Question 1 |
An antenna with a directive gain of 6\text{ dB}
is radiating a total power of \text{16 kW}. The amplitude of the electric field in free space at a distance of \text{8 km}
from the antenna in the direction of \text{6 dB}
gain (rounded off to three decimal places) is ______ \text{V/m}
.
0.244 | |
0.417 | |
1.254 | |
2.365 |
Question 1 Explanation:
\begin{aligned} G_{d}(\mathrm{~dB}) &=10 \log _{10}\left(G_{d}\right) \\ \Rightarrow\qquad 6 &=10 \log _{10} G_{d} \\ \Rightarrow\qquad G_{d} &=10^{0.6} \\ G_{d} &=3.981 \\ E_{m} &=\frac{\sqrt{60 G_{d} P_{\text {rad }}}}{r} \\ E_{m} &=\frac{\sqrt{60\left(16 \times 10^{3}\right)(3.981)}}{8 \times 10^{3}}=0.244(\mathrm{~V} / \mathrm{m}) \end{aligned}
Question 2 |
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
\lambda | |
0.5\lambda | |
0.25\lambda | |
0.75\lambda |
Question 2 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 3 |
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 _____ %.
0.86 | |
2.01 | |
2.54 | |
2.68 |
Question 3 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}
\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 4 |
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.
\theta =90^{\circ},\; \; \phi =45^{\circ} | |
\theta =45^{\circ},\; \; \phi =90^{\circ} | |
\theta =90^{\circ},\; \; \phi =135^{\circ} | |
\theta =45^{\circ},\; \; \phi =135^{\circ} |
Question 4 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}
With null points along the axis of the dipole it is \phi=45^{\circ} and 225^{\circ}
Question 5 |
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
increases by a factor of 2 | |
decreases by a factor of 2 | |
remains unchanged | |
decreases by a factor of \sqrt{2} |
Question 5 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.
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 6 |
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
\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
1.5C_{o},10dB | |
1.256C_{o},10dB | |
1.256C_{o},12dB | |
1.5C_{o},12dB |
Question 6 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 7 |
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,
\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,
Linear, Circular (clockwise), -5dB | |
Circular (clockwise), Linear, -5dB | |
Circular (clockwise), Linear, -3dB | |
Circular (anti clockwise), Linear, -3dB |
Question 7 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}
\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 8 |
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 ________
1 | |
2 | |
3 | |
4 |
Question 8 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 9 |
The directivity of an antenna array can be increased by adding more antenna elements, as a larger number of elements
improves the radiation efficiency | |
increases the effective area of the antenna | |
results in a better impedance matching | |
allows more power to be transmitted by the antenna |
Question 9 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
A_{e}=\frac{\lambda^{2}}{4 \pi} D
\Rightarrow D \uparrow A_{e} \uparrow
Question 10 |
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 _________.
30 | |
40 | |
50 | |
60 |
Question 10 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}
\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}
There are 10 questions to complete.