A $50 \Omega$ lossless transmission line is terminated with a load $\mathrm{Z}_{\mathrm{L}}$ of $(50-\mathrm{j} 75) \Omega$ . If the average incident power on the line is 10 mW , then the average power delivered to the load (in mW , rounded off to one decimal place) is __________ .

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📖 Explanation

Step 1: Characteristic Impedance Given: $Z_0 = 50\ \Omega$ (lossless line) $Z_L = 50 - j75\ \Omega$ Incident power = $P_{\text{inc}} = 10\ \text{mW}$ Step 2: Reflection Coefficient Reflection coefficient $\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0}$ First compute: $Z_L + Z_0 = (50 - j75) + 50 = 100 - j75$ $Z_L - Z_0 = (50 - j75) - 50 = -j75$ So: $\Gamma = \frac{-j75}{100 - j75}$ Multiply numerator and denominator by the complex conjugate of the denominator: $\Gamma = \frac{-j75 \cdot (100 + j75)}{(100 - j75)(100 + j75)}$ Numerator: $-j75(100 + j75) = -j7500 - j^2 75 \cdot 75 = -j7500 + 5625$ Denominator: $100^2 + 75^2 = 10000 + 5625 = 15625$ So: $\Gamma = \frac{5625 - j7500}{15625} = 0.36 - j0.48$ Step 3: Reflected Power $|\Gamma|^2 = (0.36)^2 + (0.48)^2 = 0.1296 + 0.2304 = 0.36$ So, reflected power = $P_{\text{ref}} = |\Gamma|^2 \cdot P_{\text{inc}} = 0.36 \cdot 10 = 3.6\ \text{mW}$ Step 4: Delivered Power $P_{\text{delivered}} = P_{\text{inc}} - P_{\text{ref}} = 10 - 3.6 = \boxed{6.4\ \text{mW}}$

Q2GATE 2024MCQ

Consider a lossless transmission line terminated with a short circuit as shown in the figure below. As one moves towards the generator from the load, the normalized impedances $z_{\text {\in } A}, z_{\text {\in } B}, z_{\text {\in } C}$ and $z_{\text {\in } D}$ (indicated in the figure) are _____

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📖 Explanation

$Z_{s / c}=j Z_{0} \tan \beta l$ . $\Rightarrow$ Normalized impedance,

\begin{aligned} & \bar{Z}_{S / C}=\frac{Z_{S / C}}{Z_{0}}=j \tan \beta l \\ & \bar{Z}_{\text {\in } A}=j \tan \left(\frac{2 \pi}{\lambda} \cdot \frac{\lambda}{8}\right)=j \tan (\pi / 4)=j 1 \Omega \\ & \bar{Z}_{\text {\in } B}=j \tan \left(\frac{2 \pi}{\lambda} \cdot \frac{\lambda}{4}\right)=\infty \\ & \bar{Z}_{\text {\in } C}=j \tan \left(\frac{2 \pi}{\lambda} \cdot \frac{3 \lambda}{8}\right)=-j 1 \Omega \\ & \bar{Z}_{\text {\in } D}=j \tan \left(\frac{2 \pi}{\lambda} \cdot \frac{\lambda}{2}\right)=0 \end{aligned}

Q6GATE 2023MSQ

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

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📖 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.

Q7GATE 2021MCQ

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

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📖 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, $Z_{L}=\frac{Z_{1}^{2}}{Z_{0}}=\frac{(75)^{2}}{50}=112.5$

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