# Transmission Lines

 Question 1
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   Electromagnetics
 Question 2
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   Electromagnetics
Question 2 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 3
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   Electromagnetics
Question 3 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.
 Question 4
The impedance matching network shown n the figure is to match a lossless line having characteristic impedance $Z_{0}= 50 \:\Omega$ with a load impedance $Z_{L}$. A quarter-wave line having a characteristic impedance $Z_1=75\:\Omega$ is connected to $Z_{L}$. Two stubs having characteristic impedance of $75\:\Omega$ each are connected to this quarter-wave line. One is a short-circuited ($\text{S.C}$) stub of length $0.25\lambda$ connected across PQ and the other one is an open-circuited $(\text{O.C})$ stub of length $0.5\lambda$ connected across RS.

The impedance matching in achieved when the real part of $Z_{L}$ is
 A $112.5\:\Omega$ B $75.0\:\Omega$ C $50.0\:\Omega$ D $33.3\:\Omega$
GATE EC 2021   Electromagnetics
Question 4 Explanation:
\begin{aligned} Z_{\text {in } \lambda / 4}&=\frac{Z_{O}^{2}}{Z_{L}}=\frac{(75)^{2}}{0}=\infty \quad\left[\text { for } \frac{\lambda}{4} T_{X} \text { line }\right]\\ \text{Given }Z_{L}\text{ of }\frac{\lambda}{4}\text{ line is 0}(\mathrm{SC})\\ Z_{\text {in }_{\lambda / 2}}&=Z_{L}=\infty\left[\text { for } \frac{\lambda}{2} T_{X} \text { line }\right]\\ \text{Given }Z_{L}\text{ of }\frac{\lambda}{2}\text{ line is }\infty(O . C) \end{aligned}
The input impedance of $\frac{\lambda}{4}$ transmission line, as well as $\frac{\lambda}{2}$ transmission line is 00 , and they are in parallel with main transmission line, so they are not effective for main.
Transmission line.
Final configuration of given line is

For impedance matching, $Z_{L}=\frac{Z_{1}^{2}}{Z_{0}}=\frac{(75)^{2}}{50}=112.5$
 Question 5
A transmission line of length $3\lambda /4$ and having a characteristic impedance of 50$\Omega$ is terminated with a load of 400$\Omega$. The impedance (rounded off to two decimal places) seen at the input end of the transmission line is __________ $\Omega$.
 A 5.55 B 2.25 C 4.45 D 6.25
GATE EC 2020   Electromagnetics
Question 5 Explanation:

$Z_{in}\,\, for \, (l=\lambda /4)=\frac{Z_{0}^{2}}{Z_{L}}=\frac{50^{2}}{400}=\frac{25}{4}=6.25\Omega$

There are 5 questions to complete.