Antenna Parameters Questions and Answers – Directive Gain

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

1. For an isotropic antenna, the average power Pav can be expressed in terms of radiated power Pr as ____
a) Pav=Pr/4π
b) Pav=Pr/2πr2
c) Pav=Pr/2π
d) Pav=Pr/4πr2
View Answer

Answer: d
Explanation: Average power is the total power radiated in the unit area. Here for isotropic radiation, area is spherical (say with radius r) and the area is 4πr2.
∴ Pav=Pr/4πr2
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2. Directive gain is defined as a measure of concentration of power in a particular direction.
a) True
b) False
View Answer

Answer: a
Explanation: Directive gain is the ratio of power density to the average power radiated.
\(G_d = \frac{P_{d(\theta,\emptyset)}}{P_{avg}}\)

3. What is the directive gain when the magnitude of radiation intensity equals to average radiation intensity?
a) 4π
b) ∞
c) 1
d) 0
View Answer

Answer: c
Explanation: Directive gain \(G_d = \frac{P_{d(\theta,\emptyset)}}{P_{avg}} = \frac{P_{d(\theta,\emptyset)}}{P_r/4\pi r^2} = \frac{P_{d(\theta,\emptyset)^{r^2}}}{P_r/4\pi} = \frac{U_{d(\theta,\emptyset)}}{U_{avg}} \)
∴ \(G_d = \frac{U_{(\theta,\emptyset)}}{U_{avg}}\)=1.
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4. Directive gain of antenna when radiation intensity is 5W/Steradian and radiated power 5W is ____
a) 4π
b) 1/4π
c) 25
d) 1
View Answer

Answer: a
Explanation: Given Ud(θ,∅) = 5W/steradian , Pr=5W
Directive gain \(G_d=\frac{P_{d(\theta,\emptyset)^{r^2}}}{P_r/4\pi} = \frac{U_{d(\theta,\emptyset)}}{P_r/4\pi}=4\pi\)

5. The Directive gain is ______ on input power to antenna and _____ on power due to ohmic losses.
a) Independent, independent
b) Dependent, independent
c) Independent, dependent
d) Dependent, dependent
View Answer

Answer: a
Explanation: Directive gain is the ratio of power density to the average power radiated. \(G_d = \frac{P_{d(\theta,\emptyset)}}{P_{avg}}\) So, the Directive gain is independent on both input power to antenna and power due to ohmic losses.
Power gain is dependent on input power and ohmic losses to antenna.
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6. What is the maximum directive gain of antenna with radiation efficiency 98% and maximum power gain 1?
a) 0.98
b) 1.02
c) 1.98
d) 1
View Answer

Answer: b
Explanation: Gpmaxr Gdmax where ηr is radiation efficiency
Therefore Gdmax=1/0.98=1.02

7. Which of the following expression is correct for radiation efficiency?
a) \(\eta_r=\frac{R_r}{R_l}\)
b) \(\eta_r=\frac{R_r}{R_r-R_l}\)
c) \(\eta_r=\frac{R_r}{R_r+R_l}\)
d) \(\eta_r=\frac{R_l}{R_r+R_l}\)
View Answer

Answer: c
Explanation: Radiation efficiency is defined as the ratio of power radiated to the total input power to the antenna. Total input power is the sum of the radiated power Pr and the ohmic losses Pl.
\(\eta_r = \frac{P_r}{P_{in}} = \frac{P_r}{P_r+P_l} = \frac{R_r I_{rms}^2}{I_{rms}^2 R_r+I_{rms}^2 R_l} = \frac{R_r}{R_r+R_l} \)
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8. For a lossless antenna, maximum Power gain equals to the maximum directive gain.
a) True
b) False
View Answer

Answer: a
Explanation: For a lossless antenna, ohmic losses will be zero. So, radiation efficiency will be 100%. Hence, maximum power gain will be equal to the maximum directive gain of antenna.

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Manish Bhojasia - Founder & CTO at Sanfoundry
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