Antenna Array Questions and Answers – End Fire Array

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This set of Antennas Multiple Choice Questions & Answers (MCQs) 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°
View Answer

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°.
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2. What is the phase excitation difference for an end-fire array?
a) 0
b) ±kd /2
c) π
d) ±kd
View Answer

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
View Answer

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.
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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) \(\frac{π}{λ} d \)
View Answer

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
View Answer

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.
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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) \(\frac{π}{4}\)
View Answer

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) \(\frac{9λ}{4}\)
b) \(\frac{5λ}{4}\)
c) \(\frac{5λ}{2}\)
d) \(\frac{9λ}{2}\)
View Answer

Answer: a
Explanation: 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
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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
View Answer

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°

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter