Antenna Theory Questions and Answers Part-9

1. Find the power radiated from the half wave dipole at 2km away with magnetic field at point \(\theta=\frac{\pi}{2}\) is 10μA/m ?
a) 0.576mW
b) 0.576W
c) 0.756W
d) 0.675W

Answer: b
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
\(P_{avg}=I_{rms}^2 R=(\frac{I_m}{\sqrt 2})^2 R=(\frac{0.125}{\sqrt 2})^2 ×73×=0.576W\)

2. For the same current, the power radiated by half-wave dipole is four times that of the radiation by quarter wave monopole.
a) True
b) False

Answer: b
Explanation: 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.

3. If the power radiated by a half wave dipole is 100mW then power radiated by a quarter wave monopole under same current input is _____
a) 50mW
b) 200mW
c) 100mW
d) 50W

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

4. If the power radiated by a quarter wave monopo le is 100mW, then power radiated by a half wave dipole with doubled current is ______
a) 800mW
b) 400mW
c) 200mW
d) 100mW

Answer: a
Explanation: 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

5. Angular width between the first nulls or first side lobes is called as _______
a) half power beam width
b) full null beam width
c) beam area
d) directivity

Answer: b
Explanation: Angular width between the first nulls or first side lobes is called full null beam width. Half power beam width is the angular width measured between the 3dB power points of the major lobe. Beam area is the product of HPBW in perpendicular directions. Directivity is the maximum directive gain.

6. If the HPBW is 30° then FNBW is approximately _____
a) 60°
b) 30°
c) 15°
d) 20°

Answer: a
Explanation: FNBW ≈ 2HPBW=2×30°=60°

7. If beam efficiency is 0.87 then the stray factor is ____
a) 1.87
b) 0.13
c) 1.30
d) 0.87

Answer: b
Explanation: Stray factor ∈s is the ratio of minor beam area to total beam area.
Given beam efficiency ∈B=0.87
B+∈s = 1
s = 1-0.87 = 0.13

8. If the antennas revolution time is 10 sec and 3 dB beam width duration is 150ms then the antenna beam width is _______
a) 5.4°
b) 54°
c) 15°
d) 1.5°

Answer: a
Explanation: Antenna beam width = \(\frac{Beam \,duration}{rotating \,period}×360° = \frac{150m}{10} ×360°=5.4°\)

9. Relation between beam efficiency and stray factor is given by ____
a) ∈B+∈s=1
b) ∈B-∈s=1
c) \(\frac{\in_B}{\in_s}=1\)
d) ∈B+∈s=2

Answer: a
Explanation: Stray factor ∈s is the ratio of minor beam area to total beam area. Beam efficiency ∈B is the ratio of major beam area to total beam area. Total beam area is sum of major lobe area ΩM and minor lobe area Ωm.
\(\in_B=\frac{\Omega_M}{\Omega_A}, \in_s = \frac{\Omega_m}{\Omega_A}\)
B+∈s=1

10. The Maxwell equation ∇×E=\(\frac{-\partial B}{\partial t}\) is derived from which law?
a) Amperes law
b) Faradays Law
c) Lens law
d) Gauss law

Answer: b
Explanation: Faradays law states that emf generated around a loop of wire in magnetic field is proportional to the rate of change of time-varying magnetic field through the loop.
Amperes law gives ∇×H=J
Lens law gives only the reason for the negative sign in the Faradays law of induction.
Gauss’s law states that the net flux of an electricfield in a closed surface is directly proportional to the enclosed electric charge.