Transmission Lines

Question 1
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 1 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 2
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 2 Explanation: 


Z_{in}\,\, for \, (l=\lambda /4)=\frac{Z_{0}^{2}}{Z_{L}}=\frac{50^{2}}{400}=\frac{25}{4}=6.25\Omega
Question 3
The impedances Z=jX, for all X in the range (-\infty ,\infty), map to the Smith chart as
A
a circle of radius 1 with centre at (0, 0).
B
a point at the centre of the chart.
C
a line passing through the centre of the chart
D
a circle of radius 0.5 with centre at (0.5, 0).
GATE EC 2020   Electromagnetics
Question 3 Explanation: 
For given impedance Normalized impedance is
\begin{aligned} \frac{Z}{Z_{0}}&=\frac{jX}{Z_{0}}\\ Z&=jX \\ \Rightarrow \; Z&=0+jX \end{aligned}
Normalized Resistance =0 \; \Rightarrow \; r=0
X=-\infty \text{ to } \infty
r=0 and X from -\infty \, to\, \infty is a unit circle (radius 1) and centre (0,0) on a complex reflection coeffient plane:
Question 4
A lossy transmission line has resistance per unit length R = 0.05 \Omega /m. The line is distortionless and has characteristic impedance of 50 \Omega . The attenuation constant (in Np/m, correct to three decimal places) of the line is _______.
A
0.0001
B
0.001
C
0.01
D
0.1
GATE EC 2018   Electromagnetics
Question 4 Explanation: 
For a distortionless transmission line,
\frac{L}{R}=\frac{C}{G}
Propagation constant,
\begin{aligned} \gamma &=\alpha+j \beta=\sqrt{(R+j \omega L)(G+j \omega C)} \\ &=\sqrt{R G}\left(1+j \omega \frac{L}{R}\right) \end{aligned}
Attenuation constant,
\alpha=\sqrt{R G}
Characteristic impedance,
\begin{aligned} Z_{0} &=\sqrt{\frac{(R+j \omega L)}{(G+j \omega C)}}=\sqrt{\frac{R}{G}} \\ \sqrt{G} &=\frac{\sqrt{R}}{Z_{0}} \\ \text{So,}\quad\alpha &=\sqrt{R} \cdot \frac{\sqrt{R}}{Z_{0}}=\frac{R}{Z_{0}} \\ &=\frac{0.05}{50}=\frac{0.01}{10}=0.001 \mathrm{Np} / \mathrm{m} \end{aligned}
Question 5
The points P, Q, and R shown on the Smith chart (normalized impedance chart) in the following figure represent:
A
P: Open Circuit, Q: Short Circuit, R: Matched Load
B
P: Open Circuit, Q: Matched Load, R: Short Circuit
C
P: Short Circuit, Q: Matched Load, R: Open Circuit
D
P: Short Circuit, Q: Open Circuit, R: Matched Load
GATE EC 2018   Electromagnetics
Question 5 Explanation: 
For Short circuit,
r=x=0 \quad \Rightarrow \text { Point } " P^{\prime \prime}
For Open circuit,
r=x=\infty \quad \Rightarrow \text { Point }^{\prime \prime} R^{\prime \prime}
For Matched load,
r=1 \text { and } x=0 \Rightarrow \text { Point " } Q^{\prime \prime}
P: Short Circuit, Q: Matched Load R: Open circuit
Question 6
A two - wire transmission line terminates in a television set. The VSWR measured on the line is 5.8. The percentage of power that is reflected from the television set is ______________
A
14.4
B
49.83
C
64.3
D
38
GATE EC 2017-SET-2   Electromagnetics
Question 6 Explanation: 
Given that, VSWR (or) s=5.8
\begin{aligned} s &=\frac{1+|\Gamma|}{1-|\Gamma|}=5.8 \\ \Gamma &=\text { Reflection coefficient of } \\ &\quad \text { television set } \\ \Gamma &=\frac{s-1}{s+1}=\frac{4.8}{6.8}=0.70588 \\ \frac{P_{\text {reflected }}}{P_{\text {incident }}} &=\Gamma^{2}=0.4983 \text { or } 49.83 \% \end{aligned}
Question 7
The voltage of an electromagnetic wave propagating in a coaxial cable with uniform characteristic impedance is V(\iota )=e^{-y \iota +j\omega t} volts, Where \iota is the distance along the length of the cable in meters. \gamma =(0.1+j40)m^{-1} is the complex propagation constant, and \omega = 2\pi \times 10^{9} rad/ s is the angular frequency. The absolute value of the attenuation in the cable in dB/meter is __________.
A
0.5
B
0.86
C
0.34
D
0.94
GATE EC 2017-SET-1   Electromagnetics
Question 7 Explanation: 
\begin{aligned} V(l) &=V_{O} e^{-\alpha l} e^{-j\beta l} e^{j\omega t} \\ \text { Attenuation } &=\frac{\mid \text { Input } \mid}{\mid \text { Output } \mid}=\frac{\left|V_{O}(0)\right|}{\left|V_{o}(l)\right|} \end{aligned}
Attenuation per meter =\frac{\left|V_{0}\right|}{\left|V_{0}(1 \mathrm{m})\right|}=e^{\alpha}
Attenuation in \mathrm{dB} / \mathrm{m}=\left(20 \log e^{\alpha}\right) \mathrm{dB} / \mathrm{m}
=20(0.1) \log e=0.868 \mathrm{dB} / \mathrm{m}
Question 8
A microwave circuit consisting of lossless transmission lines T_{1} \; and \; T_{2} is shown in the figure. The plot shows the magnitude of the input reflection coefficient \tau as a function of frequency f. The phase velocity of the signal in the transmission lines is 2*10^{8} m/s.

The length L (in meters) of T_2 is _______
A
0.05
B
0.1
C
0.2
D
3
GATE EC 2016-SET-2   Electromagnetics
Question 8 Explanation: 
At frequency of 1 \mathrm{GHz} the line is matched and | \Gamma |=0 as seen in the graph.


The line T_{1} is having 50 \Omega load parallel to T_{2} input impedance.
This load is 50\Omega when T_{2} input impedance is infinite.
\begin{aligned} Z_{\text {in } 2}&=\infty=Z_{L 2} \text{ (open circuit) }\\ T_{2}\text{ has to be a }\frac{\lambda}{2}\text{ line}.&\\ f &=1 \mathrm{GHz} \\ \lambda &=\frac{2 \times 10^{10}}{1 \times 10^{9}} \mathrm{cm} \\ \frac{\lambda}{2} &=L=10 \mathrm{cm}=0.1 \mathrm{m} \end{aligned}
Question 9
A lossless microstrip transmission line consists of a trace of width w. It is drawn over a practically infinite ground plane and is separated by a dielectric slab of thickness t and relative permittivity \varepsilon _{r} \gt 1. The inductance per unit length and the characteristic impedance of this line are L and Z_{0}, respectively.

Which one of the following inequalities is always satisfied?
A
Z_{0} \gt \sqrt{\frac{Lt}{\varepsilon _{0}\varepsilon_{r}w }}
B
Z_{0} \lt \sqrt{\frac{Lt}{\varepsilon _{0}\varepsilon_{r} w}}
C
Z_{0} \gt\sqrt{\frac{Lw}{\varepsilon _{0}\varepsilon_{r}t }}
D
Z_{0} \lt \sqrt{\frac{Lw}{\varepsilon _{0}\varepsilon_{r}t }}
GATE EC 2016-SET-2   Electromagnetics
Question 10
The propagation constant of a lossy transmission line is (2 +j5) m^{-1} and its characteristic impedance is (50 + j0) \Omega at \omega = 10^{6} rad s^{-1}. The values of the line constants L, C,R, G are, respectively,
A
L=200\mu H/m, C=0.1 \mu F/m, R=50\Omega /m, G=0.02 S/m
B
L=250\mu H/m, C=0.1 \mu F/m, R=100\Omega /m, G=0.04 S/m
C
L=200\mu H/m, C=0.2 \mu F/m, R=100\Omega /m, G=0.04 S/m
D
L=250\mu H/m, C=0.2 \mu F/m, R=50\Omega /m, G=0.04 S/m
GATE EC 2016-SET-1   Electromagnetics
Question 10 Explanation: 
\begin{aligned} \gamma &=\sqrt{(R+j \omega L)(G+j \omega C)} \\ Z_{0} &=\sqrt{\frac{R+j \omega L}{G+j \omega C}} \\ \gamma \cdot Z_{0}=R+j \omega L =&(2+j 5)(50+j 0)=100+j 250 \\ R &=100 \Omega / m \\ L &=\frac{250}{\omega}=\frac{250}{10^{6}}=250 \mu \mathrm{H} / \mathrm{m} \\ \frac{\gamma}{Z_{0}}=\frac{2+j 5}{50}=G &+\dot{j \omega C}=0.04+j 0.1 \\ G &=0.04 \mathrm{S} / \mathrm{m} \\ C &=\frac{0.1}{\omega}=\frac{0.1}{10^{6}}=0.1 \mu \mathrm{F} / \mathrm{m} \end{aligned}
There are 10 questions to complete.