Antenna Theory Questions and Answers Part-7

1. Find the power received by the receiving antenna if it is placed at a distance of 20m from the transmitting antenna which is radiating 50W power at a frequency 900MHz and are made-up of half-wave dipoles.
a) 23.65μW
b) 2.365μW
c) 236.5μW
d) 4.73μW

Answer: c
Explanation: given d=20m, Pt=50W and f=900MHz
Gain of half-wave dipoles is 1.64
\(λ = \frac{c}{f} = \frac{3×10^8}{900Mhz} = \frac{1}{3} m \)
\(\frac{P_r}{P_t} = \frac{G_t G_r \lambda^2}{(4\pi R)^2} = \frac{1.64×1.64×1/3^2}{(4\pi ×20)^2}\)
Pr=236.5μW

2. Let’s assume a transmitting antenna having gain 10dB is placed at a distance of 100m from the receiving antenna and radiates a power of 5W. Find the gain of the receiving antenna in dB when the received power is 150μW and transmitter frequency 500MHz?
a) 1.31dB
b) 1.19dB
c) 11.19dB
d) 13.16dB

Answer: c
Explanation:
Given Pt=5W, Pr=150μW, f=500MHz, R=100m and Gt in dB=10dB
Gt in dB=10log10 Gt=10dB
Gt=10
⇨ \(\lambda = \frac{c}{f} = \frac{3×10^8}{500Mhz} = 0.6m\)
From Friss transmission equation, \(\frac{P_r}{P_t} = \frac{G_t G_r \lambda^2}{(4\pi R)^2} \)
⇨ \(G_r=\frac{P_r (4\pi R)^2}{P_t G_t \lambda^2} = \frac{150\mu(4\pi ×100)^2}{5×10×0.6^2}=13.16\)
⇨ Gr in dB=10log10 Gr=10log10 13.16=11.19dB

3. If the distance between the transmitting and receiving antenna is decreased by a factor 2 while other factors remain same, then the new power received by the antenna _______
a) increases by factor 2
b) decreases by factor 2
c) increases by factor 4
d) decreases by factor 4

Answer: c
Explanation: From Friss transmission equation, \(P_r=P_t\frac{G_t G_r \lambda^2}{(4πR)^2}\)
\(\frac{P_{r1}}{P_{r2}} = \frac{R_2^2}{R_1^2} = \frac{(R/2)^2}{R^2} = \frac{1}{4}\)
Pr2=4Pr1.

4. Assume two similar antennas for transmitting and receiving. If the operating frequency gets reduced by 3 times then the received power gets _______
a) increases by factor 3
b) decreases by factor 3
c) increases by factor 9
d) decreases by factor 9

Answer: c
Explanation: From Friss transmission equation,
\(\frac{P_r}{P_t} = \frac{G_t G_r λ^2}{(4\pi R)^2} = \frac{G_t G_r c^2}{(4\pi Rf)^2}\)
\(\frac{P_{r1}}{P_{r2}} = \frac{f_2^2}{f_1^2} = \frac{(f/3)^2}{f^2} = \frac{1}{9}\)
Pr2=9Pr1

5. If the reflection co-efficient is ½ then emissivity is ___
a) 3/4
b) 1/4
c) 1/2
d) 3/2

Answer: a
Explanation: Emissivity in terms of reflection coefficient is given by \(\epsilon=1-\mid\Gamma_s\mid^2=1-\frac{1}{4}=\frac{3}{4}.\)

6. Overall receiver noise temperature expression if T1, T2… are amplifier 1, 2, and so on noise Temperature and G1, G2, and so on are their gain respectively is_____
a) T = \(T_1+\frac{T_2}{G_1}+\frac{T_3}{G_1 G_2}+⋯ \)
b) T = T1+T2 (1-G1)+T3(1-G1G2)+⋯
c) T = \(T_1+\frac{T_2}{(1-G_1)}+\frac{T_3}{(1-G_1 G_2)}+⋯\)
d) T = T1+T2 (G1)+T3(G1G2)+⋯

Answer: a
Explanation: Overall receiver noise temperature expression is given by T = \(T_1+\frac{T_2}{G_1}+\frac{T_3}{G_1 G_2}+⋯ \) System Temperature is one of the important factors to determine the antenna sensitivity and SNR.


7. Total noise power of the system is P=_____
a) k(TA+TR)B
b) k(TA+TR)/B
c) k(TR)B
d) kB/Tsys

Answer: a
Explanation: The overall noise temperature of the system is the sum of noise temperature of antenna TA and the receiver surrounding TR.
⇨ Total noise power of the system is P= k(TA+TR)B
⇨ K is Boltzmann’s constant and B is the bandwidth

8. What is the relation between noise temperature introduced by beam TB and the antenna temperature TA when the solid angle obtained by the noise source is greater than antenna solid angle?
a) TA= TB
b) TA > TB
c) TA < TB
d) TA « TB

Answer: a
Explanation: When the solid angle obtained by the noise source ΩB is greater than antenna solid angle ΩA, then relation between noise temperature introduced by beam TB and the antenna temperatureTA is given by
TA= TB (If Lossless antenna).
For radio astronomy, ΩBA and TA≠ TB; ΔTA=\(\frac{\Omega_B}{\Omega_A}T_B\)

9. Which expression suits best when the solid angle obtained by the noise source is less than antenna solid angle?
a) PA ΩA=PB ΩB and ΔTA=\(\frac{\Omega_B}{\Omega_A} T_B\)
b) PA ΩB=PB ΩA and ΔTA=\(\frac{\Omega_B}{\Omega_A} T_B\)
c) ΔTA=\(\frac{\Omega_A}{\Omega_B} T_B\) and PA ΩB=PB ΩA
d) ΔTA=\(\frac{\Omega_A}{\Omega_B} T_B\) and PA ΩA=PB ΩB

Answer: a
Explanation: When the solid angle obtained by the noise sourceΩB is greater than antenna solid angle ΩA, then relation between noise temperature introduced by beam TB and the antenna temperatureTA is given by
TA= TB (If Lossless antenna).
For radio astronomy, ΩBA and TA≠ TB; ΔTA=\(\frac{\Omega_B}{\Omega_A} T_B\) and PA ΩA=PB ΩB

10. Expression for noise figure F related to the effective noise temperature Te is ____
a) \(F=1+\frac{T_e}{T_o}\)
b) \(F=1+\frac{T_0}{T_e}\)
c) \(F=1-\frac{T_e}{T_o}\)
d) \(F=1-\frac{T_0}{T_e}\)

Answer: a
Explanation: The noise introduced by antenna is known as the effective noise temperature. The relation between noise figure and effective noise temperature is given by
\(F=1+\frac{T_e}{T_o}, T_o\) is the room temperature.