Question 1 |
An optical fiber is kept along the \hat{z} direction. The refractive indices for the electric fields along \hat{x} and \hat{y} directions in the fiber are n_{x}=1.5000 and n_{y}=1.5001, respectively (n_{x} \neq n_{y} due to the imperfection in the fiber cross-section). The free space wavelength of a light wave propagating in the fiber is 1.5\mum. If the light wave is circularly polarized at the input of the fiber, the minimum propagation distance after which it becomes linearly polarized, in centimeter, is ___________.
0.2 | |
0.25 | |
0.37 | |
0.45 |
Question 1 Explanation:
Initially the wave is circularly polarized. So, the
initial phase difference between field components
in \hat{a}_{x} direction and \hat{a}_{y} direction is \frac{\pi}{2}
To become linearly polarized, the wave must travel a minimum distance, such that, the phase difference at that point between the field components in \hat{a}_{x} direction and \hat{a}_{y} direction is \pi
(i.e., the travel of this minimum distance should provide an additional phase difference of \pi / 2 between \hat{a}_{x} and \hat{a}_{y} field components).
So,
\begin{aligned} z_{min}k_z&\sim z_{min}k_y=\frac{\pi}{2} \\ z_{min}\left [ \frac{\omega }{V_{px}}\sim \frac{\omega }{V_{py}} \right ]&= \frac{\pi}{2} \\ 2 \pi z_{min}\left [ \frac{f}{c}\sqrt{\varepsilon _{rx}}\sim \frac{f}{c}\sqrt{\varepsilon _{ry}}\right ]&=\frac{\pi}{2} \\ \frac{4 z_{min}}{\lambda _0}[n_x\sim n_y]&=1 \\ z_{min}&=\frac{\lambda _0}{4(n_x\sim n_y)} \\ &=\frac{1.5}{4(1.5\sim 1.5001)}\mu m \\ &= \frac{1.5}{4(0.0001)}\mu m=\frac{1.5}{4}cm\\ z_{min}&=0.375cm \end{aligned}
To become linearly polarized, the wave must travel a minimum distance, such that, the phase difference at that point between the field components in \hat{a}_{x} direction and \hat{a}_{y} direction is \pi
(i.e., the travel of this minimum distance should provide an additional phase difference of \pi / 2 between \hat{a}_{x} and \hat{a}_{y} field components).
So,
\begin{aligned} z_{min}k_z&\sim z_{min}k_y=\frac{\pi}{2} \\ z_{min}\left [ \frac{\omega }{V_{px}}\sim \frac{\omega }{V_{py}} \right ]&= \frac{\pi}{2} \\ 2 \pi z_{min}\left [ \frac{f}{c}\sqrt{\varepsilon _{rx}}\sim \frac{f}{c}\sqrt{\varepsilon _{ry}}\right ]&=\frac{\pi}{2} \\ \frac{4 z_{min}}{\lambda _0}[n_x\sim n_y]&=1 \\ z_{min}&=\frac{\lambda _0}{4(n_x\sim n_y)} \\ &=\frac{1.5}{4(1.5\sim 1.5001)}\mu m \\ &= \frac{1.5}{4(0.0001)}\mu m=\frac{1.5}{4}cm\\ z_{min}&=0.375cm \end{aligned}
Question 2 |
A radar operating at 5 GHz uses a common antenna for transmission and reception. The antenna has a gain of 150 and is aligned for maximum directional radiation and reception to a target 1 km away having radar cross-section of 3 m^{2}. If it transmits 100 kW, then the received power (in \muW) is__________
0.01 | |
0.1 | |
0.06 | |
0.6 |
Question 2 Explanation:
\begin{aligned} f&=5 \mathrm{GHz}, G=150, R=1 \mathrm{km} \\ P_{t}&=100 \mathrm{kW}, \sigma=3 \mathrm{m}^{2} \\ P_{r}&= \frac{(G \lambda)^{2} \cdot \sigma}{(4 \pi)^{3} R^{4}} \times P_{t} \\ &=\frac{\left(150 \times \frac{3 \times 10^{8}}{5 \times 10^{9}}\right)^{2} \cdot 3}{(4 \pi)^{3} \cdot 10^{12}} \times 10^{5} \\ &=\frac{81 \times 3 \times 10^{5}}{(4 \pi)^{3} \times 10^{12}} \\ &= 1.224 \times 10^{-8}=0.012 \mu \mathrm{W} \end{aligned}
Question 3 |
Light from free space is incident at an angle \theta _{i} to the normal of the facet of a step-index large core optical fibre. The core and cladding refractive indices are n _{1}=1.5 \; and \; n_{2}=1.4, respectively.
The maximum value of \theta _{i} (in degrees) for which the incident light will be guided in the core of the fibre is ________
The maximum value of \theta _{i} (in degrees) for which the incident light will be guided in the core of the fibre is ________
8.25 | |
16.36 | |
24.24 | |
32.58 |
Question 3 Explanation:
\begin{aligned} & \sin \alpha_{\max } &=\sqrt{n_{1}^{2}-n_{2}^{2}}=\sqrt{1.5^{2}-1.4^{2}} \\ \Rightarrow & \alpha_{\max } &=\sin ^{-1}(0.5385)=32.58^{\circ} \end{aligned}
Question 4 |
A two-port network has scattering parameters given by [S]=\begin{bmatrix} S_{11} &S_{12} \\ S_{21} & S_{22} \end{bmatrix}. If the
port-2 of the two port is short circuited, the S_{11}
parameter for the resultant one port network is
\frac{S_{11}-S_{11}S_{22}+S_{12}S_{21}}{1+S_{22}} | |
\frac{S_{11}+S_{11}S_{22}-S_{12}S_{21}}{1+S_{22}} | |
\frac{S_{11}+S_{11}S_{22}+S_{12}S_{21}}{1-S_{22}} | |
\frac{S_{11}-S_{11}S_{22}+S_{12}S_{21}}{1-S_{22}} |
Question 4 Explanation:
\begin{array}{l} b_{1}=s_{11} \cdot a_{1}+s_{12} \cdot a_{2} \qquad\ldots(i)\\ b_{2}=s_{21} \cdot a_{1}+s_{22} a_{2}\qquad\ldots(ii) \end{array}
When output is short, a_{2}=-b_{2}
\begin{aligned} -a_{2} &=s_{21} \cdot a_{1}+s_{22} a_{2} \\ \Rightarrow \quad a_{2} &=\frac{-s_{21} \cdot a_{1}}{1+s_{22}} \end{aligned}
Substituting in first equation,
\begin{aligned} b_{1} &=s_{11} \cdot a_{1}+s_{12}\left(\frac{-s_{21} \cdot a_{1}}{1+s_{22}}\right) \\ &=\frac{s_{11} \cdot a_{1}+s_{11} \cdot s_{22} a_{1}-s_{12} \cdot s_{21} a_{1}}{1+s_{22}} \\ \frac{b_{1}}{a_{1}} &=s_{11}=\frac{s_{11}+s_{11} \cdot s_{22}-s_{12} \cdot s_{21}}{1+s_{22}} \end{aligned}
When output is short, a_{2}=-b_{2}
\begin{aligned} -a_{2} &=s_{21} \cdot a_{1}+s_{22} a_{2} \\ \Rightarrow \quad a_{2} &=\frac{-s_{21} \cdot a_{1}}{1+s_{22}} \end{aligned}
Substituting in first equation,
\begin{aligned} b_{1} &=s_{11} \cdot a_{1}+s_{12}\left(\frac{-s_{21} \cdot a_{1}}{1+s_{22}}\right) \\ &=\frac{s_{11} \cdot a_{1}+s_{11} \cdot s_{22} a_{1}-s_{12} \cdot s_{21} a_{1}}{1+s_{22}} \\ \frac{b_{1}}{a_{1}} &=s_{11}=\frac{s_{11}+s_{11} \cdot s_{22}-s_{12} \cdot s_{21}}{1+s_{22}} \end{aligned}
Question 5 |
If the scattering matrix [S] of a two port network is
|S|=\begin{bmatrix} 0.2\angle 0^{\circ}&0.9\angle 90^{\circ} \\ 0.9\angle 90^{\circ}& 0.1\angle 90^{\circ} \end{bmatrix}
then the network is
|S|=\begin{bmatrix} 0.2\angle 0^{\circ}&0.9\angle 90^{\circ} \\ 0.9\angle 90^{\circ}& 0.1\angle 90^{\circ} \end{bmatrix}
then the network is
lossless and reciprocal | |
lossless but not reciprocal | |
not lossless but reciprocal | |
neither lossless nor reciprocal |
Question 5 Explanation:
For reciprocal networks, S_{12}=S_{21}
For symmetrical networks, S_{11}=S_{22}
For antimetrical networks, S_{11}=-S_{22}
For lossless reciprocal networks,
\left|S_{11}\right|=\left|S_{22}\right|
and \left|S_{11}\right|^{2}+\left|S_{12}\right|^{2}=1
For symmetrical networks, S_{11}=S_{22}
For antimetrical networks, S_{11}=-S_{22}
For lossless reciprocal networks,
\left|S_{11}\right|=\left|S_{22}\right|
and \left|S_{11}\right|^{2}+\left|S_{12}\right|^{2}=1
Question 6 |
In the design of a single mode step index optical fibre close to upper cut-off, the
single-mode operation is not preserved if
radius as well as operating wavelength are halved | |
radius as well as operating wavelength are doubled | |
radius is halved and operating wavelength is doubled | |
radius is doubled and operating wavelength is halved |
Question 7 |
A load of 50\Omega is connected in shunt in a 2-wire transmission line of Z_{0} = 50\Omega as shown in the figure. The 2-port scattering parameter matrix (s-matrix) of the
shunt element is
\begin{bmatrix} -\frac{1}{2} &\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix} | |
\begin{bmatrix} 0 & 1\\ 1&0 \end{bmatrix} | |
\begin{bmatrix} -\frac{1}{3} &\frac{2}{3} \\ \frac{2}{3} & -\frac{1}{3} \end{bmatrix} | |
\begin{bmatrix} \frac{1}{4} &-\frac{3}{4} \\ -\frac{3}{4} & \frac{1}{4} \end{bmatrix} |
Question 7 Explanation:
The line is terminated with 50 \Omega at the center and
so, matched on both the sides.
Question 8 |
In a microwave test bench, why is the microwave signal amplitude modulated at 1 kHz
To increase the sensitivity of measurement | |
To transmit the signal to a far-off place | |
To study amplitude modulations | |
Because crystal detector fails at microwave frequencies |
There are 8 questions to complete.