# Antenna Parameters Questions and Answers – Effective Aperture

This set of Antennas Multiple Choice Questions & Answers (MCQs) focuses on “Effective Aperture”.

1. Effective aperture is the ability of antenna to extract energy from the electromagnetic wave.
a) True
b) False

Explanation: Effective aperture is defined as the ratio of power received from load to the average power density produced at that point. So it is the ability of antenna to extract energy from EM wave.

2. Which of the following best describes the condition for Maximum effective aperture?
a) Load impedance must be equal to the antenna impedance
b) Load impedance must be equal complex conjugate to the antenna impedance
c) Receiver power should be minimum
d) Transmitter power should be minimum

Explanation: Foe effective aperture to be maximum, the receiver power should be maximum. Maximum power transfer states that the maximum received power is obtained when the impedance of network matches with the complex conjugate of the load impedance. Hence, Load impedance must be equal to the complex conjugate of the antenna impedance.

3. What is the effective aperture of Hertzian dipole antenna operating at frequency 100 MHz?
a) 1.07m2
b) 0.17m2
c) 1.7m2
d) 1.2m2

Explanation: Effective aperture for a Hertzian dipole is given by $$A_e=1.5\frac{\lambda^2}{4\pi}$$
Gain of Hertzian dipole is 1.5
$$\lambda=\frac{c}{f}=\frac{3×10^8}{100×10^6}=3m$$
$$A_e=1.5 \frac{\lambda^2}{4\pi} =1.5\frac{3^2}{4\pi} =1.07m^2$$

4. If physical aperture of antenna is 0.02m2 and aperture efficiency is 0.5, then what is the value of effective aperture?
a) 0.0004m2
b) 0.001m2
c) 0.01m2
d) 25m2

Explanation: Effective aperture Aem=Aeη=0.02×0.5=0.01 m2

5. Expression for aperture efficiency in terms of physical aperture Ae and effective aperture Aem is ____
a) $$\frac{A_e}{A_{em}}$$
b) $$\frac{A_{em}}{A_e}$$
c) $$\frac{A_e+A_{em}}{A_e-A_{em}}$$
d) $$\frac{A_e-A_{em}}{A_e+A_{em}}$$

Explanation: The aperture efficiency is defined as the ratio of effective aperture to physical aperture of antenna. So aperture efficiency $$\eta=\frac{A_{em}}{A_e}.$$

6. What is the effective aperture of a Half-wave dipole operating at 100MHz?
a) 1.07m2
b) 1.17m2
c) 1.27m2
d) 1.77m2

Explanation: The directivity of half-wave dipole is 1.64
The effective aperture of half-wave dipole is $$A_e=1.64\frac{\lambda^2}{4\pi}$$
$$\lambda=\frac{c}{f}=\frac{3×10^8}{100×10^6}=3m$$
$$A_e=1.64 \frac{\lambda^2}{4\pi} = 1.64 \frac{3^2}{4\pi}$$=1.17m2.

7. What is the relation between effective length and Effective aperture of antenna?
a) $$A_e = \frac{dL^2\eta}{4R_{rad}}$$
b) $$A_e = \frac{dL^2}{4\eta R_{rad}}$$
c) $$A_e = \frac{dL^2 R_{rad}}{4\eta}$$
d) $$A_e = \frac{dL^2 \eta ^2}{4R_{rad}}$$

Explanation: Maximum effective aperture is ratio of maximum power received to the average power density. $$A_e=\frac{P_{Rmax}}{P_{avg}}$$
The received power is maximum when load equal to complex conjugate of network resistance.
⇨ $$I_{total}=\frac{V_{oc}}{2R_{rad}}$$
⇨ $$P_{Rmax}=I_{rms}^2 R_{rad}= (\frac{I_{total}}{\sqrt 2})^2 R_{rad}=\frac{V_{oc}^2}{8R_{rad}}=\frac{\mid E_\theta\mid^2 dL^2}{8R_{rad}} and P_{avg}=\frac{\mid E_\theta\mid^2}{2\eta}$$
$$A_e=\frac{P_{Rmax}}{P_{avg}}=\frac{dL^2\eta}{4R_{rad}}$$

8. The physical aperture of an isotropic radiator is _______
a) $$\frac{4\pi \eta}{\lambda^2}$$
b) $$\frac{4\pi}{\lambda^2 \eta}$$
c) $$\frac{\lambda^2}{4\pi \eta}$$
d) $$\frac{\lambda^2\eta}{4\pi}$$

Explanation: For isotropic radiator, directivity is 1. So the effective aperture is given by $$A_{em}=D \frac{\lambda^2}{4\pi} = \frac{\lambda^2}{4\pi}$$
Then physical aperture =$$\frac{Effective\, aperture}{Aperture\, efficiency}=\frac{\lambda^2}{4\pi \eta}$$

Sanfoundry Global Education & Learning Series – Antennas.