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Correct choice is (a) 800mW

To elaborate: Average Power radiated from the half-wave dipole \(P_{AVG-hlf}=I_{RMS}^2 R_{hlf}=4I_{rms}^2 R_{hlf}\)

⇨\(\frac{P_{avg-hlf}}{P_{avg\, MONO}}=\frac{4I_{rms}^2 R_{hlf}}{I_{rms}^2 R_{mono}}=4*\frac{73}{36.5}=8\) (under same current)

Pavg-hlf=8(Pavg mono)=800mW

Correct answer is (a) 50mW

To ELABORATE: 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)

\(P_{avg \,mono}=\frac{P_{avg-hlf}}{2}=\frac{100mW}{2}=50mW.\)

The correct option is (a) Half-wave dipole

To EXPLAIN: 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.

The correct option is (b) Noise figure value lies between 0 and 1

The best I can EXPLAIN: 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 kTAW/Hz while noise power of antenna is kTAB W

Correct option is (a) \(73I_{RMS}^2\)

To explain I WOULD SAY: 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.

Right OPTION is (a) 159μA/m

To elaborate: η=\(\FRAC{E}{H}\)

⇨ 120π=60m/H

⇨ H = 159μA/m

The correct ANSWER is (b) False

For explanation I would say: The radiation resistance of a HALF wave dipole is 73Ω and that of a QUARTER wave monopole is 36.5Ω. So the POWER radiated by half-wave dipole is two TIMES that of the radiation by quarter wave monopole.

Right CHOICE is (a) 2

The best I can EXPLAIN: 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)

The correct ANSWER is (a) 2

Easy explanation: NOISE figure F=1+\(\FRAC{T_e}{T_o}\)

Te=To(F-1)

 F-1=1

 F=2.

The CORRECT OPTION is (d) 200mW

The 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

Right ANSWER is (b) 580K

To explain I would SAY: Room temperature To=290K

Noise figure F=1+\(\frac{T_e}{T_o}\)

 Te=To(F-1)=290(3-1)=580K.

Correct choice is (a) 365mW

Best 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=\((\frac{100×10^{-3}}{\sqrt 2})^2×73=365mW.\)

The correct answer is (b) 0.576W

Easy EXPLANATION: Magnetic FIELD strength \(H=\frac{I_m}{2\pi r}(\frac{cos⁡(\frac{\pi}{2}cos\theta)}{sin\theta})\)

⇨ 10×10^-6=\(\frac{I_m}{2\pi×2×10^3}(\frac{cos⁡(\frac{\pi}{2}cos\frac{\pi}{2})}{sin\frac{\pi}{2}})\)

⇨ Im=0.125A

Now Average power RADIATED \(P_{avg}=I_{rms}^2 R=(\frac{I_m}{\SQRT 2})^2 R=(\frac{0.125}{\sqrt 2})^2 ×73×=0.576W\)

The correct ANSWER is (a) \(F=1+\frac{T_e}{T_o}\)

Best 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.

The CORRECT answer is (a) Te=To (F-1)

To explain I would say: 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)

Correct option is (a) TA= TB

The explanation is: 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 temperatureTAis given by

TA= TB (If Lossless antenna).

For radio astronomy, ΩB<ΩA and TA≠ TB; ΔTA=\(\frac{\Omega_B}{\Omega_A}T_B\)

Right answer is (a) PA ΩA=PB ΩBand ΔTA=\(\frac{\Omega_B}{\Omega_A} T_B\)

To elaborate: 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 temperatureTAis given by

TA= TB (If Lossless antenna).

For radio astronomy, ΩB<ΩA and TA≠ TB; ΔTA=\(\frac{\Omega_B}{\Omega_A} T_B\) and PA ΩA=PB ΩB

The CORRECT OPTION is (a) 3/4

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

The CORRECT option is (a) TB=\((1-\mid\Gamma_s \mid^2) T_p\)

The explanation is: The relation between brightness temperature and the physical body temperature is GIVEN by TB=\((1-\mid\Gamma_s \mid^2) T_p\)

Here Γsis the reflection coefficient for a given POLARIZATION and emissivity =\(1-\mid\Gamma_s \mid^2.\)

The CORRECT option is (a) k(TA+TR)B

The explanation is: The overall noise TEMPERATURE of the system is the sum of noise temperature of antenna TAand 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

Right OPTION is (a) T = \(T_1+\frac{T_2}{G_1}+\frac{T_3}{G_1 G_2}+⋯ \)

To elaborate: 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.