Antennas Questions and Answers – Aperture Antenna – Beamwidths

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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
View Answer

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
View Answer

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
View Answer

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.
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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
View Answer

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⁡(λ/b)
b) sec-1⁡(λ/b)
c) cos-1⁡(λ/a)
d) sin-1⁡(λ/a)
View Answer

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⁡(nλ/b)
⇨ For first null n=1 θ= sin-1⁡(λ/b).

6. The first null beam width in the E-plane of a rectangular aperture of a×b is given by _______________
a) 2sin-1⁡(λ/b)
b) sin-1⁡(λ/a)
c) 2sec-1⁡(λ/b)
d) 2cos-1⁡(λ/a)
View Answer

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\(⁡(\frac{n\lambda}{b})\)
Therefore, the FNBW in E-plane is given by FNBW=2 θ = sin-1⁡\(⁡(\frac{\lambda}{b})\).

7. Larger the size of the aperture, the narrower is the Beam-widths.
a) True
b) False
View Answer

Answer: a
Explanation: The FNBW in E-plane is given by FNBW=2 θ = sin-1⁡\(⁡(\frac{n\lambda}{b})\). As the dimension of the antenna aperture increases, the FNBW will decrease. Thereby, beam-width becomes narrower.
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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
View Answer

Answer: a
Explanation: By equating the field in E-plane to half power point
\(\frac{sin⁡(0.5kbsin\theta)}{0.5kbsin\theta} = \frac{1}{\sqrt 2} => \theta = arcsin⁡(\frac{0.443
\lambda}{b})\)
Now HPBW = 2 arcsin⁡\((\frac{0.443\lambda}{b})\)≈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
View Answer

Answer: a
Explanation: HPBW = 2 arcsin⁡\((\frac{0.443\lambda}{b})\)≈0.886λ/b=0.886/2=0.443.
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10. The value at which the second null occurs in H-plane of rectangular aperture of a*b is given by ____
a) sin-1⁡(2λ/a)
b) sin-1⁡(λ/a)
c) sin-1⁡(a/2λ)
d) sin-1⁡(a/λ)
View Answer

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⁡(nλ/a)
⇨ For null n=2 θ= sin-1⁡(2λ/a).

Sanfoundry Global Education & Learning Series – Antennas.

To practice all written questions on Antennas, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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