# Miscellaneous Topics

 Question 1
An optical fiber is kept along the $\hat{z}$ direction. The refractive indices for the electric fields along $\hat{x}$ and $\hat{y}$ directions in the fiber are $n_{x}$=1.5000 and $n_{y}$=1.5001, respectively ($n_{x} \neq n_{y}$ due to the imperfection in the fiber cross-section). The free space wavelength of a light wave propagating in the fiber is 1.5$\mu$m. If the light wave is circularly polarized at the input of the fiber, the minimum propagation distance after which it becomes linearly polarized, in centimeter, is ___________.
 A 0.2 B 0.25 C 0.37 D 0.45
GATE EC 2017-SET-1   Electromagnetics
Question 1 Explanation:
Initially the wave is circularly polarized. So, the initial phase difference between field components in $\hat{a}_{x}$ direction and $\hat{a}_{y}$ direction is $\frac{\pi}{2}$
To become linearly polarized, the wave must travel a minimum distance, such that, the phase difference at that point between the field components in $\hat{a}_{x}$ direction and $\hat{a}_{y}$ direction is $\pi$
(i.e., the travel of this minimum distance should provide an additional phase difference of $\pi / 2$ between $\hat{a}_{x}$ and $\hat{a}_{y}$ field components).
So,
\begin{aligned} z_{min}k_z&\sim z_{min}k_y=\frac{\pi}{2} \\ z_{min}\left [ \frac{\omega }{V_{px}}\sim \frac{\omega }{V_{py}} \right ]&= \frac{\pi}{2} \\ 2 \pi z_{min}\left [ \frac{f}{c}\sqrt{\varepsilon _{rx}}\sim \frac{f}{c}\sqrt{\varepsilon _{ry}}\right ]&=\frac{\pi}{2} \\ \frac{4 z_{min}}{\lambda _0}[n_x\sim n_y]&=1 \\ z_{min}&=\frac{\lambda _0}{4(n_x\sim n_y)} \\ &=\frac{1.5}{4(1.5\sim 1.5001)}\mu m \\ &= \frac{1.5}{4(0.0001)}\mu m=\frac{1.5}{4}cm\\ z_{min}&=0.375cm \end{aligned}
 Question 2
A radar operating at 5 GHz uses a common antenna for transmission and reception. The antenna has a gain of 150 and is aligned for maximum directional radiation and reception to a target 1 km away having radar cross-section of 3 $m^{2}$. If it transmits 100 kW, then the received power (in $\mu$W) is__________
 A 0.01 B 0.1 C 0.06 D 0.6
GATE EC 2016-SET-3   Electromagnetics
Question 2 Explanation:
\begin{aligned} f&=5 \mathrm{GHz}, G=150, R=1 \mathrm{km} \\ P_{t}&=100 \mathrm{kW}, \sigma=3 \mathrm{m}^{2} \\ P_{r}&= \frac{(G \lambda)^{2} \cdot \sigma}{(4 \pi)^{3} R^{4}} \times P_{t} \\ &=\frac{\left(150 \times \frac{3 \times 10^{8}}{5 \times 10^{9}}\right)^{2} \cdot 3}{(4 \pi)^{3} \cdot 10^{12}} \times 10^{5} \\ &=\frac{81 \times 3 \times 10^{5}}{(4 \pi)^{3} \times 10^{12}} \\ &= 1.224 \times 10^{-8}=0.012 \mu \mathrm{W} \end{aligned}
 Question 3
Light from free space is incident at an angle $\theta _{i}$ to the normal of the facet of a step-index large core optical fibre. The core and cladding refractive indices are $n _{1}=1.5 \; and \; n_{2}=1.4$, respectively.

The maximum value of $\theta _{i}$ (in degrees) for which the incident light will be guided in the core of the fibre is ________
 A 8.25 B 16.36 C 24.24 D 32.58
GATE EC 2016-SET-2   Electromagnetics
Question 3 Explanation:
\begin{aligned} & \sin \alpha_{\max } &=\sqrt{n_{1}^{2}-n_{2}^{2}}=\sqrt{1.5^{2}-1.4^{2}} \\ \Rightarrow & \alpha_{\max } &=\sin ^{-1}(0.5385)=32.58^{\circ} \end{aligned}
 Question 4
A two-port network has scattering parameters given by $[S]=\begin{bmatrix} S_{11} &S_{12} \\ S_{21} & S_{22} \end{bmatrix}$. If the port-2 of the two port is short circuited, the $S_{11}$ parameter for the resultant one port network is
 A $\frac{S_{11}-S_{11}S_{22}+S_{12}S_{21}}{1+S_{22}}$ B $\frac{S_{11}+S_{11}S_{22}-S_{12}S_{21}}{1+S_{22}}$ C $\frac{S_{11}+S_{11}S_{22}+S_{12}S_{21}}{1-S_{22}}$ D $\frac{S_{11}-S_{11}S_{22}+S_{12}S_{21}}{1-S_{22}}$
GATE EC 2014-SET-1   Electromagnetics
Question 4 Explanation:
$\begin{array}{l} b_{1}=s_{11} \cdot a_{1}+s_{12} \cdot a_{2} \qquad\ldots(i)\\ b_{2}=s_{21} \cdot a_{1}+s_{22} a_{2}\qquad\ldots(ii) \end{array}$
When output is short, $a_{2}=-b_{2}$
\begin{aligned} -a_{2} &=s_{21} \cdot a_{1}+s_{22} a_{2} \\ \Rightarrow \quad a_{2} &=\frac{-s_{21} \cdot a_{1}}{1+s_{22}} \end{aligned}
Substituting in first equation,
\begin{aligned} b_{1} &=s_{11} \cdot a_{1}+s_{12}\left(\frac{-s_{21} \cdot a_{1}}{1+s_{22}}\right) \\ &=\frac{s_{11} \cdot a_{1}+s_{11} \cdot s_{22} a_{1}-s_{12} \cdot s_{21} a_{1}}{1+s_{22}} \\ \frac{b_{1}}{a_{1}} &=s_{11}=\frac{s_{11}+s_{11} \cdot s_{22}-s_{12} \cdot s_{21}}{1+s_{22}} \end{aligned}
 Question 5
If the scattering matrix [S] of a two port network is
$|S|=\begin{bmatrix} 0.2\angle 0^{\circ}&0.9\angle 90^{\circ} \\ 0.9\angle 90^{\circ}& 0.1\angle 90^{\circ} \end{bmatrix}$
then the network is
 A lossless and reciprocal B lossless but not reciprocal C not lossless but reciprocal D neither lossless nor reciprocal
GATE EC 2010   Electromagnetics
Question 5 Explanation:
For reciprocal networks, $S_{12}=S_{21}$
For symmetrical networks, $S_{11}=S_{22}$
For antimetrical networks, $S_{11}=-S_{22}$
For lossless reciprocal networks,
$\left|S_{11}\right|=\left|S_{22}\right|$
and $\left|S_{11}\right|^{2}+\left|S_{12}\right|^{2}=1$
 Question 6
In the design of a single mode step index optical fibre close to upper cut-off, the single-mode operation is not preserved if
 A radius as well as operating wavelength are halved B radius as well as operating wavelength are doubled C radius is halved and operating wavelength is doubled D radius is doubled and operating wavelength is halved
GATE EC 2008   Electromagnetics
 Question 7
A load of 50$\Omega$ is connected in shunt in a 2-wire transmission line of $Z_{0} = 50\Omega$ as shown in the figure. The 2-port scattering parameter matrix (s-matrix) of the shunt element is
 A $\begin{bmatrix} -\frac{1}{2} &\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix}$ B $\begin{bmatrix} 0 & 1\\ 1&0 \end{bmatrix}$ C $\begin{bmatrix} -\frac{1}{3} &\frac{2}{3} \\ \frac{2}{3} & -\frac{1}{3} \end{bmatrix}$ D $\begin{bmatrix} \frac{1}{4} &-\frac{3}{4} \\ -\frac{3}{4} & \frac{1}{4} \end{bmatrix}$
GATE EC 2007   Electromagnetics
Question 7 Explanation:
The line is terminated with $50 \Omega$ at the center and so, matched on both the sides.
 Question 8
In a microwave test bench, why is the microwave signal amplitude modulated at 1 kHz
 A To increase the sensitivity of measurement B To transmit the signal to a far-off place C To study amplitude modulations D Because crystal detector fails at microwave frequencies
GATE EC 2004   Electromagnetics
There are 8 questions to complete.