250+ TOP MCQs on Broadside Array and Answers

Antennas Multiple Choice Questions 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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 θn at which null occurs for an 8-element broadside array with spacing d.
a) (cos^{-1}⁡frac{λn}{Nd})
b) (sin^{-1}⁡frac{λn}{Nd})
c) (cos^{-1}⁡frac{2λn}{Nd})
d) (sin^{-1}⁡frac{2λn}{Nd})
Answer: a
Clarification: Nulls occurs when array factor (AF=frac{sin frac{Nφ}{2}}{frac{Nφ}{2}} = 0)
⇨ (sinfrac{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
Clarification: Directivity (D=frac{2Nd}{λ}=frac{2×5×frac{λ}{4}}{λ}=2.5)
D (dB) = 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
Clarification: 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
Clarification: 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 (cos^{-1}⁡frac{1}{2} )
d) The maxima occurs normal to the axis of array and also at other angles
Answer: b
Clarification: Since the spacing between elements is an integral multiple of λ (n=1), 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
Clarification: 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°.

250+ TOP MCQs on Corner Reflector and Answers

Antennas Multiple Choice Questions on “Corner Reflector”.

1. Corner reflector is designed for radiation in forward direction unlike in plane reflector.
a) True
b) False
Answer: a
Clarification: 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
Clarification: 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
Clarification: 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) (frac{π}{4})
b) (frac{π}{2})
c) (frac{π}{8})
d) 4π
Answer: a
Clarification: 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
Clarification: From the design equations of a corner reflector,
Aperture dimension DA= 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: The corner angle depends on value of N (current element). For a flat reflector N=1. So the corner angle is (α=frac{π}{N}=π=180°.)

250+ TOP MCQs on Radiation – Hertzian Dipole and Answers

Antennas MCQs 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
Clarification: 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
Clarification: Power radiated by a Hertzian dipole Prad=Rrad I2
Rrad = (80pi^2(frac{l}{lambda})^2=80pi^2(frac{frac{lambda}{30}}{lambda})^2=0.877Omega)
Prad = Rrad I2=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
Clarification: 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
Clarification: Since a constant current flow and there is no any charge accumulation at the ends of the dipole, the term 1/r3disappears. Therefore, electrostatic field is absent.

5.Which of the following is the radiation resistance of the Hertzian dipole?
a) (frac{eta_0 w^2 dl^2}{6pi c^2})
b) (frac{eta_0 wdl^2}{6pi c^2})
c) (frac{eta_0 w^2 dl^2}{3pi c^2})
d) (frac{eta_0 w^3 dl^2}{3pi c^3})
Answer: a
Clarification: Radiation resistance of a Hertzian dipole is (R_{rad} = 80pi^2(frac{l}{lambda})^2)
By simplifying the options given above,
(frac{eta_0 w^2 dl^2}{6pi c^2} = frac{120pi(2pi/lambda)^2 dl^2}{6pi} = 80pi^2(frac{l}{lambda})^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
Clarification: 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
Clarification: The radiation resistance of a monopole is 1/8 times the current element.
(R_{rad}=10pi^2 (l/lambda)^2)
For a monopole height h= l/2 => l= 2h
(R_{rad}=10pi^2(frac{2h}{lambda})^2=40pi^2 (frac{hf}{c})^2=40pi^2 (frac{1×100×10^6}{3×10^{10} })^2=4.38mOmega)

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
Clarification: The radiation resistance of monopole is ½ times the short dipole. But the radiation resistance of short dipole is ¼ time the current element.
(R_{rad ,mono}=frac{1}{2}×frac{1}{4}×80pi^2(frac{l}{lambda})^2=frac{1}{8}×80pi^2(frac{l}{lambda})^2=10pi^2 (frac{l}{lambda})^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
Clarification: 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
Clarification: The radiation resistance of monopole is 1/8 times the current element.
Rrad Herztian = 8×Rrad mono=8×18=144Ω

MCQs on all areas of Antennas,

250+ TOP MCQs on End Fire Array and Answers

Antennas Multiple Choice Questions 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
Clarification: 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
Clarification: 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
Clarification: 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{2π}{λ} d )
b) (frac{2π}{λ} d )
c) (-frac{π}{λ} d )
d) (frac{π}{λ} d )
Answer: a
Clarification: 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
Clarification: 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) (frac{π}{2})
b) (-frac{π}{2})
c) π
d) (frac{π}{4})
Answer: a
Clarification: 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}{4})
b) (frac{5λ}{4})
c) (frac{5λ}{2})
d) (frac{9λ}{2})
Answer: a
Clarification: For an N-element end-fire array, the overall length of the array is given by ρ=(N-1)d
⇨ ρ=(N-1)d=(10-1)(λ/4)=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
Clarification: 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°

250+ TOP MCQs on Radiation Pattern of 8-Isotropic Elements and Answers

Antenna Array Questions & Answers for Exams on “Radiation Pattern of 8-Isotropic Elements”.

1. The array factor of 8 – isotropic elements of broadside array is given by ____
a) (frac{sin(2kdcosθ)}{2kdcosθ} )
b) (frac{sin(4kdcosθ)}{4kdcosθ} )
c) (frac{sin(2kdcosθ)}{kdcosθ} )
d) (frac{cos(2kdcosθ)}{2kdcosθ} )
Answer: b
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 arrayβ=0,
ᴪ=kdcosθ+β=kdcosθ
(frac{Nᴪ}{2}=4kdcosθ)
(AF=frac{sin(Nᴪ/2)}{N frac{ᴪ}{2}}=frac{sin(4kdcosθ)}{4kdcosθ} )

2. An 8-isotropic element broadside array separated by a λ/2 distance has nulls occurring at ____
a) (cos^{-1} (±frac{n}{4}))
b) (cos^{-1} (±frac{n}{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}[-β±frac{2πn}{N}])=cos^{-1}⁡(frac{λ}{2πd} [±frac{2πn}{N}]))
⇨ (θ_n=cos^{-1}⁡(frac{λ}{2π(λ/2)} [±frac{2πn}{N}])=cos^{-1} (±frac{2n}{8})=cos^{-1} (±frac{n}{4}) )

3. An 8-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: b
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} [±frac{2πn}{N}]))
⇨ (θ_n=cos^{-1}⁡(frac{λ}{2π(λ/4)}) [±frac{2πn}{N}])=cos^{-1} (±frac{4n}{8})=cos^{-1} (±frac{n}{2}) [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(cosθ)
b) cos(sinθ)
c) sin(sinθ)
d) sinc(2cosθ)
Answer: d
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}=4kdcosθ=4(frac{2π}{λ})(frac{λ}{4})cosθ=2πcosθ)
(AF=frac{sin(Nᴪ/2)}{N frac{ᴪ}{2}}=frac{sin(2πcosθ)}{2πcosθ}=sinc(2cosθ).)

5. The array factor of 8 – isotropic elements of broadside array separated by a λ/2 is given by ____
a) sinc(4cosθ)
b) sin(2πcosθ)
c) sinc(4πsinθ)
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}=4kdcosθ=4(frac{2π}{λ})(frac{λ}{2})cosθ=4πcosθ )
(AF=frac{sin(Nᴪ/2)}{N frac{ᴪ}{2}}=frac{sin(4πcosθ)}{4πcosθ}=sinc(4cosθ).)

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
Clarification: The nulls of the N- element array is given by
(θ_n=cos^{-1}⁡(frac{λ}{2πd} [-β±frac{2πn}{N}])=cos^{-1}⁡(frac{λ}{2πd} [±frac{2πn}{N}]))
⇨ (θ_n=cos^{-1}⁡(frac{λ}{2π(λ/2)} [±frac{2πn}{N}])=cos^{-1} (±frac{2n}{8})=cos^{-1} (±frac{n}{4}) )
⇨ (n=1 (first ,null) cos^{-1} (±frac{n}{4})=cos^{-1} (±frac{1}{4})=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
Clarification: The nulls of the N- element array is given by
(θ_n=cos^{-1}⁡(frac{λ}{2πd} [-β±frac{2πn}{N}])=cos^{-1}⁡(frac{λ}{2πd} [±frac{2πn}{N}]))
(θ_n=cos^{-1}⁡(frac{λ}{2π(frac{λ}{4})}left[±frac{2πn}{N}right])=cos^{-1} (±frac{4n}{8})=cos^{-1} (±frac{n}{2})=cos^{-1} (±1/2)=60)

8. The necessary condition for maximum of the first 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: b
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{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
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)} [±frac{3}{8}π]) ) (s=1 for 1st minor lobe)
⇨ (θ_s=cos^{-1} (±frac{3}{8})=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
Clarification: The nulls of the N- element array is given by (θ_n=cos^{-1}⁡(frac{λ}{2πd} [-β±frac{2πn}{N}]))
Since its given broad side array (β=±kd=±frac{2πd}{λ}=±frac{π}{2},)
(θ_n=cos^{-1}⁡(frac{2}{π} [∓frac{π}{2}±frac{2πn}{8}]))
(=cos^{-1}⁡([∓1±frac{n}{2}]))
First null at n=1; (θ_n= =cos^{-1}⁡([1±frac{1}{2}]) (considering ,β=-frac{π}{2}) )
(θ_n =cos^{-1} (frac{1}{2}),or ,cos^{-1} (3/2) )
(θ_n =cos^{-1} (frac{1}{2})=60.)

Global Education & Learning Series – Antennas.

all exam questions on Antenna Array, here is complete set of 1000+ Multiple Choice Questions and Answers .

250+ TOP MCQs on Flat Sheet Reflector and Answers

Antennas Multiple Choice Questions 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
Clarification: 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
Clarification: For a corner reflector to become a plane reflector the corner angle must be 180°.
For a plane reflector N (current element) =1
(α=frac{π}{N}=π.)

3. Which of the following reflector antenna is formed by joining two plane reflectors at some angle (≠nπ)?
a) Corner
b) Parabolic
c) Cheese
d) Truncated
Answer: a
Clarification: 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.

all areas of Antennas,