Category: Antennas Objective Questions

250+ TOP MCQs on Baluns and Answers

Antennas Multiple Choice Questions 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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}{2} )
d) (frac{1}{4} )
Answer: a
Clarification: A Balun can be viewed as a transformer. (frac{Z_{unbal}}{Z_{bal}} = (turns ,ratio)^2=(frac{1}{2})^2=frac{1}{4} )
⇨ Zbal = 4Zunbal
So the balanced impedance is four times the unbalanced impedance.

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.)

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