Question 1 |
The following circuit(s) representing a lumped element equivalent of an infinitesimal section of a transmission line is/are
A | |
B | |
C | |
D |
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
R_{m}=100 \Omega, l_{m}=20 \mathrm{~cm} | |
R_{m}=25 \Omega, l_{m}=20 \mathrm{~cm} | |
R_{m}=100 \Omega, l_{m}=5 \mathrm{~cm} | |
R_{m}=25 \Omega, l_{m}=5 \mathrm{~cm} |
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.
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
the closed loop gain is less than 1 and the phase shift is less than 180^{\circ}. | |
the closed loop gain is greater than 1 and the phase shift is less than 180^{\circ}. | |
the closed loop gain is less than 1 and the phase shift is greater than 180^{\circ}. | |
the closed loop gain is greater than 1 and the phase shift is greater than 180^{\circ}. |
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.
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
The impedance matching in achieved when the real part of Z_{L} is
112.5\:\Omega | |
75.0\:\Omega | |
50.0\:\Omega | |
33.3\:\Omega |
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
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.
5.55 | |
2.25 | |
4.45 | |
6.25 |
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.