Antenna Theory Questions and Answers Part-8

1. 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)

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)

2. 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

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

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

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

4. 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 )

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

5. If the current input to the antenna is 100mA, then find the average power radiated from the half-wave dipole antenna?
a) 365mW
b) 0.356mW
c) 0.365mW
d) 356mW

Answer: a
Explanation: Average Power radiated from the half-wave dipole Pavg=\(I_{rms}^2 R=(\frac{I_m}{\sqrt 2})^2 R \)
Radiation resistance of a half-wave dipole is 73Ω.
Given Im=100mA => Pavg=\((\frac{100×10^{-3}}{\sqrt 2})^2×73=365mW.\)

6. The average radiated power of half-wave dipole is given by ______
a) \(73I_{rms}^2\)
b) \(36.5I_{rms}^2\)
c) \(13.25I_{rms}^2\)
d) \(146I_{rms}^2\)

Answer: a
Explanation: Radiation resistance of a half-wave dipole is 73Ω.
Average Power radiated from the half-wave dipole \(P_{avg}=I_{rms}^2 R=73I_{rms}^2\)
Radiation resistance of a quarter-wave monopole is 36.5Ω.

7. If the power radiated by a quarter-wave monopole is 100mW then power radiated by a half wave dipole under same current input is _____
a) 100W
b) 100mW
c) 200W
d) 200mW

Answer: d
Explanation: Average Power radiated from the half-wave dipole \(P_{avg-hlf}=I_{rms}^2 R_{hlf}\)
⇨ \(\frac{P_{avg-hlf}}{P_{avg\, mono}} = \frac{R_{hlf}}{R_{mono}} = \frac{73}{36.5}=2\) (under same current)
⇨ Pavg-hlf = 2Pavg mono = 2*100mW=200mW

8. Power radiated by a half wave dipole is how many times the power radiated by a quarter wave monopole under same input current to antennas?
a) 2
b) 3
c) 4
d) 1

Answer: a
Explanation: Average Power radiated from the half-wave dipole \(P_{avg-hlf}=I_{rms}^2 R_{hlf}\)
⇨ \(\frac{P_{avg-hlf}}{P_{avg\, mono}} = \frac{R_{hlf}}{R_{mono}} = \frac{73}{36.5}=2\) (under same current)

9. Find the magnetic field if the electric field radiated by the half-wave dipole is 60mV/m?
a) 159μA/m
b) 195μA/m
c) 159mA/m
d) 195mA/m

Answer: a
Explanation: η=\(\frac{E}{H}\)
⇨ 120π=60m/H
⇨ H = 159μA/m

10. In which of the following the power is radiated through a complete spherical surface?
a) Half-wave dipole
b) Quarter-wave Monopole
c) Both Half-wave dipole & Quarter-wave Monopole
d) Neither Half-wave dipole nor Quarter-wave Monopole

Answer: a
Explanation: In a half-wave dipole the power is radiated in the entire spherical surface and in quarter wave monopole the power is radiated only through a hemispherical surface. Hence its radiation resistance is also twice that of the quarter wave monopole.