# Miscellaneous Topics

 Question 1
Consider a narrow band signal, propagating in a lossless dielectric medium $\left(\epsilon_{r}=4, \mu_{r}=1\right)$, with phase velocity $v_{p}$ and group velocity $v_{g}$. Which of the following statement is true? ( $c$ is the velocity of light in vacuum.)
 A $v_{p} \gt c, v_{g} \gt c$ B $v_{p} \lt c, v_{g} \gt c$ C $v_{p} \gt c, v_{g} \lt c$ D $v_{p} \lt c, v_{g} \lt c$
GATE EC 2023   Electromagnetics
Question 1 Explanation:
Phase velocity, $V_{p}=\frac{\omega}{\beta}=\frac{\omega}{\omega \sqrt{\mu \epsilon}}=\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}, \frac{1}{\sqrt{\mu_{r} \epsilon_{r}}}=\frac{C}{\sqrt{\mu_{r} \epsilon_{r}}}$
$\therefore \quad V_{p} \lt C$
Group velocity, $V_{g}=\frac{d \omega}{d \beta}=\frac{V_{p}}{1-\frac{\omega}{V_{p}} \frac{d V_{p}}{d \omega}}$
Here, $\quad V_{p} \neq f(\omega)$
$\therefore \quad V_{g}=V_{p} \lt C$
Hence, $\quad V_{p} \lt C$
$\quad \quad V_{g} \lt C$
 Question 2
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 2 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 3
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 3 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 4
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 4 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 5
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 5 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}

There are 5 questions to complete.