Question 1 |
A transparent dielectric coating is applied to glass \left(\epsilon_{r}=4, \mu_{r}=1\right) to eliminate the reflection of red light \left(\lambda_{0}=0.75 \mu \mathrm{m}\right). The minimum thickness of the dielectric coating, in \mu \mathrm{m}, that can be used is ____ (rounded off to two decimal places).
0.02 | |
0.18 | |
0.13 | |
0.52 |
Question 1 Explanation:
For no reflection, impedance must be matched.
Hence, \eta_{2} acts like a quarter wave impedance transformer.
So,
(i) \quad \eta_{2}=\sqrt{\eta_{1} \cdot \eta_{3}} \Rightarrow \epsilon_{r_{2}}=\sqrt{\epsilon_{r_{1}} \cdot \epsilon_{r_{3}}} \Rightarrow \epsilon_{r_{2}}=2
(ii) For impedance matching,
\begin{aligned} & d=(2 n+1) \frac{\lambda}{4} ; n=0,1,2 \ldots \\ & \lambda=\frac{\lambda_{0}}{\sqrt{\epsilon_{r}}}=\frac{\lambda_{0}}{\sqrt{\epsilon_{r_{2}}}} \\ Here \;\;& \lambda=\frac{0.75 \times 10^{-6}}{\sqrt{2}}=0.53 \times 10^{-6} \end{aligned}
Hence, for minimum distance, n=0
So, d=\frac{\lambda}{4}=\frac{0.53 \times 10^{-6}}{4}=0.133 \mu \mathrm{m}
Hence, \eta_{2} acts like a quarter wave impedance transformer.
So,
(i) \quad \eta_{2}=\sqrt{\eta_{1} \cdot \eta_{3}} \Rightarrow \epsilon_{r_{2}}=\sqrt{\epsilon_{r_{1}} \cdot \epsilon_{r_{3}}} \Rightarrow \epsilon_{r_{2}}=2
(ii) For impedance matching,
\begin{aligned} & d=(2 n+1) \frac{\lambda}{4} ; n=0,1,2 \ldots \\ & \lambda=\frac{\lambda_{0}}{\sqrt{\epsilon_{r}}}=\frac{\lambda_{0}}{\sqrt{\epsilon_{r_{2}}}} \\ Here \;\;& \lambda=\frac{0.75 \times 10^{-6}}{\sqrt{2}}=0.53 \times 10^{-6} \end{aligned}
Hence, for minimum distance, n=0
So, d=\frac{\lambda}{4}=\frac{0.53 \times 10^{-6}}{4}=0.133 \mu \mathrm{m}
Question 2 |
The following circuit(s) representing a lumped element equivalent of an infinitesimal section of a transmission line is/are
A | |
B | |
C | |
D |
Question 3 |
The electric field of a plane electromagnetic wave is
E=a_{x} C_{1 x} \cos (\omega t-\beta z)+a_{y} C_{1 y} \cos (\omega t-\beta z+\theta) \mathrm{V} / \mathrm{m}
Which of the following combination(s) will give rise to a left handed elliptically polarized (LHEP) wave?
E=a_{x} C_{1 x} \cos (\omega t-\beta z)+a_{y} C_{1 y} \cos (\omega t-\beta z+\theta) \mathrm{V} / \mathrm{m}
Which of the following combination(s) will give rise to a left handed elliptically polarized (LHEP) wave?
C_{1 x}=1, C_{1 y}=1, \theta=\pi / 4 | |
C_{1 x}=2, C_{1 y}=1, \theta=\pi / 2 | |
C_{1 x}=1, C_{1 y}=2, \theta=3 \pi / 2 | |
C_{1 x}=2, C_{1 y}=1, \theta=3 \pi / 4 |
Question 3 Explanation:
Given, \quad \vec{E}=\hat{a}_{x} C_{1 x} \cos (\omega t-\beta z)+\hat{a}_{y} C_{1 y} \cos (\omega t-\beta z+\theta)
at z=0
\vec{E}=C_{1 x} \cos \omega t \hat{a}_{x}+C_{1 y} \cos (\omega t+\theta) \hat{a}_{y}
Going by options,
Option (A) \quad \vec{E}=\cos \omega t \hat{a}_{x}+\cos (\omega t+\pi / 4) \hat{a}_{y}
at t=0, \omega t=0, \vec{E}=\hat{a}_{x}+\frac{1}{\sqrt{2}} \hat{a}_{y}
at t=T / 4, \omega t=\pi / 2, \vec{E}=0-\frac{1}{\sqrt{2}} \hat{a}_{y}
\Rightarrow Hence, it is LHEP.
Option (B) \quad \vec{E}=2 \cos \omega t \hat{a}_{x}+\cos (\omega t+\pi / 2) \hat{a}_{y}
at t=0, \omega t=0, \vec{E}=2 \hat{a}_{x}
at t=T / 4, \omega t=\pi / 2, \vec{E}=-1 \hat{a}_{y}
\Rightarrow Hence, it is LHEP.
Option (C) \quad \vec{E}=\cos \omega t \hat{a}_{x}+2 \cos (\omega t+3 \pi / 2) \hat{a}_{y}
at t=0, \quad \quad \omega t=0, \vec{E}=\hat{a}_{x}
at t=T / 4, \omega t=\pi / 2, \vec{E}=2 \hat{a}_{y}
\Rightarrow Hence, it is RHEP.
Option (D) \quad \vec{E}=2 \cos \omega t \hat{a}_{x}+\cos (\omega t+3 \pi / 4) \hat{a}_{y}
at t=0, \omega t=0, \vec{E}=2 \hat{a}_{x}-\frac{1}{\sqrt{2}} \hat{a}_{y}
at t=T / 4, \omega t=\pi / 2, \vec{E}=0-\frac{1}{\sqrt{2}} \hat{a}_{y}=\frac{-1}{\sqrt{2}} \hat{a}_{y}
\Rightarrow Hence, it is LHEP.
\therefore Option (A), (B) and (D) are correct.
at z=0
\vec{E}=C_{1 x} \cos \omega t \hat{a}_{x}+C_{1 y} \cos (\omega t+\theta) \hat{a}_{y}
Going by options,
Option (A) \quad \vec{E}=\cos \omega t \hat{a}_{x}+\cos (\omega t+\pi / 4) \hat{a}_{y}
at t=0, \omega t=0, \vec{E}=\hat{a}_{x}+\frac{1}{\sqrt{2}} \hat{a}_{y}
at t=T / 4, \omega t=\pi / 2, \vec{E}=0-\frac{1}{\sqrt{2}} \hat{a}_{y}
\Rightarrow Hence, it is LHEP.
Option (B) \quad \vec{E}=2 \cos \omega t \hat{a}_{x}+\cos (\omega t+\pi / 2) \hat{a}_{y}
at t=0, \omega t=0, \vec{E}=2 \hat{a}_{x}
at t=T / 4, \omega t=\pi / 2, \vec{E}=-1 \hat{a}_{y}
\Rightarrow Hence, it is LHEP.
Option (C) \quad \vec{E}=\cos \omega t \hat{a}_{x}+2 \cos (\omega t+3 \pi / 2) \hat{a}_{y}
at t=0, \quad \quad \omega t=0, \vec{E}=\hat{a}_{x}
at t=T / 4, \omega t=\pi / 2, \vec{E}=2 \hat{a}_{y}
\Rightarrow Hence, it is RHEP.
Option (D) \quad \vec{E}=2 \cos \omega t \hat{a}_{x}+\cos (\omega t+3 \pi / 4) \hat{a}_{y}
at t=0, \omega t=0, \vec{E}=2 \hat{a}_{x}-\frac{1}{\sqrt{2}} \hat{a}_{y}
at t=T / 4, \omega t=\pi / 2, \vec{E}=0-\frac{1}{\sqrt{2}} \hat{a}_{y}=\frac{-1}{\sqrt{2}} \hat{a}_{y}
\Rightarrow Hence, it is LHEP.
\therefore Option (A), (B) and (D) are correct.
Question 4 |
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 4 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 5 |
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 5 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.
There are 5 questions to complete.