Antennas

Question 1
For an infinitesimally small dipole in free space, the electric field E_\theta in the far field is proportional to (e^{-jkr}/r) \sin \theta, where k = 2 \pi /\lambda. A vertical infinitesimally small electric dipole (\delta l \lt \lt \lambda) is placed at a distance h(h \gt 0) above an infinite ideal conducting plane, as shown in the figure. The minimum value of h, for which one of the maxima in the far field radiation pattern occurs at \theta =60^{\circ}, is
A
\lambda
B
0.5\lambda
C
0.25\lambda
D
0.75\lambda
GATE EC 2020   Electromagnetics
Question 1 Explanation: 


\begin{aligned}\left | \text{Total E }\right |&=\left | (E_{\text{single element}}) \right | \left | (A.F.) \right | \\ \left | (A.F.) \right |&=\frac{\sin\left ( N\frac{\psi }{2} \right )}{\sin (\frac{\psi }{2})}=\frac{\sin \left ( 2\frac{\psi }{2} \right )}{\sin \left ( \frac{\psi }{2} \right )} \\ & =\frac{2\sin \left ( \frac{\psi }{2} \right )\cos (\frac{\psi }{2})}{\sin \frac{\psi }{2}} \\ & =2\cos (\frac{\psi }{2})\\ \left | A.F_{N} \right | &=\frac{A.F}{A.F_{max}}=\frac{2\cos (\frac{\psi }{2})}{2} \\ &=\left | \cos \left ( \frac{\psi }{2} \right ) \right | \\ \text{where, }\psi &=\beta d\cos \theta =\frac{2\pi }{\lambda }(2h)\cos \theta \\ \left.\begin{matrix} \left | A.F_{N} \right |_{\theta =60^{\circ}} \end{matrix}\right|& =\left | \cos (\frac{2\pi }{\lambda }h\cos 60^{\circ}) \right |\\ &=\left | \cos \left ( \frac{\pi h}{\lambda } \right ) \right |\\ \cos \theta \text{ is maximum ,where } \theta &=n\pi \;\; n=0,1,2..\\ \frac{\pi h}{\lambda }&=n\pi \Rightarrow h=n\lambda \\ \Rightarrow \; \text{For } n=1, \;\; h_{min}&=\lambda \end{aligned}
Question 2
Radiation resistance of a small dipole current element of length l at a frequency of 3 GHz is 3 ohms. If the length is changed by 1%, then the percentage change in the radiation resistance, rounded off to two decimal places, is _____ %.
A
0.86
B
2.01
C
2.54
D
2.68
GATE EC 2019   Electromagnetics
Question 2 Explanation: 
Radiation resistance of a small dipole current element of length 'I' is
\begin{array}{l} R_{\mathrm{rad}}=80 \pi^{2}\left(\frac{l}{\lambda}\right)^{2} \Rightarrow R \propto l^{2} \\ \frac{R_{2}}{R_{1}}=\left(\frac{l_{2}}{l_{1}}\right)^{2} \end{array}
If length is changed by 1% then percentage change in the radiation resistance.
\frac{R_{2}}{R_{1}}=\left(\frac{1.01 l}{l}\right)^{2}=1.0201
\text { Percentage change in radiation resistance }
\begin{array}{l} =\frac{R_{2}-R_{1}}{R_{1}} \times 100 \\ =0.0201 \times 100=2.01 \% \end{array}
Question 3
A half wavelength dipole is kept in the x-y plane and oriented along 45^{\circ} from the x-axis. Determine the direction of null in the radiation pattern for 0\leq \phi \leq \pi. Here the angle \theta (0\leq \theta \leq \pi) is measured from the z-axis, and the angle \phi(0\leq \phi \leq 2\pi) is measured from the x-axis in the x-y plane.
A
\theta =90^{\circ},\; \; \phi =45^{\circ}
B
\theta =45^{\circ},\; \; \phi =90^{\circ}
C
\theta =90^{\circ},\; \; \phi =135^{\circ}
D
\theta =45^{\circ},\; \; \phi =135^{\circ}
GATE EC 2017-SET-1   Electromagnetics
Question 3 Explanation: 
x-y plane means Z=0 with Z=r \cos \theta (spherical coordinates) \theta=90^{\circ}
With null points along the axis of the dipole it is \phi=45^{\circ} and 225^{\circ}

Question 4
Consider a wireless communication link between a transmitter and a receiver located in free space, with finite and strictly positive capacity. If the effective areas of the transmitter and the receiver antennas, and the distance between them are all doubled, and everything else remains unchanged, the maximum capacity of the wireless link
A
increases by a factor of 2
B
decreases by a factor of 2
C
remains unchanged
D
decreases by a factor of \sqrt{2}
GATE EC 2017-SET-1   Electromagnetics
Question 4 Explanation: 
As per Friis free space propagation equation,
W_{r}=\frac{W_{t} A_{e r} A_{e t}}{(\lambda d)^{2}}
when A_{e r} and A_{e t} are doubled and d also doubled, W_{r} is same. Hence capacity is also same.
Question 5
The far-zone power density radiated by a helical antenna is approximated as:

\vec{W}_{rad}=\vec{W}_{average}\approx \hat{a}_{r}C_{o}\frac{1}{r^{2}}cos^{4}\theta

The radiated power density is symmetrical with respect to \phi and exists only in the upper hemisphere: 0\leq \theta \leq \frac{\pi }{2};0\leq \phi \leq 2\pi;C_{o} is a constant. The power radiated by the antenna (in watts) and the maximum directivity of the antenna, respectively, are
A
1.5C_{o},10dB
B
1.256C_{o},10dB
C
1.256C_{o},12dB
D
1.5C_{o},12dB
GATE EC 2016-SET-1   Electromagnetics
Question 5 Explanation: 
\begin{aligned} \text { Power radiated }&=\int P_{\mathrm{rad}} \cdot d s \\ =& \int_{\theta=0}^{\pi / 2} \int_{\theta=0}^{2 \pi} C_{0} \frac{1}{r^{2}} \cos ^{4} \theta \cdot r^{2} \sin \theta d \theta d \phi \\ =& 2 \pi \cdot \frac{C_{0}}{r^{2}} \cdot r^{2} \int_{\theta=0}^{\pi / 2} \cos ^{4} \theta \cdot d(-\cos \theta) \\ =&\left.2 \pi \cdot C_{0}\left(-\frac{\cos ^{5} \theta}{5}\right)\right|_{0} ^{\pi / 2} \\ =& \frac{2 \pi}{5} C_{0}=1.256 C_{0} \\ \text{Directivity}=& \frac{4 \pi \cdot U(\theta, \phi)}{\int(\theta, \phi) \cdot d \Omega} \\ =& \frac{4 \pi \cdot C_{0} \cdot \cos ^{4} \theta}{\int_{\theta=0}^{\pi / 2}\int_{\phi=0}^{2\pi}C_{0} \cdot \cos ^{4} \theta \cdot \sin \theta d \theta d \phi} \\ &=\frac{2 \cos ^{5} \theta}{1 / 5}=10 \cos ^{5} \theta \\ \max _{\operatorname{value}}=& 10 \\ \Rightarrow \max \text { value }(\text { in } \mathrm{dB}) &=10 \log _{10} 10=10 \mathrm{dB} \end{aligned}
Question 6
The electric field of a uniform plane wave travelling along the negative z direction is given by the following equation:

\vec{E}^{1}_{w}=(\hat{a}_{x}+j\hat{a}_{y})E_{0}e^{jkz}

This wave is incident upon a receiving antenna placed at the origin and whose radiated electric field towards the incident wave is given by the following equation:

\vec{E}_{a}=(\hat{a}_{x}+2\hat{a}_{y})E_{I}\frac{1}{r}e^{-jkr}

The polarization of the incident wave, the polarization of the antenna and losses due to the polarization mismatch are, respectively,
A
Linear, Circular (clockwise), -5dB
B
Circular (clockwise), Linear, -5dB
C
Circular (clockwise), Linear, -3dB
D
Circular (anti clockwise), Linear, -3dB
GATE EC 2016-SET-1   Electromagnetics
Question 6 Explanation: 
\vec{E}_{T}=\left(\hat{a}_{x}+j \hat{a}_{y}\right) E_{0} e^{j k z}
\Rightarrow Wave contains two orthogonal components and Y component leads X component leads by 90^{\circ} and also wave is travelling in negative z-direction.
\vec{E}_{R}=\left(\hat{a}_{x}+2 \hat{a}_{y}\right) E_{1} \frac{1}{r} e^{-j k r}
\Rightarrow Wave contains two orthogonal components with unequal amplitudes and both are in-phase.
\Rightarrow Linear polarization.
Polarization Loss Factor, \mathrm{PLF}=\left|\hat{E}_{T} \cdot \hat{E}_{R}\right|^{2}
\begin{aligned} Where, E_{T}&=\frac{\hat{a}_{x}+j \hat{a}_{y}}{\sqrt{2}} \\ E_{R}&=\frac{\hat{a}_{x}+2 \hat{a}_{y}}{\sqrt{5}} \\ P L&= \left|\frac{1+j 2^{2}}{\sqrt{10} }\right|=\frac{5}{10}=\frac{1}{2} \\ \Rightarrow \quad P L F(d B)&=10 \log \frac{1}{2}=-3 \mathrm{dB} \end{aligned}
Question 7
Two lossless X-band horn antennas are separated by a distance of 200\lambda. The amplitude reflection coefficients at the terminals of the transmitting and receiving antennas are 0.15 and 0.18, respectively. The maximum directivities of the transmitting and receiving antennas (over the isotropic antenna) are 18 dB and 22 dB, respectively. Assuming that the input power in the lossless transmission line connected to the antenna is 2 W, and that the antennas are perfectly aligned and polarization matched, the power ( in mW) delivered to the load at the receiver is ________
A
1
B
2
C
3
D
4
GATE EC 2016-SET-1   Electromagnetics
Question 7 Explanation: 


\begin{aligned} G_{t}&=10^{1.8}, G_{r}=10^{2.2} \\ P_{r}&=\frac{\left(1-\left|\Gamma_{t}\right|^{2}\right)\left(1-\left|\Gamma_{r}\right|^{2}\right) G_{t} G_{r}}{\left(\frac{4 \pi d}{\lambda}\right)^{2}} \cdot P_{t}\\ &=\frac{\left(1-|0.15|^{2}\right)\left(1-|0.18|^{2}\right) 10^{1.8} \cdot 10^{2.2}}{\left(\frac{4 \pi 200 \lambda}{\lambda}\right)^{2}} \times 2 \\ &=2.99476 \times 10^{-3} \mathrm{W}=2.995 \mathrm{mW} \approx 3 \mathrm{mW} \end{aligned}
Question 8
The directivity of an antenna array can be increased by adding more antenna elements, as a larger number of elements
A
improves the radiation efficiency
B
increases the effective area of the antenna
C
results in a better impedance matching
D
allows more power to be transmitted by the antenna
GATE EC 2015-SET-3   Electromagnetics
Question 8 Explanation: 
Directivity (0) is increased by adding more antenna elements in an antenna array. Effective area (A_{e}) and directivity (D) are related by
A_{e}=\frac{\lambda^{2}}{4 \pi} D
\Rightarrow D \uparrow A_{e} \uparrow
Question 9
Two half-wave dipole antennas placed as shown in the figure are excited with sinusoidally varying currents of frequency 3 MHz and phase shift of \pi/2 between them(the element at the origin leads in phase). If the maximum radiated E-field at the point P in the x-y plane occurs at an azimuthal angle of 60^{\circ}, the distance d (in meters) between the antennas is _________.
A
30
B
40
C
50
D
60
GATE EC 2015-SET-2   Electromagnetics
Question 9 Explanation: 
Both dipole antennas have isotropic pattern in \phi view or H plane view.


\psi=\beta d \cos \theta+\alpha
where, \alpha=-90^{\circ}, \beta=\frac{2 \pi}{\lambda}=\frac{2 \pi}{\frac{3 \times 10^{8}}{3 \times 10^{6}}}=\frac{2 \pi}{100}
Since maximum radiation (\psi=0) is at \theta=60^{\circ}
\begin{array}{l} \Rightarrow \quad 0=\frac{2 \pi}{100} d \cos 60^{\circ}-90^{\circ} \\ \Rightarrow \quad 90=\frac{360}{100} d \frac{1}{2} \\ \Rightarrow \quad d=50 \mathrm{m} \end{array}
Question 10
Match column A with column B.
A
1\rightarrow P \; 2\rightarrow Q \; 3\rightarrow R
B
1\rightarrow R \; 2\rightarrow P \; 3\rightarrow Q
C
1\rightarrow Q \; 2\rightarrow P \; 3\rightarrow R
D
1\rightarrow R \; 2\rightarrow Q \; 3\rightarrow P
GATE EC 2014-SET-4   Electromagnetics
Question 10 Explanation: 
Point electromagnetic source radiates in all direction Dish antenna radiates any electromagnetic energy in any particular direction with narrow beam-width and high directivity Yagi-uda antenna is high gain antenna used for T.V. reception. Its radiation is along the axis of the antenna.
There are 10 questions to complete.
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