Question 1 |
Consider the following statements regarding the complex Poynting vector \vec{P} for the power radiated by a point source in an infinite homogeneous and lossless medium. Re(\vec{P}) denotes the real part of \vec{P}, S denotes a spherical surface whose centre is at the point source, and \hat{n}
denotes the unit surface normal on S. Which of the following statements is TRUE?
Re(\vec{P}) remains constant at any radial distance from the source | |
Re(\vec{P}) increases with increasing radial distance from the source | |
\oint \oint_{S}Re(\vec{P}).\hat{n} dS remains constant at any radial distance from the source | |
\oint \oint_{S}Re(\vec{P}).\hat{n} dS decreases with increasing radial distance form the source |
Question 1 Explanation:
Power density of any point source decreases with
distance i.e. the density decreases and area of
cross-over increases with the product being
constant.
Question 2 |
A transmission line of characteristic impedance 50 W is terminated by a 50 \Omega load. When excited by a sinusoidal voltage source at 10 GHz, the phase
difference between two points spaced 2 mm apart on the line is found to \frac{\pi }{4} radians. The phase velocity of the wave along the line is
0.8 \times 10^{8} m/s | |
1.2 \times 10^{8} m/s | |
1.6 \times 10^{8} m/s | |
3 \times 10^{8} m/s |
Question 2 Explanation:
\begin{aligned} \text { Phase difference }&=\frac{2 \pi}{\lambda} \text { (path difference) } \\ \Rightarrow \frac{\pi}{4}&=\frac{2 \pi}{\lambda}\left(2 \times 10^{-3}\right) \\ \therefore \quad &=8 \times 2 \times 10^{-3} \\ &=16 \times 10^{-3} \mathrm{m} \\ f&= 10 \mathrm{GHz}=10 \times 10^{9} \mathrm{Hz} \end{aligned}
Hence the phase velocity of wave along the line is,
\begin{aligned} & v=f \lambda=10 \times 10^{9} \times 16 \times 10^{-3} \mathrm{m} / \mathrm{sec} \\ \therefore \quad v &=1.6 \times 10^{8} \mathrm{m} / \mathrm{s} \end{aligned}
Hence the phase velocity of wave along the line is,
\begin{aligned} & v=f \lambda=10 \times 10^{9} \times 16 \times 10^{-3} \mathrm{m} / \mathrm{sec} \\ \therefore \quad v &=1.6 \times 10^{8} \mathrm{m} / \mathrm{s} \end{aligned}
Question 3 |
An analog signal is band-limited to 4 kHz, sampled at the Nyquist rate and
the samples are quantized into 4 levels. The quantized levels are assumed to be
independent and equally probable. If we transmit two quantized samples per
second, the information rate is ________ bits / second.
1 | |
2 | |
3 | |
4 |
Question 3 Explanation:
Quantized levels are equiprobable; hence
\begin{aligned} H&=\log _{2} 4=2 \text { bits/sample }\\ r&=2 \text{samples/sec} \\ \end{aligned}
Hence information rate R=r \cdot H=2 \text{ samples/sec}
\times 2\text{ bits/sample}
\Rightarrow R=4 \mathrm{bits} / \mathrm{sec}
\begin{aligned} H&=\log _{2} 4=2 \text { bits/sample }\\ r&=2 \text{samples/sec} \\ \end{aligned}
Hence information rate R=r \cdot H=2 \text{ samples/sec}
\times 2\text{ bits/sample}
\Rightarrow R=4 \mathrm{bits} / \mathrm{sec}
Question 4 |
The root locus plot for a system is given below. The open loop transfer function
corresponding to this plot is given by
G(S)H(S)=k\frac{s(s+1)}{(s+2)(s+3)} | |
G(S)H(S)=k\frac{(s+1)}{(s+2)(s+3)^{2}} | |
G(S)H(S)=k\frac{1}{s(s-1)(s+2)(s+3)} | |
G(S)H(S)=k\frac{(s+1)}{(s+2)(s+3)} |
Question 4 Explanation:
From plot we can observe that one pole terminates
at one zero at position -1 and three poles
terminates to \infty. It means there are four poles and
1 zero. Pole at -3 goes on both sides. It means
there are two poles at -3.
Question 5 |
A system is defined by its impulse response h(n) =2^{n} u(n - 2). The system is
stable and causal | |
causal but not stable | |
stable but not causal | |
unstable and non-causal |
Question 5 Explanation:
h(n)=2^{n} u(n-2)
For causal system h(n)=0 for n \lt 0
Hence given system is causal.
For stability:
\sum_{n=2}^{\infty} 2^{n}=\infty, so given system is not stable.
For causal system h(n)=0 for n \lt 0
Hence given system is causal.
For stability:
\sum_{n=2}^{\infty} 2^{n}=\infty, so given system is not stable.
Question 6 |
If the unit step response of a network is (1-e^{-at}), then its unit impulse response is
\alpha e^{-at} | |
\alpha^{-1} e^{-at} | |
(1-\alpha ^{-1})e^{at} | |
(1-\alpha)e^{-at} |
Question 6 Explanation:
Unit step response s(t)=\left(1-e^{-\alpha t}\right)
So, unit impulse response is
\begin{aligned} h(t) &=\frac{d s(t)}{d t}=\frac{d}{d t}\left(1-e^{-\alpha t}\right) \\ &=\alpha e^{-\alpha t} \end{aligned}
So, unit impulse response is
\begin{aligned} h(t) &=\frac{d s(t)}{d t}=\frac{d}{d t}\left(1-e^{-\alpha t}\right) \\ &=\alpha e^{-\alpha t} \end{aligned}
Question 7 |
The output Y in the circuit below is always '1' when
two or more of the inputs P, Q, R are '0' | |
two or more of the inputs P, Q, R are '1' | |
any odd number of the inputs P, Q, R is '0' | |
any odd number of the inputs P, Q, R is '1' |
Question 7 Explanation:
Y= PO+ PR+ RO
Question 8 |
In the circuit shown below, capacitors C1 and C2 are very large and are shorts
at the input frequency. v_{i} is a small signal input. The gain magnitude |\frac{v_{0}}{v_{i}}| at 10
M rad/s is
maximum | |
minimum | |
unity | |
zero |
Question 8 Explanation:
Whenever we use a bypass capacitor in parallel R_E then it always increases voltage gain. As compare to "CE with R_E without capacitor". Hence only answer (A) is possible. and
\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{10 \times 10^{-6} \times 10^{-9}}}
=10 M rad/sec and gain is maximum ar resonance frequency.
\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{10 \times 10^{-6} \times 10^{-9}}}
=10 M rad/sec and gain is maximum ar resonance frequency.
Question 9 |
Drift current in the semiconductors depends upon
only the electric field | |
only the carrier concentration gradient | |
both the electric field and the carrier concentration | |
both the electric field and the carrier concentration gradient |
Question 9 Explanation:
\begin{aligned} j &= n\;e\;v_{d}\\ \text{Put,}\quad v_{d} &= \mu E\\ \therefore J&=n e \mu E\\ \text{Hence}\quad l&=n e \mu EA\\ \end{aligned}
So, 1 depends upon carrier concentration and electric field.
So, 1 depends upon carrier concentration and electric field.
Question 10 |
A Zener diode, when used in voltage stabilization circuits, is biased in
reverse bias region below the breakdown voltage | |
reverse breakdown region | |
forward bias region | |
forward bias constant current mode |
Question 10 Explanation:
In breakdown region only, zener diode is useful.
In reverse bias region below the breakdown voltage it will behave like an open circuit.
There are 10 questions to complete.
