Transmission Lines – GATE ECE Electromagnetics Practice Questions & PYQs – Page 2 | BattleExam | BattleExam

Q24GATE 2014NAT

The input impedance of a $\frac{\lambda }{8}$ section of a lossless transmission line of characteristic impedance 50 $\Omega$ is found to be real when the other end is terminated by a load $Z_{L}=(R+jX)\Omega$ . If X is 30 $\Omega$ , the value of R(in $\Omega$ ) is_____.

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} Z_{\text {\in }}&=Z_{0}\left(\frac{Z_{L}+j Z_{o} \tan \beta l}{Z_{o}+j Z_{L} \tan \beta l}\right)\\ \text{at }\;l&=\frac{\lambda}{8} \\ \Rightarrow \quad \beta l&=\frac{2 \pi}{\lambda} \times \frac{\lambda}{8}=\frac{\pi}{4} \\ \therefore \quad\tan \frac{\pi}{4}&=1 \\ Z_{\text {\in }}&=Z_{o}\left[\frac{Z_{L}+j 50}{Z_{o}+j Z_{L}}\right] \end{aligned}

\begin{aligned} \text{Given},\quad Z_{L}&=R+j X=R+j 30 \text{ and }Z_{0}=50 \Omega \\ 30 R &=80(50+R) \\ Z_{\text {\in }} &=50\left[\frac{R+j 30+j 50}{50+j(R+j 30)}\right] \\ &=50\left[\frac{R+j 80}{(50-30)+j R}\right]=50\left[\frac{R+j 80}{20+j R}\right] \\ &=50\left[\frac{(R+j 80)(20-j R)}{20^{2}+R^{2}}\right]\$ \end{aligned}

Since, only real part of $Z_{\text {\in }}$ exists, so imaginary part of $Z_{\text {\in }}=0$

\begin{aligned} \therefore \quad Z_{\text {\in }} &=50\left[\frac{20 R+1600 j-j R^{2}+50 R}{20^{2}+R^{2}}\right]\$ \\ \text { Zimaginary } &=0=\frac{1600-R^{2}}{20^{2}+R^{2}}\\ \therefore \quad R^{2} &=1600 \\ R &=40 \Omega \end{aligned}

Q31GATE 2010MCQ

In the circuit shown, all the transmission line sections are lossless. The Voltage Standing Wave Ration(VSWR) on the 60 $\Omega$ line is

🚀 Solve in Practice Mode

📖 Explanation

$Z_{\text {\in }}=Z_{0}\left(\frac{Z_{L}+j Z_{0} \tan \beta l}{Z_{0}+j Z_{L} \tan \beta l}\right)$ Input impedance,

\begin{aligned} Z_{1} &=30\left[\frac{0+j 30 \tan \left(\frac{2 \pi}{\lambda}\right)\left(\frac{\lambda}{8}\right)}{30+0}\right] \\ \Rightarrow Z_{1} &=j 30 \\ Z_{2} &=30 \sqrt{2}\left[\frac{30+j 30 \sqrt{2} \tan \left(\frac{2 \pi}{\lambda}\right)\left(\frac{\lambda}{4}\right)}{30 \sqrt{2}+j 30 \tan \left(\frac{2 \pi}{\lambda}\right)\left(\frac{\lambda}{4}\right)}\right] \\ &=30 \sqrt{2}\left[\frac{\frac{30}{\tan \pi / 2}+j 30 \sqrt{2}}{\frac{30 \sqrt{2}}{\tan \pi / 2}+j 30}\right] \\ &=60 \Omega \\ \text { or, } Z_{2}&=\frac{(30 \sqrt{2})^{2}}{30}=60 \Omega & \end{aligned}

Load impedance, $Z_{L}=Z_{1}+Z_{2}=j 30+6$ Magnitude of reflection coefficient,

\begin{aligned} |\Gamma|&=\left|\frac{Z_{L}-Z_{0}}{Z_{L}+Z_{0}}\right|=\left|\frac{60+j 30-60}{60+j 30+60}\right|\\ &=\left|\frac{j 30}{120+j 30}\right|=\mid \frac{j 1}{4+j \mid} \\ &=\frac{1}{\sqrt{16+1}=\frac{1}{\sqrt{17}}} \end{aligned}

VSWR on 60 \Omega line, $\text { VSWR }=\frac{1+|\Gamma|}{1-|\Gamma|}=\frac{1+\frac{1}{\sqrt{17}}}{1-\frac{1}{\sqrt{17}}}=1.64$

Explanation Figure 2