# GATE EC 2011

 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?
 A Re($\vec{P}$) remains constant at any radial distance from the source B Re($\vec{P}$) increases with increasing radial distance from the source C $\oint \oint_{S}Re(\vec{P}).\hat{n} dS$ remains constant at any radial distance from the source D $\oint \oint_{S}Re(\vec{P}).\hat{n} dS$ decreases with increasing radial distance form the source
Electromagnetics   Antennas
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
 A $0.8 \times 10^{8} m/s$ B $1.2 \times 10^{8} m/s$ C $1.6 \times 10^{8} m/s$ D $3 \times 10^{8} m/s$
Electromagnetics   Transmission Lines
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}
 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.
 A 1 B 2 C 3 D 4
Communication Systems   Digital Communications
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 $R=r \cdot H=2 \text{ samples/sec}$
$\times 2\text{ bits/sample}$
$\Rightarrow R=4 \mathrm{bits} / \mathrm{sec}$
 Question 4
The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by
 A $G(S)H(S)=k\frac{s(s+1)}{(s+2)(s+3)}$ B $G(S)H(S)=k\frac{(s+1)}{(s+2)(s+3)^{2}}$ C $G(S)H(S)=k\frac{1}{s(s-1)(s+2)(s+3)}$ D $G(S)H(S)=k\frac{(s+1)}{(s+2)(s+3)}$
Control Systems   Root Locus
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
 A stable and causal B causal but not stable C stable but not causal D unstable and non-causal
Signals and Systems   LTI Systems Continuous and Discrete
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.
 Question 6
If the unit step response of a network is $(1-e^{-at})$, then its unit impulse response is
 A $\alpha e^{-at}$ B $\alpha^{-1} e^{-at}$ C $(1-\alpha ^{-1})e^{at}$ D $(1-\alpha)e^{-at}$
Signals and Systems   Laplace Transform
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}
 Question 7
The output Y in the circuit below is always '1' when
 A two or more of the inputs P, Q, R are '0' B two or more of the inputs P, Q, R are '1' C any odd number of the inputs P, Q, R is '0' D any odd number of the inputs P, Q, R is '1'
Digital Circuits   Logic Gates
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
 A maximum B minimum C unity D zero
Analog Circuits   BJT Analysis
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.
 Question 9
Drift current in the semiconductors depends upon
 A only the electric field B only the carrier concentration gradient C both the electric field and the carrier concentration D both the electric field and the carrier concentration gradient
Electronic Devices   Basic Semiconductor Physics
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.
 Question 10
A Zener diode, when used in voltage stabilization circuits, is biased in
 A reverse bias region below the breakdown voltage B reverse breakdown region C forward bias region D forward bias constant current mode
Electronic Devices   PN-Junction Diodes and Special Diodes
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.