Antenna Parameters Questions and Answers – Effective Aperture

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This set of Antennas Multiple Choice Questions & Answers (MCQs) focuses on “Effective Aperture”.

1. Effective aperture is the ability of antenna to extract energy from the electromagnetic wave.
a) True
b) False
View Answer

Answer: a
Explanation: Effective aperture is defined as the ratio of power received from load to the average power density produced at that point. So it is the ability of antenna to extract energy from EM wave.
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2. Which of the following best describes the condition for Maximum effective aperture?
a) Load impedance must be equal to the antenna impedance
b) Load impedance must be equal complex conjugate to the antenna impedance
c) Receiver power should be minimum
d) Transmitter power should be minimum
View Answer

Answer: b
Explanation: Foe effective aperture to be maximum, the receiver power should be maximum. Maximum power transfer states that the maximum received power is obtained when the impedance of network matches with the complex conjugate of the load impedance. Hence, Load impedance must be equal to the complex conjugate of the antenna impedance.

3. What is the effective aperture of Hertzian dipole antenna operating at frequency 100 MHz?
a) 1.07m2
b) 0.17m2
c) 1.7m2
d) 1.2m2
View Answer

Answer: a
Explanation: Effective aperture for a Hertzian dipole is given by \(A_e=1.5\frac{\lambda^2}{4\pi} \)
Gain of Hertzian dipole is 1.5
\(\lambda=\frac{c}{f}=\frac{3×10^8}{100×10^6}=3m\)
\(A_e=1.5 \frac{\lambda^2}{4\pi} =1.5\frac{3^2}{4\pi} =1.07m^2\)
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4. If physical aperture of antenna is 0.02m2 and aperture efficiency is 0.5, then what is the value of effective aperture?
a) 0.0004m2
b) 0.001m2
c) 0.01m2
d) 25m2
View Answer

Answer: c
Explanation: Effective aperture Aem=Aeη=0.02×0.5=0.01 m2

5. Expression for aperture efficiency in terms of physical aperture Ae and effective aperture Aem is ____
a) \(\frac{A_e}{A_{em}}\)
b) \(\frac{A_{em}}{A_e}\)
c) \(\frac{A_e+A_{em}}{A_e-A_{em}}\)
d) \(\frac{A_e-A_{em}}{A_e+A_{em}}\)
View Answer

Answer: b
Explanation: The aperture efficiency is defined as the ratio of effective aperture to physical aperture of antenna. So aperture efficiency \(\eta=\frac{A_{em}}{A_e}.\)
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6. What is the effective aperture of a Half-wave dipole operating at 100MHz?
a) 1.07m2
b) 1.17m2
c) 1.27m2
d) 1.77m2
View Answer

Answer: b
Explanation: The directivity of half-wave dipole is 1.64
The effective aperture of half-wave dipole is \(A_e=1.64\frac{\lambda^2}{4\pi}\)
\(\lambda=\frac{c}{f}=\frac{3×10^8}{100×10^6}=3m\)
\(A_e=1.64 \frac{\lambda^2}{4\pi} = 1.64 \frac{3^2}{4\pi}\)=1.17m2.

7. What is the relation between effective length and Effective aperture of antenna?
a) \(A_e = \frac{dL^2\eta}{4R_{rad}}\)
b) \(A_e = \frac{dL^2}{4\eta R_{rad}}\)
c) \(A_e = \frac{dL^2 R_{rad}}{4\eta}\)
d) \(A_e = \frac{dL^2 \eta ^2}{4R_{rad}}\)
View Answer

Answer: a
Explanation: Maximum effective aperture is ratio of maximum power received to the average power density. \(A_e=\frac{P_{Rmax}}{P_{avg}}\)
The received power is maximum when load equal to complex conjugate of network resistance.
⇨ \(I_{total}=\frac{V_{oc}}{2R_{rad}}\)
⇨ \(P_{Rmax}=I_{rms}^2 R_{rad}= (\frac{I_{total}}{\sqrt 2})^2 R_{rad}=\frac{V_{oc}^2}{8R_{rad}}=\frac{\mid E_\theta\mid^2 dL^2}{8R_{rad}} and P_{avg}=\frac{\mid E_\theta\mid^2}{2\eta} \)
\(A_e=\frac{P_{Rmax}}{P_{avg}}=\frac{dL^2\eta}{4R_{rad}}\)
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8. The physical aperture of an isotropic radiator is _______
a) \(\frac{4\pi \eta}{\lambda^2}\)
b) \(\frac{4\pi}{\lambda^2 \eta}\)
c) \(\frac{\lambda^2}{4\pi \eta}\)
d) \(\frac{\lambda^2\eta}{4\pi}\)
View Answer

Answer: c
Explanation: For isotropic radiator, directivity is 1. So the effective aperture is given by \(A_{em}=D \frac{\lambda^2}{4\pi} = \frac{\lambda^2}{4\pi}\)
Then physical aperture =\(\frac{Effective\, aperture}{Aperture\, efficiency}=\frac{\lambda^2}{4\pi \eta}\)

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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