# Electromagnetics

 Question 1
A transparent dielectric coating is applied to glass $\left(\epsilon_{r}=4, \mu_{r}=1\right)$ to eliminate the reflection of red light $\left(\lambda_{0}=0.75 \mu \mathrm{m}\right)$. The minimum thickness of the dielectric coating, in $\mu \mathrm{m}$, that can be used is ____ (rounded off to two decimal places).
 A 0.02 B 0.18 C 0.13 D 0.52
GATE EC 2023      Basics of Electromagnetics
Question 1 Explanation:
For no reflection, impedance must be matched.
Hence, $\eta_{2}$ acts like a quarter wave impedance transformer.

So,
$(i) \quad \eta_{2}=\sqrt{\eta_{1} \cdot \eta_{3}} \Rightarrow \epsilon_{r_{2}}=\sqrt{\epsilon_{r_{1}} \cdot \epsilon_{r_{3}}} \Rightarrow \epsilon_{r_{2}}=2$

(ii) For impedance matching,
\begin{aligned} & d=(2 n+1) \frac{\lambda}{4} ; n=0,1,2 \ldots \\ & \lambda=\frac{\lambda_{0}}{\sqrt{\epsilon_{r}}}=\frac{\lambda_{0}}{\sqrt{\epsilon_{r_{2}}}} \\ Here \;\;& \lambda=\frac{0.75 \times 10^{-6}}{\sqrt{2}}=0.53 \times 10^{-6} \end{aligned}
Hence, for minimum distance, $n=0$
So, $d=\frac{\lambda}{4}=\frac{0.53 \times 10^{-6}}{4}=0.133 \mu \mathrm{m}$
 Question 2
The following circuit(s) representing a lumped element equivalent of an infinitesimal section of a transmission line is/are

 A A B B C C D D
GATE EC 2023      Transmission Lines

 Question 3
The electric field of a plane electromagnetic wave is

$E=a_{x} C_{1 x} \cos (\omega t-\beta z)+a_{y} C_{1 y} \cos (\omega t-\beta z+\theta) \mathrm{V} / \mathrm{m}$

Which of the following combination(s) will give rise to a left handed elliptically polarized (LHEP) wave?
 A $C_{1 x}=1, C_{1 y}=1, \theta=\pi / 4$ B $C_{1 x}=2, C_{1 y}=1, \theta=\pi / 2$ C $C_{1 x}=1, C_{1 y}=2, \theta=3 \pi / 2$ D $C_{1 x}=2, C_{1 y}=1, \theta=3 \pi / 4$
GATE EC 2023      Uniform Plane Waves
Question 3 Explanation:
Given, $\quad \vec{E}=\hat{a}_{x} C_{1 x} \cos (\omega t-\beta z)+\hat{a}_{y} C_{1 y} \cos (\omega t-\beta z+\theta)$
at $z=0$
$\vec{E}=C_{1 x} \cos \omega t \hat{a}_{x}+C_{1 y} \cos (\omega t+\theta) \hat{a}_{y}$
Going by options,

Option (A) $\quad \vec{E}=\cos \omega t \hat{a}_{x}+\cos (\omega t+\pi / 4) \hat{a}_{y}$
at $t=0, \omega t=0, \vec{E}=\hat{a}_{x}+\frac{1}{\sqrt{2}} \hat{a}_{y}$
at $t=T / 4, \omega t=\pi / 2, \vec{E}=0-\frac{1}{\sqrt{2}} \hat{a}_{y}$
$\Rightarrow$ Hence, it is LHEP.

Option (B) $\quad \vec{E}=2 \cos \omega t \hat{a}_{x}+\cos (\omega t+\pi / 2) \hat{a}_{y}$
at $t=0, \omega t=0, \vec{E}=2 \hat{a}_{x}$
at $t=T / 4, \omega t=\pi / 2, \vec{E}=-1 \hat{a}_{y}$
$\Rightarrow$ Hence, it is LHEP.

Option (C) $\quad \vec{E}=\cos \omega t \hat{a}_{x}+2 \cos (\omega t+3 \pi / 2) \hat{a}_{y}$
at $t=0, \quad \quad \omega t=0, \vec{E}=\hat{a}_{x}$
at $t=T / 4, \omega t=\pi / 2, \vec{E}=2 \hat{a}_{y}$
$\Rightarrow$ Hence, it is RHEP.

Option (D) $\quad \vec{E}=2 \cos \omega t \hat{a}_{x}+\cos (\omega t+3 \pi / 4) \hat{a}_{y}$
at $t=0, \omega t=0, \vec{E}=2 \hat{a}_{x}-\frac{1}{\sqrt{2}} \hat{a}_{y}$
at $t=T / 4, \omega t=\pi / 2, \vec{E}=0-\frac{1}{\sqrt{2}} \hat{a}_{y}=\frac{-1}{\sqrt{2}} \hat{a}_{y}$
$\Rightarrow$ Hence, it is LHEP.

$\therefore$ Option (A), (B) and (D) are correct.
 Question 4
The standing wave ratio on a $50 \Omega$ lossless transmission line terminated in an unknown load impedance is found to be 2.0. The distance between successive voltage minima is $30 \mathrm{~cm}$ and the first minimum is located at $10 \mathrm{~cm}$ from the load. $Z_{L}$ can be replaced by an equivalent length $l_{m}$ and terminating resistance $R_{m}$ of the same line. The value of $R_{m}$ and $l_{m}$, respectively, are

 A $R_{m}=100 \Omega, l_{m}=20 \mathrm{~cm}$ B $R_{m}=25 \Omega, l_{m}=20 \mathrm{~cm}$ C $R_{m}=100 \Omega, l_{m}=5 \mathrm{~cm}$ D $R_{m}=25 \Omega, l_{m}=5 \mathrm{~cm}$
GATE EC 2023      Transmission Lines
Question 4 Explanation:
Given $S=2, Z_{\min }=10 \mathrm{~cm}, Z_{0}=50 \Omega$
As we know that, $\quad|\Gamma|=\frac{S-1}{S+1}=\frac{2-1}{2+1}=\frac{1}{3}$
Now, distance between successive voltage minima $=30 \mathrm{~cm}$
$\Rightarrow \quad \frac{\lambda}{2}=30 \mathrm{~cm}$
$\Rightarrow \quad \lambda=60 \mathrm{~cm}$

Also, for minima,
$2 \beta Z_{\min }=(2 n+1) \pi+\theta_{\Gamma}$

At $n=0,1 \mathrm{st}$ minima, $Z_{\text {min }}=10 \mathrm{~cm}$
$\frac{4 \pi}{\lambda} Z_{\min }=\pi+\theta_{\Gamma}$
$\Rightarrow \quad \frac{4 \pi}{60} * 10=\pi+\theta_{\Gamma}$
$\Rightarrow \quad \frac{2 \pi}{3}-\pi=\theta_{\Gamma}$
$\Rightarrow \quad \theta_{\Gamma}=\frac{-\pi}{3} \quad \therefore \Gamma=\frac{1}{3} \angle-60^{\circ}$

Now, $\Gamma=\frac{Z_{L}-Z_{0}}{Z_{L}+Z_{0}}$

$\begin{array}{ll}\Rightarrow \quad & Z_{L}=Z_{0}\left[\frac{1+\Gamma}{1-\Gamma}\right] \\ \Rightarrow & Z_{L}=50\left[\frac{1+0.33 e^{-j \frac{\pi}{3}}}{1-0.33 e^{-j \frac{\pi}{3}}}\right]\end{array}$
$\Rightarrow \quad Z_{L}=67.97 \angle-32.67^{\circ}$
Now $Z_{\text {in }}=Z_{0}\left[\frac{Z_{L}+j Z_{0} \tan \beta l}{Z_{0}+j Z_{L} \tan \beta l}\right]$
$\Rightarrow \quad Z_{\text {in }}=50\left[\frac{R_{m}+j 50 \tan \beta l_{m}}{50+j R_{m} \tan \beta l_{m}}\right]$
Here, $\quad Z_{\text {in }}=Z_{L}=67.97 \angle-32.67^{\circ}$
Going through options,
$\left.\begin{array}{l}R_{m}=100 \Omega \text { and } L_{m}=5 \mathrm{~cm} \\ R_{m}=25 \Omega \text { and } L_{m}=20 \mathrm{~cm}\end{array}\right\}$ satisfy this identity, hence option (B) and (C) are correct.
 Question 5
A cascade of common-source amplifiers in a unity gain feedback configuration oscillates when
 A the closed loop gain is less than 1 and the phase shift is less than $180^{\circ}$. B the closed loop gain is greater than 1 and the phase shift is less than $180^{\circ}$. C the closed loop gain is less than 1 and the phase shift is greater than $180^{\circ}$. D the closed loop gain is greater than 1 and the phase shift is greater than $180^{\circ}$.
GATE EC 2023      Transmission Lines
Question 5 Explanation:
For oscillation,
Loop gain magnitude $\geq 1$.
Phase of loop gain $=360^{\circ}$
So, correct answer is option (D).
In options, closes loop gain is mentioned technically it should be loop gain.

There are 5 questions to complete.