1/8 times the current element.

R rad Herztian = 8×R rad mono =8×18=144Ω

This set of Antennas Multiple Choice Questions & Answers  focuses on “Baluns”.

1. What is a Balun?

a) It is used to balance the unbalanced systems

b) It unbalances the balanced systems

c) A twisted wire

d) Main beam of antenna with large beam width

Answer: a

Explanation: A Balun is a device which connects a balanced two –conductor line to an unbalanced coaxial line. It eliminates field mismatch. The current distribution is present in the inner conductor and is zero at the outer conductor.

2. A Balun is used to make the current along the outer side of the outer conductor along the coaxial cable _______

a) Zero

b) Maximum

c) Maximum or minimum depends on power supply

d) Infinity

Answer: a

Explanation: When a coaxial cable is connected to a half wave dipole, the current distribution along the outer side of the outer conductor may also present. It results in loss and field mismatch. So a Balun is used to make the current at the outer side of the outer conductor zero. Current distribution is present at the inner side of the outer conductor.

3. The process of forcing the current at the outer side of the outer conductor to be zero is called _____

a) Current distribution

b) Current chokes

c) Field effect

d) Impedance chokes

Answer: a

Explanation: A Balun is a device which connects a balanced two –conductor line to an unbalanced coaxial line. It eliminates field mismatch. It forces the current at the outer side of outer conductor to be zero. This is called current choke.

4. A Balun joins a balanced line and an unbalanced line.

a) True

b) False

Answer: a

Explanation: A Balun joins a balanced line which is usually a two conductor’s twisted cable with equal current distribution to an unbalanced line with one conductor with unequal current distribution and vice-versa. It acts like a transformer.

5. All Baluns provide impedance transformation.

a) True

b) False

Answer: b

Explanation: A 1:1 Balun doesn’t provide any impedance transformation. Some Baluns provide impedance transformation like 1: 4 or 9:1.

6. For a 4: 1 Balun, what is the unbalanced impedance if the balanced impedance is 2KΩ?

a) 8K

b) 0.5K

c) 4K

d) 2K

Answer: b

Explanation: A 4: 1 Balun implies its impedance ratio. \(\frac{Z_{bal}}{Z_{unbal}} = \frac{4}{1} \)

Unbalanced impedance = 2K/4=0.5K.

7. Among current Balun and voltage Balun, which works better?

a) Current Balun

b) Voltage Balun

c) Both work equally

d) Depends the power supply

Answer: a

Explanation: A current Balun offers better balance and can tolerate load impedances and balance variations better compared to Voltage Balun. In a current Balun the output terminal voltage can be of any value to make the currents equal in the feed line.

8. Which of the following is false regarding Transformer Balun?

a) It has a narrow band frequency

b) It has infinitely wide band frequency

c) Ideally it provides zero insertion loss

d) Impedance matching is adjusted by its turn ratio

Answer: a

Explanation: A transformer Balun has ideally infinite wide band frequency and provides zero insertion loss. By changing the number of turns we can modify the impedance only when the input and output impedances are only resistive.

9. The phase difference between the outputs of Balun in frequency domain is ____

a) 180°

b) 120°

c) 60°

d) 90°

Answer:a

Explanation: A Balun can be viewed as a three port device, with matched input and a differential output. The differential outputs are equal and opposite. So, they are 1800 out of phase with respect to each other.

10. For an unbalanced to balanced signal conversion, if the turn’s ratio in a Balun is 1:2 then the balanced impedance is ______ times the unbalanced impedance.

a) 4

b) 2

c) \

 

 \(\frac{1}{4} \)

Answer: a

Explanation: A Balun can be viewed as a transformer. \

 

^2=

 

^2=\frac{1}{4} \)

⇨ Z bal = 4Z unbal

So the balanced impedance is four times the unbalanced impedance.

This set of Antennas Question Bank focuses on “Types of Baluns”.

1. Which of the following statement is false regarding Type-1 Sleeve Balun?

a) It is shorted at the base

b) It has a λ/4 sleeve

c) It has a λ/2 sleeve

d) Ideally it provides infinite impedance at top

Answer: c

Explanation: Sleeve Balun has a λ/4 sleeve which is shorted at base electrically and it provides infinite impedance at the top.

2. Consider a 9:1 Balun ; if the unbalanced impedance is RΩ then the balanced impedance will be _____Ω.

a) 9R

b) 3R

c) 0.11R

d) 0.45R

Answer: a

Explanation: A 1:9 Balun represents that the balanced impedance is 9 times the unbalanced impedance.

⇨ Balanced impedance =9×R=9R

3. Which of the following Balun changes the shape of the unbalanced transmission line to that of a balanced transmission line?

a) Sleeve Balun

b) Folded Balun

c) Tapered Balun

d) Infinite Balun

Answer: c

Explanation: Tapered Balun changes the shape of the unbalanced transmission line to that of a balanced transmission line. Sleeve Balun uses a λ/4 sleeve which is shorted at base electrically and it provides infinite impedance at the top. Infinite Balun uses the current flowing outside the conductor as a part of antenna.

4. In which of the following Balun we don’t choke the current?

a) Sleeve Balun

b) Folded Balun

c) Tapered Balun

d) Infinite Balun

Answer: d

Explanation: Infinite Balun uses the current flowing outside the conductor as a part of antenna. In this we don’t choke the current, but we use it as a part of antenna, where we want the current flow in it.

5. Which of the following Balun contains a sliding short circuit bar for frequency adjustment?

a) Type – 1

b) Sleeve Balun

c) Type – 2

d) Type – 3

Answer: d

Explanation: Type – 3 Baluns has a sliding short circuit bar for frequency adjustment. Sleeve Balun or Type -1 Balun has a λ/4 sleeve which is shorted at base electrically.

6. Type II Balun has two Type I balloons in series providing more bandwidth and load impedances at all frequencies.

a) True

b) False

Answer: a

Explanation: Type II Balun has 2 Type I Baluns in series providing more bandwidth and load impedances at all frequencies. Type -1 Balun has a λ/4 sleeve which is shorted at base electrically.

7. Reciprocity applies to which of the following devices?

a) Isolator

b) Circulator

c) Balun

d) RF amplifiers

Answer: c

Explanation: In a Balun, a 3 port network, contains one matched input and two differential outputs. It is used to convert the unbalanced signal to balanced and vice-versa. So it is reciprocal. Isolator, Circulator, RF amplifiers are Non reciprocal devices.

8. For a Balun to work effectively, it must have ____ impedance for common mode currents and ____ impedance for differential mode current.

a) high, low

b) low, high

c) high, high

d) low, low

Answer: a

Explanation: Since Balun have to minimize the common mode currents it should have High impedance.

For differential mode currents it must have low impedance.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Types of Polarization”.

1. Which of the following polarization is used in monopole antenna?

a) Right-hand Circular

b) Linear

c) Depends on the feed

d) Left-hand Circular

Answer: b

Explanation: The linear polarization takes place in the monopole dipole. In the Dipole also a linear polarization takes place. Parabolic reflectors take the polarization of the feed. In helical, circular loop antennas we can find the circular polarization.

2. Which type of polarization is found in parabolic reflector?

a) Right-hand Circular

b) Linear

c) Depends on the feed

d) Left-hand Circular

Answer: c

Explanation: The parabolic antennas take the polarization of the feed. The linear polarization takes place in the monopole dipole. In the Dipole also a linear polarization takes place. In helical, circular loop antennas we can find the circular polarization.

3. In which of the following polarization the electric field components are perpendicular to each other and have equal magnitude?

a) Linear

b) Vertical

c) Circular

d) Elliptical

Answer: c

Explanation: In linear components are in the same plane. Vertical and horizontal are types in linear polarization. Circular polarization has electric field components are perpendicular to each other and have equal magnitude. In elliptic, the electric field components are perpendicular to each other and have unequal magnitude.

4. In which of the following polarization the electric field components are perpendicular to each other and have unequal magnitudes?

a) Linear

b) Vertical

c) Circular

d) Elliptical

Answer: d

Explanation: In linear components are in the same plane. Vertical and horizontal are types in linear polarization. Circular polarization has electric field components are perpendicular to each other and have equal magnitude. In elliptic, the electric field components are perpendicular to each other and have unequal magnitude.

5. Tilt angle of the elliptic polarization with respect to horizontal is ____________

a) τ=\

 

\)

b) τ=\

 

 τ=\

 

\)

d) τ=\

 

\)

Answer: a

Explanation: The two properties of ellipse that relates to the polarization are — eccentricity and tilt or inclination angle with respect to horizontal. Tilt angle of the elliptic polarization is given by

τ=\

 

\)

6. The locus traced by the extremity of the time-varying field vector at a fixed observational point is called ________

a) polarization

b) gain

c) directivity

d) height

Answer: a

Explanation: The locus traced by the extremity of the time-varying field vector at a fixed observational point is called polarization. Gain is the output to input power ratio. Directivity is the amount of power radiated in the desired direction.

7. Which of the following is true for the circular polarization?

a) E x =E y , and ∅=π/2

b) E x ≠E y , and ∅=π/2

c) E x ≠E y , and ∅=π/4

d) E x =E y , and ∅=π/4

Answer: a

Explanation: The locus traced by the extremity of the time-varying field vector at a fixed observational point is called polarization. Circular polarization has electric field components are perpendicular to each other and have equal magnitude. E x =E y , and ∅=π/2.

8. Which of the following is true for the elliptical polarization?

a) E x =E y , and ∅=π/2

b) E x ≠E y , and ∅=π/2

c) E x ≠E y , and ∅=π/4

d) E x =E y , and ∅=π/4

Answer: b

Explanation: The locus traced by the extremity of the time-varying field vector at a fixed observational point is called polarization. In elliptical polarization, the electric field components are perpendicular to each other and have unequal magnitude.

E x ≠E y ,and ∅=π/2

9. The transmission mode polarization vector and receiving mode polarization vector of antenna polarization are ___________

a) always equal

b) conjugate to each other

c) negative of conjugate of other

d) inverse of Conjugate of other vector

Answer: b

Explanation: The antenna polarization is defined by the polarization vector it transmits. In a common coordinate system, the transmission mode polarization vector is the conjugate of its receiving mode polarization vector.

10. The linear and circular polarizations are special cases of elliptical polarization.

a) True

b) False

Answer: a

Explanation: In Elliptical polarization, if the amplitude is made equal, then it becomes a circular polarization. If the phase difference of the two linear components is zero or nπ then it is becomes a linear polarization. Therefore linear and circular polarization is a special case of elliptical polarization.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Polarization Mismatch”.

1. What is the polarization loss factor due to mismatch if the two linear polarized antennas are rotated by an angle ∅?

a) tan⁡∅

b) cos⁡∅

c) cos 2 ∅

d) cos -1 ⁡∅

Answer: c

Explanation: The polarization loss factor describes the power loss due to polarization mismatch. The polarization loss factor due to mismatch if the two linear polarized antennas are rotated by an angle ∅ is given by PLF =cos 2 ∅.

2. What is the polarization loss factor when a linear polarized antenna receives a circular polarized?

a) 0.5

b) 1

c) 0

d) 2

Answer: a

Explanation: Circular polarization is combination of two orthogonal linear polarized waves at 90° phase difference. The linear component just selects one in-phase component from this so ∅=45

⇨ PLF = cos 2 ∅.=1/2=0.5

3. What is the polarization loss factor when two antennas are rotated by an angle 30°?

a) 0.75

b) 0.25

c) 1

d) 0

Answer: a

Explanation: The polarization loss factor due to mismatch if the two linear polarized antennas are rotated by an angle ∅ is given by PLF = cos 2 ∅

⇨ PLF = cos 2 30=\(\frac{3}{4}=0.75\)

4. Which of the following is best described by the polarization loss factor?

a) Power loss due to mismatch

b) Height of antenna

c) Directivity

d) Power gain

Answer: a

Explanation: The polarization loss factor describes the power loss due to polarization mismatch. The polarization loss factor due to mismatch if the two linear polarized antennas are rotated by an angle ∅ is given by PLF =cos 2 ∅.

5. The polarization of received antenna is not same as incident wave polarization is termed as ___

a) polarization mismatch

b) polarization loss factor

c) directivity

d) transmitter

Answer: a

Explanation: The polarization of received antenna is not same as incident wave polarization is termed as Polarization mismatch. The polarization loss factor describes the power loss due to polarization mismatch. PLF =cos 2 ∅.

6. Which of the following is the condition for no power loss?

a) PLF=1

b) PLF=0

c) PLF=0.5

d) PLF=0.75

Answer: a

Explanation: Polarization loss factor describes the power loss. When PLF = 1 = 0dB indicates total power incident is received by the antenna. Therefore the condition for the no power loss is PLF=1

7. Which of the following limit is correct for the polarization loss factor?

a) 0≤PLF≤1

b) -1≤PLF≤1

c) -1≤PLF≤0

d) 0≤PLF≤∞

Answer: a

Explanation: The polarization loss factor due to mismatch if the two linear polarized antennas are rotated by an angle ∅ is given by PLF = cos 2 ∅.

⇨ Max value of PLF = 1

⇨ Min value of PLF = 0

⇨ 0≤PLF≤1

8. Which of the following holds true for complete polarization mismatch?

a) PLF=1

b) PLF=0

c) PLF=0.5

d) PLF=0.75

Answer: b

Explanation: Polarization loss factor describes the power loss. When PLF=0=∞ dB, the receiver antenna doesn’t receive any incident power so there will be a complete mismatch.

9. Polarization loss factor describes the amount of power loss due to polarization mismatch at the receiving antenna from incident wave.

a) True

b) False

Answer: a

Explanation: Polarization loss factor describes the amount of power loss due to polarization mismatch at the receiving antenna from incident wave. The polarization loss factor due to mismatch if the two linear polarized antennas are rotated by an angle ∅ is given by PLF = cos 2 ∅.

10. Polarization loss factor in dB when two linear polarized antennas are rotated by an angle 45° is ___

a) 3dB

b) -3dB

c) 0.5dB

d) -0.5dB

Answer: b

Explanation: The polarization loss factor due to mismatch if the two linear polarized antennas are rotated by an angle ∅ is given by PLF = cos 2 ∅.

⇨ PLF = cos 2 ∅=cos 2 45=0.5=-3dB.

This set of Antennas Questions and Answers for Campus interviews focuses on “Folded Dipole Antenna – Types”.

1. Which of the following statement is not correct?

a) Folded dipole antenna has less impedance than half dipole

b) Folded dipole is a balanced antenna

c) Folded dipole antenna is a dipole antenna with its end folded back forming a loop

d) Balun is used at the feeder when unbalanced feed is used in the folded dipole

Answer: a

Explanation: The impedance of folded dipole is four times the impedance of the half dipole. So its impedance is higher than the half dipole. Since folded dipole is a balanced antenna we use a balanced feeder so Balun is used.

2. Folded dipole antenna belongs to which type of antenna?

a) Reflector

b) Aperture

c) Lens

d) Wire

Answer: d

Explanation: Folded dipole antenna belongs to wire antenna. It is a dipole antenna with two ends folded back and connected to each other forming a loop.

3. What is the input impedance of a half wave folded dipole?

a) 73Ω

b) 292Ω

c) 146Ω

d) 36.5Ω

Answer: b

Explanation: The input impedance of a half wave folded dipole is four times the half wave dipole.

So Z= 4×73=292Ω

4. A half wave folded dipole of 3 wires has the impedance _____Ω.

a) 675

b) 657

c) 219

d) 292

Answer: b

Explanation: For a half wave folded dipole of 3 wires the impedance = n 2 ×73=9×73=657

5. In which of the following type of folded dipole the number of conductors is varied?

a) Multi conductor folded dipole

b) Unequal conductor folded dipole

c) Both multi conductor and unequal conductor

d) Neither multi conductor nor unequal conductor

Answer: a

Explanation: In Multi conductor folded dipole, the two or more conductors are connected together. This increases the impedance and also the bandwidth.

6. In unequal conductor folded dipole, the diameter of the conductor of feed arm and the non feed arm are unequal.

a) True

b) False

Answer: a

Explanation: In unequal conductor folded dipole, the diameters of the two conductors connecting together are unequal. If the radii of two conductors are r1 , r2  and separated by a distance, then the impedance transformation for folded dipole is given by

\Missing or unrecognized delimiter for \right^2.\)

7. If the radii of two conductors are r1 , r2  and separated by a distance,then the impedance transformation for a half folded dipole is given by _______

a) \Missing or unrecognized delimiter for \right^2\)

b) \Missing or unrecognized delimiter for \right^2\)

c) \Missing or unrecognized delimiter for \right^2\)

d) \Missing or unrecognized delimiter for \right^2\)

Answer: b

Explanation: If the radii of two conductors are r1 , r2  and separated by a distance, then the impedance transformation for folded dipole is given by

\Missing or unrecognized delimiter for \right^2\)

For a half dipole Z’=73Ω

∴ \Missing or unrecognized delimiter for \right^2\)

8. For a multi conductor folded dipole with n conductors, the impedance is ____ Ω.

a) 2 n Z

b) nZ

c) n 2 Z

d) Z/n

Answer: c

Explanation: For a multi conductor folded dipole with n conductors, the impedance is n 2 Z.

Where value of radiation resistance Z depends on the length of the dipole used.

9. Which of the following causes the shortening effect on multi conductor folded dipole?

a) Only thickness of conductors

b) Space between conductors

c) Radiation resistance of the dipole

d) Both the radii and radiation resistance

Answer: a

Explanation: The shortening effect on the multi conductors depends on the thickness of the conductor whereas in wires it depends on the length and the frequency of operation.

10. On which of the following factors does the overall impedance of the unequal conductor folded dipole depends?

a) Only thickness of conductors

b) Space between conductors

c) Radiation resistance of the dipole

d) Radii, radiation resistance and space between conductors.

Answer: d

Explanation: If the radii of two conductors are r1 , r2  and separated by a distance, then the impedance transformation for folded dipole is given by

\Missing or unrecognized delimiter for \right^2\) and Z’ is the radiation resistance of the dipole used .

So the overall impedance depends on radii, radiation resistance of the dipole and space between conductors.

This set of Antennas Assessment Questions and Answers focuses on “Yagi – Uda Antenna”.

1. The directivity of Yagi-Uda antenna is increased by adding ______

a) reflectors

b) driven element

c) directors

d) boom

Answer: a

Explanation: Addition of directors leads to focus the beam in the forward direction. So, directors will increase the gain of antenna. Folded dipole acts like a feed or driven element. Reflectors will increase the directivity of antenna by reflecting all energy towards radiation direction of antenna. Boom is a center rod on which elements are mounted.

2. Directors are used to increase ______ of the Yagi-Uda antenna.

a) Directivity

b) Gain

c) Back lobe

d) Reflection away from the radiation

Answer: b

Explanation: Reflectors will increase the directivity of antenna by reflecting all energy towards radiation direction of antenna. Addition of directors leads to focus the beam in the forward direction. So, directors will increase the gain of antenna.

3. What is the radiation pattern of a Yagi-Uda antenna?

a) Broad-side

b) End-fire

c) Collinear

d) Both Broadside and End-fire

Answer: b

Explanation: Radiation pattern of a Yagi-Uda antenna is end-fire. It has its main beam parallel to the axis of antenna . The addition of directors will increase the gain of antenna while reflectors will increase the directivity of antenna.

4. A narrow beam-width is obtained through a large number of directors.

a) True

b) False

Answer: a

Explanation: Large number of directors increases the direction of radiation towards the desired direction and a narrow beam-width is obtained. It also helps in increasing the directivity of antenna.

5. The dipole to which the power is applied directly from the feeder in the Yagi-Uda antenna is called as _____

a) Director

b) Reflector

c) Driven element

d) Boom

Answer: c

Explanation: The dipole to which the power is applied directly from the feeder in the Yagi-Uda antenna is called driven element. Directors add the field of the driven element and will excite the next parasitic element. Reflectors will increase the directivity of antenna.

6. Folded dipole is used than a single dipole in Yagi-Uda to obtain wider frequency range.

a) True

b) False

Answer: a

Explanation: Folded dipole produces flatter impedance v/s frequency compared to single dipole. So, folded dipole is used in Yagi-Uda to obtain wider frequency range.

7. In which of the following bands Yagi-Uda antenna operates?

a) HF-UFH

b) VLF-MF

c) LF-HF

d) UHF-EHF

Answer: a

Explanation: Yagi-Uda antenna operates mostly in the HF to UFH band frequency. IT ranges from 3MHz to 3GHz.

VLF-MF: 3 kHz to 3MHz

LF-HF: 30 kHz to 30MHz

UHF –EHF: 300MHz to 300GHz.

8. A Yagi-Uda antenna is ____

a) Only a super directive antenna

b) Only a super gain antenna

c) Both super directive and super gain

d) Neither super directive nor super gain

Answer: c

Explanation: A Yagi-Uda antenna provides high directivity by increasing reflectors and gain due to the directors. Directors will increase the forward gain of the antenna. So it is both super directive and super gain antenna.

9. In order to convert the bidirectional dipole to unidirectional system, we use _______

a) Active element

b) Driven element

c) Parasitic element

d) Isolator

Answer: c

Explanation: We use reflectors and directors which are passive elements also known as the parasitic elements to increase the directivity and gain if the antenna.

10. Which of the following will add the field of the driven element and will excite the next parasitic element to increase the gain of the antenna?

a) Director

b) Reflector

c) Active element

d) Boom

Answer: a

Explanation: Directors add the field of the driven element and will excite the next parasitic element. Directors will increase the gain of the antenna in the forward direction. Reflectors will add fields of the driven element in the direction from reflector to driven element.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Construction of Yagi – Uda”.

1. The length of the director compared to the driven element is ______

a) Greater

b) Smaller

c) Independent to each other

d) Depends on the type of driven element

Answer: b

Explanation: The length of the director is less than the driven element. It is shorter than half wavelength of the dipole. Director is used to increase the gain of antenna.

2. The length of the reflector compared to the driven element is ______

a) Greater

b) Smaller

c) Independent to each other

d) Depends on the type of driven element

Answer: a

Explanation: Reflector is used to reflect all the energy towards the radiation direction. It increases the front to back lobe ratio. So always the reflector length is greater than the driven element length.

3. The distance between directors of Yagi-Uda antenna is ____

a) 0.2λ

b) 0.49λ

c) 0.52λ

d) 0.62λ

Answer: a

Explanation: The distance between directors of Yagi-Uda antenna is 0.2λ. The directors will increase the gain.

4. What is the approximate distance between directors for a Yagi-Uda antenna operating at frequency 150MHz?

a) 0.4m

b) 1.6m

c) 2m

d) 4.8m

Answer: a

Explanation: λ = \

 

 

⇨=0.4m

5. If length of the reflector, driven element, director isl r , l de , l d , respectively then the order in the increasing order of their lengths is _____

a) l r , l de , l d

b) l d , l de , l r

c) l de , l r , l d

d) l r , l de , l de

Answer: b

Explanation: The lengths of reflector, driven element, director is in the orderl r > l de > l d .

So, the increasing order is l d , l de , l r .

6. For a feeder length of 0.5λ, which among the following would be the reflector length?

a) 0.525λ

b) 0.425λ

c) 0.23λ

d) 0.35λ

Answer: a

Explanation: The reflector length is approximately 5% greater than the driven element length.

So, l = 0.525λ is correct as it is the only option which is greater than 0.5λ.

7. If the input impedance of a half wave dipole Yagi-Uda is 73Ω, then input impedance of a half wave folded dipole as driven element is ____

a) 73Ω

b) 146Ω

c) 36.5Ω

d) 292Ω

Answer: d

Explanation: Half wave folded dipole impedance is 4 times the half wave dipole.

⇨ Z = 4 × 73 = 292Ω.

8. What is the reflector length  for a three element Yagi-Uda array, operating at a frequency 500MHz?

a) 2

b) 0.2

c) 1

d) 0.1

Answer: c

Explanation: For a 3 element Yagi-Uda array,

Reflector length  = 500/f  = 500/500 = 1 feet

9. What is the driven length  for a three element Yagi-Uda array, operating at a frequency 500MHz?

a) 0.95

b) 0.2

c) 1

d) 2

Answer: a

Explanation: For a 3 element Yagi-Uda array,

Driven element length  = 475/f  = 475/500 = 0.95 feet.

10. What is the Director length  for a three element Yagi-Uda array, operating at a frequency 500MHz?

a) 0.91

b) 1

c) 2

d) 0.5

Answer: a

Explanation: For a 3 element Yagi-Uda array,

Director element length  = 455/f  = 455/500 = 0.91 feet.

11. For a feeder length of 0.5λ, which among the following would be the director length?

a) 0.525λ

b) 0.455λ

c) 0.493λ

d) 0.55λ

Answer: b

Explanation: The director length is approximately 5% less than the driven element length.

⇨ 0.5-) = 0.475

So, l = 0.455λ is correct as it is the only option which is less than 0.475λ.

This set of Antennas Question Paper focuses on “Radiation from Rectangular Aperture”.

1. Which of the following antenna belongs to rectangular aperture?

a) Horn antenna

b) Helical antenna

c) Parabolic antenna

d) Conical antenna

Answer: a

Explanation: The aperture of antenna at the end determines its shape. Horn antenna has a rectangular aperture. Helical, Parabolic, conical antenna have circular apertures. Helical antenna belongs to frequency independent antenna.

2. The radiation pattern of rectangular is similar to line source integrated in two directions

a) True

b) False

Answer: a

Explanation: The radiation pattern for the rectangular distributions is similar to the line source distributions. In this the patterns is calculated by integration in two directions as rectangle have two different lengths in different direction.

3. The total pattern function for rectangular aperture f if f and f are separable is given by ____

a) f=f f

b) f=f+f

c) f=f/f

d) f=f-f

Answer: a

Explanation: The radiation pattern for the rectangular aperture is likely relatable to the line source distributions. If the functions f and f are separable, then total pattern will be the product of the two functions. f=ff.

4. The first-level of the side lobe occurs at ______ dB for a uniform rectangular aperture antenna.

a) -13.26

b) -6.63

c) 3

d) 8.5

Answer: a

Explanation: The aperture of antenna at the end determines its shape. If the field is uniform in amplitude and phase along the rectangular aperture then it is called a uniform rectangular aperture antenna. The first side-lobe occurs at -13.26dB.

5. A rectangular aperture a ×b is placed in xy-plane, The HPBW in H-plane is given by _____

a) 0.886λ/a

b) 0.443λ/a

c) 0.5λ/b

d) λ/b

Answer: a

Explanation: By equating the field in H-plane to half power point

\

 

 

 

\)

Now HPBW = \

 

 \approx 0.886\lambda/a.\)

6. The relation between directivity and the effective aperture of the uniform aperture antenna is given by _____

a) \

 

 \

 

 \

 

 \(D = \frac{4\pi\lambda^2}{A_{eff}}\)

Answer: a

Explanation:For a uniform aperture antenna, the physical and effective apertures are equal.

The relation between directivity and the effective aperture of the antenna is given by

\(D = \frac{4\pi}{\lambda^2}A_{eff}\).

7. Which of the following is used to reduce side lobe levels in aperture antenna?

a) Tapering

b) Increasing the power

c) Using repeaters

d) Reducing power

Answer: a

Explanation: The uniform aperture produces the high SLL under a constant phase amplitude excitation. To reduce this SLL effect, tapering is done. Tapering is done maximum at center and reduces to zero at the edges for anequivalent source distribution.

8. The principle plane pattern of the E-plane of rectangular aperture of a×b is given by F = ______

a) \

 

 \

 

 \

 

 \(\frac{1-sin⁡}{0.5kbsin\theta}\)

Answer: a

Explanation: The principle plane patterns for the uniform rectangular aperture antenna is given by

F = \(\frac{sin⁡}{0.5kbsin\theta}\) and main lobe occurs when θ =0.

9. Tapering is done in order to reduce the side lobe level.

a) True

b) False

Answer: a

Explanation: The uniform aperture produces the high SLL under a constant phase amplitude excitation. To reduce this SLL effect, tapering is done. Tapering is done maximum at center and reduces to zero at the edges for an equivalent source distribution.

10. If the aperture antenna is tapered only in H-plane then which of the following is true compared to uniform non-tapered aperture antenna?

a) Principle patterns in E-plane and H-plane are same in both cases

b) Principle patterns in E-plane and H-plane are different in both cases

c) Principle patterns in E-plane is same and H-plane is different

d) Principle patterns in E-plane is different and H-plane is same

Answer: c

Explanation: Tapering is done to reduce the SLL effect. Since the E-plane is not tapered, its principle pattern is same but in H-plane as the aperture is tapered principle pattern will be different from the uniform aperture antenna.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Uniform Aperture Field”.

1. For aperture antenna to be efficient and have high directivity, its area should be ____________

a) ≥ λ 2

b) ≥ 1/λ

c) ≤ λ 2

d) ≤ λ

Answer: a

Explanation: Antenna with an aperture at the end is known as aperture antenna. Example is waveguide. For aperture antenna to have high directivity its area should be ≥ λ 2 . These antennas usually operated at UHF and above frequencies.

2. At which of the following frequencies aperture antennas are operated?

a) UHF and EHF

b) MF and HF

c) HF and UHF

d) LF and MF

Answer: a

Explanation: Antenna with an aperture at the end is known as aperture antenna. Example is waveguide. These antennas usually operated at UHF and EHF frequencies .

LF – 30 kHz – 300 kHz

MF – 300k-3MHz

HF – 3MHz -30MHz

3. Which of the following does not belong to the aperture antenna?

a) Half-Dipole

b) Horn Antenna

c) Waveguide antenna

d) Slot antenna

Answer: a

Explanation: Antenna with an aperture at the end is known as aperture antenna. Horn antenna, Waveguide antenna and slot antenna are examples of aperture antenna. Half-dipole is a wire antenna.

4. Which of the following principle is used for analysis of aperture antennas?

a) Equivalence principle

b) Friss Equation

c) Reflectivity

d) Diffraction

Answer: a

Explanation: Equivalence principle follows the uniqueness theorem. It provides a unique solution for the boundary conditions. So this is used in the analysis of the aperture antennas. Friss transmission is used to relate the distance and power radiation between the antennas. Reflectivity and diffraction are not the principles used for analysis of aperture antenna.

5. Equivalence Principle follows which f the following theorem?

a) Uniqueness Theorem

b) Poynting theorem

c) Friss Theorem

d) Gauss theorem

Answer: a

Explanation: Equivalence principle follows the uniqueness theorem. It provides a unique solution for the boundary conditions. Uniqueness theorem is defined from the pointing theorem. Friss transmission is used to relate the distance and power radiation between the antennas. Gauss theorem states that total electric flux enclosed by charge is equal to net positive charge.

6. Which of the following condition is true for the electric conductor equivalent?

a) J s =n×H=0;M s =-n×E

b) J s =n×H;M s =-n×E=0

c) J s =n×H≠0;M s =-n×E

d) J s =n×H≠0;M s =-n×E=0

Answer: a

Explanation: The equivalent surface currents J s , M s radiates fields H, E respectively. The conditions for the electric conductor equivalent is J s =n×H=0;M s =-n×E.

7. Which of the following condition is true for the magnetic conductor equivalent?

a) J s =n×H=0;M s =-n×E

b) J s =n×H;M s =-n×E=0

c) J s =n×H≠0;M s =-n×E

d) J s =n×H≠0;M s =-n×E=0

Answer: b

Explanation: The equivalent surface currents J s , M s radiates fields H, E. The conditions for the electric conductor equivalent is J s =n×H;M s =-n×E=0. This is one of the equivalence principle modes.

8. Equivalence principle is mainly used for far filed analysis of the antenna in outer region.

a) True

b) False

Answer: a

Explanation: Equivalence principle follows the uniqueness theorem. It provides a unique solution for the boundary conditions. So this is used in the far field analysis of the aperture antennas in the outer regions.

9. The gain of the aperture antenna increases with square of the frequency.

a) True

b) False

Answer: a

Explanation: One of the distinguishing features of the aperture antenna is the increase in gain with operating frequency. The gain of the aperture antenna increases with square of the frequency if its aperture efficiency is kept constant with respect to the frequency.

10. Huygens principle in mathematical form is referred to as equivalence principle for aperture antennas.

a) True

b) False

Answer: a

Explanation: The equivalence principle replaces the aperture antenna with surface currents and thereby fields. It is derived from the uniqueness theorem and provides a unique solution for the boundary conditions for the fields. This is mainly used for the far field analysis of the aperture antenna.

This set of Antennas written test Questions & Answers focuses on “Aperture Antenna – Beamwidths”.

1. Half-power Beamwidth is given by ____

a) 70λ/D

b) 70D/λ

c) 35λ/D

d) 35D/λ

Answer: a

Explanation: The area the power is radiated is given by Beamwidth. Half power Beamwidth is the area at which the power is radiated 50% of peak power. The half-power beamwidth is given by70λ/D.

2. If the antenna dimension is two times the wavelength of the signal then the half power beam width will be _____

a) 35

b) 140

c) 70

d) 280

Answer: a

Explanation: The half-power beamwidth is given by70λ/D.

⇨ \(\frac{70\lambda}{D}=\frac{70}{2}=35.\)

3. For a circular aperture the FNBW is ______

a) 140λ/D

b) 70λ/D

c) 140D/λ

d) 70D/λ

Answer: a

Explanation: The area the power is radiated is given by Beam-width. The beam-width between the first nulls is the FNBW. For circular aperture the FNBW is given by 140λ/D. The half-power beam width is given by 70λ/D.

4. If the antenna dimension is two times the wavelength of the signal then the First null beam width will be _____

a) 35

b) 140

c) 70

d) 280

Answer: c

Explanation: The first null beam-width is given by 140λ/D.

⇨ \(\frac{140\lambda}{D} = \frac{140}{2}=70.\)

5. For a rectangular aperture of a*b the first null in E-plane occur at _______

a) sin -1 ⁡

b) sec -1 ⁡

c) cos -1 ⁡

d) sin -1 ⁡

Answer: a

Explanation: The area the power is radiated is given by Beam-width. The beam-width between the first nulls is the FNBW. \(\frac{kb}{2}\) sinθ=nπ

⇨ θ= sin -1 ⁡

⇨ For first null n=1 θ= sin -1 ⁡.

6. The first null beam width in the E-plane of a rectangular aperture of a×b is given by _______________

a) 2sin -1 ⁡

b) sin -1 ⁡

c) 2sec -1 ⁡

d) 2cos -1 ⁡

Answer: a

Explanation: The area the power is radiated is given by Beam-width. The beam-width between the first nulls is the FNBW. \(\frac{kb}{2}\) sinθ=nπ

θ= sin -1 \

 

\)

Therefore, the FNBW in E-plane is given by FNBW=2 θ = sin -1 ⁡\

 

\).

7. Larger the size of the aperture, the narrower is the Beam-widths.

a) True

b) False

Answer: a

Explanation: The FNBW in E-plane is given by FNBW=2 θ = sin -1 ⁡\

 

\). As the dimension of the antenna aperture increases, the FNBW will decrease. Thereby, beam-width becomes narrower.

8. Half-power Beam width in E-plane for a rectangular aperture antenna of a×b is given by ____

a) 0.886λ/b

b) 0.443λ/b

c) 0.5λ/b

d) λ/b

Answer: a

Explanation: By equating the field in E-plane to half power point

\

 

 

 

\)

Now HPBW = 2 arcsin⁡\

 

\)≈0.886λ/b.

9. Find the HPBW of the uniform rectangular aperture antenna with 4λ×2λ in the E-plane?

a) 0.443

b) 0.886

c) 0.25

d) 0.5

Answer: a

Explanation: HPBW = 2 arcsin⁡\

 

\)≈0.886λ/b=0.886/2=0.443.

10. The value at which the second null occurs in H-plane of rectangular aperture of a*b is given by ____

a) sin -1 ⁡

b) sin -1 ⁡

c) sin -1 ⁡

d) sin -1 ⁡

Answer: a

Explanation: The area the power is radiated is given by Beam-width. The beam-width between the first nulls is the FNBW. For H-plane \(\frac{ka}{2}\) sinθ=nπ

⇨ θ= sin -1 ⁡

⇨ For null n=2 θ= sin -1 ⁡.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Principles”.

1. Which of the following type does horn antenna belongs?

a) Wire Antenna

b) Array Antenna

c) Aperture Antenna

d) Lens Antenna

Answer: c

Explanation: Horn antenna belongs to Aperture antenna. Flaring done is done at the aperture of the rectangular waveguide gives different types of horn antenna. Dipole belongs to Wire antenna. Yagi-Uda is an array antenna. Convex-plane is example for lens antenna.

2. Which of the following antenna is mainly used for broadband signals?

a) Marconi antenna

b) Horn antenna

c) Wire antenna

d) Yagi-Uda antenna

Answer: b

Explanation: Horn antenna is used for broadband signals due to its flared nature. It is used to properly match the waveguide to a large radiating aperture by shaping transition gradually. It is used in 300MHz to 30GHz frequency range.

3. In which of the following bands the horn antenna operates?

a) HF and VHF

b) UHF and LF

c) UHF and SHF

d) LF and VHF

Answer: c

Explanation: Horn antenna is used in the frequency range 300MHz to 30GHZ. So, it is used in UHF and SHF frequency bands.

HF- VHF: 3MHz to 300MHz

LF-UHF: 30kHz to 3GHz

LF-VHF: 30kHz to 300MHz

4. In a horn antenna, with increase in aperture, the directivity is ____ and diffraction is ______

a) increased, decreased

b) decreased, increased

c) increased, increased

d) decreased, decreased

Answer: a

Explanation: Horn antenna is designed to improve the directivity and reduce the diffractions. It is used to properly match the waveguide to a large radiating aperture by shaping transition gradually. So, reflections are also decreased.

5. In Horn antennas impedance matching is provided by _______

a) flaring

b) increasing Power

c) decreasing axial length

d) Balun

Answer: a

Explanation: In horn antennas, we use Flaring technique for impedance matching. The sides of the waveguide are flared to match the impedance and improve the radiation efficiency.

6. Principle used in Horn antenna is Huygens principle.

a) True

b) False

Answer: a

Explanation: Horn antenna uses Huygens principle where aperture field is considered as a second source which gives rise to far field. It states that each point on a primary wave front can be considered to be a new source of a secondary spherical wave and that a secondary wave front can be constructed as the envelop of these secondary spherical waves.

7. Which of the following doesn’t apt for flaring?

a) Improve Gain

b) Impedance matching

c) Decreases side lobes

d) Improve Standing waves

Answer: d

Explanation: Flaring is done to provide the impedance matching. So, the reflections are minimized so there will be no standing waves. Side lobes will be reduced so the forward gain is also improved.

8. The radiation pattern of H-plane Sectoral horn is narrower than the E-plane Sectoral horn.

a) True

b) False

Answer: a

Explanation: The radiation pattern of H-plane Sectoral horn is narrower than the E-plane Sectoral horn. It is because of the dimension of the horn antenna in that direction.

9. The small the flare angle, ___ is the directivity and ____ is the beam width.

a) high, low

b) low, high

c) low, low

d) high, high

Answer: a

Explanation: The smaller flare angle means the beam is narrower. The Beamwidth and directivity are inversely proportional to each other. If beamwidth increases then directivity will decrease. So, for a small fare angle, the directivity is more and beamwidth is less.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Types”.

1. For a horn antenna, in which flaring is done only in one direction is ________

a) Conical antenna

b) Sectoral antenna

c) Pyramidal horn antenna

d) Exponential horn antenna

Answer: b

Explanation: In Sectoral horn antenna, flaring is done only in one direction. Depending on the flaring direction with respect to field propagation, it is divided into E-plane or H-plane horn antenna. Conical horn flared cross section is in shape of a cone.Pyramidal horn flared cross section is in shape of a four-sided pyramid. In exponential horn, the separation of sides increases as a function of length.

2. If flaring is done in the boarder direction of the rectangular waveguide then it is called ______

a) E-plane horn

b) H-plane horn

c) Conical horn

d) Pyramidal horn

Answer: a

Explanation: If the flaring is done in the electric field that means in the boarder direction of the rectangular waveguide, then it is called E-plane horn antenna.

3. If flaring is done in the magnetic field direction of the rectangular waveguide then it is called ______

a) E-plane horn

b) H-plane horn

c) Conical horn

d) Pyramidal horn

Answer: b

Explanation: In Sectorial horn antenna, flaring is done only in one direction.If the flaring is done in the magnetic field direction, then it is called H-plane horn antenna.

4. Which of the following antenna has the parallel slots along the inside surface of the horn?

a) E-plane horn

b) Pyramidal horn

c) Exponential horn

d) Corrugated horn

Answer: d

Explanation: A Corrugated horn has parallel slots or grooves which are small in size compared to the wavelength. These are present along the inside surface of the horn and are transverse to the axis.These corrugated antennas have wider bandwidths and minimizes the side lobes.

5. In which of the following antennas the separation of sides increases as a function of length?

a) Conical horn

b) Exponential tapered pyramidal horn

c) Pyramidal horn

d) Sectoral horn

Answer: b

Explanation: In exponentially tapered pyramidal horn, the transition region is gradually tapered exponentially to minimize the reflections. These are mainly used when reflections in the waveguide are critical.

6. If the walls of the circular waveguide are flared out, then it is called _____

a) Pyramidal horn

b) E-plane horn

c) H-plane horn

d) Conical horn

Answer: d

Explanation: In conical horn, the walls of the circular waveguide are flared out. For pyramidal horn, the walls of the rectangular waveguide are flared out in both directions.

7. If the all the walls of rectangular waveguide are flared out, then it is called ______

a) Pyramidal horn

b) Conical horn

c) E-plane horn

d) H-plane horn

Answer: a

Explanation: For pyramidal horn, all the walls of the rectangular waveguide are flared out. E-plane and H-plane are Sectoral horn antennas which are flared in only one direction.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Design – 1”.

1. If the flare angle of the horn increases, then its beam width ______

a) Decreases

b) Increases

c) Depends on aperture of horn

d) Independent of flare angle

Answer: b

Explanation: The beam width of the horn antenna increases as the flare angle increases. The reflections are decreases as the flare angle increases. But at 900, the antenna approximates to an open end waveguide and its gain decreases.

2. Find the width of aperture of the E-plane horn operating at frequency 800MHz with the slant length 5cm?

a) 1.936 cm

b) 19.36 cm

c) 0.194 cm

d) 1.936 m

Answer: b

Explanation: Aperture width \( = \sqrt{2λL}=\sqrt{\frac{2c}{f} L}=\sqrt{\frac{2×3×10^{10}}{800×10^6}×5}=19.36 cm\)

3. Find the aperture width of the H-plane horn operating at frequency 900MHz with the slant length 5cm?

a) 2.236 cm

b) 22.36 cm

c) 0.224 cm

d) 2.236 m

Answer: b

Explanation: Aperture width \( = \sqrt{3λL}=\sqrt{\frac{3c}{f} L}=\sqrt{\frac{3×3×10^{10}}{900×10^6}×5}=22.36 cm\)

4. Find the diameter of the conical horn operating at frequency 900MHz with the slant length 1.5cm?

a) 1.224 cm

b) 12.24 cm

c) 0.122 cm

d) 1.224 m

Answer: b

Explanation: Diameter \( = \sqrt{3λL}=\sqrt{\frac{3c}{f} L}=\sqrt{\frac{3×3×10^10}{900×10^6}×1.5}=12.24 cm \)

5. Find the directivity of a pyramidal horn antenna with E-plane aperture width 10λ and H-plane aperture width 15λ?

a) 30.15dB

b) 30.51dB

c) 3.051dB

d) 35.01dB

Answer: a

Explanation: Directivity \( = 10log_{10}\frac{7.5A_e}{λ^2} = 10log_{10}\frac{7.5×10λ×15λ}{λ^2} =30.51dB

\)

6. Find the length of the E-plane horn with aperture 12λ and path travel difference 0.4λ?

a) 3.5 λ

b) 45 λ

c) 54 λ

d) 9 λ

Answer: b

Explanation: Length of the E-plane horn \(ρ =\frac{h^2}{8δ}=\frac{^2}{8×0.4λ}=45 λ\)

7. Find the flare angle of the H-plane horn with aperture 12λ and path travel difference 0.4λ?

a) 15.18°

b) 7.09°

c) 30.18°

d) 1.518°

Answer: a

Explanation: Length of the H-plane horn \(ρ =\frac{h^2}{8δ}=\frac{^2}{8×0.4λ}=45 λ\)

Flare angle of the H-plane horn = \(2tan^{-1}⁡\frac{h}{2ρ}=2×tan^{-1}⁡\frac{12λ}{2×45λ}=15.18°\)

8. Which of the following is the correct form of design equations of a horn antenna?

a) Length \

 

 

 Length \

 

 

 Length \

 

 

 Length \(ρ=\frac{h^2}{8δ}, flare\, angle=tan^{-1}⁡\frac{h}{2ρ}\)

Answer: a

Explanation: Design equations of a horn antenna are:

Length \(ρ=\frac{h^2}{8δ},flare\, angle\, θ=2 tan^{-1}⁡\frac{h}{2ρ}\)

9. What are the length and the flare angle of the H-plane horn antenna with aperture 10 λ and path difference δ=0.2λ?

a) Length=31.8λ,flare angle=4.15°

b) Length=62.5λ,flare angle=9.14°

c) Length=31.8λ,flare angle=9.14°

d) Length=62.5λ,flare angle=4.15°

Answer: b

Explanation: Length of the H-plane horn \(ρ=\frac{h^2}{8δ}=\frac{^2}{8×0.2λ}=62.5λ\)

Flare angle of the H-plane horn \(= 2tan^{-1}⁡\frac{h}{2ρ}=2×tan^{-1}⁡\frac{10λ}{2×62.5λ}=9.14° \)

10. Find the optimum aperture angle for H-plane horn with aperture 12λ and path travel difference 0.4λ?

a) 75.9°

b) 1.59°

c) 7.59°

d) 15.18°

Answer: c

Explanation: Optimum aperture angle of the H-plane horn = \(tan^{-1}⁡\frac{h}{2ρ}=tan^{-1}⁡\frac{12λ}{2×45λ}=7.59°\)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Design – 2”.

1. HPBW of E-plane horn is  _____

a) \

 

 \

 

 \

 

 \(\frac{λ}{67h_e} \)

Answer: a

Explanation: HPBW of E-plane horn is \(\frac{56λ}{h_e} \) where h e the aperture dimension of E-plane horn.

2. HPBW of H-plane horn is  _____

a) \

 

 \

 

 \

 

 \(\frac{λ}{67h_h} \)

Answer: b

Explanation: HPBW of H-plane horn is \(\frac{67λ}{h_h} \) where h h the aperture dimension of H-plane horn.

3. HPBW of E-plane horn with aperture dimension 10 λ in degrees is ____

a) 56

b) 67

c) 5.6

d) 6.7

Answer: c

Explanation: HPBW of E-plane horn is \(\frac{56λ}{h_e} = \frac{56λ}{10λ}=5.6\).

4. HPBW of E-plane horn with aperture dimension 10 λ in degrees is ____

a) 56

b) 67

c) 5.6

d) 6.7

Answer: d

Explanation: HPBW of H-plane horn is \(\frac{67λ}{h_h} = \frac{67λ}{10λ}=6.7\).

5. What is the physical area of the rectangular horn antenna with E-plane aperture 5 λ and H-plane aperture 4 λ?

a) 20λ 2

b) 10λ 2

c) 0.8

d) 9 λ

Answer: a

Explanation: The physical area of rectangular horn is A = h h ×h e =5λ×4λ=20 λ 2

6. Physical area of a conical horn antenna of radius 5cm is____

a) 25π

b) 6.15π

c) 2.5π

d) 25

Answer: a

Explanation: Physical area of conical horn A=πr 2 =π×5 2 =25π.

7. Flare angle of horn antenna when axial length l and path difference δ is given is _____

a) \

 

 \

 

 \

 

 \(θ= cos^{-1}⁡\frac{l}{l-δ} \)

Answer: a

Explanation: Flare angle of horn antenna when axial length l and path difference d is given is

\(θ= 2 cos^{-1}⁡\frac{l}{l+δ} \).

8. Flare angle  of horn antenna when axial length 10λ and path difference δ = 0.3λ is given is _____

a) θ = 1.386

b) θ = 2.772

c) θ = 27.72

d) θ = 13.86

Answer: c

Explanation: Flare angle of horn antenna when axial length l and path difference d is given is

\(θ = 2 cos^{-1}⁡\frac{l}{l+δ}=2 cos^{-1}⁡\frac{10λ}{10.3λ}=27.72\)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Basics”.

1. Which of the following refers to the pattern of reflector in the reflector antenna?

a) Primary pattern

b) Secondary pattern

c) Reflector pattern

d) Feed pattern

Answer: b

Explanation: In a reflector antenna, primary pattern is the feed pattern and secondary pattern is the pattern of reflector. Reflector antennas are high gain antenna and are used in RADARs and for some communication purpose.

2. Reflector antenna operates on the Geometric optics principle?

a) True

b) False

Answer: a

Explanation: Reflector antenna is high gain antenna and works on the principle of the Geometric optics. Geometrical optics shows that if a beam of parallel rays is incident on a reflector antenna whose geometrical shape is a parabola, then the array beams will converge at the focal point.

3. Which of the following is a dual reflector antenna?

a) Cassegrain antenna

b) Parabolic antenna

c) Offset reflector antenna

d) Wire antenna

Answer: a

Explanation: A dual reflector antenna consists of two reflectors and one feed antenna. Cassegrain antenna is the best example of dual reflector antenna since it contains main reflector as parabolic and sub-reflector as hyperbola.

4. Which of the following combination forms a Cassegrain antenna?

a) The main reflector is parabolic and sub-reflector is hyperbolic

b) The main reflector is parabolic and sub-reflector is concave

c) The main reflector is hyperbolic and sub-reflector is parabolic

d) The main reflector is hyperbolic and sub-reflector is convex

Answer: a

Explanation: A Cassegrain antenna is a dual reflector antenna and is axis symmetry. It consists of two reflectors and one primary feed. The main reflector is parabolic and sub-reflector is hyperbolic .

5. Which of the following efficiency is used to measure the power-loss at the feed pattern which is intercepted by reflector?

a) Spillover

b) Illumination

c) Taper

d) Aperture

Answer: a

Explanation: When the feed pattern exceeds beyond reflectors rim, not all the energy is redirected by the reflector. This power loss is measure results in spillover efficiency. Illumination efficiency is the combination of both taper and spillover efficiency. Aperture efficiency is the ratio of effective aperture to physical aperture. Taper efficiency gives the directivity measure of the antenna.

6. What is the value of magnification of the Cassegrain antenna if its sub-reflector eccentricity is 2?

a) 3

b) 2

c) 1/3

d) 1/2

Answer: a

Explanation: Given eccentricity e=2

Magnification \(M=\frac{e+1}{e-1}=3 \)

7. Which of the following is false regarding a reflector antenna?

a) Reflector antennas are high gain antennas with two antennas

b) Both the primary and secondary antennas are excited

c) The pattern of the reflector in the reflector antenna is the Secondary pattern

d) A dual reflector contains two reflectors and one primary feed

Answer: b

Explanation: In a reflector antenna the horn or a dipole acts as a feed and the antenna which is excited is the primary antenna. Reflector is the secondary antenna. Reflector antenna is a high gain antenna consisting of primary and secondary antennas.

8. Which of the following is not a reflector antenna?

a) Convex-convex

b) Corner

c) Gregorian

d) Cassegrain

Answer: a

Explanation: Corner, Cassegrain and Gregorian belong to reflector antenna. Convex-convex is a type of lens antenna. Gregorian is a dual reflector antenna with a concave sub-reflector.

9. When a reflector is placed at the foci along the feed it is called ____ antenna

a) Dual reflector antenna

b) Plane antenna

c) Wire antenna

d) Convex-Convex

Answer: a

Explanation: Usually a feed forward is used in reflector antenna and feed antenna is placed at focus of the reflector. When a sub-reflector is placed at the focus along with the feed antenna it is called dual reflector antenna. Example of dual reflector antenna: Cassegrain antenna.

Convex-Convex is a lens antenna.

10. Which of the following is used as a secondary antenna in the reflector antenna?

a) Horn

b) Feed antenna

c) Parabolic

d) Dipole

Answer: c

Explanation: A parabolic reflector is used as a secondary antenna. Horn and dipole are used for feed antenna. Feed antenna of the reflector antenna is called Primary antenna.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Flat Sheet Reflector”.

1. Which of the following statements is false for flat reflector?

a) Plane reflector is used to reduce the backward radiation

b) Increases the gain in forward direction

c) A large flat sheet placed in front of dipole increases the directivity

d) Decreases the gain in forward direction

Answer: d

Explanation: A plane or a flat reflector is used to radiate energy in desired direction by placing a feed in front of it. It reduces the backward radiation and increases the gain in the forward direction. A reflector antenna is a high gain antenna.

2. A corner reflector is converted to plane reflector when the corner angle is ______

a) 90°

b) 180°

c) 45°

d) 60°

Answer: b

Explanation: For a corner reflector to become a plane reflector the corner angle must be 180°.

For a plane reflector N  =1

\(α=\frac{π}{N}=π.\)

3. Which of the following reflector antenna is formed by joining two plane reflectors at some angle ?

a) Corner

b) Parabolic

c) Cheese

d) Truncated

Answer: a

Explanation: By joining two plane sheet reflectors at some angle we get the corner antenna. A plane or a flat reflector is used to radiate energy in desired direction by placing a feed in front of it. Cheese and Truncated are types of parabolic antenna.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Corner Reflector”.

1. Corner reflector is designed for radiation in forward direction unlike in plane reflector.

a) True

b) False

Answer: a

Explanation: The plane reflector allows back and side radiation. In order to avoid this it is modifies to a corner reflector which consists of two plane reflectors joined at a corner to allow radiation in forward direction only.

2. In a corner reflector, included angle α refers to _______

a) angle at which two plane reflectors are joined

b) angle between vertex and the feed radiator

c) angle between major axis to the main beam

d) angle between vertex and the main beam axis

Answer: a

Explanation: A corner reflector consists of two plane reflectors joined at a corner to allow radiation in forward direction only. Included angle is the angle at which two reflectors are joined, measured at the corner.

3. In order to achieve good system efficiency in a corner reflector, the spacing between the vertex and the feed must be ______ as included angle decreases.

a) Increased

b) Decreased

c) Constant

d) Either increases or decreases

Answer: a

Explanation: The system efficiency in a corner reflector depends on the spacing between vertex of the corner and the feed. It is adjusted depending on the included angle. As the included decreases, spacing must be increased.

4. Corner angle of a corner reflector with 4 current elements is ______

a) \

 

 \

 

 \

 

 4π

Answer: a

Explanation: Corner angle of a corner reflector with 4 current elements is given by \(α=\frac{π}{N}\)

Given N=4

⇨ \(α=\frac{π}{N}=\frac{π}{4}.\)

5. Find the aperture dimension of the corner reflector having reflector sheet side length 1m?

a) 1.414

b) 2.828

c) 2

d) 1

Answer: a

Explanation: From the design equations of a corner reflector,

Aperture dimension D A = 1.414l

Where l is side length of reflector sheet and l=2d where d is distance between feed and vertex of the reflector.

6. In a corner reflector antenna, if the spacing between vertex of reflector and feed is 2m then side length of reflector sheet is _____cm.

a) 4

b) 2

c) 400

d) 200

Answer: c

Explanation: In a corner reflector antenna, side length of reflector sheet and l=2d

Where, d is distance between feed and vertex of the reflector.

Given d=2m so l=2*2=4m=400cm.

7. Which of the following statements is false for a corner reflector antenna?

a) Increases directivity in the forward direction

b) Back radiation is not reduced compare to the plane reflector

c) System efficiency depends on spacing between reflector vertex and the feed

d) Angle between the reflecting plates is called included angle

Answer: b

Explanation: Compared to the plane reflector antenna, back and side radiations are reduced in the corner reflector. A corner reflector consists of two plane reflectors joined at a corner to allow radiation in forward direction only. As the included angle decreases, spacing will increases.

8. The number of images, polarity and position in the analysis of the radiation field of corner reflector depends on what?

a) Only included angle

b) Polarization of feed element

c) Both included angle & Polarization of feed element

d) Neither included angle nor Polarization of feed element

Answer: c

Explanation: The number of images, polarity and position in the analysis of the radiation field of corner reflector depends on both included angle & Polarization of feed element with perpendicular polarization. Angle between the reflecting plates is called included angle.

9. Corner angle of a square corner reflector antenna is _____

a) 90°

b) 180°

c) 60°

d) 45°

Answer: a

Explanation: For a square reflector angle N=2 and corner angle is given by \(α=\frac{π}{N}=\frac{π}{2}=90°.\) A corner reflector consists of two plane reflectors joined at a corner to allow radiation in forward direction only. This is one of the most practically used antennas.

10. What is the corner angle of a flat reflector antenna?

a) 90°

b) 180°

c) 60°

d) 45°

Answer: b

Explanation: The corner angle depends on value of N . For a flat reflector N=1. So the corner angle is \(α=\frac{π}{N}=π=180°.\)

This set of Reflector Antenna online test focuses on “Parabolic Reflector Antenna”.

1. Which of the following wave conversion mechanism is performed in a parabolic reflector antenna?

a) Plane to spherical

b) Spherical to plane

c) Performs both plane to spherical and spherical to plane

d) Elliptic polarization

Answer: b

Explanation: In a parabolic reflector antenna, the wave conversion mechanism used is spherical to plane. It works on the principle of geometric optics. The reflected plane waves travel parallel o the major axis of the reflector.

2. The power gain of the parabolic reflector with circular aperture of diameter 10λ?

a) 100π

b) 10π

c) 600π

d) 360π

Answer: c

Explanation: The power gain of the parabolic reflector with circular aperture of diameter d is given by

\(G_p = 6\frac{d^2}{λ^2} = 600π.\)

3. Find the aperture ratio of the paraboloid with aperture diameter 1m at 1.5GHz frequency?

a) 5

b) 0.2

c) 1.5

d) 7.5

Answer: a

Explanation: Aperture ratio \(= \frac{d}{λ}\)

\(λ=\frac{c}{f}=\frac{3×10^8}{1.5×10^9}=0.2m\)

Aperture ratio\( = \frac{1}{0.2}=5.\)

4. BWFN of a paraboloid antenna with circular aperture assuming feed is isotropic is ____

a) \

 

 \

 

 \

 

 \(\frac{40λ}{d} \)

Answer: a

Explanation: A paraboloid antenna with a circular aperture has

BWFN  = \

 

For a rectangular aperture \

 

.

5. Find the BWFN of a paraboloid with a circular aperture of diameter 10λ?

a) 14 degrees

b) 28 degrees

c) 11.5 degrees

d) 41 degrees

Answer: a

Explanation: A paraboloid antenna with a circular aperture has

BWFN  = \

 

⇨ \(BWFN =\frac{140λ}{10λ}=14 degrees.\)

6. What is the ratio of focal length to diameter for practical applications in a parabolic reflector?

a) 0.25 to 0.5

b) < 0.25

c) 0.125 to 0.3

d) 0.5 to 1

Answer: a

Explanation: The ratio of focal length  to diameter  f/d < \(\frac{1}{4}\) indicates radiation away from the parabolic surface of the reflector. So for practical applications it lies between 0.125 and 0.5.

7. Which of the following is false regarding a paraboloid antenna?

a) Spill over decreases due to back lobe of primary radiator

b) Feed placed at the focus is used to improve the beam pattern

c) Pill box provides wide beam in one plane and narrow beam in other plane

d) At lower frequencies parabolic antennas are not used frequently

Answer: a

Explanation: Spill over occurs due to the non-captured radiation by the reflector. From the primary radiators also some of forward radiation gets added up with the desired parallel beams. This is called back lobe radiation. It increases due to the back lobe of the primary radiator.

8. Which of the following paraboloid reflector is formed by cutting some part of paraboloid to meet requirements?

a) Truncated paraboloid

b) Cassegrain

c) Corner

d) Pill box

Answer: a

Explanation: When a portion of the paraboloid reflector is cut off or truncated it is called as truncated paraboloid. Pill box provides wide beam in one plane and narrow beam in other plane. Cassegrain is a dual reflector antenna and Corner reflector is also a type of reflector antenna.

9. Which of the following is used to produce wide beam in one plane and narrow beam in other plane?

a) Pill box

b) Truncated paraboloid

c) Cassegrain

d) Paraboloid with rectangular aperture

Answer: a

Explanation: Pill box is short parabolic right cylinder enclosed by parallel plates. It produces a wide beam in one plane and narrow beam in other plane. Cassegrain is a dual reflector. Truncated is a type of paraboloid.

10. In a paraboloid antenna, all rays leaving the focal point are collimated along the reflector’s axis after reflection.

a) True

b) False

Answer: a

Explanation: According to the geometry of the paraboloid reflector,

⇨ All rays leaving the focal point are collimated along the reflector’s axis after reflection.

⇨ All overall ray path lengths  are the same and equal to 2F.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Feeding Systems”.

1. Which of the following is called as the source placed at the focus?

a) Feed radiator

b) Reflector

c) Secondary radiator

d) Primary / reflector

Answer: a

Explanation: The source placed at the focus is called as feed radiator. It is also known as primary radiator. The reflector is called as the secondary radiator. A reflector antenna contains primary and secondary radiators.

2. In which of the following cases, a parabolic reflector primary radiator is said to be ideal feed?

a) The entire reflector is illuminated with no radiation in unwanted direction when feed radiated entire energy towards it

b) Some part of reflector is illuminated with no radiation in unwanted direction when feed radiated entire energy towards it

c) Some part of reflector is illuminated with radiation in unwanted direction when feed radiated entire energy towards it

d) The entire reflector is illuminated with small radiation in unwanted direction when feed radiated entire energy towards it

Answer: a

Explanation: A parabolic reflector primary radiator is said to be ideal feed, when the feed radiates entire energy, the reflector is fully illuminated and there is no energy radiation in unwanted direction.

3. To obtain maximum beam pattern, the primary radiator is placed ______ in a reflector antenna.

a) between Focus and Directrix

b) after the focus

c) at the focus point

d) can be placed anywhere

Answer: c

Explanation: Maximum beam pattern is obtained only when the feed is placed at the focus of the parabola. Reflector antenna is a high gain antenna. So this is one of the important points to obtain the maximized gain.

4. When the feed is moved along the main axis in a reflector antenna what happens to the beam pattern?

a) It broadens

b) It deteriorates

c) Beam remains unchanged

d) Side lobes are increased

Answer: a

Explanation: Maximum beam pattern is obtained only when the feed is placed at the focus of the parabola. When the feed is moved along the main axis, beam gets broadened.

5. The beam gets deteriorated when it is moved along a line perpendicular to the main axis passing through focus?

a) True

b) False

Answer: a

Explanation: Beam pattern is maximum only when the feed is placed at the focus of the parabola. When the feed is moved along the main axis, beam gets broadened. When it is moved along perpendicular line beam gets deteriorated.

6. In a Cassegrain feed system; the feed is placed at _____

a) Focus

b) Vertex

c) Directrix

d) Anywhere between vertex and focus

Answer: b

Explanation: Cassegrain is a dual reflector antenna. In this the feed is placed at the vertex of the parabolic reflector. The focus of the sub-reflector coincides with the focus of parabolic reflector and it illuminates the reflector.

7. Which of the following coincides with the focus of the parabolic reflector in a Cassegrain antenna?

a) Feed

b) Parabolic reflector

c) Focus of hyperboloid reflector

d) Focus of primary radiator

Answer: c

Explanation: The focus of the sub-reflector coincides with the focus of parabolic reflector and it illuminates the reflector. Since it is a Cassegrain, the sub-reflector is a hyperboloid reflector.

8. Which of the following regarding the Cassegrain feed system is false?

a) Spill over is reduced

b) It is a dual reflector antenna

c) Minor lobe radiation increases

d) A convex sub-reflector is used

Answer: c

Explanation: Spill over is the power loss when the reflector fails to redirect the energy. With the use of sub-reflector, this is reduced. Thus the minor lobe radiation also decreased. A convex hyperboloid is used as a sub-reflector. It is a dual reflector antenna with primary and two secondary radiators.

9. When a dipole with a parasitic reflector is used as a feed system, the distance between them is _____

a) 0.125λ

b) 0.4λ

c) 1λ

d) 0.625λ

Answer: a

Explanation: In a parabolic reflector antenna, a dipole along a parasitic reflector serves well as a feed radiator. The distance between parasitic reflector and the dipole is 0.125λ. The distance between dipole to dipole is approximately 0.4λ.

10. Which of the following uses a concave sub-reflector in a dual reflector antenna?

a) Cassegrain

b) Gregorian

c) Both Gregorian &Cassegrain

d) Neither Gregorian nor Cassegrain

Answer: b

Explanation: Cassegrain and Gregorian are dual reflector antennas. Gregorian antenna has a concave sub-reflector which is elliptic. Cassegrain uses a hyperboloid reflector which is convex towards the feed. The sub-reflectors are used to reduce the spillover efficiency.

This set of Reflector Antenna online quiz focuses on “Aperture Blockage”.

1. Which of the following is used to avoid the aperture blockage due to secondary reflector?

a) Offset feed system

b) Cassegrain

c) Plane reflector

d) Parabolic antenna

Answer: a

Explanation: Offset feed system is used to avoid the aperture blocking effect due to the dependence of secondary reflector. In this a feed radiator is placed at the focus in an inclined position such that all rays are collimated without formation of blockage region. In Cassegrain, plane reflector there is a blocking effect.

2. Which of the following causes aperture blockage in a Cassegrain antenna?

a) Main reflector

b) Sub-reflector

c) Feed radiator

d) Noise from surroundings

Answer: b

Explanation: The sub-reflector causes the aperture blockage in a Cassegrain antenna. It can be reduced by reducing the size of the sub-reflector.

3. In which of the following the secondary reflector faces the primary reflector?

a) Cassegrain

b) Gregorian

c) Corner reflector

d) Plane reflector

Answer: b

Explanation: In Gregorian antenna, sub-reflector is ellipsoidal and has its concave side faces the primary reflector. In Cassegrain, sub-reflector is hyperboloid with its convex side facing the feed. Corner and plane reflectors are normal reflector antennas while Cassegrain and Gregorian are dual reflector antennas.

4. Which of the following statement is true regarding a Gregorian antenna?

a) Sub-reflector ellipsoidal lies closer to the focus of the paraboloid

b) Sub-reflector is a hyperboloid with its convex facing feed

c) Sub-reflector is an ellipsoidal with its concave facing reflector

d) Sub-reflector is an ellipsoidal with its concave facing feed

Answer: c

Explanation: Gregorian antenna is a dual reflector antenna having a concave secondary reflector. The secondary reflector also known as sub-reflector is a ellipsoidal. It lies beyond the focus of the paraboloid.

5. In which of the following the aperture blocking is reduced by permitting antenna to operate in a single polarization?

a) Truncated reflector

b) Cassegrain

c) Polarization-twist reflector

d) Plane reflector

Answer: c

Explanation: Aperture blocking is reduced by permitting the antenna to operate with a single polarization in Polarization-twist reflector. Truncated reflector is a type of parabolic reflector. Cassegrain is a dual reflector.

6. In which of the following a sub-reflector called trans-reflector is present?

a) Polarization-twist reflector

b) Plane reflector

c) Cassegrain

d) Gregorian

Answer: a

Explanation: It is present in Polarization-twist reflector. Sub-reflector consists of a horizontal grating of wires. Cassegrain contains a hyperboloid sub-reflector. Gregorian contains an ellipsoidal sub-reflector.

7. Which of the following is not used for aperture blocking?

a) Reducing size of sub-reflector

b) Offset feed system

c) High directive feed

d) Reducing spillover

Answer: c

Explanation: A high directive feed means a large feed. It partially shadows the primary reflector and becomes an obstacle for blockage. Aperture blocking can be avoided by reducing size of sub-reflector, thereby spillover and by using Offset feed system.

8. Which of the following condition holds good for minimum total aperture blocking in Cassegrain?

a) Area of sub-reflector and projected area of feed are equal

b) Area of primary reflector and projected area of feed are equal

c) Area of sub-reflector greater than area of primary reflector

d) Area of sub-reflector is less than projected area of feed

Answer: a

Explanation: Minimum total aperture blocking occurs when the area of sub-reflector and projected area of feed are approximately equal. In other conditions, the main ray back radiated ray gets added up with collinear rays from the primary reflector.

9. In Polarization-twist reflector, sub-reflector consists of a horizontal grating of wires.

a) True

b) False

Answer: a

Explanation: In Polarization-twist reflector, sub-reflector consists of a horizontal grating of wires called trans-reflector. In this the antenna is permitted to operate only at a single polarization. The polarized radiation is rotated by 90° by sub-reflector at the primary reflector with the help of twist reflector.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Application of Parabolic Reflector”.

1. Which one of the following antennas is mostly used in TV Dish?

a) Parabolic reflector

b) Lens antenna

c) Log periodic

d) Rhombus antenna

Answer: a

Explanation: Parabolic reflector antenna also known as Dish antenna is widely used in the TV dishes. It is a high gain antenna used in VHF. Due to its narrow beamwidth it is widely used as dish antenna.

2. Which of the following antenna has a shape of letter ‘C’?

a) Orange peel antenna

b) Pill box

c) Yagi-Uda antenna

d) Wire dipole

Answer: a

Explanation: Orange peel antenna is one of the types of parabolic antenna with its shape in form of letter C. It is mainly used in the search radars. Pill box antenna gives wide and narrow beams in different planes. Yagi-Uda antenna is formed with the help of dipoles and reflectors placed at some distances to get a unidirectional beam with good directivity.

3. Which of the following statements is false about Array-fed antenna?

a) It is widely used to create a downlink radiation in satellite communication

b) An array of feed horns are clustered at focal point to produce arbitrary shaped beam

c) It uses only a single horn feed and a multiple array of passive reflectors

d) It can also be used with secondary reflector antennas

Answer: c

Explanation: Array-fed antenna uses a cluster of horn feeds instead of a single horn to produce an arbitrary beam. It is mainly used in direct satellite communication for downlink radiation. It is also used with secondary reflectors like Cassegrain.

4. Which type of feed in parabolic antenna blocks the beam limiting aperture efficiency to 55%?

a) Axial feed

b) Offset feed

c) Cassegrain

d) Gregorian

Answer: a

Explanation: In Axial feed, the feed is located at the focal point along the beam axis infront of dish and is pointed back and blocks some part of beam. Offset is used to avoid the aperture blocking. In Cassegrain and Gregorian the aperture efficiency is approximately 70%.

5. Which of the following reflector antenna doesn’t require a directive feed?

a) Parabolic

b) Hyperbolic

c) Corner

d) Truncated

Answer: c

Explanation: Corner reflector doesn’t require a directional feed because the direct and reflected waves are combined properly and also it doesn’t require any specific focal point. In Parabolic, hyperbolic the feed is directed towards the focal point.

6. In which of the following secondary reflector is concave?

a) Gregorian

b) Cassegrain

c) Parabolic

d) Corner

Answer: a

Explanation: Cassegrain and Gregorian are dual reflector antennas having primary and secondary reflectors. In Gregorian concave ellipsoidal is used as secondary reflector whereas in Cassegrain convex hyperboloid. Corner and parabolic are reflector antennas and secondary reflectors are not used widely.

7. In which of the following secondary reflector is convex?

a) Gregorian

b) Cassegrain

c) Parabolic

d) Corner

Answer: b

Explanation: Dual reflector antennas have both primary and secondary reflectors. Cassegrain and Gregorian antennas are dual reflector antennas. Secondary reflector is convex for Cassegrain antenna. In Gregorian concave ellipsoidal is used as secondary reflector. Parabolic and corner antennas are normal reflector antennas.

8. Cassegrain antennas cannot be used in radio telescopes.

a) True

b) False

Answer: b

Explanation: One of the advantages of Cassegrain antenna is that its feed is not placed in front of the dish or at the focal point. So, it can be used with bulky feeds which requires in radio telescopes and large satellite communications.

9. Which of the following antenna is a combination of pillbox and parabolic cylinder?

a) Cheese antenna

b) Orange peel antenna

c) Corner antenna

d) Cassegrain

Answer: a

Explanation: A pillbox alone sometimes is called as a Cheese antenna as it is made from the parabolic cylinder. It is a thin slice of the parabolic cylinder cross-section covered with plates. Orange peel antenna consists of array of feeds and is C shaped. Cassegrain is a dual reflector antenna. Corner antenna is made from joining two plane sheet reflectors.

10. Which of the following is formed by moving parabolic counter parallel to itself?

a) Parabolic cylinder

b) Pill box

c) Cheese antenna

d) Truncated antenna

Answer: a

Explanation: A parabolic cylinder is formed by moving the parabolic counter parallel to itself. It consists of a line of focus unlike the focal point in basic parabolic antenna. Pillbox or cheese antenna is obtained from the parabolic cylinder reflector. Truncated is obtained by cutting some part of parabolic antenna.

11. Large f/D ratio is used for deep-dish reflectors.

a) True

b) False

Answer: b

Explanation: The ratio of focal length to the aperture size is known as the f/D ratio. Large f/D ratio is used for shallow-dish reflectors. Small f/D ratio is used for deep-dish reflectors. As the f/D ratio increases, the spillover efficiency of antenna reduces.

12. In which of the following feed there is no impedance mismatch?

a) Offset feed

b) Cassegrain feed

c) Front feed

d) Gregorian feed

Answer: a

Explanation: In Front feed, due to the reflections from the dishes impedance mismatch occurs. Cassegrain and Gregorian are front feed dual reflectors. Offset feed there is no aperture blocking and no impedance mismatch.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Types”.

1. Which of the following type does the slot antenna belongs?

a) Aperture Type

b) Wire Type

c) Lens Type

d) Reflector Type

Answer: a

Explanation: Slot antenna belongs to the aperture antenna and operates at 300MHz to 30GHz. Dipole is an example of Wire type. Reflector type example is Parabolic. Lens type: Convex-plane.

2. In which of the following band the slot antenna operates?

a) UHF and SHF

b) SHF and VLF

c) VLF and MF

d) UHF and EHF

Answer: a

Explanation: Slot antenna operates in the range of 300MHz to 30GHz.

VLF-3 KHz – 30 KHz

MF- 300 KHz – 3MHz

UHF- 300 MHz-3GHz

SHF- 3 GHz -30 GHz

EHF- 30GHz-300GHz

So slot antenna operates at UHF and SHF range.

3. Which of the following antenna is obtained by removing a small area of metal from an infinite ground plane?

a) Slot antenna

b) Plane reflector

c) Dipole

d) Yagi-Uda

Answer: a

Explanation: Slot antenna belongs to the aperture antenna and operates at 300MHz to 30GHz. It is obtained by removing a small area of metal from an infinite ground plane.

4. What is the principle used in slot antennas?

a) Babinet’s principle

b) Archimedes principle

c) Geo-Optics

d) Image Theory

Answer: a

Explanation: Babinet’s principle is used in slot antenna. It relates the fields obtained directly by slot antenna with corresponding complementary strip dipoles. It relates transmission/reception characteristics like radiation pattern, gain, and input impedance of those two complementary antennas. Geo-optics is used in reflectors and Image theory in Corner reflectors.

5. Microstrip slot antennas are complementary to the Microstrip dipole antennas

a) True

b) False

Answer: a

Explanation: Microstrip slot antenna uses a Babinet’s principle. It relates transmission/reception characteristics like radiation pattern, gain, and input impedance of those two complementary antennas. Slot antenna belongs to the aperture antenna and operates at 300MHz to 30GHz.

6. Which of the following slot antenna radiation can be modified electronically by tunable capacitor?

a) Annular slot antenna

b) Tapered slot antenna

c) Both annular and tapered

d) Neither annular nor tapered

Answer: a

Explanation: Annular slot antenna radiation can be tuned electronically by using tunable capacitor and is used for mobile communications. Tapered slot produces end-fire radiation.

7. Which of the following type of slot antenna gives good end-fire radiation?

a) Annular slot antenna

b) Tapered slot antenna

c) All slot antennas gives good end-fire radiation

d) None of slot antenna gives good end-fire radiation

Answer: b

Explanation: Microstrip slot antennas are usually poor end-fire radiators except tapered slot antenna. Slot antenna belongs to the aperture antenna and operates at 300MHz to 30GHz.

8. What is the impedance of 2- element slot antenna with single slot antenna resistance =500Ω?

a) 125Ω

b) 500Ω

c) 1000Ω

d) 250Ω

Answer: a

Explanation: impedance of N-element slot antenna \(Z_{in,N} = \frac{Z_{slot}}{N^2} \)

Given N=2, Now \(Z_{in,N} = \frac{Z_{slot}}{N^2} = \frac{500}{4} = 125\Omega \)

9. Which of the following cannot be used to reduce the radiation resistance in slot antenna for a given slot size?

a) Center feed

b) Off-center feed

c) Inclined slot

d) Stub tuning

Answer: a

Explanation: A center feed slot has high radiation resistance and requires a matching network to match the antenna. Off-center, Inclined slot and stub-tuning are some techniques used to reduce the radiation resistance of slot antenna.

10. The bi-directional radiation disadvantage in slot antenna can be rectified with metallic cavity

a) True

b) False

Answer: a

Explanation: The slot antenna has low cross polarization compared to patch antenna but its disadvantage is that it has bi-directional radiation which is corrected with the metallic cavity or reflector.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Patch Types”.

1. Which of the following is the most common version of the printed antenna?

a) Horn antenna

b) Microstrip antenna

c) Wire antenna

d) Lens antenna

Answer: b

Explanation: Microstrip antenna is a printed antenna. It is fabricated using photolithography technique. It consists of a radiation patch on one side of dielectric and a ground plane on other side. The patch can be of any shape like circular or rectangular.

2. In which of the following type of antenna a patch is created during fabrication?

a) Microstrip antenna

b) Horn antenna

c) Wire antenna

d) Lens antenna

Answer: a

Explanation: In Microstrip antenna patch is created on the dielectric substrate using photo-etching. The shape and size of patch may vary the return loss. Horn, Wire and Lens doesn’t have any patch. Horn is a waveguide antenna and a wire antenna is made of dipoles.

3. In Microstrip antenna, the patch and feed line is photo etched on dielectric substrate

a) True

b) False

Answer: a

Explanation: Microstrip antenna is a printed antenna which is fabricated using photolithography technique. Its design process consists of a ground plane and a dielectric substrate in which patch is created. The source is connected to it through a feed line by photo etching process.

4. The shape of the patch in rectangular Microstrip antenna is ____

a) Rectangular

b) Circular

c) Cylindrical

d) Elliptical

Answer: a

Explanation: The antenna is named under the shape of the patch. For a rectangular Microstrip antenna, the patch shape is rectangle. For a general Microstrip antenna patch can be any shape like rectangle, circular, elliptic, triangle.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Feed Methods”.

1. Which of the following is used to excite to radiate the antenna?

a) Feed line

b) Ground plane

c) Patch

d) Substrate

Answer: a

Explanation: The feedline is sued to excite to radiate the antenna in direct or indirect way. There are mainly four different types of feeding in Microstrip antenna: Microstrip line feed, coaxial probe, aperture coupling, proximity coupling.

2. Which of the following fed line is connected to patch just like a conducting strip?

a) Microstrip line feed

b) Coaxial feed

c) Aperture coupling

d) Proximity Coupling

Answer: a

Explanation: Microstrip line feed acts as an extent to patch and is connected to patch just like a conducting strip. In coaxial feed, inner conductor is attached to patch and outer conductor to ground plane. In Aperture coupling, two different substrates are separated by ground plane. In Proximity coupling, the length of stub and length to width ratio of patch controls the match.

3. Which of the following feed line contains two different substrates separated by a ground plane?

a) Microstrip line feed

b) Coaxial feed

c) Aperture coupling

d) Proximity Coupling

Answer: c

Explanation: In Aperture coupling, the ground plane separates two different substrates. In Microstripline feed, the feed is connected to the patch like a conducting strip. The coaxial feed consists an inner conductor attached to the patch and ground plane to the outer conductor. The length of stub and length to width ratio of patch is used for matching in proximity coupling.

4. Which of the following is true regarding the coaxial coupling feed in Microstrip antenna?

a) Inner conductor is connected to patch and outer conductor to ground plane

b) Outer conductor is connected to patch and inner conductor to ground plane

c) It is connected to patch just like a conducting strip

d) It contains two different substrates separated by a ground plane

Answer: a

Explanation: A coaxial feed consists of inner and outer conductors to which the patch and ground planes are attached respectively. In a Microstrip feed line the patch is attached to the feed like a conducting strip. Different substrates are used in aperture coupling.

5. Which of the following feed line uses the length of stub and L/W ratio to control the match?

a) Microstrip line feed

b) Coaxial feed

c) Aperture coupling

d) Proximity Coupling

Answer: d

Explanation: In Proximity coupling, the length of stub and length to width ratio of patch controls the match. Microstrip line feed acts as an extent to patch and is connected to patch just like a conducting strip. In coaxial feed, inner conductor is attached to patch and outer conductor to ground plane. In Aperture coupling, ground plane separates the different substrates.

6. Which of the following is the disadvantage of Microstrip line feeding?

a) Spurious feed radiation increases with increase in substrate thickness

b) Spurious feed radiation decreases with increase in substrate thickness

c) There is no Bandwidth limit

d) Low spurious radiation

Answer: a

Explanation: Microstrip line feed acts as an extent to patch and is connected to patch just like a conducting strip. So the substrate thickness may increases leading to the increase in spurious radiation and this limits the bandwidth.

7. Which of the following is the disadvantage of the coaxial feeding?

a) Fabrication is very difficult

b) Low spurious radiation

c) Narrow Bandwidth

d) No generation of higher order modes

Answer: c

Explanation: In coaxial feed, inner conductor is attached to patch and outer conductor to ground plane. Coaxial feeding fabrication is simple and has low spurious radiation which is its advantage. Due to asymmetries, higher modes are generated and produces cross polarization radiation.

8. Which of the following allows independent optimization of feed mechanism?

a) Microstrip line feed

b) Coaxial feed

c) Aperture coupling

d) Proximity Coupling

Answer: c

Explanation: In Aperture coupling, two different substrates are separated by ground plane. This type of arrangement optimizes feed mechanism and radiating element.In Proximity coupling, the length of stub and length to width ratio of patch controls the match. Microstrip line feed acts as an extent to patch and is connected to patch just like a conducting strip. In coaxial feed, inner conductor is attached to patch and outer conductor to ground plane.

9. Which of the following feeding has largest bandwidth?

a) Microstrip line feed

b) Coaxial feed

c) Aperture coupling

d) Proximity Coupling

Answer: d

Explanation: Proximity coupling feed provides largest bandwidth. Microstrip line feed limits the bandwidth due to increase in substrate thickness. Coaxial feed has narrow bandwidth.

10. In aperture coupling, ground plane is used to minimize the inference due to spurious radiation.

a) True

b) False

Answer: a

Explanation: In Aperture coupling, two different substrates are separated by ground plane. This type of arrangement optimizes feed mechanism and radiating element. The ground plane is between the substrates and isolates feed from the radiating element and minimizes the interference due to spurious radiation.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Applications”.

1. In mobiles, which of the following antenna is widely used?

a) Microstrip antenna

b) Horn antenna

c) Yagi-Uda antenna

d) Lens antenna

Answer: a

Explanation: Microstrip antenna is a printed antenna. It is fabricated using photolithography technique. It is small in size and is used in mobile communication widely. Horn, Yagi-Uda and Lens antennas are large in size.

2. One of the advantages of the Microstrip antenna compared to conventional microwave antenna is _________

a) Small size

b) Low gain

c) No surface wave excitation

d) High gain

Answer: a

Explanation: Compared to conventional microwave antennas like wire, lens, reflector antennas, the Microstrip antenna is small in size. It has low gain and has surface wave excitation which is a disadvantage.

3. In Microstrip antennas, the feed line and matching networks cannot be fabricated separated separately

a) False

b) True

Answer: a

Explanation: Microstrip antenna is a printed antenna. It consists of a radiation patch on one side of dielectric and a ground plane on other side. The patch is connected to the feed line. In this at fabrication only, the feed line and matching networks are fabricated simultaneously.

4. Which of the following is the application of Microstrip antenna in a telemedicine industry?

a) Wireless Body Area Network

b) Detection of moving targets

c) Rectenna application

d) WiMax

Answer: a

Explanation: A 2.4 GHz Wearable Microstrip antenna is used for the Wireless Body Area Network. Detection of moving targets comes under radar application. Rectenna is a special rectifying antenna used to converts directly the microwave energy to the DC power. WiMax is an IEEE 802.16 standard used for communication.

5. Rectenna Application of Microstrip antenna converts _______________

a) Microwave energy to DC power

b) Microwave energy to AC power

c) Microwave energy to solar energy

d) Inductive to capacitive

Answer: a

Explanation: Rectenna is a special rectifying antenna used to converts directly the microwave energy to the DC power. This is used in the long distance links. It is a combination of antenna, rectification filters and rectifiers.

6. One of the disadvantages of the Microstrip antenna is excitation of surface waves

a) True

b) False

Answer: a

Explanation: Microstrip antenna is a printed antenna. It consists of a radiation patch on one side of dielectric and a ground plane on other side. It has low gain and has surface wave excitation which is a disadvantage.

This set of Advanced Microstrip Antenna Questions and Answers focuses on “Numerical Tool for Antenna Analysis”.

1. Which of the following is not a technique to analyze the Microstrip antenna?

a) FEM

b) MoM

c) FDTD

d) PID

Answer: d

Explanation: FEM, MoM, FDTD are techniques used to know the Microstrip antenna performance. PI, PD, PID are control system mechanisms. MoM  is used to solve the mixed potential integrated equations obtained from the electromagnetic properties.

2. Which of the following is used to solve the mixed potential integrated equations?

a) MoM

b) FEM

c) FDTD

d) Jordan’s method

Answer: a

Explanation: MoM  is used to solve the mixed potential integrated equations obtained from the electromagnetic properties. FEM  and FDTD  solve the partial differential equations.

3. Which of the following technique in Microstrip antennas is used to solve the partial differential equations?

a) MoM but not FEM

b) FEM and MoM

c) FDTD and MoM

d) FEM and FDTD

Answer: d

Explanation: FEM  and FDTF  techniques are used in the Microstrip antenna to solve the partial differential equations. MoM  is used to solve the mixed potential integrated equations obtained from the electromagnetic properties.

4. Acronym of HFSS is for _______

a) High frequency structural stimulator

b) High frequency synthesis software

c) Hardware for stimulating system

d) High frequency system software

Answer: a

Explanation: HFSS is an Ansys tool used for the analysis of the antenna properties like radiation and designing antenna. HFSS is acronym used for High frequency structural stimulator. HFSS is a commercial FEM solver for the electromagnetic structure of the antenna.

5. Which of the following is a versatile technique to analyze the complex structure of Microstrip antenna in HFSS?

a) MoM

b) FEM

c) FDTD

d) Cavity Method

Answer: b

Explanation: HFSS is a High Frequency Structural Stimulator used as a commercial FEM solver for the electromagnetic properties of the antenna. FEM and FDTD solve the partial differential equations while MoM is used to solve the mixed potential integrated equations obtained from the electromagnetic properties.

6. In FEM, the region of interest is divided into a number of finite layers elements to analyze it

a) True

b) False

Answer: a

Explanation: FEM  and FDTF  solve the partial differential equations. HFSS is a commercial FEM solver for the electromagnetic structure of the antenna. Those finite layers can be rectangles or triangles. FEM is used for complex structural elements also.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Introduction”.

1. Which of the following is false regarding Antenna array?

a) Directivity increases

b) Directivity decreases

c) Beam width decreases

d) Gain increases

Answer: b

Explanation: A single antenna provides low gain and less directivity. To increase the directivity antenna arrays are used. With the antenna arrays, directivity and gain increases and beam width decreases.

2. Electrical size of antenna is increased by which of the following?

a) Antenna Array

b) Decreasing the coverage area

c) Increasing the coverage area

d) Using a single antenna

Answer: a

Explanation: To increase the directivity antenna arrays are used. With the antenna arrays, directivity and gain increases and beam width decreases. The electrical size of the antenna is increased by placing an array antenna together to achieve high directivity.

3. For long distance communication, which of the property is mainly necessary for the antenna?

a) High directivity

b) Low directivity

c) Low gain

d) Broad beam width

Answer: a

Explanation: Long distance communication requires antenna with high directivity. To increase the directivity antenna arrays are used. With the antenna arrays, directivity and gain increases and beam width decreases.

4. Which of the following is false about the single antenna for long distance communication?

a) Enlarging may create side lobes

b) No side lobes

c) High directivity is required

d) High Gain is required

Answer: b

Explanation: High directive antennas are required for the long distance communications. The array of antennas is used to increase the directivity. The directivity can be increased by increasing the dimensions of antenna but it creates side lobes.

5. The electrical size of antenna is increased by antenna array to avoid size lobes compared to single antenna.

a) True

b) False

Answer: a

Explanation: Increasing the dimensions of antennas may lead to the appearance of the side lobes. So by placing a group of antennas together the electrical size of antenna can be increased. With the antenna arrays, directivity and gain increases and beam width decreases.

6. A uniform linear array contains _____________

a) N elements placed at equidistance and fed currents of equal magnitude and progressive phase shift

b) N elements at non-equidistance and fed currents of equal magnitude and progressive phase shift

c) N elements at equidistance and fed currents of unequal magnitude and progressive phase shift

d) N elements at equidistance and fed currents of unequal magnitude and equal phase shift

Answer: a

Explanation: An array is said to be linear if N elements are spaced equally long the line and is a uniform array if the current is fed with equal magnitude to all elements and progressive phase shift along the line. High directivity can be obtained by antenna array.

7. Total resultant field obtained by the antenna array is given by which of following?

a) Vector superposition of individual field from the element

b) Maximum field from individual sources in the array

c) Minimum field from individual sources in the array

d) Field from the individual source

Answer: a

Explanation: The total resultant field is obtained by adding all the fields obtained by the individual sources in the array. An Array containing N elements has the resultant field equal to the vector superposition of individual field from the elements.

8. If the progressive shift in antenna array is equal to zero then it is called _________

a) Broad side

b) End-fire

c) Yagi-uda

d) Fishbone antenna

Answer: a

Explanation: The total phase difference of the fields is given by Ѱ=kdcosθ+β

Here β is the progressive phase shift

⇨ β=0, array is a uniform broadside array

⇨ β=180, array is a uniform end-fire array

Yagi-uda antenna, fishbone antenna are end-fire antenna array.

9. What is the progressive phase shift of the end-fire array?

a) 0

b) 90

c) 180

d) 60

Answer: c

Explanation: The progressive phase shift of the end-fire array is 180°. It is a linear array whose direction of radiation is along the axis of the array. For a broadside array it is 0°.

10. Which of the following statement about antenna array is false?

a) Field pattern is the product of individual elements in array

b) Field pattern is the sum of individual elements in array

c) Resultant field is the vector superposition of the fields from individual elements in array

d) High directivity can be achieved for long distance communications

Answer: b

Explanation: The total resultant field is obtained by adding all the fields obtained by the individual sources in the array. Radiation pattern is obtained by multiplying the individual pattern of the element. Field pattern is the product of individual elements in array. Antenna arrays are used to get high directivity with less side lobes.

This set of Antennas Multiple Choice Questions & Answers  focuses on “N-element Linear Array”.

1. Which of the following statement is true?

a) As the number of elements increase in array it becomes more directive

b) As the number of elements increase in array it becomes less directive

c) Point to point communication is not possible with more number of array elements

d) There is no uniform progressive phase shift in linear uniform array

Answer: a

Explanation: To get a single beam for the point to point communication more number of array elements is used. It increases the directivity of the antenna. An array is said to be uniform if the elements are excited equally and there is a uniform progressive phase shift.

2. Normalized array factor of N –element linear array is ________

a) \

ᴪ

ᴪ

 

 \

ᴪ

ᴪ

 

 \

ᴪ

ᴪ

 

 \(N\frac{cosᴪ}{Nᴪ/2} \)

Answer: a

Explanation: The N-element linear uniform array, having a constant phase difference will have the array factor \(AF = ∑_{n=1}^Ne^{jᴪ}\)

Normalized array factor is given by \(\frac{sinᴪ}{Nᴪ/2}. \)

3. Which of the following expression gives the nulls for the N- element linear array?

a) \

 

 

\)

b) \

 

 

\)

c) \

 

 

\)

d) \

 

 

\)

Answer: a

Explanation: To determine Null points the array factor is set equal to zero.

\

ᴪ

 

ᴪ

 

 

⇨

ᴪ

 

=0 \)

⇨ \

ᴪ

 

⇨

 

⇨

 

 

.\)

4. Maximum value of array factor for N-element linear array occurs at ______

a) \

 

\)

b) \

 

\)

c) \

 

\)

d) \

 

\)

Answer: a

Explanation: Normalized array factor is given by \

ᴪ

ᴪ

 

 

⇨

ᴪ

 

=0 \)

⇨ \

ᴪ

 

⇨⇨

 

.\)

5. Find the maximum value of array factor when elements are separated by a λ/4 and phase difference is 0?

a) θ m =cos -1 ⁡

b) θ m =sin -1 ⁡

c) θ m =cos -1 ⁡

d) θ m =sin -1 

Answer: a

Explanation: The maximum value of array factor is \

 

.\)

\

 

.\)

θ m =cos -1 ⁡.

6. Find the Nulls of the N-element array in which elements are separated by λ/4 and phase difference is 0?

a) \

 

\)

b) \

 

\)

c) \

 

\)

d) \

 

\)

Answer: a

Explanation: The nulls of the N- element array is given by \

 

 

\)

⇨ \

 

 

\)

⇨ \

 

\)

7. Find the Nulls of the 8-element array in which elements are separated by λ/4 and phase difference is 0?

a) \

 

\)

b) \

 

\)

c) \

 

\)

d) \

 

\)

Answer: a

Explanation: The nulls of the N- element array is given by \

 

 

\)

⇨ \

 

 

\)

⇨ \

 

\)

⇨ \

 

.\)

8. The radiating pattern of single element multiplied by the array factor simply gives the ___________

a) Pattern multiplication

b) Normalized array factor

c) Beamwidth of the array

d) Field strength of the array

Answer: a

Explanation: The radiation pattern of the single array antenna is multiplied by the antenna factor then it is called pattern multiplication. Array factor is the function of antenna positions in the array and its weights. Total array field is the field generated by the sum of the individual elements in array.

9. Condition for the half power width of the Array factor is given by ___________

a) \

ᴪ

 

 \

ᴪ

 

 \

ᴪ

 

 \(\frac{Nᴪ}{2}=±1\)

Answer: a

Explanation: Array factor is the function of antenna positions in the array and its weights. Half power beamwidth is also known as the 3 decibel points. The half power beam widths of the array factor will occur at \(\frac{Nᴪ}{2}=±1.391. \)

10. The maximum of the first minor lobe of array factor occurs at 13.46 dB down the maximum major lobe.

a) True

b) False

Answer: a

Explanation: The maximum of 1st minor lobe occurs at \(\frac{Nᴪ}{2}=±3π/2\)

⇨ \(AF = \frac{sin

ᴪ

 

}{\frac{Nᴪ}{2}}=0.212= -13.46dB.\)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Factors”.

1. The normalized array factor of a two element array antenna is given by ___________

a) \

 

\)

b) \

 

\)

c) \

 

\)

d) \

 

\)

Answer: a

Explanation: Array factor is the function of antenna positions in the array and its weights. The array factor for a two element array antenna is given by \

 

\)

The normalized array factor is given by \

 

 

.\)

2. Which of the following is a function of position of antennas in array and the weights?

a) Array Factor

b) Field pattern

c) Total array field

d) Beamwidth

Answer: a

Explanation: Array factor is the function of antenna positions in the array and its weights. The normalized array factor is given by \

 

 

\). Field pattern is multiplication of single element with the array factor. Total array field is the field generated by the sum of the individual elements in array.

3. Find the normalized Array factor when two antenna elements are separated by a distance of λ/4 and phase difference is 0 and θ=0?

a) \

 

\)

b) \

 

\)

c) \

 

\)

d) \

 

\)

Answer: a

Explanation: The normalized array factor is given by \

 

 

\)

⇨ \

 

=cos⁡

 

 

=cos⁡

 

\)

4. Which of the following pattern represents the array factor of a two element array separated by a distance of λ/4 and phase difference is 0?

a) 

b) 

c) 

d) 

Answer: a

Explanation: The normalized array factor is given by \

 

 

\)

⇨ AF n =0

⇨ \

 

=0\)

⇨ ⁡\

 

 

=\frac{π}{2}\)

⇨ cosθ=2

No Nulls will occur for the given pattern. So the pattern is represented as

5. Which of the following is true for uniform linear array elements, to obtain the total field?

a) The single element field is multiplied by the array factor

b) The single element field is multiplied by the normalized array factor

c) The single element field is multiplied by the beamwidth

d) The single element field is multiplied by the directivity

Answer: a

Explanation: Total array field is the field generated by the sum of the individual elements in array and is given simple by multiplying the field due to single element by the array factor. Array factor is the function of antenna positions in the array and its weights. Multiplying the normalized field with the normalized array factor gives the pattern multiplication.

6. Multiplying the normalized field with the normalized array factor gives ___________

a) pattern multiplication

b) array factor

c) beamwidth

d) null

Answer: a

Explanation: The radiation pattern of the single array antenna is multiplied by the antenna factor then it is called pattern multiplication. So multiplying the normalized field with the normalized array factor gives the pattern multiplication. Array factor is the function of antenna positions in the array and its weights. Nulls are known by equating array factor to zero.

7. Find the angle at which nulls occur for the two element array antenna with separation λ/4 and phase difference is π/2?

a) 0

b) π/2

c) π/4

d) π

Answer: a

Explanation: The normalized array factor is given by \

 

 

\)

⇨ AF n =0

⇨ \

 

=0\)

⇨ ⁡\(\frac{

 

cosθ+\frac{π}{2}}{2}=\frac{π}{2}\)

⇨ Cosθ=1

⇨ θ=0

So Nulls occur at 0°.

8. Find the angle at which nulls occur for the two element array antenna with separation λ/4 and phase difference is -π/2?

a) 0

b) π/2

c) π/4

d) π

Answer: d

Explanation: The normalized array factor is given by \

 

 

\)

⇨ AF n =0

⇨ \

 

=0\)

⇨ \(\frac{

 

 

cosθ-\frac{π}{2}}{2}=±\frac{π}{2}\)

⇨ Cosθ=-1

⇨ θ=180 or π

So Nulls occur at 180°.

9. Find the angle at which nulls occur for the two element array antenna with separation λ/4 and phase difference is 0?

a) Doesn’t exist

b) 0

c) π/2

d) π/4

Answer: a

Explanation: The normalized array factor is given by \

 

 

\)

⇨ AF n =0

⇨ \

 

=0\)

⇨ \(\frac{

 

cosθ+0}{2}=\frac{π}{2}\)

⇨ cosθ=2

No Nulls will occur for the given pattern.

10. For N- element linear uniform array, the normalized array factor is represented as ______________

a) \

ᴪ

ᴪ

 

 \

ᴪ

ᴪ

 

 \

ᴪ

ᴪ

 

 \(N\frac{cosᴪ}{Nᴪ/2}\)

Answer: a

Explanation: The N-element linear uniform array, having a constant phase difference will have the array factor AF = \(∑_{n=1}^Ne^{jᴪ}\)

Normalized array factor is given by \(\frac{sinᴪ}{Nᴪ/2}\)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Pattern Multiplications”.

1. Pattern multiplication is the multiplication of single array radiation pattern with ___

a) Array Factor

b) Beamwidth

c) Total Field

d) Directivity

Answer: a

Explanation: The radiation pattern of the single array antenna is multiplied by the antenna factor then it is called pattern multiplication. Array factor is the function of antenna positions in the array and its weights.Total array field is the field generated by the sum of the individual elements in array.

2. Which of the following expression represents the pattern multiplication?

a) E total =E θ × Array Factor

b) E total =E θ × Directivity

c) E total =E all elements × Array Factor

d) E total =E θ × Beamwidth

Answer: a

Explanation: The pattern multiplication is the product of the radiation pattern of the single array antenna by the antenna factor. This helps to sketch the radiation pattern of the entire array. An Array factor is defined as the function of antenna positions in the array and its weights.

3. All the elements must be identical to apply pattern multiplication principle.

a) True

b) False

Answer: a

Explanation: This pattern multiplication principle works only for the identical elements in the array. This is one of the disadvantages of the pattern multiplication. This is used to sketch the radiation pattern of the entire array.

4. Which of the following is the resultant pattern obtained by pattern multiplication principle for the figure shown below?

a) 

b) 

c) 

d) 

Answer: a

Explanation:The radiation pattern of the single array antenna is multiplied by the antenna factor then it is called pattern multiplication. At nulls the final pattern will also have same nulls from bath patterns.

5. Which of the following statements is false?

a) Pattern multiplications gives use the radiation pattern of the array

b) All elements must be identical to apply pattern multiplication principle

c) The radiation pattern of the single array antenna is multiplied by the antenna factor then is pattern multiplication

d) Pattern multiplication is also applicable for array with different unequal elements

Answer: d

Explanation: The pattern of the individual array element is multiplied by the array factor. This is pattern multiplication. To apply this principle all elements must be identical. This shows the radiation pattern of the entire array.

6. Which of the following is the resultant pattern obtained by pattern multiplication principle for the figure shown below?

a) 

b) 

c) 

d) 

Answer: a

Explanation: The radiation pattern of the single array antenna is multiplied by the antenna factor then it is called pattern multiplication. At nulls the final pattern will also have same nulls from bath patterns.

7. Which of the following doesn’t applicable for pattern multiplication?

a) It is the product of the individual radiation pattern of the element with the array factor of the array

b) All elements must be identical

c) All elements need not be identical

d) This gives the radiation pattern of the entire array

Answer: c

Explanation: The pattern of the individual array element is multiplied by the array factor. This is pattern multiplication. To apply this principle all elements must be identical. This shows the radiation pattern of the entire array.

8. For two elements array find the radiation pattern of array separated by λ/4 and phase difference is 0?

a) 

b) 

c) 

d) 

Answer: a

Explanation: Consider the far field pattern, equation the field pattern to zero to get nulls.

E total =E θ ×Array Factor=0

E θ =0 =>cosθ=0,θ=±π/2

Or

⇨ Array Factor=0

⇨ AF n =0

⇨ \

 

=0\)

⇨ \(\frac{

 

cosθ+0}{2}=\frac{π}{2}\)

⇨ cosθ=2

9. The total radiation pattern of a two element array, elements separated by a distance λ/4 and phase difference \

 

 ____________

a) 0, ±π/2

b) π, ±π/2

c) π, ±π/4

d) π/4, ±π/2

Answer: a

Explanation: The normalized array factor is given by \

 

 

\)

⇨ AF n =0

⇨ \

 

=0\)

⇨ \(\frac{

 

cosθ+\frac{π}{2}}{2}=\frac{π}{2}\)

⇨ Cosθ=1

⇨ θ=0

So Nulls occur at 0°.

For far field, the E θ =0 => cosθ=0, θE θ =0 cosθ=0, θ=±π/2

Therefore nulls occur atθ=±\(\frac{π}{2}, \) 0.

10. The isotropic element array radiation pattern depends on the nulls of the array factor only.

a) True

b) False

Answer: a

Explanation: For isotropic elements it radiated n all directions. The radiation pattern of the array is the pattern of the individual array element is multiplied by the array factor. So its total radiation pattern depends on the nulls of the array factor only.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Radiation Pattern for 4-Isotropic Elements”.

1. The array factor of 4- isotropic elements of broadside array is given by ____________

a) \

 

 \

 

 \

 

 \(\frac{cos}{2kdcosθ} \)

Answer: a

Explanation: Normalized array factor is given by

\(AF=\frac{sinᴪ}{N \frac{ᴪ}{2}}\)

And ᴪ=kdcosθ+β

Since its given broad side array β=0,

ᴪ=kdcosθ+β=kdcosθ

\(\frac{Nᴪ}{2}=2kdcosθ\)

\(AF=\frac{sinᴪ}{N \frac{ᴪ}{2}}=\frac{sin}{2kdcosθ} \)

2. A 4-isotropic element broadside array separated by a λ/2 distance has nulls occurring at ____________

a) \

 

 \)

b) \

 

 \)

c) \

 

 \)

d) \

 

 \)

Answer: a

Explanation: The nulls of the N- element array is given by

\

 

 

=cos^{-1}⁡

 

 

\)

⇨ \

 

 

=cos^{-1} 

 

=cos^{-1} 

 

 \)

3. A 4-isotropic element broadside array separated by a λ/4 distance has nulls occurring at ____________

a) cos -1 

b) \

 

\)

c) \

 

\)

d) sin -1 

Answer: a

Explanation: The nulls of the N- element array is given by

\

 

 

=cos^{-1}⁡

 

 

\)

⇨ \

 

 

=cos^{-1} 

 

=cos^{-1}  \left[n=1,2,3 \,and \,n≠N,2N…\right]\)

4. The array factor of 4- isotropic elements of broadside array separated by a λ/4 is given by ____________

a) sinc

b) cos

c) sin

d) sin

Answer: a

Explanation: Normalized array factor is given by \

ᴪ

ᴪ

 

 

ᴪᴪ

ᴪ

 

 

 

cosθ=πcosθ \)

\

ᴪ

ᴪ

 

 

 

.\)

5. The array factor of 4- isotropic elements of broadside array separated by a λ/2 is given by ____________

a) sinc

b) sin

c) sinc

d) sin

Answer: a

Explanation: Normalized array factor is given by \

ᴪ

ᴪ

 

 

ᴪᴪ

ᴪ

 

 

 

cosθ=2πcosθ \)

\

ᴪ

ᴪ

 

 

 

.\)

6. What is the direction of first null of broadside 4-element isotropic antenna having a separation of λ/2?

a) 60°

b) 30°

c) 180°

d) 0°

Answer: a

Explanation: The nulls of the N- element array is given by

\

 

 

=cos^{-1}⁡

 

 

\)

⇨ \

 

 

=cos^{-1} 

 

=cos^{-1}

 

 \)

⇨ \ cos^{-1} 

 

=cos^{-1} 

 

=60° or 120°. \)

7. What is the direction of first null of broadside 4-element isotropic antenna having a separation of λ/4?

a) 0

b) 60

c) 30

d) 120

Answer: a

Explanation: The nulls of the N- element array is given by

\

 

 

=cos^{-1}⁡

 

 

\)

\

 

 

=cos^{-1} 

 

=cos^{-1}=0\)

8. The necessary condition for maximum of the second side lobe of n element array is __________

a) \

ᴪ

 

 

 \

ᴪ

 

 

 \

ᴪ

 

 

 \( \frac{Nᴪ}{2}=±\frac{4π}{2}\)

Answer: a

Explanation: The secondary maxima occur when the numerator of the array factor equals to 1.

⇨ \

ᴪ

 

=±1\)

⇨ \(\frac{Nᴪ}{2} =±\frac{2s+1}{2}π\)

⇨ \(\frac{Nᴪ}{2}=±\frac{5π}{2}\) [s=2 for second minor lobe].

9. The direction of the first minor lobe of 4 element isotropic broadside array separated by λ/2 is ___________

a) 41.4°

b) 30°

c) 60°

d) 90°

Answer: a

Explanation: The direction of the secondary maxima  occur at θ s

\

 

 

\)

⇨ \

 

 

\) (s=1 for 1 st minor lobe)

⇨ \

 

=41.4°\)

10. A 4-isotropic element end-fire array separated by a λ/4 distance has first null occurring at ____________

a) 60

b) 30

c) 90

d) 150

Answer: c

Explanation: The nulls of the N- element array is given by \

 

 

\)

Since its given broad side array \

 

 

 

 

 

\)

=cos -1 

First null at n=1; θ n =cos -1 ⁡ 

 

 \)

θ n = cos -1 ⁡  or cos -1 ⁡

θ n = cos -1 ⁡ =90.

This set of Antenna Array Questions & Answers for Exams focuses on “Radiation Pattern of 8-Isotropic Elements”.

1. The array factor of 8 – isotropic elements of broadside array is given by ____

a) \

 

 \

 

 \

 

 \(\frac{cos}{2kdcosθ} \)

Answer: b

Explanation: Normalized array factor is given by \(AF=\frac{sinᴪ}{N \frac{ᴪ}{2}}\)

And ᴪ=kdcosθ+β

Since its given broad side arrayβ=0,

ᴪ=kdcosθ+β=kdcosθ

\(\frac{Nᴪ}{2}=4kdcosθ\)

\(AF=\frac{sinᴪ}{N \frac{ᴪ}{2}}=\frac{sin}{4kdcosθ} \)

2. An 8-isotropic element broadside array separated by a λ/2 distance has nulls occurring at ____

a) \

 

\)

b) \

 

\)

c) \

 

\)

d) \

 

\)

Answer: a

Explanation: The nulls of the N- element array is given by

\

 

 

=cos^{-1}⁡

 

 

\)

⇨ \

 

 

=cos^{-1} 

 

=cos^{-1} 

 

 \)

3. An 8-isotropic element broadside array separated by a λ/4 distance has nulls occurring at ____

a) cos -1 

b) \

 

\)

c) \

 

\)

d) \

\)

Answer: b

Explanation: The nulls of the N- element array is given by

\

 

 

=cos^{-1}⁡

 

 

\)

⇨ \

 

 [±\frac{2πn}{N}])=cos^{-1} 

 

=cos^{-1} 

 

 [n=1,2,3 \,and\, n≠N,2N…]\)

4. The array factor of 8- isotropic elements of broadside array separated by a λ/4 is given by ____

a) sinc

b) cos

c) sin

d) sinc

Answer: d

Explanation: Normalized array factor is given by

\

ᴪ

ᴪ

 

 

ᴪᴪ

ᴪ

 

 

 

cosθ=2πcosθ\)

\

ᴪ

ᴪ

 

 

 

.\)

5. The array factor of 8 – isotropic elements of broadside array separated by a λ/2 is given by ____

a) sinc

b) sin

c) sinc

d) sin

Answer: a

Explanation: Normalized array factor is given by \

ᴪ

ᴪ

 

 

ᴪᴪ

ᴪ

 

 

 

cosθ=4πcosθ \)

\

ᴪ

ᴪ

 

 

 

.\)

6. What is the direction of first null of broadside 8-element isotropic antenna having a separation of λ/2?

a) 60°

b) 75.5°

c) 37.5°

d) 57.5°

Answer: b

Explanation: The nulls of the N- element array is given by

\

 

 

=cos^{-1}⁡

 

 

\)

⇨ \

 

 

=cos^{-1} 

 

=cos^{-1} 

 

 \)

⇨ \ cos^{-1} 

 

=cos^{-1} 

 

=75.5°. \)

7. What is the direction of first null of broadside 8-element isotropic antenna having a separation of \frac{λ}{4}?

a) 0

b) 60

c) 30

d) 120

Answer: b

Explanation: The nulls of the N- element array is given by

\

 

 

=cos^{-1}⁡

 

 

\)

\

 

 

 

=cos^{-1} 

 

=cos^{-1} 

 

=cos^{-1} =60\)

8. The necessary condition for maximum of the first side lobe of n element array is ______

a) \

ᴪ

 

 

 \

ᴪ

 

 

 \

ᴪ

 

 

 \(\frac{Nᴪ}{2}=±\frac{4π}{2}\)

Answer: b

Explanation: The secondary maxima occur when the numerator of the array factor equals to 1.

⇨ \

ᴪ

 

=±1\)

⇨ \(\frac{Nᴪ}{2}=±\frac{2s+1}{2} π \)

⇨ \(\frac{Nᴪ}{2}=±\frac{3π}{2}\) [s=1 for first minor lobe].

9. The direction of the first minor lobe of 8 element isotropic broadside array separated by λ/2 is ___

a) 41.4°

b) 76.6°

c) 67.7°

d) 90°

Answer: b

Explanation: The direction of the secondary maxima  occur at θ s

\

 

 

\)

⇨ \

 

 

 \) (s=1 for 1 st minor lobe)

⇨ \

 

=67.7°\)

10. An 8-isotropic element end-fire array separated by a λ/4 distance has first null occurring at ____

a) 60

b) 30

c) 90

d) 150

Answer: a

Explanation: The nulls of the N- element array is given by \

 

 

\)

Since its given broad side array \

 

 

 

 

 

\)

\

 

\)

First null at n=1; \

 

 

 

 \)

\

 

\,or \,cos^{-1}  \)

\

 

=60.\)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Broadside Array”.

1. In Broadside array the maximum radiation is directed with respected to the array axis at an angle____

a) 90°

b) 45°

c) 0°

d) 180°

Answer: a

Explanation: In a Broadside array the maximum radiated is directed towards the normal to the axis of the array. So it is at angle 90°. In the end-fire array maximum radiation is along the axis of the array.

2. What is the phase excitation difference for a broadside array?

a) 0

b) π/2

c) π

d) 3π/2

Answer: a

Explanation: The maximum array factor occurs when \(\frac{sin \frac{Nφ}{2}}{\frac{Nφ}{2}}\) maximum that is \(\frac{Nφ}{2}=0.\) And φ=kdcosθ+β

=> kdcosθ+β=0 For a broadside maximum radiation is normal to axis of array so θ=90

=> β=0

3. Which of the following statements is false regarding a broadside array?

a) The maximum radiation is normal to the axis of the array

b) Must have same amplitude excitation but different phase excitation among different elements

c) The spacing between elements must not equal to the integral multiples of λ

d) The phase excitation difference must be equal to zero

Answer: b

Explanation: Since the phase excitation difference is zero it means that all are equally excited with same phase. In a Broadside array the maximum radiated is directed towards the normal to the axis of the array. The spacing between elements is not equal to integral multiples of λ to avoid grating lobes.

4. Which of the following cannot be the separation between elements in a broadside array to avoid grating lobes?

a) 4λ/2

b) λ/2

c) 3λ/2

d) 5λ/2

Answer: a

Explanation: The spacing between elements should not equal to integral multiples of λ to avoid grating lobes. The option 4λ/2=2λ

So when d=2λ grating lobes occurs which means maxima are found at other angles also. So this is not a desired spacing.

5. Find the value θ n at which null occurs for an 8-element broadside array with spacing d.

a) \

 

 \

 

 \

 

 \(sin^{-1}⁡\frac{2λn}{Nd}\)

Answer: a

Explanation: Nulls occurs when array factor \(AF=\frac{sin \frac{Nφ}{2}}{\frac{Nφ}{2}} = 0\)

⇨ \(sin\frac{Nφ}{2} = 0 =>\frac{Nφ}{2}=±nπ \,and\, φ=kdcosθ+β=kdcosθ_n=\frac{2π}{λ} dcosθ_n\)

⇨ Null occurs at \(θ_n=cos^{-1}⁡\frac{λn}{Nd}\)

6. What would be the directivity of a linear broadside array in dB consisting 5 isotropic elements with element spacing λ/4?

a) 9.37

b) 3.97

c) 6.53

d) 3.79

Answer: b

Explanation: Directivity \

 

 

 

 = 10log2.5=3.97 dB

7. In a broadside array all the elements must have equal ______ excitation with similar amplitude excitations to get maximum radiation.

a) Phase

b) Frequency

c) Voltage

d) Current

Answer: a

Explanation: Since the phase excitation difference is zero it means that all are equally excited with same phase. So in order to get maximum radiation it should have equal phase excitations along with similar amplitude excitations.

8. The directivity of a linear broadside array with half wave length spacing is equal to _____

a) Unity

b) Zero

c) Half of the number of elements present in array

d) Number of elements present in array

Answer: d

Explanation: The directivity of N isotropic elements with spacing d is given by

Directivity \(D=\frac{2Nd}{λ} \)

⇨ \(D=\frac{2Nd}{λ}=\frac{2Nλ/2}{λ}=N\)

9. Which of the following is false regarding a linear broadside array with 2 elements and spacing λ?

a) Directivity = 6.02 dB

b) No grating lobes are present

c) Nulls occur at \

 

 The maxima occurs normal to the axis of array and also at other angles

Answer: b

Explanation: Since the spacing between elements is an integral multiple of λ , grating lobes occurs.

Directivity \(D=\frac{2Nd}{λ}=4=6.02dB\)

⇨ Null occurs at \(θ_n=cos^{-1}⁡\frac{λn}{Nd}=cos^{-1}⁡\frac{1}{2} \)

10. What is the radiation pattern of a broadside array when array element axis coincides with the 0° line?

a) 

b) 

c) 

d) 

Answer: a

Explanation: In a Broadside array the maximum radiated is directed towards the normal to the axis of the array. Broadside array is a bidirectional antenna and when axis coincides with 00 then maximum radiation is at 90°.

This set of Antenna Array Questions and Answers for Aptitude test focuses on “Array of N-Isotropic Sources”.

1. The direction of nulls for broadside array of N –Isotropic sources is given by _____

a) \

 

\)

b) \

 

\)

c) \

 

\)

d) \

 

\)

Answer: a

Explanation: The nulls of the N- element array is given by \

 

 

\)

Given it’s a broadside array so β=0

\

 

 

= cos^{-1}⁡

 

.\)

2. The direction of first null of the broadside array of N-Isotropic sources is _____

a) \

 

\)

b) \

 

\)

c) \

 

\)

d) \

 

\)

Answer: a

Explanation: The nulls of the N- element array is given by \

 

 

\)

Given it’s a broadside array so β=0 and n=1 for first null

\

 

 

= cos^{-1}⁡

 

=cos^{-1}⁡

 

\)

3. The direction of nulls for end-fire array of N –Isotropic sources separated by λ/4 is given by ____

a) \

 

\)

b) \

 

\)

c) \

 

\)

d) \

 

\)

Answer: a

Explanation: The nulls of the N- element array is given by \

 

 

\)

Since its given broad side array \

 

 

 

 

 

\)

\

 

\)

4. The necessary condition for the direction of maximum side lobe level of the N-element isotropic array is _______

a) \ᴪ

 

 \ᴪ

 

 \ᴪ

 

 \(ᴪ=±\frac{2}{N} π\)

Answer: a

Explanation: The secondary maxima occur when the numerator of the array factor equals to 1.

⇨ \

ᴪ

 

=±1\)

⇨ \(\frac{Nᴪ}{2}=±\frac{2s+1}{2} π \)

⇨ \(ᴪ=±\frac{2s+1}{N} π .\)

5. The necessary condition for the direction of maximum first side lobe level of the 8-element isotropic array is _______

a) \

 

 \

 

 \

 

 \(\frac{5}{8} π\)

Answer: a

Explanation: The secondary maxima occur when the numerator of the array factor equals to 1.

⇨ \

ᴪ

 

=±1\)

⇨ \(\frac{Nᴪ}{2}=±\frac{2s+1}{2} π \)

⇨ \(ᴪ=±\frac{2s+1}{N}π=\frac{2+1}{8} π=\frac{3}{8} π.\)

6. The necessary condition for the direction of maximum second side lobe level of the 4-element isotropic array is _______

a) \

 

 \

 

 \

 

 \(\frac{5}{8} π\)

Answer: a

Explanation: The secondary maxima occur when the numerator of the array factor equals to 1.

⇨ \

ᴪ

 

=±1\)

⇨ \(\frac{Nᴪ}{2}=±\frac{2s+1}{2} π \)

⇨ \(ᴪ=±\frac{2s+1}{N}π=\frac{2+1}{4} π=\frac{5}{4} π.\)

7. The Half-power beam width of the N-element isotropic source array can be known when _____

a) \ᴪ

 

 \ᴪ

 

 \ᴪ

 

 \(ᴪ=\frac{3}{N}\)

Answer: a

Explanation: Normalized array factor is given by \(AF=\frac{sinᴪ}{N \frac{ᴪ}{2}}=\frac{1}{√2} \)

⇨ \(\frac{Nᴪ}{2}=1.391\)

⇨ \(ᴪ=\frac{2.782}{N} \)

8. Which of the following is the necessary condition to avoid grating lobes in N-Isotropic element array?

a) \

 

 

 \

 

 

 \

 

 

 \(\frac{λ}{d}=\frac{1}{1+|cosθ_m |} \)

Answer: a

Explanation: Grating lobes are the minor and unnecessary lobes other than the major lobe.

To avoid grating lobes, kd(cosθ-cosθ m ) ≤ 2π

θ m – Direction of maximum radiation

⇨ \

 

≤2π\)

\(\frac{d}{λ}≤\frac{1}{cosθ-cosθ_m} \)

\(\frac{d}{λ}≤\frac{1}{1+|cosθ_m |} \)

9. Which of the following is the necessary condition to avoid grating lobes in N-Isotropic element broadside array?

a) d < λ

b) d > λ

c) d=λ

d) d < 2λ

Answer: a

Explanation: Grating lobes are the minor and unnecessary lobes other than the major lobe.

To avoid grating lobes, kd(cosθ-cosθ m )≤2π

\(\frac{d}{λ}≤\frac{1}{1+|cosθ_m |} \)

For broadside to avoid grating lobes (θ m =90)

⇨ \(\frac{d}{λ}\) < 1

⇨ d < λ

10. Which of the following is the necessary condition to avoid grating lobes in N-Isotropic element end-fire array?

a) d < λ/2

b) d < λ

c) d > λ/2

d) d=λ

Answer: a

Explanation: Grating lobes are the minor and unnecessary lobes other than the major lobe.

To avoid grating lobes, kd(cosθ-cosθ m )≤2π

\(\frac{d}{λ}≤\frac{1}{1+|cosθ_m|} \)

For broadside to avoid grating lobes (θ m =0)

⇨ \(\frac{d}{λ}\) < 1/2

⇨ d < λ/2

This set of Antennas Multiple Choice Questions & Answers  focuses on “End Fire Array”.

1. The direction of maximum radiation in end-fire array is ______ with respect to the array axis.

a) 0° or 180°

b) 90°

c) 45°

d) 270°

Answer: a

Explanation: In an End-fire array the maximum radiation is along the axis of the array. So it is at either 0° or 180°. In broad-side array the maximum radiation is perpendicular to the axis of array that is at 90°.

2. What is the phase excitation difference for an end-fire array?

a) 0

b) ±kd /2

c) π

d) ±kd

Answer: d

Explanation: The maximum array factor occurs when \(\frac{sin \frac{Nφ}{2}}{\frac{Nφ}{2}}\) maximum that is \(\frac{Nφ}{2}=0. \)

And φ=kdcosθ+β

=> kdcosθ+β=0 For an end-fire array maximum radiation is along the axis of array so

θ=0° or 180°

=> β=kd when θ=180°

=> β=-kd when θ=0°

3. The phase excitation difference is zero in end-fire array.

a) True

b) False

Answer: b

Explanation: In end-fire array the phase excitation difference is ±kd and their phase vary progressively and get unidirectional maximum radiation finally. In broadside side array the phase excitation difference is zero.

4. What is the phase excitation difference in end-fire array with array spacing d at θ=0°?

a) \

 

 \

 

 \

 

 \(\frac{π}{λ} d \)

Answer: a

Explanation: In end-fire array the phase excitation difference is –kd for θ=0°.

kdcosθ+β=0

β=-kd

\(β=-\frac{2π}{λ} d \)

5. Which of the following statements is true regarding end-fire array?

a) The necessary condition of an ordinary end-fire array is β=±kd+nπ

b) The phase excitation difference is zero

c) Same input current is fed through the array, but the phase excitation is varies progressively

d) Maximum radiation occurs at normal to the axis of array

Answer: c

Explanation: For end-fire array:

Phase excitation difference β=±kd

Maximum radiation occurs along the axis of the array that is at θ=0° or 180°.

Even though same input current s fed to the arrays of equal magnitude, the phase vary progressively along the line to get the unidirectional pattern.

6. What is the progressive phase excitation of an end-fire array with element spacing λ/4 at θ=180°?

a) \

 

 \

 

 π

d) \(\frac{π}{4}\)

Answer: a

Explanation: At θ=180°, for end-fire array progressive phase excitation=\(kd=\frac{2π}{λ} d=\frac{2π}{λ} \frac{λ}{4}=\frac{π}{2}\)

Therefore \(β=\frac{π}{2}\)

7. Find the overall length of an end-fire array with 10 elements and spacing λ/4.

a) \

 

 \

 

 \

 

 \(\frac{9λ}{2}\)

Answer: a

Explanation: For an N-element end-fire array, the overall length of the array is given by ρ=d

⇨ ρ=d==9 λ/4

8. Which of the following is the correct condition of an ordinary end-fire array?

a) β=±kd

b) β=kd

c) β > kd

d) β < ±kd

Answer: a

Explanation: For an end-fire array maximum radiation is along the axis of array so

θ=0° or 180°

=> β=kd when θ=180°

=> β=-kd when θ=0°

This set of Antennas Multiple Choice Questions & Answers  focuses on “Phased Array – Basics”.

1. In which of the following a single radiator source is used to feed antenna with different phase shifts?

a) Phased array

b) FMCW radar

c) Monopulse radar

d) CW radar

Answer: a

Explanation: In phased array radar a single radiator is used to feed antenna with different phase shifts and thus the desired beam width is obtained. The beam of phased array radar is a synthesis of the multiple antennas used at different phase shifts. In FMCW, Monopulse, CW radar multiple antennas are not usually used at different phase shifts.

2. Which of the following principle is used in the phased array radar?

a) Interference

b) Diffraction

c) Reflection

d) Refraction

Answer: a

Explanation: In phased array radar a single radiator is used to feed antenna array with different phase shifts. When the signals are in-phase the superposition occurs and out of phase cancellation takes place. Thus it is able to steer the beam to required direction.

3. Electronic scanning is more advantageous compared to mechanical scanning

a) True

b) False

Answer: a

Explanation: Electronic scanning is flexible and also requires low maintenance. Electronic scanning is composed of phase shifters, electrical circuits that steer the beam. So Electronic Scanning is better compared to mechanical scanning.

4. What is the phase shift of the array is the scan angle is 30° and spacing distance is λ/2?

a) \

 

 π

c) 2π

d) \(\frac{π}{4} \)

Answer: a

Explanation: The phase \(ᴪ=\frac{2π}{λ} dsinθ=\frac{2π}{λ} \frac{λ}{2} sin30=\frac{π}{2} \)

5. Which of the following equation gives the incremental phase shift of the phase array radar with d element spacing and scan angle θ?

a) \ᴪ

 

 \ᴪ

 

 \ᴪ

 

 \(ᴪ=\frac{π}{d} λsinθ\)

Answer: a

Explanation: The incremental phase shift of the phase array radar with d element spacing and scan angle θ is given by \(ᴪ=\frac{2π}{λ} dsinθ\). The phase shifters are electronically tuned to adjust from 0 to 2π.

6. In Frequency scanning, the phase of all radiators is constant for the frequency tuned at that instant

a) True

b) False

Answer: a

Explanation: In frequency scanning, for one frequency all radiators will be in phase. As the frequency is changed then a corresponding phase shift occurs and the beam is scanned.

7. In which of the following type of scanning a heterodyne is used?

a) IF Scanning

b) Frequency scanning

c) Phase scanning

d) Digital beam-forming

Answer: a

Explanation: At receiving the signals received are heterodyned to IF . The remaining methods of scanning like beam forming are applicable. In IF scanning the heterodyne is used. In frequency scanning the frequency is varied while in phase scanning the phase of the radiators is varied.

8. In which of the following scanning the phase of the radiator is constant at one frequency?

a) Phase scanning

b) Frequency scanning

c) Both Phase and frequency scanning

d) Neither Frequency nor Phase scanning

Answer: b

Explanation: In frequency scanning, for one frequency all radiators will be in phase. As the frequency is changed then a corresponding phase shift occurs and the beam is scanned. In phase scanning the phase shifters are used and radiators have different phases.

9. Which of the following is mainly used in electronic scanning?

a) Phase shifters

b) Amplifiers

c) Repeaters

d) Limiters

Answer: a

Explanation: Electronic scanning is flexible and also requires low maintenance. Electronic scanning is composed of phase shifters, electrical circuits that steer the beam. For the beam to change its direction, the phase of the radiators has to be changed and so the phase shifters used.

10. In which of the following scanning the phase of the radiators varies incrementally for scan?

a) Phase scanning

b) Frequency scanning

c) IF scanning

d) Beam forming

Answer: a

Explanation: In phased array radar a single radiator is used to feed antenna with different phase shifts and thus the desired beam width is obtained. In phase scanning the phase shifters are used and radiators have different phases. The incremental phase shift of the phase array radar with d element spacing and scan angle θ is given by \(ᴪ=\frac{2π}{λ} dsinθ. \)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Phased Array – Fed Types”.

1. Phase array with lens array come under which feeding system?

a) Optical-feed system

b) Constrained feed system

c) Sub array feed

d) Phase shifter

Answer: a

Explanation: Phased array with lens or reflector arrays comes under the optical feed system. Proper aperture illumination is obtained by the optical feed system. In constrained the power is divided into many elements under different steps.

2. Which of the following statement is true?

a) Lens array has different input and output radiators while same radiator is used for reflector array

b) There is no surface matching required for the lens array

c) In lens and reflect array same radiator is used for input and output

d) Lens and reflect array belong to constrained feed systems

Answer: a

Explanation: In Lens array, different input and output radiators are used and hence there two surface matching is required. In reflect array the same radiator is used to radiate and collect.Phased array with lens or reflector arrays comes under the optical feed system.

3. In which of the following feed system, the power division among elements on aperture is done in several steps?

a) Constrained feed systems

b) Optical feed systems

c) Lens array

d) Reflect array

Answer: a

Explanation: The power is divided into many elements under different steps in constrained feed systems. Phased array antennas with lens or reflector arrays belong to the optical feed system. Proper aperture illumination is obtained through the optical feed system and systems power is divided in one step.

4. Which of the following series feed has same bandwidth similar to parallel feed network?

a) End fed array

b) Center fed array

c) Both end fed and Center fed array

d) Neither End fed nor center fed array

Answer: b

Explanation: The Center fed array in the series feed network has same bandwidth as the parallel feed networks. The end fed array has several bandwidth restrictions.

5. In which of the following the feed is provided at the end of the array?

a) Series end fed

b) Series center fed

c) Parallel feed

d) Optical feed systems

Answer: a

Explanation: In series end fed, the radiator is placed at the end of the array and its electrical path length depends on the frequency. In center fed, radiator is provided at the center of the array and has bandwidth similar to the parallel feed network. In parallel feed, a number of radiators are combined into sub-arrays and then to elements in series or center fed.

6. In which of the following radiators are combined into sub-arrays and then fed to elements in end or center fed?

a) Parallel feed

b) Series feed

c) Optical feed system

d) Series end feed

Answer: a

Explanation: In parallel feed, a number of radiators are combined into sub-arrays and then to elements in series or center fed. In series end fed, the radiator is placed at the end of the array and its electrical path length depends on the frequency.In optical feed systems power is divided in one step.

7. In which of the following the power is divided in single step among the elements?

a) Optical feed system

b) Constrained feed system

c) Parallel feed system

d) Sub-array

Answer: a

Explanation: Proper aperture illumination is obtained by the optical feed system. In constrained the power is divided into many elements under different steps. In optical feed systems power is divided in one step. In parallel feed, a number of radiators are combined into sub-arrays and then to elements in series or center fed.

8. The power divider used to connect array elements to single port is called _____

a) array feed

b) phase shifter

c) frequency shifter

d) repeater

Answer: a

Explanation: The power divider used to connect array elements to single port is called array feed. Array feeds are classified into three types: Constrained feeds, Optical feed/ space feed, parallel plate feed.

9. Which of the following feed techniques the reflect array used particularly?

a) Space feed

b) Constrained feed

c) Parallel plate feed

d) Sub array

Answer: a

Explanation: In space feed, the energy is distributed to the lens array in a manner analogous to point feed illuminating a reflect antenna. In constrained the power is divided into many elements under different steps. In optical feed systems power is divided in one step. Parallel plate feed uses microwave structure principles and provides efficient power division.

10. A folded pillbox antenna is used in which of the following feeding system?

a) Parallel plate feed

b) Space feed

c) Optical feed system

d) Constrained feed system

Answer: a

Explanation: Folded pillbox antenna is used to provide the power distribution to antenna elements. So this is used in the parallel plate feed system. Lens and reflect array used in the space/optical feed systems. Constrained feed provides a series and parallel feed combination.

11. Parallel plate feed is a cross of constrained and space feed.

a) True

b) False

Answer: a

Explanation: Parallel plate feed uses microwave structure principles and provides efficient power division.In constrained the power is divided into many elements under different steps. In optical feed systems power is divided in one step. In space feed, the energy is distributed to the lens array.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Adaptive Array”.

1. In which of the following the noise interference is detected and is suppressed further?

a) Adaptive arrays

b) Phased array

c) Planar array

d) Sub-array

Answer: a

Explanation: The adaptive arrays are extensively used to detect the noise interference in the desired signal and suppress them. It is a combination of array sensors and real time adaptive signal receiver processor.

2. The noise detection and suppression is possible with adaptive arrays.

a) True

b) False

Answer: a

Explanation: The noise can be detected and can be suppressed in desired direction using the adaptive arrays. Adaptive arrays are combination of array sensors along real time adaptive signal receiver processor.

3. Adaptive arrays provide best results compared to switching beams under more interference.

a) True

b) False

Answer: a

Explanation: Switching beams provide intra cellular handoffs but also works best only at less or nil interference. The adaptive arrays can be used under more interference. These are used to detect the noise interference in the desired signal and suppress them. Adaptive array consists of array sensors and real time adaptive signal receiver processor.

4. In which of the following the nulls are introduced effectively when different other users interferes the desired user?

a) Adaptive array

b) Switching beam

c) Phased array

d) Sub-array

Answer: a

Explanation: The advanced digital signal processing in the adaptive arrays helps to reduce the interference compared to switch beams. These adaptive arrays detect the noise interference and also used to suppress it in desired direction. Phased array and sub array don’t provide this.

5. ___ is used to guide the beam in desired direction and place nulls in unwanted direction/interference.

a) Adaptive array

b) Switching beams

c) Phased radar

d) Pulse radar

Answer: a

Explanation: The adaptive arrays are extensively used to detect the noise interference in the desired signal and suppress them. The advanced digital signal processing in the adaptive arrays helps to reduce the interference compared to switch beams. Pulse radars are used to know the distance of target. Phased array radars steer the beam in desired direction.

6. Which of the following statements is false?

a) The intra cellular handoffs are less in adaptive array compared to switching beams

b) Noise interference cannot be detected in adaptive arrays effectively compared to switching beams

c) Advanced digital signal processors are used in adaptive arrays

d) Algorithms used in adaptive array is more complex than switching beams

Answer: b

Explanation: Adaptive array is used to guide the beam in desired direction and place nulls in unwanted direction/interference. More complex algorithms are used in adaptive arrays and also advance digital signal processors are used to different the desired user and the unwanted signals/users.

7. Which of the following is not a benefit of the adaptive array technology?

a) Co-channel interference reduction

b) Reduction in multipath interference

c) Increase in multipath interference

d) Reduction in handoffs

Answer: c

Explanation: Adaptive array place nulls in unwanted direction/interference and are used to guide the beam in desired direction. These are used to suppress the noise interference in desired direction. So, the handoffs, co-channel interference, multipath interference are reduced with the help of the advanced digital signal processors and the complex algorithms.

8. In which of the following a real time adaptive signal processor is used?

a) Adaptive array

b) Phased array

c) Pulse radar

d) CW radar

Answer: a

Explanation: In adaptive array a real time adaptive signal processor is used to detect and nullify the noise interference. Phased array is used to steer the beam. Pulse radar is used to know the target distance and CW radar is used to the velocity of the target.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Adaptive Array – Antenna Synthesis”.

1. Antenna synthesis is inverse process to antenna analysis.

a) True

b) False

Answer: a

Explanation: Antenna synthesis determines the distribution of input or output sources from the obtained radiation pattern. Antenna analysis is determining the radiation pattern for the given input distribution.

2. In which of the following methods nulls are introduced in the desired direction of pattern?

a) Schelkunoff

b) Taylor

c) Woodward

d) Fourier

Answer: a

Explanation: In Schelkunoff method, pattern is produced with null in specified directions. Woodward method is suited for beam shaping and Fourier is used when the complete pattern and excitation distributions locations are given. Taylor series have a constant inner minor lobes level and decreasing lobes for other.

3. In which of the following method is popularly used for beam shaping in antenna synthesis?

a) Woodward

b) Schelkunoff

c) Fourier

d) Taylor

Answer: a

Explanation: Woodward method is suited for beam shaping and the complete pattern and excitation distributions locations are given by the Fourier method. The nulls in specified pattern are produced by the Schelkunoff method. The Taylor synthesis has a constant inner minor lobes level.

4. Which of the following synthesis method have a constant inner minor lobe level and remaining one decreasing gradually?

a) Woodward

b) Schelkunoff

c) Fourier

d) Taylor

Answer: d

Explanation: Taylor synthesis has a constant inner minor lobes level and gradually decreasing lobes for other. Nulls are produced in a specified direction for the pattern using Schelkunoff method. The Woodward method is used for beam shaping in antenna synthesis while Fourier is used when the complete pattern and excitation distributions locations are given.

5. Antenna synthesis is analogous to _____

a) Antenna array

b) Antenna analysis

c) Antenna radiation

d) Schelkunoff synthesis

Answer: b

Explanation: Schelkunoff is one of the methods of antenna synthesis. Antenna synthesis determines the distribution of input or output sources from the obtained radiation pattern. Antenna analysis is determining the radiation pattern for the given input distribution.

6. Which of the following is not a technique of antenna synthesis?

a) Woodward

b) Schelkunoff

c) Fourier

d) Thinning

Answer: d

Explanation: Thinning is a process to reduce the array elements with degrading the performance. Woodward-Lawson, Schelkunoff, Fourier, Taylor are techniques of antenna synthesis.Antenna synthesis determines the distribution of input or output sources from the obtained radiation pattern.

7. Which of the following method is used for antenna design with no side lobes?

a) Binomial array

b) Dolph Chebyshev

c) Schelkunoff

d) Woodward

Answer: a

Explanation: Dolph Chebyshev is a special case  of Binomial array. Dolph Chebyshev is used to design array at desired side lobes. In Schelkunoff method, pattern is produced with null in specified directions. Woodward method is suited for beam shaping.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Adaptive Array – Chebyshev Arrays”.

1. Which of the following is the key feature of the Chebyshev array?

a) Have constant side lobes and elements are placed uniformly

b) Have varying side lobes and elements are placed uniformly

c) Minor lobes are present at |x| > 1

d) Major lobes are present at |x| < 1

Answer: a

Explanation: Chebyshev array exhibits the properties of the Chebyshev polynomial having constant amplitude 1-≤x≤1. Minor lobes are present at |x| < 1 andMajor lobes are present at |x| > 1.

2. In Chebyshev array, its array factor is approximated to Chebyshev polynomial.

a) True

b) False

Answer: a

Explanation: One of the properties exhibited by the Chebyshev array is having a constant amplitude range of 1-≤x≤1. Its array factor is approximated to Chebyshev polynomial. This array has constant side lobes and elements are placed uniformly.

3. _____________ is used to obtain array with any desired side lobe level.

a) Chebyshev

b) Schelkunoff

c) Woodward-Lawson

d) Binomial

Answer: a

Explanation: Chebyshev technique is used to obtain array with any desired side lobe level. In Schelkunoff method, pattern is produced with null in specified directions. Woodward method is suited for beam shaping. Binomial array is used when no side lobes are desired.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Adaptive Array – Chebyshev Polynomials Fundamentals”.

1. Which of the following is holds true for the Chebyshev polynomial?

a) T m  =  m T m 

b) T m  =  m T m-1 

c) T m  =  m T m 

d) T m  =  m T m-1 

Answer: a

Explanation:The Chebyshev polynomial is given by

T 0 =1 m=0

T 1 =x m=1

T 2 =2x 2 -1 m=2

T 3 =4x 3 -3x m=3

There fore T m = m T m  holds true.

2. The recurrence relation for the Chebyshev polynomial is __________

a) T m  = 2zT m-1  – T m-2 

b) T m  = T m-1  – 2zT m-2 

c) T m  = 2zT m  – T m-2 

d) T m  = 2zT m  – T m-1 

Answer: a

Explanation:The Chebyshev polynomial of any order m can be derived from the recursive formula. This is one of the main features of the Chebyshev polynomial. The recurrence relation for the Chebyshev polynomial is

T m  = 2zT m-1  – T m-2 .

3. The value of T m  for every odd value of m is _____________

a) 0

b) 1

c) -1

d) m

Answer: a

Explanation:The Chebyshev polynomial is given by

T 0  = 1 m=0

T 1  = x m=1

T 2  = 2x 2 -1 m=2

T 3  = 4x 3 -3x m=3

For m=odd; T m  = 0

For m=even; T m  =  m/2

4. What is the value of T m  when m is an even number?

a) 0

b)  m

c) -1

d)  m/2

Answer: d

Explanation: The Chebyshev polynomial is given by

T 0  = 1 m=0

T 1  = x m=1

T 2  = 2x 2 -1 m=2

T 3  = 4x 3 -3x m=3

For m=even; T m  =  m/2

5. The value of T 0  is _____

a) 1

b) 100

c) 10000

d) 200

Answer: a

Explanation:The Chebyshev polynomial is given by

T 0 =1 m=0

⇨ T 0 =1.

6. The value of T m =_______

a)  m

b)  m-1

c)  m/2

d)  2m

Answer: a

Explanation:The Chebyshev polynomial is given by

T 0  = 1 m=0

T 1  = x m=1

T 2  = 2x 2 -1 m=2

T 3  = 4x 3 -3x m=3

ThereforeT m = m .

7. The value of T m  in the range between -1 to 1 is ____

a) -1 to 1

b) 1

c) 0

d) 0 to infinity

Answer: a

Explanation: The polynomial oscillates between -1 to 1 with amplitude varying from -1 to 1. Hence this is used to get the constant desired side lobe for any order within the desired range. The array factor of the Chebyshev array is approximated to the Chebyshev polynomial.

This set of Antennas Questions and Answers for Entrance exams focuses on “Adaptive Array – Chebyshev Polynomials Properties”.

1. Which of the following statement is true about the Chebyshev function T m ?

a) It is a continuously increasing function after x=1

b) It is a continuously decreasing function after x=1

c) It is a continuously increasing function after x=0

d) It is a continuously decreasing function after x=0

Answer: a

Explanation: T 0  = 1

T 1  = x

T 2  = 2x 2 -1

⇨ T 3  = 4x 3 -3x

This is the chebyshev polynomial and it increases continuously after x=1

2. How many times the polynomial T 5  crosses the x-axis between [-1, 1]?

a) 5

b) 4

c) 2

d) 6

Answer: a

Explanation: The polynomial T m  crosses the x axis m times between -1 and 1. Given polynomial is T 5 .

M=5 therefore it crosses the axis 5 times between [-1, 1].

3. Which of the following statements is true?

a) The polynomials are unstable at interval [-1, 1]

b) The polynomials are marginally stable at interval [-1, 1]

c) The polynomial doesn’t oscillate at interval [-1, 1]

d) The polynomials crosses the axis m-1 times at [-1, 1]

Answer: b

Explanation: The polynomials oscillate between -1 and 1 interval. So they are either stable or marginally stable. The polynomial crosses the axis m times in the interval [-1, 1].

4. What is the possible level from the following for the minor lobe when the main beam level is at 50db and SLL at 10 db according to Chebyshev?

a) 40dB

b) 45dB

c) 50dB

d) 80dB

Answer: a

Explanation: The minor lobes are present below the main beam level at a value 1/SLL. SLL is the side lobe level.

Possible level for the minor lobe is 50-10=40dB

5. The condition for the existence of the main lobe according to the Chebyshev is _________

a) |x| > 1

b) |x| < 1

c) |x| = 0

d) 2|x| > 1

Answer: a

Explanation: The condition for the existence of the main lobe according to the Chebyshev is |x| > 1.

The condition for the existence of the minor lobe according to the Chebyshev is |x| < 1.

6. The condition for the existence of the main lobe according to the Chebyshev is _________

a) |x| > 1

b) |x| < 1

c) |x| = 0

d) 2|x| > 1

Answer: a

Explanation: The condition for the existence of the main lobe according to the Chebyshev is |x| > 1.

The condition for the existence of the minor lobe according to the Chebyshev is |x| < 1.

7. Which of the following statements regarding Chebyshev polynomial is true?

a) The polynomial T m  is symmetric for m = even

b) The polynomial T m  is symmetric for m = odd

c) The polynomial T m  is anti-symmetric for m = even

d) The polynomial T m  is symmetric for m = even and odd

Answer: a

Explanation: The polynomial T m  is symmetric for m=even and is anti-symmetric for m=odd.

For m=even, at x=0 it is 1. For m=odd, at x=0 it is 0.

8. Which of the following properties of Chebyshev polynomial is false?

a) The minor lobes have unequal amplitudes

b) The polynomial T m  is symmetric for m = even

c) The polynomial T m  crosses the x axis m times between -1 and 1

d) Minor lobes exists for |x| < 1

Answer: a

Explanation: The minor lobes have equal amplitudes. The polynomial T m  crosses the x axis m times between -1 and 1.

Minor lobes exist for |x| < 1 and major lobes exist for |x| > 1.

9. All the polynomials of the order m pass through the point ____________

a) 

b) 

c) 

d) 

Answer: a

Explanation: All the polynomials of the order m of the chebyshev pass through the point (x, T m )

= .

T 0  = 1

T 1  = x

T 2  = 2x 2 -1

T 3  = 4x 3 -3x

10. All the nulls occur at  in the Chebyshev polynomial.

a) True

b) False

Answer: b

Explanation: All the nulls in the Chebyshev polynomial occur in the range -1≤ x≤1.

 here is representing a point. So it is false.

11. As the order of the polynomial increases, the slope becomes steeper.

a) True

b) False

Answer: a

Explanation: The Chebyshev polynomial is given by

T 0  = 1 m=0

T 1  = x m=1

T 2  = 2x 2 -1 m=2

T 3  = 4x 3 -3x m=3

As the order of the polynomial increases, the slope becomes steeper.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Adaptive Array – Dolph Pattern Method”.

1. Which of the following a method proposed to design array with any side lobes?

a) Dolph-Chebyshev

b) Binomial array

c) Taylor

d) Broad-side Array

Answer: a

Explanation: Dolph- chebyshev method is used to design array with desired side lobes. It uses the approximation of the Chebyshev polynomials. Binomial array is used to synthesis the patterns without side lobes. Taylor and broad-side array have side lobes.

2. Dolph pattern uses which of the following polynomial approximation?

a) Chebyshev

b) Binomial

c) Pascal’s

d) Taylor

Answer: a

Explanation: An array with desired side lobes can be designed by using the Dolph-chebyshev method. This method uses the approximation of the Chebyshev polynomials. Binomial array uses the Pascal’s triangle for amplitude excitations.

3. What should be the side lobe level to make Dolph Chebyshev to binomial array?

a) -∞ dB

b) 0 dB

c) 1 dB

d) 10dB

Answer: a

Explanation: Dolph –Chebyshev will become binomial array when there are no any side lobes.

So side lobes a =0

Side lobe level = major lobe level / minor lobe max =0

In dB = 20 log  = -∞ dB.

4. Which of the following represents the chebyshev recursive formula?

a) T m  = 2zT m-1  – T m-2 

b) T m  = zT m-1  – 2T m-2 

c) T m  = 2T m-1  – zT m-2 

d) T m  = T m-1  – 2zT m-2 

Answer: a

Explanation: The Chebyshev polynomial of any order m can be derived from the recursive formula. This is one of the main features of the Chebyshev polynomial. It is given by

T m  = 2zT m-1  – T m-2 .

5. What is the HPBW of the Dolph Chebyshev array if the broadening factor is 2 and HPBW of uniform array is 30°?

a) 30°

b) 60°

c) 90°

d) 120°

Answer: b

Explanation: HPBW of the Dolph Chebyshev array= broadening factor×HPBW of uniform array

=2×30°=60°.

6. In Chebyshev array which of the following is approximated to the Chebyshev polynomial?

a) Array factor

b) HPBW

c) Side lobe level

d) Range equation

Answer: a

Explanation: The array factor is approximated to the Chebyshev polynomial and the desired side lobe level is maintained similar to the polynomial where constant amplitude is maintained between -1 to 1. Dolph- chebyshev method is used to design array with desired side lobes.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Binomial Array”.

1. Which of the following array is used to synthesize pattern without side lobes?

a) Binomial array

b) Linear array

c) End-fire array

d) Chebyshev array

Answer: a

Explanation: Binomial array is used to synthesize patterns without side lobes. In Chebyshev there are side-lobes present below the major beam level. The excitation amplitudes of the binomial array follow a Pascal’s triangle.

2. The excitation amplitudes of the binomial array follow ____________

a) Pascal’s triangle

b) Bermuda triangle

c) Right angled triangle

d) Natural squares

Answer: a

Explanation: The excitation amplitudes of the binomial array follow a Pascal’s triangle.

   1

  1 1

 1 2 1  

1 3 3 1

The array factor AF =  N-1 and N is number of elements.

3. Which of following is the disadvantage of the binomial array?

a) It has low amplitude variations

b) It has high range of amplitude variations

c) It has a high HPBW

d) It has low HPBW

Answer: a

Explanation: Binomial array follows a Pascal’s triangle type of amplitude variations. So there are high variations in the amplitudes. Binomial array has high HPBW and hence more wide beams are present.

4. Binomial array has high HPBW compared to Dolph-chebychev.

a) True

b) False

Answer: a

Explanation: Binomial array has wide beams and so it has high HPBW. Its HPBW is high compared to Dolph-Chebyshev array. Binomial array is used to synthesize patterns without side lobes.

5. Which of the following can be the excitation amplitude of binomial array with 4 elements?

a) 1 3 3 1

b) 1 2 3 1

c) 2 1 1 2

d) 1 4 6 4 1

Answer: a

Explanation: The excitation amplitudes of the binomial array follow a Pascal’s triangle.

     1             N=1

    1 1          N=2 

   1 2 1        N=3  

  1 3 3 1     N=4

The array factor AF =  N-1 and N is number of elements.

6. Dolph- Chebyshev becomes a Binomial array when there are no side lobes.

a) True

b) False

Answer: a

Explanation: Binomial array is used to synthesize patterns without side lobes. Dolph- chebyshev method is used to design array with desired side lobes. When the side lobe of the Dolph drops to -∞ dB, it behaves like a binomial array.

7. What will be the array factor in polynomial form for the 3 element binomial array if Z=e^ᴪ?

a) 1+2z+z 2

b) 1+3z+3z 2 +z 3

c) 1+2z+3z 2 +z 3

d) 1+3z+z 2

Answer: a

Explanation: The array factor AF =  N-1 and N is number of elements.

AF =  3-1 =  2 = 1+2z+z 2

Binomial array follows the Pascal’s triangle coefficients

     1             N=1

    1 1          N=2 

   1 2 1        N=3  

  1 3 3 1     N=4.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Frequency Independent Antenna – Rumsey Principle”.

1. Which of the following is Rumsey’s principle?

a) The impedance & pattern of antenna is frequency independent if antennasize is specified in terms of angle

b) The impedance & pattern of antenna is frequency dependent if antennasize is specified in terms of angle

c) The impedance & pattern of antenna is frequency dependent if antennasize is specified in terms of radiated power

d) The impedance & pattern of antenna is frequency dependent if antennasize is variable

Answer: a

Explanation: The frequency independent antennas are fixed in size and operate over wide bandwidth with constant impedance, polarization and pattern. Rumsey’s principle states that the impedance & pattern of antenna is frequency independent if antennasize is specified in terms of angle.

2. Rumsey’s principle is related to which type of antennas?

a) Frequency independent antennas

b) Frequency dependent antennas

c) Reflector antennas

d) Both Frequency independent and independent antennas

Answer: a

Explanation: Rumsey’s principle is applicable to the frequency independent antennas where the antenna size is specified in terms of angle and thus making the impedance and pattern independent of frequency.

3. Which of the following is truefor a frequency independent antenna?

a) Dimensions of antenna are specified in terms of angles

b) Dimensions of antenna are specified in terms of varying wavelengths

c) The antennas have variable impedance, pattern over operating wide band frequency

d) It does not follow Rumsey’s principle

Answer: a

Explanation: Rumsey’s principle states that the impedance & pattern of antenna is frequency independent if antennasize is specified in terms of angle. The Rumsey’s principle is applicable for the frequency independent antennas.The frequency independent antennas are fixed in size and operate over wide bandwidth with constant impedance, polarization and pattern.

4. The impedance and pattern are frequency independent if antenna shape is specified in terms of angle.

a) True

b) False

Answer: a

Explanation: Rumsey’s principle is the working principle of the frequency independent antennas. According to it the impedance & pattern of antenna is frequency independent if antenna size is specified in terms of angle. The antennas have fixed size and operate over a wide bandwidth having constant impedance and polarization.

5. Which of the following property doesn’t meet a true frequency independent antenna?

a) Fixed physical size

b) Constant impedance, pattern

c) Operates over wide bandwidth

d) Variable physical size

Answer: d

Explanation: Rumsey’s principle is used for the frequency independent antenna. A true frequency independent antenna must have a fixed physical size and operates on instantaneous basics over a wide bandwidth at a constant impedance, pattern.

6. Impedance of self-complementary antenna is ____

a) Z o /2

b) Z o /4

c) Z o

d) 2Z o

Answer: a

Explanation: The self complementary antennas have constant impedance which is half of their intrinsic impedance at all frequencies. These have a metal area congruent to open area. Impedance of self-complementary antenna = Z o /2.

7. Which of the following principle is mainly used in frequency independent antennas?

a) Rumsey’s principle

b) Fabry perot

c) Brewster law

d) Archimedes principle

Answer: a

Explanation: Rumsey’s principle is used in the frequency independent antennas states that the impedance & pattern of antenna is frequency independent if antennasize is specified in terms of angle. Fabry perot is used with optical waves where rays are passed only when they are in resonance. Brewster law gives the relation to maximum polarization of light.

8. Which of the following does not belongs to frequency independent antenna?

a) Spiral antenna

b) Log periodic antenna

c) Helical antenna

d) Parabolic antenna

Answer: d

Explanation: Parabolic antenna is a reflector antenna. Spiral, log periodic,helical antenna are examples of frequency independent antennas. A true frequency independent antenna must have a fixed physical size and operates on instantaneous basics over a wide bandwidth at a constant impedance, pattern.

9. Constant impedance of a self complementary antenna is Z o /4.

a) True

b) False

Answer: b

Explanation: The self complementary antennas have constant impedance which is half of their intrinsic impedance at all frequencies. These have a metal area congruent to open area. Impedance of self-complementary antenna = Z o /2.

10. Value of self complementary antenna when Z o =120π?

a) 120π

b) 60π

c) 240π

d) 30π

Answer: b

Explanation: One of the properties of the self-complementary antennas is constant impedance. Its impedance is half of the intrinsic impedance at all frequencies. Impedance of self-complementary antenna = \(\frac{Z_o}{2} = 60π.\)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Frequency Independent Antenna – Helical Antenna”.

1. Which of the following modes does the condition πD << λ is satisfied in helical antenna?

a) Axial mode

b) Normal mode

c) Conical mode

d) Axial, Normal and Conical

Answer: b

Explanation: The circumference of the helical antenna C= πD varies with the wavelength in different modes as follows

In Normal mode: πD << λ

Axial mode: C ≈ λ

Conical mode: C ≈n λ

2. What is the pitch angle of the helical antenna to become a loop antenna?

a) 90

b) 0

c) 45

d) 60

Answer: b

Explanation: C is the circumference and L is the length, s is the spacing between turns.

When pitch angle α=0, the separation becomes zero and length will be equal to circumference of the loop. Hence it acts as a loop antenna.

3. What is the pitch angle if the separation of turns is at 2 cm and the circumference is of 4 cm?

a) 26.56

b) 62.65

c) 30

d) 25.34

Answer: a

Explanation: From the figure, \(α=tan^{-1}⁡\frac{S}{C}=tan^{-1}\frac{2}{4}=tan^{-1}⁡\frac{1}{2}=26.56\)

4. What is the necessary condition for a helical antenna to operate in the axial mode?

a) πD << λ

b) πD ≈ λ

c) πD ≈ nλ

d) πD=0

Answer: b

Explanation: In Normal mode: πD << λ

Axial mode: C ≈ λ

Conical mode: C ≈n λ

D=0 then there is no loop and is no longer helical.

5. What is the pitch angle of the helical antenna to become a linear antenna?

a) 90

b) 0

c) 45

d) 60

Answer: a

Explanation: C is the circumference and L is the length, s is the spacing between turns.

When pitch angle α=90, the separation between turns becomes parallel to length  hence it acts as a linear antenna.

6. What is the Input impedance of the axial feed helical antenna with circumference per unit wavelength is 1.2?

a) 168Ω

b) 186Ω

c) 198Ω

d) 158Ω

Answer: a

Explanation: Input impedance of helical antenna with axial feed is given by \(Z=140 \frac{C}{λ}\)

\(Z=140 \frac{C}{λ}=140×1.2=168\Omega\)

7. What is the total axial length of the helical antenna with 5 turns and circumference per wavelength is 1.2 and separation between turns is 2 cm?

a) 10cm

b) 20cm

c) 6cm

d) 12cm

Answer: a

Explanation: The total axial length of the helical antenna with N turns and separation distance between the turns S is given by A= NS

A=NS=5 * 2 = 10cm

8. The Axial ratio of the helical antenna with 3 turns is _____

a) 1.11

b) 0.89

c) 6

d) 0.98

Answer: a

Explanation: The axial ratio of helical antenna is given by \(AR = \frac{2N+1}{2N} \)

\(AR = \frac{2N+1}{2N} = \frac{7}{6} = 1.11\)

9. Find the number of turns in the helical antenna with directivity 120 and C λ =1 and pitch angle 6°.

a) 100

b) 50

c) 120

d) 12

Answer: a

Explanation: Directivity of helical antenna D = 12C λ nS λ

Pitch angle \(α=tan^{-1}⁡\frac{S_λ}{C_λ}\)

⇨ S λ =C λ tan α=1*0.1=0.1

D = 12C λ nS λ

120=12×1×n×0.1

n=100 turns

10. Which of the following type does helical antenna belongs to?

a) Wire type

b) Aperture

c) Reflector

d) Array

Answer: a

Explanation: Helical antenna belongs to wire type antenna. It is in the form of the helix and operates in the VHF and UHF. It is mainly a circular polarized antenna.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Frequency Independent Antenna – Modes of Radiation”.

1. Which of the following modes of radiation, the radiation of helical antenna is along the end-fire direction?

a) Axial mode

b) Normal mode

c) Perpendicular mode

d) Beam mode and Normal mode

Answer: a

Explanation: In Axial mode of radiation, the radiation is in end-fire direction along the helical axis. In Axial mode of radiation, the dimensions of antenna are approximately equal to the wavelength and spacing space of λ/4. Axial mode is also known as beam mode or end-fire mode. In normal mode, the radiation is perpendicular to axis.

2. In which of the following mode, the radiation is perpendicular to the axis of helical antenna?

a) Normal mode

b) Axial mode

c) Beam mode

d) End-fire mode

Answer: a

Explanation: In normal mode, the radiation is perpendicular to axis. It is also known as perpendicular mode. In this the antenna dimensions are smaller compared to the wavelength. Axial mode is also known as beam mode or end-fire mode. In Axial mode of radiation, the radiation is in end-fire direction along the helical axis.

3. In which of the following mode the dimensions of the antenna must be smaller compared to the wavelength?

a) Normal mode

b) Axial mode

c) Beam mode

d) End-fire mode

Answer: a

Explanation: In normal mode, the radiation is perpendicular to axis. It is also known as perpendicular mode. In this the antenna dimensions are smaller compared to the wavelength.Axial mode is also known as beam mode or end-fire mode. In Axial mode of radiation, the radiation is in end-fire direction along the helical axis.

4. Drawbacks of the normal mode of radiation in helical antenna are _______

a) low radiation efficiency and narrow bandwidth

b) high radiation efficiency and narrow bandwidth

c) low radiation efficiency and Broad bandwidth

d) high radiation efficiency and broad bandwidth

Answer: a

Explanation: In normal mode, the radiation is perpendicular to axis. The radiation pattern of a helical antenna is a combination of small dipole and loop antenna. In normal mode of radiation, it provides narrow bandwidth and low radiation efficiency.

5. Which of the following is true regarding the axial mode of radiation in helical antenna?

a) Radiation is broad, directional along axis and produces minor lobes

b) Radiation pattern is narrow and no minor lobes are produced

c) Minor lobes are produced only along the direction of helical axis

d) The dimensions of antenna are smaller than the wavelength

Answer: a

Explanation: In Axial mode of radiation, the radiation is in end-fire direction along the helical axis. The radiation pattern of the antenna is broad, directional along axis and produces minor lobes at oblique angles. The dimensions of antenna are approximately in the order of the wavelengths.

6. In which of the following mode, the helical antenna has narrow bandwidth?

a) Normal mode

b) Axial mode

c) Beam mode

d) End-fire mode

Answer: a

Explanation: The radiation pattern of a helical antenna is a combination of small dipole and loop antenna. In normal mode of radiation, it provides narrow bandwidth and low radiation efficiency. Axial mode is also known as beam mode or end-fire mode.

7. In which of the following mode, there are no minor lobes at oblique angles in radiation pattern?

a) Normal mode

b) Axial mode

c) Conical mode

d) Beam mode

Answer: a

Explanation: In normal mode of radiation, it provides narrow bandwidth and low radiation efficiency. In Axial mode of radiation, the radiation is in end-fire direction along the helical axis. The radiation pattern of the antenna is broad, directional along axis and produces minor lobes at oblique angles.

8. In which of the following modes, the dimension of the helical antenna circumference is greater than the wavelength?

a) Normal mode

b) Axial mode

c) Conical mode

d) Beam mode

Answer: c

Explanation: In Axial mode of radiation, the dimensions of antenna are approximately in the order of the wavelengths. In normal mode, dimensions of antenna are smaller compared to wavelength. In conical mode, the circumference of the antenna is greater than the wavelength.

9. The radiation pattern of the helical antenna is a combination of small dipole and loop antenna.

a) True

b) False

Answer: a

Explanation: The helical antenna radiation pattern is a combination of small dipole and loop antenna.If the diameter of the helical antenna is zero  then it is linear antenna and when spacing length becomes zero  then it acts like a loop antenna.

10. If the spacing between turns of the helical antenna is zero then it acts as a ________

a) Loop antenna

b) Linear dipole

c) Parabolic antenna

d) Log periodic antenna

Answer: a

Explanation: The radiation pattern of a helical antenna is a combination of small dipole and loop antenna. The helical antennas behave like a loop antenna when spacing between turns becomes zero. It becomes a linear dipole when the diameter is zero that means itch angle is zero. Parabolic is a reflector antenna. Helical antenna is a wire antenna.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Frequency Independent Antenna – Normal Mode of Radiation”.

1. In which of the following the radiation of helical antenna is broadside?

a) Normal mode

b) Axial mode

c) Conical mode

d) Beam mode

Answer: a

Explanation: In normal mode of radiation, the radiation is broadside direction. In Axial mode of radiation, it is in end-fire direction. Axial mode is also known as Beam mode.

2. Which of the following is the condition for the normal mode of radiation?

a) The dimensions of antenna must be smaller compared to the wavelength

b) The dimensions of antenna must be approximately equal to the wavelength

c) The dimensions of antenna must be greater compared to the wavelength

d) The dimensions of antenna must be approximately equal to or greater than the wavelength

Answer: a

Explanation: In normal mode, the radiation is normal to the axis of the antenna hence it is broadside directed. The dimensions of the antenna are small compared to the wavelength in normal mode. In axial mode, it is approximately equal to the wavelength. In conical mode, it is greater than wavelength.

3. In which of the following mode of radiation low efficiency and narrow bandwidth is observed?

a) Normal mode

b) Axial mode

c) Conical mode

d) Beam mode

Answer: a

Explanation: The radiation of broadside antenna is normal to the axis line. The radiation of antenna in normal mode is perpendicular to the axis of the antenna and hence it is broadside directed. The drawbacks of the normal mode are its low radiation efficiency and narrow bandwidth.

4. In normal mode of radiation, far field pattern is independent of number of turns in helical antenna.

a) True

b) False

Answer: a

Explanation: In normal mode, the dimensions of the antenna are small compared to the wavelength. So the current magnitude and phase is constant overall the length. The far field pattern is independent of the number of turns. It considers as a single loop.

5. Which of the following is false regarding normal mode of radiation of helical antenna?

a) The dimensions of the antenna are small compared to the wavelength

b) The far field pattern is independent of the number of turns

c) The current is assumed to vary along the length of the antenna

d) The radiation is normal to the direction of the axis

Answer: c

Explanation: For a helical antenna operating in normal mode, the radiation is normal to the direction of the axis. The antenna dimensions are small compared to the wavelength. So all over the length, the current magnitude and phase is constant. The pattern of far field is independent of the number of turns and it considers as a single loop.

6. Normal mode of radiation is limited in use because _____

a) it has narrow bandwidth and dependency on small dimensions of the antenna compared to wavelength

b) it has broad bandwidth and independent of antenna dimensions

c) it has narrow bandwidth and is independent of antenna dimensions

d) it has broad bandwidth and dependency on large dimensions of the antenna compared to wavelength

Answer: a

Explanation: The normal mode of radiation depends on the size of the antenna. Its critical dependency on dimensions of antenna which has to be less than the wavelength is difficult and so it causes for narrow bandwidth and low radiation efficiency.

7. In normal mode, the radiation is directed along ______

a) broadside

b) end-fire

c) along the axis

d) in all directions

Answer: a

Explanation: The normal mode of radiation radiates in the direction normal to the axis of the antenna. So radiation is directed in broadside. In Axial mode, it is directed along the end-fire .

8. For a helical antenna radiation, the pitch angleα should be _____

a) 0 < α < 90

b) α=0

c) α=90

d) α=0 or 90

Answer: a

Explanation: The radiation pattern of the helical antenna is a combination of the small dipole and loop antenna. Pitch angle is the tangent of ratio between the spacing and the circumference of the antenna. α=0, then helical antenna becomes loop antenna. α=90, the helical antenna becomes linear antenna.

9. Which of the following is true about normal mode of radiation?

a) The dimensions of the antenna are small compared to the wavelength

b) The radiation is along the axis of the antenna

c) It has high radiation efficiency

d) It has broad bandwidth

Answer: a

Explanation: The normal mode of radiation depends on the size of the antenna. Its critical dependency on dimensions of antenna which has to be less than the wavelength is difficult and so it causes for narrow bandwidth and low radiation efficiency. Radiation is normal to the direction of axis.

10. In normal mode of operation, the radiation is along the end-fire direction.

a) True

b) False

Answer: b

Explanation: In normal mode of radiation, radiation is normal to the direction of axis and is directed in the broadside. The dimensions of the antenna are small compared to the wavelength. In Axial mode, it is end-fire directed.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Frequency Independent Antenna – Log Periodic Antenna”.

1. Which of the following is a narrow band antenna?

a) Rhombic antenna

b) Yagi-Uda antenna

c) Log periodic antenna

d) Horn antenna

Answer: b

Explanation: Yagi-Uda antenna is a narrow band antenna. Rhombic, Log periodic and horn are wide band antennas. More number of channels is used in the Log periodic antenna compared to the Yagi-Uda in TV reception.

2. Log periodic antenna uses which range of frequencies?

a) VHF and UHF

b) VHF and SHF

c) MF and VHF

d) HF and VHF

Answer: a

Explanation: Log periodic antenna uses around 30MHz to 3GHz frequency range.

MF – Medium Frequency – 300-3000 KHz

HF – High Frequency – 3-30 MHz

VHF – Very high frequency – 30-300 MHz

UHF – Ultra high frequency – 300MHz to 3GHz

So it uses VHF and UHF.

3. Log periodic antenna is frequency dependent.

a) True

b) False

Answer: b

Explanation: Log periodic antenna is frequency independent. The geometrical structure of the antenna changes according to the wavelength. It is also called as Frequency Independent antenna. Its electrical performance is dependent on logarithmic of frequency only.

4. Which of the following antenna is known as Frequency Independent antenna?

a) LPDA

b) Dipole antenna

c) Rhombic antenna

d) Yagi-Uda antenna

Answer: a

Explanation: LPDA means Log periodic Dipole Array antenna. It’s all characteristics such as impedance, radiation pattern are frequency independent. So it is also called as Frequency independent antenna.

5. What is the approximate gain of LPDA for best performance?

a) 1 dB to 5 dB

b) 4 dB to 18 dB

c) 8 dB to 20 dB

d) 7 dB to 12 dB

Answer: d

Explanation: For best performance of the LPDA antenna, the gain lies between 7 to 12dB. It is a broadband antenna. LPDA is a wideband antenna and provides gain and directivity combined over a wide band of frequencies.

6. In order to get more number of channels in TV reception, we prefer Yagi-Uda than LPDA.

a) True

b) False

Answer: b

Explanation: LPDA is a broadband antenna while Yagi-Uda is a narrow band antenna. To increase the more number of channels for TV reception we prefer LPDA. It is also used for long distance communication.

7. Which of the following statements is false?

a) Log periodic antenna is a broadband antenna

b) The active region doesn’t change with the change in frequency in Log Periodic antenna

c) The geometry structure of the dipole changes proportional to the wavelength in LPDA

d) Impedance is function of logarithmic of frequency

Answer: b

Explanation: In Log period antenna, the active region changes with the change in frequency. There are three regions of operation in the LPDA.Impedance is function of logarithmic of frequency in LP so it is called log periodic antenna.

8. Which of the following best suits to describe the impedance of a log periodic antenna?

a) It is function of anti-logarithmic of frequency

b) It is dependent on frequency directly

c) It is a function of logarithmic of inverse frequency

d) It is a function of logarithmic of frequency

Answer: d

Explanation: Impedance is function of logarithmic of frequency. So it is called log periodic antenna. Both the impedance and standing wave ratio are functions of logarithmic of frequency.

9. The geometrical design of the Log periodic depends on _____

a) Scaling factor

b) Gain

c) Impedance

d) Stability

Answer: a

Explanation: Scaling factor is the ratio of the lengths and spacing’s of the dipoles. It is also called as periodicity factor. If r n denotes the length of the nth dipole then the scaling factor is given by

Scaling factor = \(\frac{r_n}{r_{n+1}}\)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Frequency Independent Antenna – Log Periodic Antenna Introduction”.

1. Frequency range of LPDA is ________

a) 3 MHz to 30 MHz

b) 30 KHz to 30 GHz

c) 30 MHz to 3 GHz

d) 300 KHz to 300GHz

Answer: c

Explanation: Log periodic antenna uses around 30MHz to 3GHz frequency range.

MF – Medium Frequency – 300-3000 KHz

HF – High Frequency – 3-30 MHz

VHF – Very high frequency – 30-300 MHz

UHF – Ultra high frequency – 300MHz to 3GHz

So it uses VHF and UHF.

2. Which of the following band of frequency the LPDA operates?

a) VHF & UHF

b) UHF & MF

c) MF & HF

d) HF & UHF

Answer: a

Explanation: Log periodic antenna uses VHF and UHF band.

MF and UHF – 300 kHz to 3GHz

MF and HF – 300 kHz to 30MHz

HF and UHF – 3MHz to 3GHz

LPDA operates at 30MHz to 3GHz frequency range.

3. In frequency independent antennas, the antennas are defined in terms of _________

a) Angles

b) Wavelengths

c) Range

d) Frequency

Answer: a

Explanation: If the antennas are defined only in terms of angles, then they are frequency independent antennas. Its impedance and radiation pattern are independent of the frequency. Example of frequency independent antennas are log periodic antennas, spiral antennas.

4. The relation between the dipole lengths and the scaling factor τ in LPDA is given by _________

a) \

 

 \

 

 

 \

 

 \(\frac{L_{N+2}}{L_N} =τ\)

Answer: a

Explanation: The electrical properties of the log periodic antenna are repeated periodically in terms of logarithmic frequency. The relation between the antenna dipole adjacent lengths and the scaling factor is given as \(\frac{L_N}{L_{N+1}} =τ. \)

5. Which of the following statement is false?

a) LPDA is a frequency independent antenna

b) In LPDA the lengths of the dipoles increases from the apex of the feed line toward other end

c) The included angle varies as the length of the dipole changes from the apex of feed line

d) The spacing between adjacent dipoles and their lengths are in same ratio

Answer: c

Explanation: The electrical properties of the LPDA are periodically repeated in terms of logarithmic frequency. The relation between the antenna dipole spacing lengths or adjacent lengths and the scaling factor is given as\(\frac{L_N}{L_{N+1}} =\frac{S_N}{S_{N+1}} = τ.\) LPDA is a frequency independent antenna and the lengths of the dipoles increases from the apex of the feed line toward other end by maintaining constant included angle.

6. The ratio of adjacent spacing of dipoles and its lengths in the LPDA are not equal.

a) True

b) False

Answer: b

Explanation: The ratio of adjacent spacing of dipoles and its lengths in the LPDA are equal. The electrical properties of the log periodic antenna are repeated periodically in terms of logarithmic frequency. The relation between the antenna dipole adjacent lengths, spacing and the scaling factor is given as\(\frac{L_N}{L_{N+1}} =\frac{S_N}{S_{N+1}} =τ.\)

7. The relation between the dipole spacing and the scaling factor τ in LPDA is given by _________

a) \

 

 \

 

 

 \

 

 \(\frac{S_{N+2}}{S_N} = τ\)

Answer: a

Explanation: The electrical properties of the log periodic antenna are repeated periodically in terms of logarithmic frequency. The relation between the antenna dipole adjacent spacing and the scaling factor is given as \(\frac{S_N}{S_{N+1}} = τ. \)

8. Which of the following expression is correct?

a) \

 

 

 \

 

 

 \

 

 

 

 \(\frac{L_N}{L_{N+1}} = \frac{S_N}{S_{N+1}} = \frac{1}{τ} \)

Answer: a

Explanation: The ratio of adjacent spacing of dipoles and its lengths in the LPDA are equal. The electrical properties of the log periodic antenna are repeated periodically in terms of logarithmic frequency. The relation between the antenna dipole adjacent lengths, spacing and the scaling factor is given as \(\frac{L_N}{L_{N+1}} = \frac{S_N}{S_{N+1}} = τ.\)

This set of Antennas Questions and Answers for Freshers focuses on “Frequency Independent Antenna – Working of LPDA”.

1. Which of the following is the condition for the Log periodic antenna to be in active region?

a) \

 

 \

 

 \

 

 \(L = \frac{λ}{4} \)

Answer: a

Explanation: Based on the lengths of the dipoles in LPDA, operation region is classified into three types:

Inactive transmission line region \(L < \frac{λ}{2} \)

Active region \(L=\frac{λ}{2} \)

Inactive stop region \(L > \frac{λ}{2}. \)

2. Which of the following is the condition for the Log periodic antenna to be inactive stop region?

a) \

 

 \

 

 \

 

 \(L = \frac{λ}{4} \)

Answer: b

Explanation: Based on the lengths of the dipoles in LPDA, operation region is classified into three types:

Inactive transmission line region, Active region and Inactive stop region. The region below \(L < \frac{λ}{2} \) belongs to Inactive transmission region. The region above \(L > \frac{λ}{2} \) is Inactive stop region.

3. Which of the following is the condition for the Log periodic antenna to be inactive transmission line region?

a) \

 

 \

 

 \

 

 \(L = \frac{λ}{4} \)

Answer: c

Explanation: The LPDA has dipoles of varying length along the axis. The dipole lengths of the LPDA less than half the wavelength belongs to the Inactive transmission line region. \(L < \frac{λ}{2}. \)

4. The dipoles in the inactive transmission region offer ________ impedance.

a) Capacitive

b) Resistive

c) Inductive

d) Characteristic

Answer: a

Explanation: Depending on the length of the dipoles, in different regions they offer different resistances.

In Inactive transmission line region \(L < \frac{λ}{2}\) offers Capacitive Impedance

Active region \(L=\frac{λ}{2} \) offers Resistive Impedance

Inactive stop region \(L > \frac{λ}{2} \) offers Inductive impedance.

5. The dipoles in the active region offer ________ impedance.

a) Capacitive

b) Resistive

c) Inductive

d) Characteristic

Answer: b

Explanation: The different regions of operation in the LPDA offer different kind of reactance. In Inactive transmission line region it offers Capacitive Impedance, in active region it offers Resistive Impedance and Inactive stop region it offers Inductive impedance.

6. The dipoles in the inactive stop region offer ________ impedance.

a) Capacitive

b) Resistive

c) Inductive

d) Characteristic

Answer: c

Explanation: The varying length of the dipoles in LPDA forms different regions and they offer different resistances. The Inactive transmission line region at \(L < \frac{λ}{2}\) offers Capacitive Impedance, Active region at \(L=\frac{λ}{2} \) offers Resistive Impedance and Inactive stop region \(L > \frac{λ}{2} \) offers Inductive impedance.

7. In which of the following region in LPDA, the current leads the base voltage?

a) Inactive transmission line region

b) Active region

c) Inactive stop region

d) Reflective region

Answer: a

Explanation: In LPDA, the length of dipole \(L < \frac{λ}{2}\) comes under inactive transmission region. These region dipoles offer capacitive impedance. So the current leads the base voltage in the inactive transmission line region. Active region \(L=\frac{λ}{2} \) offers Resistive Impedance and Inactive stop region \(L > \frac{λ}{2} \) offers Inductive impedance.

8. In which of the following region of the log periodic array, maximum radiation takes place?

a) Inactive transmission line region

b) Active region

c) Inactive stop region

d) Reflective region

Answer: b

Explanation: In LPDA, the length of dipole \(L=\frac{λ}{2} \) comes under active region. These region dipoles offer resistive impedance. Here currents are large and are in phase with the base voltage. Inactive stop region is also called as reflective region.In Inactive transmission line region \(L < \frac{λ}{2}\) offers Capacitive Impedance and Inactive stop region \(L > \frac{λ}{2} \) offers Inductive impedance.

9. Which of the following region is also known as reflective region in LPDA?

a) Inactive transmission line region

b) Active region

c) Inactive stop region

d) Log periodic array

Answer: c

Explanation: Inactive stop region \(L > \frac{λ}{2} \) offers Inductive impedance. Any small incident wave gets reflected easily by large inductive impedance hence this region is also known as reflective region.

10. Reflective region in LPDA offers the inductive impedance.

a) True

b) False

Answer: a

Explanation: The Reflective region is also known as the Inactive stop region. The Inactive stop region \(L > \frac{λ}{2}\) offers Inductive impedance. This region is called reflective region because any small incident wave gets reflected easily by large inductive impedance and thus offers inductive impedance.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Frequency Independent Antenna – Design of LPDA”.

1. What is the ratio of maximum frequency to minimum frequency if the scaling factor is 0.5 for 5 elements LPDA?

a) 16

b) 2

c) 4

d) 8

Answer: a

Explanation: The relation between frequency ratio and the scaling factor is given by

\(\frac{f_{max}}{f_{min}} = \frac{1}{τ^{N-1}} \)

\(\frac{f_{max}}{f_{min}} =\frac{1}{τ^{N-1}} =\frac{1}{^{5-1}}=2^4=16. \)

2. The value of periodicity factor in LPDA is _____

a) < 1

b) >1

c) ≥1

d) =0

Answer: a

Explanation: The periodicity factor also known as the scaling factor is the ratio of the adjacent lengths of the dipole \(\frac{L_N}{L_{N+1}} = τ \,and\, L_N < L_{N+1}\) where L is the length of the dipole.

3. The longest and shortest dipole lengths are taken in the form of the wavelength of the operating frequencies in LPDA.

a) True

b) False

Answer: a

Explanation: The longest length of the dipole is taken as the λ u /2 where λ u is the wavelength corresponding to upper frequency and shortest dipole length is taken as λ l /2. In LPDA, the adjacent dipole spacing ratios and their length ratios are equal.

4. The relation between the spacing factor σ and the scaling factor τ is given by ____

a) \

 

 \)

b) \

 

 \)

c) \

 

 \)

d) \

 

 \)

Answer: a

Explanation: In LPDA, the ratio of successive spacing of elements is equal to the ratio of adjacent dipole lengths. The spacing factor is \(σ=\frac{d_n}{2L_n} \,and\, d_n =\) spacing betwwen adjacent elements ‘n’ and ‘n+1’ and L n is the length of n th dipole. The relation between the spacing factor σ and the scaling factor τ is \

 

. \)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Antenna Ranges”.

1. Which of the following antenna range is used for small antennas?

a) Indoor ranges

b) Outdoor ranges

c) Reflection range

d) Slant range

Answer: a

Explanation: Indoor ranges are used for the small antenna as the far field region criteria is achieved in limited space. Space restriction is found in indoor range measurement. Outdoor range is used for larger antenna and in open environment. Reflection range, Slant range and elevation range comes under outdoor ranges.

2. Which of the following doesn’t come under indoor ranges?

a) Slant range

b) Anechoic chamber

c) Tampered chamber

d) Compact ranges

Answer: a

Explanation: For small antennas whose far filed criterion is obtained within in a limited range, indoor ranges are applicable. Reflection range, Slant range and elevation range belongs to outdoor ranges which is applicable for the larger antennas in open environment. Anechoic, tampered chambers and compact ranges come under indoor ranges.

3. What is the frequency of antennas used for Reflection range?

a) 300MHz to 16GHz

b) 3 MHz to 16MHz

c) 16MHz to 300MHz

d) 3GHz to 30GHz

Answer: a

Explanation: Reflection range, Elevation range and Slant range comes under outdoor ranges. The reflection range is an outdoor antenna range test where the antennas are operated at UHF and 16GHz. Ground acts as a reflector.

4. Which of the following doesn’t come under outdoor ranges?

a) Reflection range

b) Anechoic chamber

c) Slant range

d) Elevation range

Answer: b

Explanation: Reflection range, Slant range and elevation range belongs to outdoor ranges. Anechoic, tampered chambers and compact ranges are part of indoor ranges. Indoor ranges are used for the small antenna as the far field region criteria is achieved in limited space and outdoor ranges for larger antennas.

5. Height criteria required for mounting AUT under elevated range test is ______

a) h r ≤ 4D

b) h r ≥ 4D

c) h r ≥ 2D

d) h r ≤ 2D

Answer: b

Explanation: Elevated range test is one of the outdoor ranges. The height required to mount the AUT for the elevation range test is given by h r ≥ 4D, where D is the dimension of AUT.

6. If the dimension of antenna under test  is 2m the minimum height required to mount it for elevation range is _______

a) 8m

b) 8cm

c) 4m

d) 2m

Answer: a

Explanation: The minimum height required to mount antenna under test for the elevation range is given by a relation h r ≥ 4D, where D is the dimension of AUT.

Minimum height required h r =4D=4*2=8m

7. For AUT to encounter uniform illumination which of the following is true?

a) Height of AUT should be equal to or more than four times the dimension of antenna

b) Height of AUT should be less than four times the dimension of antenna

c) Height of AUT should be equal to or more than two times the dimension of antenna

d) Height of AUT should be less than two times the dimension of antenna

Answer: a

Explanation: The amplitude variation over AUT should be more than 0.2dB to get AUT encounter uniform illumination. To satisfy that the height criteria of AUT should be equal to or more than four times the dimension of antenna.

8. Which of the following antenna ranges are used to avoid limitations due to electromagnetic interference?

a) Anechoic Chambers

b) Slant height

c) Outdoor ranges

d) Reflection range

Answer: a

Explanation: Indoor ranges are used to avoid limitations due to electromagnetic interference and also ignore the environmental changes in surroundings. Indoor ranges have space restrictions unlike outdoor ranges. Anechoic, tampered chambers and compact ranges come under indoor ranges. Reflection range, Slant range and elevation range comes under outdoor ranges.

9. Antenna range refers to the measurement of antenna parameters.

a) True

b) False

Answer: a

Explanation: The volume, in which the antenna parameters are measured, is called the antenna range. They are of two types. Indoor ranges are used to avoid limitations due to electromagnetic interference. Outdoor ranges are used for larger antenna in environmental surroundings.

10. Which of the following is false regarding elevated range test?

a) To get AUT encounter uniform illumination, the amplitude variation should be less than 0.2dB

b) The height criteria of AUT should be equal to or more than four times the dimension of antenna

c) Range diffraction fences are used to reduce the antenna reflections.

d) To get AUT encounter uniform illumination, the amplitude variation should be more than 0.2dB

Answer: d

Explanation: To get AUT encounter uniform illumination, the amplitude variation should not be more than 0.2dB. The height required to mount the AUT for the elevation range test is given by h r ≥ 4D, where D is the dimension of AUT. Range diffraction fences are used to reduce the antenna reflections.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Measurement of Gain”.

1. In which of the following type of measurement prior gain knowledge of antennas used in measurements is not necessary?

a) Absolute gain

b) Gain transfer method

c) Gain compromise method

d) Both gain transfer and absolute gain method

Answer: a

Explanation: In absolute gain method, the prior knowledge of gain of antennas used in the measurements is not necessary. Gain transfer or Gain compromise method requires the standard gains of the antennas used for measurements to compare the AUT.

2. Friss transfer formula is given by _____

a) \

 

^2 \)

b) \

 

\)

c) \

 

 

 \

 

\)

Answer: a

Explanation: Friss transmission is basis for the absolute gain measurement. The Friss transfer equation is given by \

 

^2 \,where\, G_r,G_t \) are the gains of receiving and transmitting antenna respectively.

3. Friss equation provided the basis for the absolute gain measurement.

a) True

b) False

Answer: a

Explanation: Friss transmission is the method used for the absolute gain measurement. The Friss transfer formula is given by \

 

^2 \,where\, G_r,G_t \) are the gains of receiving and transmitting antenna respectively.

4. Which of the following is used mostly for the calibration of standard gain of antenna?

a) Absolute gain

b) Gain transfer method

c) Gain compromise method

d) Both gain transfer and absolute gain method

Answer: a

Explanation: In absolute gain method, the prior knowledge of gain of antennas used in the measurements is not necessary and it is used for standard gain calibration. Gain transfer or Gain compromise method requires the standard gains of the antennas used for measurements to compare the AUT.

5. In which of the following method does not require standard gain of antennas for gain measurements to compare AUT?

a) Absolute gain

b) Gain transfer method

c) Gain compromise method

d) Both gain transfer and absolute gain method

Answer: a

Explanation: Gain transfer and gain compromise methods require the standard gain calibration of antennas for gain measurements to compare the antennas under test. In Absolute gain method, gain of the antennas which are to be compared are known priory and this is used for the standard gain calibration

6. Antenna efficiency of a lossless isotropic antenna is dB is ____

a) 0

b) 1

c) 3

d) -3

Answer: a

Explanation: For a lossless antenna the whole peak power is radiated towards the receiver without any losses. So the antenna efficiency will be 100% so it is 0dB.

7. Relation between Antenna gain G and directivity D is ______

a) G = ∈ r D

b) G = ∈ r /D

c) G = D/∈ r

d) G = 1/∈ r D

Answer: a

Explanation: Antenna gain refers to the amount of radiation intensity in the desired direction to that of would have radiated when the isotropic antenna radiates. The relation between Antenna gain and Directivity is G = ∈ r D, ∈ r is the antenna efficiency.

8. What is the antenna efficiency if the gain equals to the directivity of the antenna?

a) 1

b) 0

c) 3

d) 2

Answer: a

Explanation: The relation between Antenna gain and Directivity is G=∈ r D, ∈ r is the antenna efficiency.

⇨ ∈ r = \(\frac{G}{D}=1\)

9. Which of the following method is employed for the gain transfer method for antenna outdoor range?

a) Anechoic chamber

b) Tampered chamber

c) Reflection range

d) Compact range

Answer: c

Explanation: Free space or elevation or Reflection range can be employed for the gain transfer method. Anechoic, tampered chambers and compact ranges come under indoor ranges. In gain transfer method the unknown gain antennas are compared with the standard gain antennas.

10. Which of the following equation holds good for gain transfer method?

a) G test antenna = G std antenna ) + \

 

 \)

b) G test antenna = G std antenna ) – \

 

 \)

c) G test antenna + G std antenna ) = \

 

 \)

d) G test antenna / G std antenna ) = \

 

 \)

Answer: a

Explanation: In gain transfer method the unknown gain antennas are compared with the standard gain antennas. The relation between gains of test and standard antenna to the powers received by them are given by the equation

G test antenna = G std antenna ) + \

 

. \)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Standing Wave Ratio Method”.

1. Which of the following statement is true?

a) Standing waves occur when mismatches occurs

b) Standing waves occur when no mismatches occurs

c) Minimum value of SWR is 0

d) Maximum value of SWR is 1

Answer: a

Explanation: When there is mismatch, reflections occur and this causes for the occurrence of the standing waves. The standing wave ratio is defined as the maximum voltage to minimum voltage.

\(S=\frac{V_{max}}{V_{min}} =\frac{1+ρ}{1-ρ}\) And reflection coefficient \(ρ=\frac{P_{ref}}{P_{inc}}. \) Therefore S ranges from 1 to infinity.

2. Relation between SWR ‘S’ and reflection coefficient ‘ρ’ is given by _________

a) \

 

 \

 

 \

 

 \(S=\frac{ρ}{1-ρ}\)

Answer: a

Explanation: Standing waves occur when there is mismatch or reflection due to uneven surfaces. The standing wave ratio is defined as the maximum voltage to minimum voltage and its relation with the reflection coefficient is given as

\(S=\frac{V_{max}}{V_{min}} =\frac{1+ρ}{1-ρ}.\)

3. If there are no any reflections, then the value of the SWR will be ________________

a) 1

b) 0

c) ∞

d) 2

Answer: a

Explanation: The standing wave ratio is defined as the maximum voltage to minimum voltage.

\

 

 

 

.

\(S=\frac{1+ρ}{1-ρ}=1\)

4. The value of the SWR when the reflection power is equal to the incident power is ____________

a) 1

b) 0

c) ∞

d) 2

Answer: c

Explanation: The Reflection coefficient is \

 

. The SWR S is given by

\(S=\frac{1+ρ}{1-ρ}=\frac{1}{0}=∞\)

5. Which of the following level can be measurable by the Low VSWR?

a) 15

b) 11

c) 9

d) 20

Answer: c

Explanation: The low VSWR measurement is able to measure the readings of the VSWR not exceeding 10. For High VSWR measurements  there is a special method called double minimum method.

6. The empirical formula for the measurement of the high VSWR is give by ____

a) \

 

 \

 

 \

 

 \(VSWR =\frac{2λ_g}{

}\)

Answer: a

Explanation: High VSWR measurements  there is a special method called double minimum method. \(VSWR =\frac{λ_g}{π

}\) where d 2 , d 1 are the readings noted on the slotted line where dips  is observed.

7. Which of the following is not used in the VSWR measurement?

a) Reflective Klystron

b) Slotted line

c) Frequency meter

d) Spectrum analyzer

Answer: d

Explanation: Spectrum analyzer frequency domain instrument and is contrast to the oscilloscope which displays in time domain. The Klystron acts as a microwave source and the by moving the slotted line the maximum reading on the meter is observed. For high VSWR the frequency of operation is noted using the frequency meter.

8. The relation between the λ g , λ c , λ 0 is given by ____

a) \

 

 

 

 \

 

 

 

 \

 

 

 

 \(\frac{1}{λ_g}=\frac{1}{λ_0^2} +\frac{1}{λ_c^2} \)

Answer: a

Explanation: The relation between the λ g , λ c , λ 0 is given by \(\frac{1}{λ_g^2} =\frac{1}{λ_0^2} -\frac{1}{λ_c^2} \)

⇨ \(\frac{1}{λ_0} =\sqrt{\frac{1}{λ_g^2} +\frac{1}{λ_c^2}}\)

and \(λ_0=\frac{c}{f}.\)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Spectrum Analyzer”.

1. Spectrum Analyzer is time domain instrument.

a) True

b) False

Answer: b

Explanation: Spectrum analyzer is a frequency domain instrument. It gives the plot of signal amplitude versus the frequency. The amplitude is the Fourier transform of the signal. It acts as a diagnostic tool for the RF and EMI measurements.

2. Which of the following instrument measures the amplitude and phase of the signal?

a) Network analyzer

b) Spectrum analyzer

c) Oscilloscope

d) Klystron

Answer: a

Explanation: Network analyzer is used to measure the amplitude of signal and phase over a frequency. Spectrum analyzer which is a frequency domain instrument gives the plot of signal amplitude versus the frequency. Oscilloscope is a time domain instrument. Klystron acts as a source for the microwave signal.

3. Which of the following is not present in the spectrum analyzer?

a) Swept local Oscillator

b) RF amplifier

c) Sweep voltage generator

d) Slotted line

Answer: d

Explanation: The slotted line is not used in the spectrum analyzer. Spectrum analyzer is a frequency domain instrument. It gives the plot of signal amplitude versus the frequency.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Near Field and Far Field”.

1. The region greater than \Missing open brace for subscript Far field region

b) Near filed region

c) Radiative Field

d) Reactive Field

Answer: a

Explanation: The characteristics of EM fields vary with distance from the antenna. It is divided into two regions. The region next to the antenna is the near field region and the region Far field region > \(\frac{2D^2}{λ}\).

2. Which of the following is known as Non Radiative region?

a) Far field region

b) Near filed region

c) Radiative Field

d) Reactive Field

Answer: d

Explanation: Radiative and reactive field regions are the parts of the near field region. The Far field region is greater than \(\frac{2D^2}{λ}\)and near field region is less than \(\frac{2D^2}{λ}\). Reactive region is also known as non- Radiative region and is next to antenna. The Radiative region is also known as Fresnel region.

3. Which of the following is true?

a) Radiative Field < 0.62\

 

 Reactive Field < 0.62\

 

 Fresnel Field << 0.62\

 

 Near field > \(\frac{2D^2}{λ}\)

Answer: b

Explanation: The characteristics of EM fields vary with distance from the antenna. It is divided into two regions. The region next to the antenna is the near field region and the region Far field region > \(\frac{2D^2}{λ}\). Reactive field < 0.62\(\sqrt{\frac{D^3}{λ}} < \,Radiative\, field\, < \frac{2D^2}{λ}\).

4. Radiative and Non Radiative fields are parts of which field region?

a) Near field

b) Fresnel field

c) Reactive field

d) Far field

Answer: a

Explanation: The Near field region is less than \(\frac{2D^2}{λ}\) the Far field region is greater than\(\frac{2D^2}{λ}\). Radiative and reactive field regions are the parts of the near field region. Reactive region is also known as non- Radiative region and is next to antenna. The Radiative region is also known as Fresnel region.

5. Near field is also known as inductive field.

a) True

b) False

Answer: a

Explanation: The region next to the antenna is the near field region and it is less than \(\frac{2D^2}{λ}\). Near field has inductive effect hence it is also known as the inductive field. The far field is known as the radiating field as radiation is high in that region.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Radio Wave Propagation – Modes of Propagation”.

1. Up to which frequency the ground wave propagation is used?

a) 2MHz

b) 2GHz

c) 30MHz

d) 30GHz

Answer: a

Explanation: Ground wave propagation also known as surface wave propagation is used up to 2MHz. sky wave propagation is used at 2MHz to 30MHz.

2. In a ground wave propagation, which component of electric field is short circuited when it’s in contact by earth?

a) Horizontal

b) Vertical

c) Both horizontal and vertical

d) Neither horizontal nor vertical

Answer: a

Explanation: Any horizontal component of electric field which is in contact with earth is short circuited by earth. Usually ground wave propagation is done by vertical antennas so it is vertically polarized.

3. During ground wave propagation earth behaves like a __________

a) Leaky capacitor

b) Leaky Inductor

c) Series combination of capacitor and inductor

d) Parallel combination of capacitor and inductor

Answer: a

Explanation: Any horizontal component of electric field which is in contact with earth is short circuited by earth. So earth behaves like a leaky capacity. It forms a resistor in shunt with a capacitor.

4. Sky wave propagation reflects the frequencies ___________

a) 2MHz

b) 2 MHz to 30MHz

c) 2 GHz to 30 GHz

d) 30 GHz to 50GHz

Answer: b

Explanation: Ground wave propagation also known as surface wave propagation is used up to 2MHz. sky wave propagation is used at 2MHz to 30MHz.

5. At what distance the sky wave propagation is present from the earth surface?

a) 50 to 400km

b) Below 50 km

c) 600 to 750km

d) 50 to 400 m

Answer: a

Explanation: Sky wave propagation also known as ionosphere propagation reflects the waves with frequency 2 to 30MHz. It is present at 50 to 400km from earth surface.

6. Space wave propagation reflects the waves with frequencies _________

a) Below 2 GHz

b) 2 to 30MHz

c) Above 30GHz

d) Above 30MHz

Answer: d

Explanation: Ground wave propagation also known as surface wave propagation is used up to 2MHz. sky wave propagation is used at 2MHz to 30MHz. Space wave propagation reflects frequencies above 30MHz.

7. Space wave propagates at which frequency band?

a) VHF

b) HF

c) MF

d) EHF

Answer: a

Explanation: Space wave propagation reflects frequencies at 30 to 300MHz range. So it propagates at VHF band.

MF- 300 KHz – 3MHz

EHF- 30GHz-300GHz

8. Communication through LOS can be increased by decreasing the height of antenna.

a) True

b) False

Answer: b

Explanation: Line of sight provides a direct communication link from transmitter to receiver. If the height of antennas is increased then the LOS is also improved.

9. In which of the following mode of propagation the waves are guided along the surface of the earth?

a) Ground wave

b) Sky wave

c) LOS

d) Space wave

Answer: a

Explanation: In ground wave or surface wave propagation the waves are guided along the surface of the earth. In sky wave they are reflected at different layers in the ionosphere. Space wave uses either direct or indirect method of propagation from transmitter to receiver directly.

10. In which of the following modes of propagation the ionosphere acts as the reflecting surface for the waves?

a) Ground wave

b) Sky wave

c) Space wave

d) LOS

Answer: b

Explanation: In Sky wave or Ionospheric wave propagation waves are reflected from the ionosphere layers depending upon different frequencies.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Radio Wave Propagation – Structure of Atmosphere”.

1. Which of the following is the lowest layer of atmosphere?

a) Troposphere

b) Stratosphere

c) Ionosphere

d) Outer atmosphere

Answer: a

Explanation: Troposphere is the lowest layer in the atmosphere ranging up to 15km. stratosphere lies at 50 to 90km and up to 400km ionosphere. Above 400km is the outer atmosphere.

2. What is the range of the region of calm from the earth surface?

a) 20- 70km

b) 2-15km

c) 150-400km

d) Above 400km

Answer: a

Explanation: Region of calm is also known as stratopause. It is at 20 to 70km. Troposphere is up to 15km and above 400km is the outer atmosphere.

3. In which of the following layer of atmosphere free electrons and ions are present?

a) Troposphere

b) Stratosphere

c) Ionosphere

d) Outer atmosphere

Answer: c

Explanation: In ionosphere ionization occurs so there are free electrons, positive and negative ions are also present. It is at 110- 400km range from earth surface.

4. In stratosphere, the excess refractive index is given as ________ in terms of refractive index μ.

a) N=×10 -6

b) N=×10 6

c) N=×10 -6

d) N=×10 6

Answer: a

Explanation: The excess refractive index over unity in millionths is given by N=×10 -6

Where refractive index μ=0.0003% greater than unity for troposphere.

5. G region is also known as _______

a) Outer atmosphere

b) Appleton layer

c) Kennelly Heaviside layer

d) Absorption layer

Answer: a

Explanation: Outer atmosphere is also known as G region. It lies above 400km from the earth surface. E layer is called as Kennelly Heaviside layer and F layer is called as Appleton layer. D layer is also known as absorption layer for short wave signal at HF.

6. Which of the following is the top region during night hours in ionosphere?

a) D layer

b) E layer

c) F2 layer

d) F layer

Answer: d

Explanation: During night time D layer disappears and F1, F2 will combine to form a single F layer. So F layer is top in the Ionosphere during night hours and F2 is top at day time.

7. Ionospheric reflections in the shape of M –type occurs between E s and D layers

a) True

b) False

Answer: b

Explanation: Ionospheric reflections in the shape of M –type occurs between E s  and higher F layers if the layers are smaller compared to the wavelengths. These M-type reflections take a single path in E layer.

8. Which of the following layer is used mostly for the long distance communication?

a) Troposphere

b) Stratosphere

c) Ionosphere

d) D layer during night hours

Answer: c

Explanation: Ionosphere is most widely used for long distance communication and it lies at 90 to 400km from earth surface. D layer is not present at night hours.

9. Stratosphere is present at _____ from earth surface.

a) 20- 70km

b) 2-15km

c) 150-400km

d) Above 400km

Answer: a

Explanation: Stratosphere is present at 20 to 70km from earth surface. Troposphere is up to 15km and above 400km is the outer atmosphere.

10. During night hours, Ionosphere consists of how many layers?

a) 2

b) 3

c) 4

d) 5

Answer: a

Explanation: During night time D layer is not present. F1 and F2 layers get combined to form an F layer during night time. Therefore only two layers namely E and F are present at night time.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Radio Wave Propagation – Structure of Troposphere”.

1. What is the frequency used for tropospheric scatter propagation?

a) Above 30GHz

b) Above 30MHz

c) Above 300MHz

d) Above 300GHz

Answer: c

Explanation: Tropospheric scatter propagation is used for UHF and microwaves. So it is used at frequencies above 300MHz. Sky wave propagation is used at frequencies above 30MHz .

2. Which of the following scattering occurs through the fine layers in the troposphere?

a) Forward Scatter propagation

b) Ionosphere

c) Space wave

d) LOS

Answer: a

Explanation: Forward scatter propagation or Tropospheric propagation occurs at frequencies above 300MHz through the fine layers or blobs in the troposphere. UFH and microwaves propagate much beyond the LOS through forward scattering in tropospheric irregularities.

3. Height of the troposphere ranging from earth surface is up to _____

a) 15m

b) 15km

c) 50m

d) 50km

Answer: b

Explanation: The Troposphere portion of earth extends up to 15km from the earth surface. Ionosphere from 50 to 400km and Outer atmosphere extends above 400km.

4. At which region of the troposphere the temperature remains constant throughout the narrow belt?

a) Tropopause

b) Region of change

c) Region of calm

d) Stratopause

Answer: a

Explanation: At Tropopause starts after the top of troposphere and ends at stratosphere. Above critical height called Tropopause the temperature remains constant throughout belt and increases thereby. Entire belt of troposphere is called Region of change.

5. The lowest layer in the structure of atmosphere extends to a distance of _____km from earth surface

a) 15m

b) 15km

c) 50m

d) 50km

Answer: b

Explanation: The lowest layer of the atmosphere is troposphere. So it ranges up to 15km from the surface of the earth. The gas components remain almost constant in this region.

6. The actual height of troposphere is least at ____

a) poles

b) equator

c) both equator and poles

d) other composition other than poles and equator

Answer: a

Explanation: The actual height of troposphere is least at poles and maximum at equator. At other compositions it is almost remains same. The entire belt of troposphere is called Region of change.

7. Which of the following is a property of troposphere?

a) Temperature decreases with increase in height

b) Temperature increases with increase in height

c) Gas components don’t remain in constant percentage with increase in height

d) Water vapor components remain same with height

Answer: a

Explanation: The property of troposphere is that temperature decreases with increase in height. Gas components remains in constant percentage with increase in height but water vapor components decrease with increase in height.

8. Which of the following is the nearest region from the earth surface in the atmosphere?

a) Troposphere

b) Stratosphere

c) Ionosphere

d) Outer atmosphere

Answer: a

Explanation: Troposphere is the lowest layer and nearest to earth surface in the atmosphere ranging up to 15km. Stratosphere lies at 50 to 90km and up to 400km ionosphere. Above 400km is the outer atmosphere.

9. The region between the top of troposphere and start of stratosphere is called ____

a) tropopause

b) stratopause

c) ionosphere

d) region of calm

Answer: a

Explanation: The region between the top of troposphere and start of stratosphere is called Tropopause. Stratopause is also known as region of calm and ranges from 20 to 70km.

10. Troposphere scatter propagation is used for point to point communication

a) True

b) False

Answer: a

Explanation: Since the troposphere propagation takes at 2- 15km from earth surface there is a possibility for great attenuation. It is used for point to point communication.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Radio Wave Propagation – Structure of Ionosphere”.

1. At what height the Ionosphere lies above the earth surface?

a) 70-400km

b) 2-15km

c) 20-70km

d) Above 400km

Answer: a

Explanation: Troposphere is the lowest layer in the atmosphere ranging up to 15km. stratosphere lies at 50 to 90km and from 70- 400km ionosphere. Above 400km is the outer atmosphere.

2. Which of the following is the nearest Ionospheric layer to the earth surface?

a) D layer

b) E layer

c) F1 layer

d) F2 layer

Answer: a

Explanation: D layer is present at 70km from the earth surface and is the nearest layer to earth surface. E layer lies at 110km, f1 layer at 220km and F2 at 250 to 400km. Based on the density of ions different layers are present in this layer.

3. Which of the following layer disappears during night time in ionosphere?

a) D layer

b) E- layer

c) F1 layer

d) F2 layer

Answer: a

Explanation: During night time D layer disappears and F1, F2 layers combine together to form an F layer. D layer is present at 70km from the earth surface and is the nearest layer to earth surface. E layer lies at 110km, f1 layer at 220km and F2 at 250 to 400km.

4. Which of the following is called as Kennelly Heaviside layer?

a) E layer

b) F1 layer

c) F2 layer

d) D layer

Answer: a

Explanation: E layer is called as Kennelly Heaviside layer and F layer is called as Appleton layer. E layer lies at 110km, f1 layer at 220km and F2 at 250 to 400km.

5. In which of the following layers the electron density is high?

a) E layer

b) F1 layer

c) F2 layer

d) D layer

Answer: c

Explanation: The electron /ion density increases with the increase in height in ionosphere. F2 layer has 3×10 5 to 2×10 6 electron density . F2 layer = 2×10 5 to 4.5×10 5

6. Which of the following follows Chapman’s law of variation?

a) D layer

b) F1 layer

c) F2 layer

d) Both F1 and F2 layer

Answer: b

Explanation: Only F1 layer follows the Chapman’s law of variation where F1 region behaves like an E-region. F1 layer is formed by the ionization of O 2 atoms. F2 layer does not follow Chapman’s law of variation.

7. Which of the following regions are present in the night time?

a) D and E layers

b) E and F layers

c) F1 and F2 layers

d) D and F layers

Answer: b

Explanation: During night time D layer is not present. F1 and F2 layers get combined to form an F layer during night time. So E and F layers are present at night time.

8. Which of the following layer is used for long distance sky wave propagation during night hours?

a) F layer

b) E layer

c) D layer

d) All layers can be used

Answer: a

Explanation: F layer is the top layer in the ionosphere and is also present at the night time. So it is used for the long distance sky wave propagation and also its ionization density is high. D layer is not present during night hours.

9. Which of the following is formed by the ionization of UV and corpuscular radiations?

a) E layer

b) F1 layer

c) F2 layer

d) D layer

Answer: c

Explanation: F2 layer is formed by the ionization of the UV, x-rays and corpuscular radiations. F1 is formed by the ionization of O 2 atoms. F layer is the top layer in the ionosphere and has highest electron density. D layer is due to photo ionization of O 2 atoms at first level.

10. Which of the following layer is known as absorption layer for short waves?

a) E layer

b) F1 layer

c) F2 layer

d) D layer

Answer: d

Explanation: D layer is also known as absorption layer for short wave signal at HF. It reflects signals up to VLF and LF. Its absorption increases with solar activity.

11. Which of the following layer is known as F layer?

a) Appleton layer

b) Kennelly Heaviside layer

c) Absorption layer

d) Sir Appleton

Answer: a

Explanation: E layer is called as Kennelly Heaviside layer and F layer is called as Appleton layer. D layer is also known as absorption layer for short wave signal at HF.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Radio Wave Propagation – Ground Wave Propagation”.

1. Ground wave is always __________ polarized.

a) Vertically

b) Horizontally

c) Either vertical or horizontal

d) Neither horizontal nor vertical

Answer: a

Explanation: If the wave is horizontally polarized, then the electric field is short circuited by the earth. So the ground wave is always vertically polarized and vertical antennas are used.

2. Ground wave propagation is used for signals up to frequency __________

a) 2MHz

b) 2GHz

c) 30MHz

d) 30GHz

Answer: a

Explanation: Ground wave propagation is used for signals up to frequency of 2MHz. Ground waves are vertically polarized and transmitting and receiving antennas are placed closely and the wave follows the curvature of the earth.

3. The broadcast signals received at low frequencies during day-time are due to _________

a) Ground wave

b) Space wave

c) Sky wave

d) Tropospheric waves

Answer: a

Explanation: Ground wave propagation is used for signals up to frequency of 2MHz. It is useful for the broadcast and low frequency signals. Space waves are also known as tropospheric waves useful for FM reception. Sky wave propagation is used for long distance communication.

4. Ground wave propagation is also known as _________

a) Surface wave

b) Tropospheric wave

c) Ionospheric wave

d) Stratospheric waves

Answer: a

Explanation: Ground waves are also known as Surface waves as the wave propagates close to the surface of earth. Space waves are known as tropospheric waves and Sky waves as Ionospheric waves.

5. The ground wave propagation uses horizontal polarized antennas

a) True

b) False

Answer: b

Explanation: If the wave is horizontally polarized, then the electric field is short circuited by the earth. So the ground wave is always vertically polarized and vertical antennas are used.

6. Which of the following is particularly used for VLF?

a) Surface wave

b) Tropospheric wave

c) Ionospheric wave

d) Stratospheric waves

Answer: a

Explanation: Ground waves are also known as Surface waves are used for low frequencies and broadcasting. Tropospheric waves are used for Mf and HF signals. Ionospheric waves are used for long distance communications.

7. Ground wave field strength is given by E= _________

a) AE o /d

b) A(E o ) 2 /d

c) dE o

d) AE o /d 2

Answer: a

Explanation: The ground wave field strength at a point \(E=\frac{AE_o}{d} V/m \)

Where E o is field strength of wave at unit distance from transmitting antenna

A is factor of ground losses

D is distance of point from Transmitting antenna.

8. Which of the following propagates by gliding over the surface of earth?

a) Surface wave

b) Tropospheric wave

c) Ionospheric wave

d) Stratospheric waves

Answer: a

Explanation: Ground waves are also known as Surface waves. Ground waves are vertically polarized and transmitting and receiving antennas are placed closely and the wave follows the curvature of the earth.

9. How the ground wave losses vary with high frequencies?

a) Increases

b) Decreases

c) Does not depend on frequency

d) Increase or decrease

Answer: a

Explanation: Ground wave propagation is used at below 2MHz frequency. Ground wave propagation is used for short distance. By using sufficient power and low frequencies it can be used effectively. Ground wave losses increase rapidly with frequencies.

10. The electric field of the component increases in ground wave when it tilts more at the surface.

a) True

b) False

Answer: b

Explanation: As the wave front tilts more and more towards the surface then the electric field gets short circuited. As it gets reduced with tilt, at some point it is completely attenuated.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Radio Wave Propagation – Field strength”.

1. The field strength due to space wave propagation is E=____

a) \

 

 \

 

 \

 

 \(\frac{4πh_t h_r}{λd^4}E_0\)

Answer: a

Explanation: Space wave propagation reflects frequencies above 30MHz. The field strength due to space wave propagation is \(E=\frac{2E_0}{d} sin⁡ \frac{2πh_t h_r}{λd} ≈ \frac{4πh_t h_r}{λd^2}E_0\)

Here E 0 is the field strength due to LOS at a unit distance which depends on transmits power, h t and h r is the height of transmitting and receiving antenna and d is the distance between two antennas.

2. The path difference given by two-ray model is ___

a) \

 

 \

 

 \

 

 \(\frac{2h_t h_r}{d}\)

Answer: d

Explanation: The path difference given by two-ray model is \(\frac{2h_t h_r}{d}.\) and its phase difference is given as \(∅=\frac{2π}{λ} ∆x=\frac{4πh_t h_r}{d}.\)

3. The field strength of the wave at a unit distance from transmitting antenna depends on _____

a) only the power radiated by the transmitting antenna

b) only on the power received by the receiving antenna

c) only on directivity of antenna in vertical and horizontal planes

d) both on the power radiated by the transmitting antenna & directivity in vertical and horizontal planes

Answer: d

Explanation: The field strength E 0 of the wave at a unit distance from transmitting antenna depends on both on the power radiated by the transmitting antenna & directivity in vertical and horizontal planes. The overall field strength is calculated from this. The field strength of the surface wave of flat earth is given by E=AE 0 /d.

4. What is the condition of roughness of earth for earth to be electrically smooth?

a) R < 0.1

b) R > 10

c) R > 0.1

d) R > 1

Answer: a

Explanation: Earth is classified as electrically smooth and electrically rough.

If the roughness of earth R < 0.1, then it is electrically smooth.

If the roughness of earth R > 10, then it is electrically rough.

5. Power density of the receiving antenna is given by _________

a) \

 

 \

 

 \

 

 \(P_D=\frac{P_t G_t}{^2}\)

Answer: a

Explanation: Power density is the power per unit area at a distance d from the transmitter and is given by \(P_D=\frac{P_t G_t}{4πd^2}\) where Pt is the transmitter power and Gt is the transmitter antenna gain.

Power received by the antenna is given by \(P_r=\frac{P_t G_t G_r}{^2}\)

6. Spatial attenuation coefficient is given by ____

a)  2

b)  2

c) 4πλ/d

d) 4πd/λ

Answer: a

Explanation: Spatial attenuation coefficient reflects the decrease in the power density due to the spherical spread of the wave. It is also known as the loss factor. Expression for loss factor is given by L s = 2

7. The field strength of the surface wave of flat earth is given by ____

a) E=AE 0 /d

b) E=AE 0 /λ

c) E=dE 0 /A

d) E=λE 0 /A

Answer: a

Explanation: The field strength of the surface wave of flat earth is given by E=AE 0 /d, here A is the ground attenuation factor, d is the distance and E 0 is the field strength per unit distance. The field strength of the wave at a unit distance from transmitting antenna depends on both on the power radiated by the transmitting antenna & directivity in vertical and horizontal planes

8. Dissipation factor for the dielectric is D f = ____

a) \

 

 \

 

 \

 

 \(1.8×10^3 \frac{σ}{f}\)

Answer: a

Explanation: Dissipation factor for the dielectric is D f = \

 

 

\)

9. The effective radius of earth is how many times the actual radius of earth at standard atmospheric conditions?

a) 3/4

b) 4/3

c) 1/4

d) 1/3

Answer: a

Explanation: Effective radius of earth is the equivalent radius of earth used to correct atmospheric refraction as the refractive index varies linearly with height. It is 4/3 times the geometrical radius of the earth.

10. What is the equivalent radius of the earth when the ray appears to be a straight line over a flat earth surface?

a) 0

b) Infinity

c) Actual radius

d) 4/3 times actual radius

Answer: b

Explanation: When the ray appears to be a straight line over a flat earth surface horizontally, its refractive index is not changing with respect to height. \(\frac{dM}{dh}=0\)

Effective radius is k times the actual radius and \(k=0.048/\frac{dM}{dh}=∞\)

This set of Antennas Problems focuses on “Radio Wave Propagation – Reflection Factor for Parallel Propagation”.

1. The reflection factor is given by the ratio of ____________

a) Reflected wave to incident wave

b) Incident wave to reflected wave

c) Sum of reflected and incident waves to difference of reflected and incident waves

d) Difference of reflected and incident waves to sum of reflected and incident waves

Answer: a

Explanation: The reflection factor is given by the ratio of Reflected wave to incident wave.

\(ρ=\frac{reflected \,wave}{incident\, wave}=\frac{E_r}{E_i} \)

ρ=1 It means that surface is perfect reflecting medium.

2. What is the reflection coefficient for a perfect reflecting surface?

a) 1

b) 0

c) <1

d) >1

Answer: a

Explanation: The reflection factor is defined as the ratio of Reflected wave to incident wave.

\(ρ=\frac{reflected \,wave}{incident \,wave}=\frac{E_r}{E_i} \)

For a perfect reflecting surface, total incident waves are reflected completely. E r = E i

\(ρ=\frac{E_r}{E_i}=1.\)

3. What is the reflection coefficient if there are no any reflections?

a) 1

b) 0

c) <1

d) >1

Answer: b

Explanation: The reflection factor is given by the ratio of Reflected wave to incident wave.

\(ρ=\frac{reflected \,wave}{incident \,wave}=\frac{E_r}{E_i} \)

No reflections means E r = 0 =>ρ = 0

4. Which of the following holds true for Brewster angle?

a) It is about the angle of reflection

b) It is the angle of incident at which no reflections occur

c) It is the angle of incident at which maximum reflections occur

d) It is about the angle of refraction

Answer: b

Explanation: Brewster angle is defined as the angle of incident at which no reflections occurs.

\(θ=tan^{-1}⁡\sqrt{\frac{ϵ_{r2}}{ϵ_{r1}}} \)

5. Reflection coefficient of plane waves with normal incidence in terms of impedances of medium is ____

a) \

 

 \

 

 \

 

 \(γ_r=\frac{η_2+η_1}{η_2-η_1} \)

Answer: a

Explanation: The reflection factor is given by the ratio of Reflected wave to incident wave.

\(ρ=\frac{reflected\, wave}{incident\, wave}=\frac{E_r}{E_i} \)

Reflection coefficient of plane waves with normal incidence in terms of impedances of medium is \(γ_r=\frac{η_2-η_1}{η_2+η_1} \)

6. Relation between transmission coefficient γ t and reflection coefficient is γ r is

a) γ r = γ t -1

b) γ r = γ t +1

c) γ r = γ t -2

d) γ r = (γ t -1) / γ t +1

Answer: a

Explanation: Relation between transmission coefficient γ t and reflection coefficient is γ r is γ r = γ t – 1

Transmission coefficient is the ratio of transmitted to incident wave. The reflection factor is given by the ratio of Reflected wave to incident wave.

7. Transmission coefficient of a plane wave with normal incidence is _________

a) \

 

 \

 

 \

 

 \(γ_r=\frac{η_2-η_1}{η_2+η_1} \)

Answer: a

Explanation: Transmission coefficient is the ratio of transmitted to incident wave. Relation between transmission coefficient γ t and reflection coefficient is γ r is γ r = γ t -1.

So the Transmission coefficient of a plane wave with normal incidence is \(γ_t=\frac{2η_2}{η_2+η_1} \)

8. Complex dielectric constant of earth is given by ____

a) \

 

 \

 

 \

 

 \(ϵ’=ϵ+\frac{σϵ}{jw} \)

Answer: a

Explanation: The earth is neither a good conductor nor a good dielectric. So the dielectric constant of earth is complex and is given by \(ϵ’=ϵ+\frac{σ}{jw} \)where σ is the conductivity and w is angular frequency.

9. Which of the following is the expression for tilt angle?

a) \

 

\)

b) \

 

\)

c) \

 

\)

d) \

 

\)

Answer: a

Explanation: The wave tilt of the surface wave depends on the conductivity of earth, permittivity and angular frequency of the waves. Expression for tilt angle is \

 

. \)

10. Find the tilt angle of the radio wave with 1MHz frequency and conductivity of earth is 5 ×10 -3 S/m and relative permittivity 10?

a) 41.8°

b) 83.6°

c) 48.1°

d) 38.6°

Answer: a

Explanation: Expression for tilt angle is \

 

. \)

\

 

 

≈41.8°\)

11. Brewster angle doesn’t exist for perpendicular polarization.

a) True

b) False

Answer: a

Explanation: Brewster angle is defined as the angle of incident at which no reflections occurs.

\(θ=tan^{-1}⁡\sqrt{\frac{ϵ_{r2}}{ϵ_{r1}}}\)

At θ =90, tan θ=∞, which doesn’t exist in reality.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Space Wave Propagation – Line of Sight Distance”.

1. Which of the following statement is defined as line of sight distance?

a) Distance covered by a direct space wave from transmitting to receiving antenna

b) Distance covered by an indirect space wave from transmitting to receiving antenna

c) Distance covered by a direct sky wave from transmitting to receiving antenna

d) Distance covered by an indirect sky wave from transmitting to receiving antenna

Answer: a

Explanation: LOS  is defined for space wave propagation. It is the distance covered by a direct space wave from transmitting to receiving antenna. It depends on the height of the transmitting and receiving antennas and effective earth’s radius factor k.

2. On which of the following factors does the LOS distance depends?

a) Height of receiving antenna alone

b) Height of transmitting antenna alone

c) Only on height of transmitting and receiving antenna

d) On height of transmitting and receiving antenna and effective earths radius factor k

Answer: d

Explanation: LOS  is the distance covered by a direct space wave from transmitting to receiving antenna. The LS distance depends on the effective earth’s radius factor k and height of the transmitting and receiving antennas. The LOS distance is given by \

\) in km.

3. Which of the following order is correct?

a) LOS > Radio horizon > Optical horizon

b) Radio horizon < LOS < Optical horizon

c) Radio horizon > Optical horizon > LOS

d) Optical horizon > Radio horizon > LOS

Answer: a

Explanation: Radio horizon is the point to which a space wave can reach maximum and is greater than optical horizon. LOS is thedistance covered by a direct space wave from transmitting to receiving antenna. If distance between antennas is less than LOS then it is referred as Optical horizon and is approximated to LOS.

4. Expression for the LOS distance is _____ 

a) \

\)

b) \

\)

c) \

\)

d) \

\)

Answer: a

Explanation: LOS is thedistance covered by a direct space wave from transmitting to receiving antenna.Expression for the LOS distance is \

\) in km.

Expression for the Radio horizon distance is \

\) in km.

5. Expression for radio horizon in km is ____

a) \

\)

b) \

\)

c) \

\)

d) \

\)

Answer: c

Explanation: Radio horizon is the point to which a space wave can reach maximum. Expression for radio horizon in km is \

\)

Expression for the LOS distance is \

\) in km.

6. The value of k at which LOS equals to the radio horizon is ___

a) 1

b) 0

c) 3/4

d) -3/4

Answer: a

Explanation: Distance between transmitter and receiver is \(d=d_t+d_r=\sqrt{2r_e h_t}+\sqrt{2r_e h_r} \)

And r e =k*6370km, k is the effective radius factor.

Expression for radio horizon in km is \

,\) LOS depends on k also so d = \

.\) At k=1 both will be equal.

7. If the heights of transmitting and receiving antenna are equal then LOS distance is ___ in km.

a) 8.24√h

b) 4.82√h

c) 4.12√h

d) 2.06√h

Answer: a

Explanation: Given heights of transmitting and receiving antenna are equal h t = h r = h

Expression for the LOS distance is \

\) in km.

⇨ D=4.12  = 8.24√h.

8. Radio horizon is less than LOS distance.

a) True

b) False

Answer: a

Explanation: Expression for radio horizon in km is \

,\) Expression for the LOS distance is \

\) in km. Under same height conditions, radio horizon is less than LOS. But at effective radius factor k=1, both will be equal.

9. The radio horizon can be equal to the LOS distance if same height antennas are used.

a) True

b) False

Answer: b

Explanation: The radio horizon and LOS become equal when k=1. LOS depends on the height of the transmitting and receiving antennas and effective earth’s radius factor k. Standard value of k is 4/3.

10. What is the value of the effective radius factor k of earth if the radius of curvature and the earth radius equals?

a) 1

b) 0

c) Infinity

d) 4/3

Answer: c

Explanation: Effective radius factor k in terms of radius of curvature r c and radius of earth r e is

\(k=\frac{1}{1-\frac{r_e}{r_c}}=\frac{1}{0}=∞.\)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Space Wave Propagation – Duct Propagation”.

1. Duct propagation is due to which layer?

a) Temperature Inversion layer

b) Higher atmospheric layer

c) Ionospheric layer

d) Surface water

Answer: a

Explanation: Duct propagation is due to temperature inversion layer. Normal atmospheric temperature reduces at 6.5°C/km where as to provide duct wave propagation temperature must increase within 15-50m from earth surface. This layer occurs due to super refraction and at lower atmospheric layers.

2. Temperature inversion layer occurs due to _____

a) Super refraction

b) Reflection

c) Diffraction

d) Increase in altitude

Answer: a

Explanation: Around 50m of height from earth surface temperature increases with height, at this level the EM waves tend to refract continuously than to reflect into ionosphere. This is termed as super refraction. Reflection and diffraction don’t affect the temperature inversion much.

3. In which of the following propagation temperature inversion layer is present?

a) Duct wave propagation

b) Sky wave propagation

c) Tropospheric scattering wave propagation

d) Space wave propagation

Answer: a

Explanation: Duct propagation is due to temperature inversion layer. This layer occurs due to super refraction and at lower atmospheric layers. Around 50m of height from earth surface temperature increases with height, at this level the EM waves tend to refract continuously than to reflect into ionosphere. This is termed as super refraction.

4. How long the waves due to duct propagation travel following the earth curvature?

a) 1000km

b) 50m

c) 15m

d) 5000km

Answer: a

Explanation: Duct wave propagation is possible for microwave frequency and is due to temperature inversion layer. The waves follow the curvature of earth up to 1000km long. This occurs at a height of 15- 50m above the earth surface.

5. Refractive index of temperature inversion layer is ____________

a) μ = ×10 6

b) μ = ×10 6

c) μ = ×10 6

d) μ = ×10 6

Answer: a

Explanation: The modified refractive index for a temperature inversion layer is μ = ×10 6

Where n is the Refractive index, h is height and r is radius of earth.

6. Which of the following represents the plot of height v/s refractive index of elevated surface duct?

a) 

b) 

c) 

d) 

Answer: d

Explanation: Curves are used to determine the presence of duct propagation  and \(\frac{dμ}{dh}\) is -ve

 This is duct propagation with no elevation

 Duct propagation with elevation

 Normal atmosphere

7. Condition for duct propagation is _________

a) \

 

 \

 

 \

 

 \(\frac{dμ}{dh}\) is unity

Answer: a

Explanation:

The height v/s the refractive index graph of duct propagation is as follows

To determine the presence of duct propagation  and \(\frac{dμ}{dh}\) is -ve.

8. The wavelength of the EM signal propagation in the T.I.L for duct propagation is _________

a) \

 \

 \

 \(λ=25h_d \sqrt{∆μ×10^{-6}} \)

Answer: a

Explanation: The wavelength of the EM signal propagation in the T.I.L for duct propagation is

\(λ=2.5h_d \sqrt{∆μ×10^{-6}} \)

Here ∆μ is change in the refraction index in height h d .

9. For duct propagation the necessary condition is \

 

 False

b) True

Answer: a

Explanation: To determine the presence of duct propagation  and \(\frac{dμ}{dh}\) is -ve

This occurs at a height of 15- 50m above the earth surface and at lower atmospheric layers.

10. Duct propagation occurs at __________

a) lower atmospheric layers

b) higher atmospheric layers

c) ionospheric layers

d) any part of the atmospheric layer

Answer: a

Explanation: Duct propagation occurs at lower atmospheric layers at 15 – 50m above the earth surface. It occurs due to temperature inversion layer. The waves follow the curvature of earth up to 1000km long.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Space Wave Propagation – Refractivity”.

1. Which of the following varies with the refractive index in the atmosphere?

a) Radio horizon

b) Optical horizon

c) Line of sight 

d) Both radio horizon and LOS

Answer: a

Explanation: LOS is the distance covered by a direct space wave from transmitting to receiving antenna. It depends on the height of the transmitting and receiving antennas and effective earth’s radius factor k. Radio horizon varies with the refractive index in the atmosphere. Optical horizon is less than LOS.

2. The expression for refractivity in terms of the Pressure and temperature coefficient is given by ____________

a) \

 

 

 \

 

 

 \

 

 

 \(\frac{77.6}{T}P+4810 \frac{T}{e} \)

Answer: a

Explanation: The expression used to calculate refractivity is given by

\(N=\frac{77.6}{T} P+4810 \frac{e}{T} \)

Where p is the total pressure in millibars, e is the partial pressure of water vapor and T is the absolute temperature.

3. Gradient of refractive index is ____________

a) Change in the refractivity with respect to height

b) Change in the height with respect to refractivity

c) Change in the refractivity with respect to Pressure

d) Change in the refractivity with respect to temperature

Answer: a

Explanation: Change in the refractivity with respect to height is known as the gradient of refractive index. Radio horizon varies with the refractive index in the atmosphere.

4. What is the standard value for the radius of curvature r c to be equal to the radius of earth r e ?

a) r c =4r e

b) r c =\

 

 r c =r e

d) r c =\(\frac{1}{4}\) r e

Answer: a

Explanation: The radius of curvature is four times the radius of earth. r c =4r e

The effective radius factor of earth is 4/3 times actual radius of earth.

5. Expression for effective radius factor k in terms of radius of curvature r c and radius of earth r e ?

a) \

 

 

 \

 

 

 \

 

 \(k= 1-\frac{r_c}{r_e} \)

Answer: b

Explanation: Effective radius factor k in terms of radius of curvature r c and radius of earth r e is

\(k=\frac{1}{1-\frac{r_e}{r_c}} \)

The radius of curvature is four times the radius of earth. r c =4r e

The effective radius factor of earth is 4/3 times actual radius of earth.

6. What is the effective radius factor when radius of curvature is 4 times the radius of earth?

a) 4

b) 4/3

c) 3

d) 3/4

Answer: b

Explanation: Effective radius factor k in terms of radius of curvature r c and radius of earth r e is

\(k=\frac{1}{1-\frac{r_e}{r_c}} \)

The radius of curvature is four times the radius of earth. r c =4r e

⇨ K=1/) =4/3

The effective radius factor of earth is 4/3 times actual radius of earth.

7. Expression for refractivity is given by ___________

a) N=×10 6

b) N=×10 -6

c) N=1/

d) N=\(\frac{1}{n-1}\)×10 -6

Answer: a

Explanation: Refractivity is to observe the change in refractive index due to the smallest change in the relative dielectric constant.

N=×10 6

8. On which of the following the refractivity depends on?

a) Air pressure

b) Water pressure

c) Temperature

d) Air pressure, water pressure and temperature

Answer: d

Explanation: The expression used to calculate refractivity is given by

\(N=\frac{77.6}{T} P+4810 \frac{e}{T}\)

Where p is the total pressure in millibars, e is the partial pressure of water vapor and T is the absolute temperature.

9. Refractive index is directly proportion to ___________

a) \

 \

 ϵ r

d) 1/ϵ r

Answer: a

Explanation: Refractive index is ratio of velocity of light in air to velocity of light in medium

n=c/v

\(n=\sqrt{ϵ_r μ} \)

10. Relation between gradient of refractive index with height and dielectric constant gradient is ____________

a) \

 

 

 \

 

 

 \

 

 

 \(\frac{dϵ_r}{dH}=-2n \frac{dn}{dH}\)

Answer: a

Explanation: Refractive index is \

\)

Differentiating with respective heights treating n, ϵ_r as dependent variables

\(\frac{dn}{dH}=\frac{1}{2} \frac{1}{\sqrt{ϵ_r}} \frac{dϵ_r}{dH}\)

\(\frac{dϵ_r}{dH}=2n \frac{dn}{dH}\)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Space Wave Propagation – Tropospheric Scatter”.

1. What is the frequency at which tropospheric scatter occurs?

a) Above 30MHz

b) Below 30 MHz

c) < 3MHz

d) > 3 MHz and < 30MHz

Answer: a

Explanation: Tropospheric scatter is also known as forward scatter propagation. It occurs at a frequency above 30MHz. It belongs to the UHF and microwave range.

2. What is the range of frequency at which tropospheric scatter occurs?

a) UHF and Microwave range

b) HF

c) MF and VHF

d) MF and HF

Answer: a

Explanation: Tropospheric scatter occurs at a frequency above 30MHz. It belongs to the UHF and microwave range.

MF- 300 KHz – 3MHz

HF – 3 to 30 MHz

VHF -30 to 300MHz.

3. Tropospheric propagation is also known as forward scatter propagation.

a) True

b) False

Answer: a

Explanation: Tropospheric scatter is also known as forward scatter propagation, occurs at a frequency above 30MHz. It belongs to the UHF and microwave range. Its scattering occurs above radio horizon and travels beyond the LOS.

4. The tropospheric scattering occurs at _________

a) Beyond the LOS

b) In ground wave propagation

c) In sky wave propagation

d) Below the radio horizon

Answer: a

Explanation: Tropospheric scattering occurs at a frequency above 30MHz. So, this occurs in space wave propagation. Signal scatters in forward direction above the radio horizon and it travels beyond the LOS. Ground wave propagation occurs at less the 2MHz and sky wave at 2 to 30 MHz frequency.

5. Tropospheric scattering, if falls under low frequencies then it leads to Ionosphere scattering.

a) True

b) False

Answer: a

Explanation: The tropospheric scattering occurs at frequency above 30MHz. Under low frequencies it falls below 30MHz and sometimes thus leads to the Ionospheric propagation.

6. The turbulences in the atmosphere lead to ___________

a) tropospheric scattering

b) ground wave propagation

c) sky wave propagation

d) constant velocity of signal

Answer: a

Explanation: The main cause of the tropospheric scattering is the turbulences in the atmosphere. When it meets the turbulences, then there will be an abrupt change in velocity leading to the tropospheric scattering.The tropospheric scattering occurs at frequency above 30MHz.

7. When the signal take-off angle increases, the height of scatter volume ________

a) increases

b) decreases

c) it remains constant

d) may increase or decrease

Answer: a

Explanation: The signal take-off angle determines the height of the scatter volume. As the signal take-off angle increases, the height of scatter volume also increases. A low take-off produces low scatter volume.

8. Which of the following statements regarding tropospheric scattering is false?

a) It occurs in the region near to the mid-point of the transmitter and receiver

b) It occurs above the radio horizon

c) It travels beyond the line of sight distance

d) It occurs below the radio horizon

Answer: d

Explanation: Signal scatters in forward direction above the radio horizon and it travels beyond the LOS. It occurs in the region near to the mid-point of the transmitter and receiver.

9. In tropospheric scattering, rapid fading occurs due to _______ and long-term fading occurs due to ______

a) multipath propagation, turbulences in atmosphere

b) low signal strength, signal travelling below horizon

c) signal travelling below horizon, low signal strength

d) daily and seasonal variations, multi-path propagation

Answer: a

Explanation: Rapid fading occurs due to the multipath propagation. As the turbulent conditions changes constantly, the path length and signal strength levels also change. The turbulence in the atmosphere gives rise to daily and seasonal variations in signal strength so results in long-term fading.

10. Which of the following is used for communication in rugged terrain where normal propagation methods fail?

a) Tropospheric Scattering

b) Ground wave

c) LOS

d) Radio Horizon

Answer: a

Explanation: Signal scatters in forward direction above the radio horizon and it travels beyond the LOS. The signal strength is decreased as only a small amount of it is forward scattered and so high power amplifiers are equipped at the receivers.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Virtual Height”.

1. The effective height of a layer of ionized gas in the atmosphere by which the radio waves are reflected around earth’s curvature is called Virtual height.

a) True

b) False

Answer: a

Explanation: The height to which a short pulse of energy is sent vertically upward and traveling with the speed of light would reach taking the same two-way travel time as does original pulse reflected from ionosphere layer is called the virtual height.

2. Relation between skip distance d, virtual height d and MUF f MUF is ______

a) \

 

 \

 

 \

 

\)

d) \

 

\)

Answer: a

Explanation: Relation between skip distance d, virtual height d and MUF f MUF is given by

skip distance \(d=2h\sqrt{[\frac{f_{MUF}^2}{f_c^2}-1]}\)

3. Find the skip distance when the angle of incidence is 200 and virtual height is 50km?

a) 13.22m

b) 13.22km

c) 36.33km

d) 36.33m

Answer: c

Explanation: Skip distance \(d=2h\sqrt{[\frac{f_{MUF}^2}{f_c^2}-1]}=2h\sqrt{[secθ_i^2-1]} =2×50×\sqrt{[^2-1]}\)

d=36.33km

4. The differences in the virtual height and actual height are affected by the electron density in the ionosphere region.

a) True

b) False

Answer: a

Explanation: The difference occurs due to the exchange of energy between wave and the electrons present in the ionosphere which changes the velocity of propagation of wave. So the difference between virtual and actual height are influenced by the electron distribution.

5. Find the virtual height h when the angle of incidence is 600 and distance of separation 120km.

a) 60km

b) 64.34km

c) 34.64km

d) 72.42km

Answer: c

Explanation: Virtual height \(h=\frac{d/2}{\sqrt{secθ_i^2-1}}=34.64km\)

6. The height at a point above the earth’s surface at which the wave bends down to the earth is called ______

a) Actual height

b) Virtual height

c) Skip distance

d) Distance of separation between transmitter and antenna

Answer: a

Explanation: The height at a point above the earth’s surface at which the wave bends down to the earth is called actual height or true height. Virtual height is greater than actual height. Skip distance is the distance the wave travels from the transmitter to the receiver without touching the ground.

7. Skip zone is distance between the point where the ground wave reception becomes zero and the sky wave returns for the first time.

a) True

b) False

Answer: a

Explanation: The Skip zone is the region between the point where the ground wave reception becomes null and the sky wave returns for the first time. It entirely depends on the ground wave coverage and the skip distance.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Critical Frequency”.

1. When a wave is incident normally then the acceptable highest frequency at which signal can be returned is the ______

a) critical frequency

b) LUF

c) optimum frequency

d) dominating frequency

Answer: a

Explanation: When a wave is incident normally then the acceptable highest frequency at which signal can be returned is the critical frequency. Beyond critical frequency, wave is penetrated into another region. When the frequency is greater than critical frequency, still some part is returned back by varying the angle of incidence and this frequency is called MUF.

2. What is the critical frequency when the electron density in the F 1 layer is 20×10 10 /cm 3 ?

a) 40.2GHz

b) 4.02 MHz

c) 40.2 MHz

d) 40.2Hz

Answer: b

Explanation: Critical frequency \(f_c=9\sqrt{N\, max} \)

\(f_c =9\sqrt{20×10^{10}}=4.02MHz \)

3. What is the electron density of the layer if critical frequency is 3 MHz?

a) 11×10 12 /cm 3

b) 0.11×10 12 /cm 3

c) 1.1×10 12 /cm 3

d) 0.11×10 10 /cm 3

Answer: b

Explanation: \(f_c=9\sqrt{N\, max} \)

\(N_{max}=\frac{f_c^2}{81}=\frac{

^2}{81}=0.11×10^{12}/cm^3\)

4. For a regular layer, the critical frequency is proportional to the ______ of electron density.

a) Square

b) Inverse

c) Square root

d) Inverse square root

Answer: c

Explanation: For a regular layer, the critical frequency is proportional to the square root of N max electron density. \(f_c=9\sqrt{N \,max} \)

5. Find the critical frequency when the refractive index of the layer is 0.54 and MUF is 9MHz?

a) 0.84MHz

b) 7.57MHz

c) 0.75MHz

d) 8.4MHz

Answer: b

Explanation: Refractive index \(n=\sqrt{

 

}\)

⇨ \(N_{max}=\frac{

f^2}{81}=\frac{

^2}{81}=0.708 ×10^{12}/cm^3\)

⇨ critical frequency \(f_c=9\sqrt{N \,max} =9\sqrt{0.708 ×10^{12}}=7.57Mhz\)

6. The value of refractive index when the MUF is equal to the critical frequency is ______

a) 1

b) 0

c) 0.5

d) 0.29

Answer: b

Explanation: Refractive index \(n=\sqrt{

 

}\)

Critical frequency \(f_c=9\sqrt{N \,max} \)

When \(f_c=f_{MUF}, n=\sqrt{

 

}=1-1=0\)

7. For f MUF ≥ f c , the wave will reflects back irrespective of the angle of incidence.

a) True

b) False

Answer: b

Explanation: For f MUF ≤ f c , the wave will reflects back irrespective of the angle of incidence. For f MUF ≥ f c , the wave will reflects back depending of the angle of incidence .

8. Which of the following frequency is greater than the critical frequency?

a) MUF

b) LUF

c) Optimum frequency

d) VLF

Answer: a

Explanation: According to the secant law, f MUF =f c secθ i

So MUF is greater than or equal to the critical frequency depending on the secθ i

9. When the frequency decreases below _____ frequency, the signal reception becomes too weak and the Noise increases.

a) MUF

b) LUF

c) Optimum frequency

d) Critical frequency

Answer: b

Explanation: When the frequency is lowered below the lowest usable frequency, the signal gets absorbed. At low frequencies the noise level is also more and signal reception will be also difficult.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Maximum Usable Frequency”.

1. The maximum possible frequency for which the wave is reflected back for a given distance of propagation in the ionosphere layer is called as_______

a) Maximum usable frequency

b) Critical frequency

c) Resonance frequency

d) Dominating frequency

Answer: a

Explanation: The maximum possible frequency for which the wave is reflected back for a given distance of propagation in the ionosphere layer is called as maximum usable frequency . For a specified angle there will be a maximum frequency for which the wave is reflected back. If wave exceeds MUF then it’s not reflected back.

2. If wave exceeds the MUF then it is not reflected back.

a) True

b) False

Answer: a

Explanation: Maximum usable frequency  is defined as the maximum possible frequency for which the wave is reflected back for a given distance of propagation in the ionosphere layer. If the wave exceeds MUF then it’s not reflected back and is transmitted to other upper layers and signal is lost.

3. The frequency below which the entire power gets absorbed is called as ____

a) MUF

b) LUF

c) Critical frequency

d) Optimum frequency

Answer: b

Explanation: The frequency below which the entire power gets absorbed is called as LUF . The maximum possible frequency for which the wave is reflected back for a given distance of propagation in the ionosphere layer is called as maximum usable frequency . The frequency at which there is optimum return of the wave is the optimum frequency.

4. Suppose a ray is incident normally in the ionosphere region with electron density 36×10 10 /cm 3 , then the critical frequency is _____

a) 5GHz

b) 54MHz

c) 5.4MHz

d) 324GHz

Answer: c

Explanation: \(f_c=9\sqrt{N \,max}\) where N max = electron density and angle of incidence is ἰ = 0 

∴ \(f_c =9\sqrt{36×10^{10}} = 5.4MHz\)

5. Relation between MUF and critical frequency is ______

a) f c = f MUF secθ i

b) f MUF = f c sec 2 θ i

c) f c = f MUF sinθ i

d) f MUF = f c secθ i

Answer: d

Explanation: The f MUF in terms of critical frequency is given by f MUF = f c secθ i for a given angle of incidence between two locations.

6. Skip distance is the _____

a) Minimum distance at which wave returns back at the lowest possible frequency

b) Maximum distance at which wave returns back at the critical frequency

c) Minimum distance at which wave returns back at the critical angle

d) Maximum distance at which wave returns back at the lowest possible frequency

Answer: c

Explanation: The minimum distance at which the wave returns back at the critical angle is called skip distance.

7. What is the value of maximum usable frequency when the incident angle is 60° and the critical frequency is 4.5MHz?

a) 4.5MHz

b) 2.25MHz

c) 9MHz

d) 18MHz

Answer: c

Explanation: MUFf MUF = f c secθ i = 4.5MHz ×sec60=9MHz.

8. Which of the following is true when a ray is incident normally in an Ionosphere region?

a) MUF is equal to critical frequency

b) MUF is greater than critical frequency

c) MUF is less than critical frequency

d) MUF is zero

Answer: a

Explanation: When a ray is incident normally in an Ionosphere region, θ i =0

⇨ f MUF =f c secθ i =f c

Therefore, MUF is equal to critical frequency.

9. Which of the following statements is false?

a) MUF is always greater than or equal to critical frequency depending on the incident angle

b) Optimum frequency is the frequency at which optimum reflection of wave takes place

c) Beyond the MUF, the entire wave gets reflected back

d) Below LUF, the entire power of wave gets absorbed

Answer: c

Explanation: Beyond MUF,  the rays get penetrated into the region and no part of it is reflected back.

10. Suppose a ray is incident normally in the ionosphere region with electron density 25×10 10 /cm 3 , then the Maximum usable frequency is _____

a) 5GHz

b) 45MHz

c) 4.5MHz

d) 125GHz

Answer: c

Explanation: \(f_c=9\sqrt{N \,max}\) where N max = electron density and angle of incidence is ἰ = 0 

∴ \(f_c = 9\sqrt{25×10^{10}} = 4.5MHz\)

⇨ And f MUF = f c secθ i = f c = 4.5MHz

11. What is the refractive index of the region operating at 16MHz frequency with the electron density 49×10 10 /cm 3 ?

a) 0.518

b) 0.919

c) 0.155

d) 0.845

Answer: b

Explanation: Refractive index \(n=\sqrt{

 

}=\sqrt{

 

}=\sqrt{}=\sqrt{0.845}\)

n=0.919

12. Find the MUF for the wave operating at a critical frequency 6MHz and having the skip distance d as 25km and virtual height of 100km in the ionosphere layer.

a) 6.05MHz

b) 1.25MHz

c) 1.025MHz

d) 3.25MHz

Answer: a

Explanation: \(f_{MUF}=f_c \sqrt{

 

^2+1)}=6×10^6×\sqrt{

 

^2+1)}=6.05MHz\)

This set of Antennas Multiple Choice Questions & Answers  focuses on “Multi Hop Propagation”.

1. The propagation of wave from transmitter to receiver without touching the ground is called as ___

a) Single hop distance

b) Virtual height

c) Actual height

d) Multi-hop

Answer: a

Explanation: If wave travels from the transmitter to receiver without touching the ground is called the single hop distance. If it touches the ground in between, then it is called multi hop propagation.

2. The propagation of wave from transmitter to receiver by touching ground in between them and goes through different layers is called _____

a) Multi-hop single layer

b) Single hop multi layer

c) Multi hop multi layer

d) Single hop single layer

Answer: c

Explanation: Multi hopping means the wave touches the ground once or more than once while travelling from transmitter to receiver. If it propagates through the multiple layers then it is called multi-hop multi layer. In single hopping the wave doesn’t touches the ground.

3. The take-off angle for the curved earth surface is given by ____

a) β = 90 – θ i – 57.3d/2R

b) β = 180 – θ i

c) β = 180 – θ i – 57.3d/2R

d) β = 90 – θ i

Answer: a

Explanation: The take-off angle for the curved earth surface is β = 90 – θ i – 57.3d/2R

Take-off angle for the flat earth surface is β = 90 – θ i

4. Calculate the skip distance for the flat earth when a wave is reflected in an ionosphere at a height of 100km at a take-off angle 10°.

a) 172km

b) 35.26km

c) 17.25km

d) 35.26m

Answer: b

Explanation: Take-off angle for the flat earth surface is β = 90 – θ i

θ i =90-β=90-10=80

skip distance \(d=2h\sqrt{[

^2-1]}=2h\sqrt{[^2-1]}=35.26km \)

5. Find the take-off angle for the flat earth surface with θ i = 75.

a) 15

b) 30

c) 105

d) 75

Answer: a

Explanation: Take-off angle for the flat earth surface is β = 90 – θ i = 90 – 75 = 15

6. Which of the following propagation is used when the receiver is beyond the skip distance?

a) Single hop

b) Single hop multi layer

c) Multi hop

d) Ground wave

Answer: c

Explanation: When the receiver is beyond the skip distance, single hop is prevented from reaching the receiver. It requires more than one hop to travel to the desired receiver. So Multi Hop propagation with single or multi layer is used.

7. The Upper ray is stronger than the lower ray and is mostly preferred for communication.

a) True

b) False

Answer: b

Explanation: The Upper ray is weaker than the lower ray in terms of energy contents and it spreads more in the electron density region compared to the lower ray. So the lower ray is mostly preferred for communication.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Smart Antennas – Benefits”.

1. Which of the following statements is false regarding smart antenna for 802.11 applications?

a) Coverage area is increased

b) Signal paths are reduced

c) Probability of collisions is increased

d) Interference is reduced

Answer: c

Explanation: Smart antenna allows improved SNR so coverage area is also increased. The signal paths are reduced so the multi-path reduction, the transmitted RF energy is confined to a desired direction so interference is reduced. Probability of collisions is reduced and network capacity is increased.

2. Which of the following is not the advantage of smart antenna?

a) Range is increased

b) Secured transmission

c) Design of trans-receiver is simple

d) Probability of collisions is reduced

Answer: c

Explanation: The design of trans-receiver for a smart antennas is quite complex than a traditional base station trans-receiver. Secured transmission, increase in range, reduction in probability of collisions is benefits of smart antenna.

3. Which of the following is used to distinguish the selected signal and the multipath signal in smart antenna architecture?

a) DSP procedure

b) Switched beam array

c) Range gates

d) Delay Cancellers

Answer: a

Explanation: Adaptive array system change the sign patterns with respective to RF settings. DSP procedure is used to differentiate the selected signal from the multipath signals. Switched beam is used for forming multi rays in specific orders. Range gates and delay line cancellers find applications in MTI radars to get the accurate target information.

4. Which of the following is mostly used to generate multiple fixed beams by augmentation in specific orders?

a) Switched Beam array

b) Range gates

c) Conical Scanning

d) FMCW radar

Answer: a

Explanation: Switched beam is used for forming multi rays in specific orders. Range gates and delay line cancellers find applications in MTI radars to get the accurate target information. In conical scanning continuous scanning is performed by beam.

5. In which of the following type of smart antenna, the performance of the multi-channel fading is limited?

a) Diversity system

b) Omni-directional system

c) Sectored system

d) Directional system

Answer: a

Explanation: In Diversity system, multi-path fading, co-channel interference, delay spread impairments are greatly reduced in smart antennas. Directional system uses sectored antennas reference to provide the energy to intended direction.

6. Which of the following is not the benefit of smart antenna?

a) It increases the SIR

b) It decreases the SIR

c) It increases the SNR

d) Probability of collisions is reduced

Answer: b

Explanation: Probability of collisions is reduced so the inference is also reduced in smart antennas. Thus the Signal to interference  increases in this. The SNR is also high as the multipath fading is also reduced n diversity type system.

7. In Smart antennas, signal to interference ratio is low.

a) True

b) False

Answer: b

Explanation: Probability of collisions is reduced so the inference is also reduced in smart antennas. Thus the Signal to interference  increases in this by increasing the signal strength and reducing the interference.

8. Which of the following suffers the co-channel interference most?

a) Smart antennas

b) Sectored antennas

c) Omni-directional

d) Both Smart antenna and Omni-direction

Answer: c

Explanation: Omni-directional suffers more co-channel interference compared to the sectored and smart antennas. Probability of collisions is reduced so the inference is also reduced in smart antennas. Thus, the Signal to interference  increases.

9. Smart antennas can be used for location-specific service.

a) True

b) False

Answer: a

Explanation: The spatial detection nature of the smart antenna enables to locate humans in emergencies. It is used for location-specific services.

10. Which of the following is benefit of smart antenna?

a) It is easy to tap the connection by the intruder apart from the user

b) Co-Channel interference is less compared to the Omni-directional

c) Design of Trans-receiver is complex

d) It cannot be used for location-specific services

Answer: b

Explanation: The spatial detection nature of the smart antenna enables for location specific services. Probability of collisions is reduced so the inference is also reduced in smart antennas. Thus the Signal to interference  increases in this. Smart antennas provide security so it is difficult for the intruder to get the user connection.

This set of Antennas Multiple Choice Questions & Answers  focuses on “Smart Antennas – Drawbacks”.

1. Which of the following is the drawback of smart antenna?

a) Design of trans-receiver

b) Spacing between base stations

c) Probability of collisions being reduced

d) Focused to an intended direction

Answer: a

Explanation: Design of trans-receiver is a complex when compared to the traditional base station trans-receivers because separate trans-receiver chains are needed for each array elements. Probability of collision reduction, focusing on desired direction is the benefits of smart antenna.

2. Which of the following statement regarding smart antenna is false?

a) Design of transceiver is complex

b) Smart antennas is more directive

c) Smart antenna is more expensive

d) Numeric processors and control systems are not needed in smart antennas

Answer: d

Explanation: Compared to the traditional base station trans-receivers, design of trans-receiver is a complex because there is a need of separate trans-receiver chains for each array elements. Smart antennas are more directive and costly. The base stations need very powerful numeric processors and control systems.