Category: Antennas Objective Questions

250+ TOP MCQs on Radiation – Basic Maxwell Equations and Answers

Antennas Multiple Choice Questions & Answers on “Radiation – Basic Maxwell Equations”.

1. The Maxwell equation ∇×E=(frac{-partial B}{partial t}) is derived from which law?
a) Amperes law
b) Faradays Law
c) Lens law
d) Gauss law
Answer: b
Clarification: 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
Clarification: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=(-frac{partial phi}{partial t})
c) ∇×E=(frac{-partial B}{partial t})
d) ∇×D=ρv
Answer: a
Clarification:

  • Faradays Law                : ∇×E=(-frac{partial B}{partial t})
  • Amperes Law                : ∇×H=J+(frac{partial D}{partial t})
  • Gauss Law for electric field     : D=ρv
  • Gauss law for magnetic field  : ∇.B=0

4. Gauss for the Magnetic Field is given by ______
a) ∇.B=0
b) ∇×B=ρv
c) ∇×B=0
d) ∇.B=ρv
Answer: a
Clarification: 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=ρv
c) ∇×D=0
d) ∇.D=ρv
Answer: d
Clarification: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=ρv
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=-(-frac{partial B}{partial t})
b) ∇×H=J
c) ∇.D=ρv
d) ∇.B=0
Answer: d
Clarification: 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=ρv

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
Clarification: 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_vrho_v dv )
  • Gauss law magnetic field : (oint_sB.ds =0)
  • Faradays law : (int_cE.dl =-int_sfrac{partial B}{partial t}.dS)
  • Amperes law : (int_cH.dl =int_s(frac{partial D}{partial t} + J).dS)

8. Divergence of Magnetic field is ______
a) volume charge density ρv
b) zero
c) infinite
d) dependent on magnetic field vector
Answer: b
Clarification: 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=ρv.

9. Which of the following Maxwell equation is correct for a non-conducting and lossless medium?
a) ∇.D=ρv
b) ∇.D=0
c) ∇×D=ρv
d) ∇×E=0
Answer: b
Clarification: Since it is given non-conducting medium, the charge density ρv=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 6.1×107s/m and relative permittivity 1.
a) 0.00091m
b) 0.9113μm
c) 0.319μm
d) 0.1913μm
Answer: b
Clarification: The skin depth is given by (δ = sqrt{frac{1}{pi fμσ}})
Given f=5GHz
Conductivity σ= 6.1×107 s/m
And μr = 1 =>μ=4π ×10-7
⇨ (δ = sqrt{frac{1}{pi fμσ}}=0.9113mu m.)

250+ TOP MCQs on Antenna Parameters Basics and Answers

Antennas Multiple Choice Questions on “Basics”.

1. The relation between vector magnetic potential and current density is given by ______
a) ∇.A=J
b) ∇×A=H
c) ∇2A=-μJ
d) ∇2A=∇×H

Answer: c
Clarification: 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= -∇2A and B= μH , ∇×H=J
⇨ ∇×μH= -∇2A
⇨ ∇2A=-μJ.

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
Clarification: For an Hertzian dipole, equating the magnitudes of maximum induction and radiation fields we get,
(mid overline{E_θ}mid_{max⁡Radiation} =mid overline{E_θ}mid_{max⁡Induction} )
(frac{I_m dl}{4piepsilon} (frac{ω}{v^2r})=frac{I_mdl}{4piepsilon} (frac{1}{r^2v}))
(r=frac{v}{ω}=frac{lambda f}{2pi 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
Clarification: 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^2frac{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
Clarification: For a lossless antenna, the radiation efficiency ecd=1.
Overall efficiency of antenna is given by eo =ecd (1-(midgamma^2mid))=1×(1-(0.152))=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
Clarification: 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
Clarification: 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
Clarification: Radiation resistance of a Hertzian dipole of length l is
R=80π2((frac{l}{lambda})^2=80pi^2 (frac{lambda/8}{lambda})^2=12.33Omega)

8. Relation between directivity and effective area of transmitting and receiving antenna is ________
a) Dt At=Dr Ar
b) Dt Ar=Dr At
c) At Dt=∈Dr Ar
d) Dt At=∈Dr Ar

Answer: b
Clarification: The power collected by the receiving antenna is
(P_r = frac{P_tD_tA_r}{4pi R^2} => D_tA_r = frac{P_r}{P_t}4pi R^2) = Dr At
∴ Dt Ar=Dr At

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
Clarification: 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
Clarification: Since radiation resistance of quarter-wave monopole (l=λ/4) is 36.56Ω, then for a half-wave dipole (l=λ/2) 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
Clarification: 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.

250+ TOP MCQs on Radiation Pattern for 4-Isotropic Elements and Answers

Antennas Multiple Choice Questions on “Radiation Pattern for 4-Isotropic Elements”.

1. The array factor of 4- isotropic elements of broadside array is given by ____________
a) (frac{sin(2kdcosθ)}{2kdcosθ} )
b) (frac{sin(kdcosθ)}{2kdcosθ} )
c) (frac{sin(2kdcosθ)}{kdcosθ} )
d) (frac{cos(2kdcosθ)}{2kdcosθ} )
Answer: a
Clarification: Normalized array factor is given by
(AF=frac{sin(Nᴪ/2)}{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{sin(Nᴪ/2)}{N frac{ᴪ}{2}}=frac{sin(2kdcosθ)}{2kdcosθ} )

2. A 4-isotropic element broadside array separated by a λ/2 distance has nulls occurring at ____________
a) (cos^{-1} (±frac{n}{2}) )
b) (cos^{-1} (±frac{4n}{2}) )
c) (sin^{-1} (±frac{n}{2}) )
d) (sin^{-1} (±frac{n}{4}) )
Answer: a
Clarification: The nulls of the N- element array is given by
(θ_n=cos^{-1}⁡(frac{λ}{2πd} left[-β±frac{2πn}{N}right])=cos^{-1}⁡(frac{λ}{2πd} left[±frac{2πn}{N}right]))
⇨ (θ_n=cos^{-1}⁡(frac{λ}{2π(λ/2)}left[±frac{2πn}{N}right])=cos^{-1} (±frac{2n}{4})=cos^{-1} (±frac{n}{2}) )

3. A 4-isotropic element broadside array separated by a λ/4 distance has nulls occurring at ____________
a) cos-1 (±n)
b) (cos^{-1} (±frac{n}{2}))
c) (sin^{-1} (±frac{n}{2}))
d) sin-1 (±n)
Answer: a
Clarification: The nulls of the N- element array is given by
(θ_n=cos^{-1}⁡(frac{λ}{2πd} left[-β±frac{2πn}{N}right])=cos^{-1}⁡(frac{λ}{2πd} left[±frac{2πn}{N}right]))
⇨ (θ_n=cos^{-1}(⁡frac{λ}{2π(λ/4)}left[±frac{2πn}{N}right])=cos^{-1} (±frac{4n}{4})=cos^{-1} (±n) 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(cosθ)
b) cos(sinθ)
c) sin(sinθ)
d) sin(cosθ)
Answer: a
Clarification: Normalized array factor is given by (AF=frac{sin(Nᴪ/2)}{N frac{ᴪ}{2}})
And ᴪ=kdcosθ+β
Since its given broad side array β=0,
ᴪ=kdcosθ+β=kdcosθ
(frac{Nᴪ}{2}=2kdcosθ=2(frac{2π}{λ})(frac{λ}{4})cosθ=πcosθ )
(AF=frac{sin(Nᴪ/2)}{N frac{ᴪ}{2}}=frac{sin(πcosθ)}{πcosθ}=sinc(cosθ).)

5. The array factor of 4- isotropic elements of broadside array separated by a λ/2 is given by ____________
a) sinc(2cosθ)
b) sin(2πcosθ)
c) sinc(2πsinθ)
d) sin(2sinθ)
Answer: a
Clarification: Normalized array factor is given by (AF=frac{sin(Nᴪ/2)}{N frac{ᴪ}{2}})
And ᴪ=kdcosθ+β
Since its given broad side array β=0,
ᴪ=kdcosθ+β=kdcosθ
(frac{Nᴪ}{2}=2kdcosθ=2(frac{2π}{λ})(frac{λ}{2})cosθ=2πcosθ )
(AF=frac{sin(Nᴪ/2)}{N frac{ᴪ}{2}}=frac{sin(2πcosθ)}{2πcosθ}=sinc(2cosθ).)

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
Clarification: The nulls of the N- element array is given by
(θ_n=cos^{-1}⁡(frac{λ}{2πd} left[-β±frac{2πn}{N}right])=cos^{-1}⁡(frac{λ}{2πd} left[±frac{2πn}{N}right]))
⇨ (θ_n=cos^{-1}⁡(frac{λ}{2π(λ/2)} left[±frac{2πn}{N}right])=cos^{-1} (±frac{2n}{4})=cos^{-1}(±frac{n}{2}) )
⇨ (n=1 (first ,null) cos^{-1} (±frac{n}{2})=cos^{-1} (±frac{1}{2})=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
Clarification: The nulls of the N- element array is given by
(θ_n=cos^{-1}⁡(frac{λ}{2πd} left[-β±frac{2πn}{N}right])=cos^{-1}⁡(frac{λ}{2πd} left[±frac{2πn}{N}right]))
(θ_n=cos^{-1}⁡(frac{λ}{2π(λ/4)}left[±frac{2πn}{N}right])=cos^{-1} (±frac{4n}{4})=cos^{-1}(±n)=0)

8. The necessary condition for maximum of the second side lobe of n element array is __________
a) ( frac{Nᴪ}{2}=±frac{5π}{2})
b) ( frac{Nᴪ}{2}=±frac{3π}{2})
c) ( frac{Nᴪ}{2}=±frac{π}{2})
d) ( frac{Nᴪ}{2}=±frac{4π}{2})
Answer: a
Clarification: The secondary maxima occur when the numerator of the array factor equals to 1.
⇨ (sin(frac{Nᴪ}{2})=±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
Clarification: The direction of the secondary maxima (minor lobes) occur at θs
(θ_s=cos^{-1} (frac{λ}{2πd} left[-β±frac{(2s+1)}{N} πright]))
⇨ (θ_s=cos^{-1} (frac{λ}{2π(λ/2)} left[±frac{3}{4} πright])) (s=1 for 1st minor lobe)
⇨ (θ_s=cos^{-1} (±frac{3}{4})=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
Clarification: The nulls of the N- element array is given by (θ_n=cos^{-1}⁡(frac{λ}{2πd} left[-β±frac{2πn}{N}right]))
Since its given broad side array (β=±kd=±frac{2πd}{λ}=±frac{π}{2},)
(θ_n=cos^{-1}⁡(frac{2}{π} left[∓frac{π}{2}±frac{2πn}{4}right]))
=cos-1([∓1±n])
First null at n=1; θn=cos-1⁡([1±1) (considering β=(-frac{π}{2}) )
θn = cos-1⁡ (0) or cos-1⁡(2)
θn = cos-1⁡ (1)=90.

250+ TOP MCQs on Antennas Basics and Answers

Antennas Multiple Choice Questions 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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 (convex).

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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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.

 

250+ TOP MCQs on First Null Beam Width and Answers

Antennas Multiple Choice Questions 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
Clarification: 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
Clarification: FNBW ≈ 2HPBW=2×30°=60°

3. Total beam area is sum of major and minor lobe areas.
a) True
b) False
Answer: a
Clarification: Total beam area is sum of major lobe area ΩM and minor lobe area Ωm.
ΩA = ΩM+Ωm

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
Clarification: 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
Clarification: 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) (frac{in_B}{in_s}=1)
d) ∈B+∈s=2
Answer: a
Clarification: 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 ΩM and minor lobe area Ωm.
(in_B=frac{Omega_M}{Omega_A}, in_s = frac{Omega_m}{Omega_A})
∈B+∈s=1

all areas of Antennas,

250+ TOP MCQs on Antenna Characteristics and Answers

Antennas Multiple Choice Questions 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
Clarification: 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
Clarification: Power radiated Pr=I2 Rr=100×32=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
Clarification: Front-to-Back ratio (FBR) =(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
Clarification: 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 r2?
a) Far field
b) Near field
c) Radiation field
d) Electrostatic field
Answer: b
Clarification: Induction field is also known as ‘Near field’, varies inversely with r2. Electrostatic field varies inversely with r3. 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) 1.28m2
b) 2.18m2
c) 0.128m2
d) 12.8m2
Answer: a
Clarification: 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
Clarification: End-fire array has its main beam parallel to the axis of antenna (θ=0° or 180°). 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
Clarification: 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
Clarification: As beam width of antenna increases its area coverage broadens, thereby directivity decreases. Beam area and directivity are inversely proportional. (D=frac{4pi}{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
Clarification: 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.