# Antenna Parameters Questions and Answers – Basics

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

1. The relation between vector magnetic potential and current density is given by ______
a) ∇.A=J
b) ∇×A=H
c) ∇2 A=-μJ
d) ∇2 A=∇×H

Explanation: Magnetic Flux density B is expressed as B=∇×A
Taking curl on both sides, we get ∇×B=∇×(∇×A ) = ( ∇.A)∇-(∇.∇)A
From Maxwell’s equation, ∇.B=0 =>∇.A=0
⇨ ∇×B= -∇2 A and B= μH , ∇×H=J
⇨ ∇×μH= -∇2A
⇨ ∇2 A=-μJ.

2. The induction and radiation fields are equal at a distance of _______
a) λ/4
b) λ/6
c) λ/8
d) λ/2

Explanation: For an Hertzian dipole, equating the magnitudes of maximum induction and radiation fields we get,
$$\mid \overline{E_θ}\mid_{max⁡Radiation} =\mid \overline{E_θ}\mid_{max⁡Induction}$$
$$\frac{I_m dl}{4\pi\epsilon} (\frac{ω}{v^2r})=\frac{I_mdl}{4\pi\epsilon} (\frac{1}{r^2v})$$
$$r=\frac{v}{ω}=\frac{\lambda f}{2\pi f}=\frac{\lambda}{6}.$$

3. The ratio of radiation intensity in a given direction from antenna to the radiation intensity over all directions is called as ________
a) Directivity
c) Gain of antenna
d) Array Factor

Explanation: Directivity of antenna is defined as the ratio of radiation intensity in a given direction from antenna to the radiation intensity over all directions. $$D = \frac{U_{max}}{U_0}.$$
Radiation Intensity is power radiated from an antenna for unit solid angle. $$U_0= w_r.r^2\frac{watts}{steradians}.$$
Gain of antenna is ratio of radiation intensity in given direction to the radiation intensity of isotropic radiation. Array factor is a function of geometry of array and the excitation phase.

4. What is the overall efficiency of a lossless antenna with reflection coefficient 0.15?
a) 0.997
b) 0.779
c) 0.669
d) 0.977

Explanation: For a lossless antenna, the radiation efficiency ecd=1.
Overall efficiency of antenna is given by eo =ecd (1-$$\mid\gamma^2\mid)$$=1×(1-(0.152))=0.977.

5. The equivalent area when multiplied by the instant power density which leads to free radiation of power at antenna is called as _______
a) Loss area
b) Scattering area
c) Captured area
d) Effective area

Explanation: Scattering area is the equivalent area when multiplied by the instant power density which leads to free radiation of power. Loss area leads to power dissipation and captured area leads to total power collection by the antenna. The relation among them is given by,
Captured area= effective area + loss area + scattering area.

6. Equivalent circuit representation of an antenna is ______
a) Series R, L, C
b) Parallel R, L, C
c) Series R, L parallel to C
d) Parallel R, C series to L

Explanation: Antenna is represented by a series R, L, C equivalent circuit. Antenna is used for impedance matching and acts like a transducer.

7. Radiation resistance of a Hertzian dipole of length λ/8 is ________
a) 12.33Ω
b) 8.54Ω
c) 10.56Ω
d) 13.22Ω

Explanation: Radiation resistance of a Hertzian dipole of length l is
R=80π2$$(\frac{l}{\lambda})^2=80\pi^2 (\frac{\lambda/8}{\lambda})^2=12.33\Omega$$

8. Relation between directivity and effective area of transmitting and receiving antenna is ________
a) Dt At=Dr Ar
b) Dt Ar=Dr At
c) At Dt=∈Dr Ar
d) Dt At=∈Dr Ar

Explanation: The power collected by the receiving antenna is
$$P_r = \frac{P_tD_tA_r}{4\pi R^2} => D_tA_r = \frac{P_r}{P_t}4\pi R^2$$ = Dr At
∴ Dt Ar=Dr At

9. The axis of back lobe makes an angle of 180° with respect to the beam of an antenna.
a) True
b) False

Explanation: The axis of back lobe is opposite to the main lobe. So it makes 180° with beam of antenna. It is also a side lobe which is at 180 to main lobe.

10.Radiation resistance of a half-wave dipole is ______
a) 36.56Ω
b) 18.28Ω
c) 73.12Ω
d) 40.24Ω

Explanation: Since radiation resistance of quarter-wave monopole (l=λ/4) is 36.56Ω, then for a half-wave dipole (l=λ/2) it is given by 36.56×2 = 73.12Ω. Hertzian dipole is an ideal dipole of infinitesimal dipole.

11. The radiation efficiency for antenna having radiation resistance 36.15Ω and loss resistance 0.85Ω is given by ________
a) 0.977
b) 0.799
c) 0.997
d) 0.779

Explanation: The radiation efficiency $$e_{cd}=\frac{R_r}{R_l+R_r}=\frac{36.15}{36.15+0.85}=0.977.$$