Antenna Parameters Questions and Answers – Antenna Noise Temperature

«
»

This set of Antennas Multiple Choice Questions & Answers (MCQs) focuses on “Antenna Noise Temperature”.

1. Relation between brightness temperature TB and physical body temperatureTp is ____
a) TB=\((1-\mid\Gamma_s \mid^2) T_p\)
b) TB=\(T_p/(1-\mid\Gamma_s \mid^2)\)
c) TB=\((1-\mid\Gamma_s \mid)T_p\)
d) TB=\((1-\mid\Gamma_s \mid)^2 T_p\)
View Answer

Answer: a
Explanation: The relation between brightness temperature and the physical body temperature is given by TB=\((1-\mid\Gamma_s \mid^2) T_p\)
Here Γs is the reflection coefficient for a given polarization and emissivity =\(1-\mid\Gamma_s \mid^2.\)
advertisement

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

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

3. 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)+⋯
View Answer

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.
advertisement
advertisement

4. 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
View Answer

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

5. 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
View Answer

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=\(\frac{\Omega_B}{\Omega_A}T_B\)
advertisement

6. 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
View Answer

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

7. 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) \(F=1-\frac{T_0}{T_e}\)
View Answer

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.
advertisement

8. Effective noise temperature Te in terms of noise figure is ____
a) Te=To (F-1)
b) Te=To/(F-1)
c) Te=To/(F+1)
d) Te=To (F+1)
View Answer

Answer: a
Explanation: The relation between noise figure and effective noise temperature is given by F=1+\(\frac{T_e}{T_o}\)
⇨ F-1=\(\frac{T_e}{T_o}\)
⇨ Te=To (F-1)

9. Which of the following statement is false?
a) Noise power of antenna depends on the antenna temperature as well as the noise due to the receiver surroundings
b) Noise figure value lies between 0 and 1
c) Any object with physical temperature greater than 0K radiates energy
d) Noise power per unit bandwidth is kTA W/Hz
View Answer

Answer: b
Explanation: The relation between noise figure and effective noise temperature is given by F=1+\(\frac{T_e}{T_o}\)
And object with physical temperature greater than 0K radiates energy. So \(\frac{T_e}{T_o}\) > 0 and F > 1
Noise power per unit bandwidth of antenna is kTA W/Hz while noise power of antenna is kTAB W
advertisement

10. Find the effective noise temperature if noise figure is 3 at room temperature (290K)?
a) 290K
b) 580K
c) 289K
d) 195K
View Answer

Answer: b
Explanation: Room temperature To=290K
Noise figure F=1+\(\frac{T_e}{T_o}\)
Te=To(F-1)=290(3-1)=580K.

11. What should be the noise figure value at which the effective noise temperature equals to room temperature?
a) 2
b) 1
c) 0
d) 1/T_(o )
View Answer

Answer: a
Explanation: Noise figure F=1+\(\frac{T_e}{T_o}\)
Te=To(F-1)
F-1=1
F=2.

Sanfoundry Global Education & Learning Series – Antennas.

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

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
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