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
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
2. Directive gain is defined as a measure of concentration of power in a particular direction.
a) True
b) False
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
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.
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
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
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.
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
Explanation: Gpmax=ηr 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) \(\eta_r=\frac{R_l}{R_r+R_l}\)
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.
\(\eta_r = \frac{P_r}{P_{in}} = \frac{P_r}{P_r+P_l} = \frac{R_r I_{rms}^2}{I_{rms}^2 R_r+I_{rms}^2 R_l} = \frac{R_r}{R_r+R_l} \)
8. For a lossless antenna, maximum Power gain equals to the maximum directive gain.
a) True
b) False
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.
9. The ratio of power radiated in a particular direction to the total input power of antenna is called as _____
a) Directive gain
b) Power gain
c) Directivity
d) Partial directivity
Explanation: The ratio of power radiated in a particular direction to the actual power input to antenna is called Power gain. \(G_p=\frac{P_{d(\theta,\emptyset)}}{P_{in}}\). Directive gain is the ratio of power radiated in desired direction to the average power radiated from the antenna. Partial directivity is the part of radiation intensity in a particular polarization to radiation intensity in all directions.
10. What is the maximum power gain of antenna with radiation efficiency 98% and directive gain 1?
a) 0.98
b) 1.02
c) 1.98
d) 1
Explanation: Gpmax=ηr Gdmax where ηr is radiation efficiency
Gpmax=0.98×1=0.98.