# GATE EE 2011

 Question 1
Roots of the algebraic equation $x^{3}+x^{2}+x+1=0$ are
 A (+1, +j, -j) B (+1, -1, +1) C (0, 0, 0) D (-1, +j, -j)
Engineering Mathematics   Calculus
Question 1 Explanation:
$-1$ is one of the root since
$(-1)^3+(-1)^2+(-1)+1=0$
By polynomial division
$\frac{x^3+x^2+x+1}{(x-(-1))}=x^2+1 \Rightarrow \;\; x^3+x^2+x+1=(x^2+1)(x+1)$
So root are (-1,+j,-j)
 Question 2
With K as a constant, the possible solution for the first order differential equation $\frac{dy}{dx}=e^{-3x}$ is
 A $-\frac{1}{3}e^{-3x}+K$ B $-\frac{1}{3}e^{3x}+K$ C $-\frac{1}{3}e^{-3x}+K$ D $-3e^{-x}+K$
Engineering Mathematics   Differential Equations
Question 2 Explanation:
\begin{aligned} \frac{dy}{dx}&=e^{-3x}\\ \int dy&=\int e^{-3x}dx\\ y&=\frac{e^{-3x}}{-3}+K\\ y&=-\frac{1}{3}e^{-3x}+K \end{aligned}
 Question 3
The r.m.s value of the current i(t) in the circuit shown below is
 A $\frac{1}{2}A$ B $\frac{1}{\sqrt{2}}A$ C 1A D $\sqrt{2}A$
Electric Circuits   Steady State AC Analysis
Question 3 Explanation:
$V_s=1 \sin t\equiv V_m \sin \omega t$
$V_m=1 V$ and $\omega =1$ rad/sec
Impedance of the branch containing inductor and capacitor
$Z=j(X_L-X_C)$
$\;\;=j\left ( \omega L-\frac{1}{\omega C} \right )$
$\;\;=j\left ( 1 times 1 -\frac{1}{1 times 1} \right )=0$
So, this branch is short circuit and the whole current flow through it
$i(t)=\frac{1.0 \sin t}{1}=1.0 \sin t$
rms value of the current $=\frac{1}{\sqrt{2}}A$
 Question 4
The fourier series expansion $f(t)=a_{0}+\sum_{n=1}^{\infty }a_{n}cosn\omega t+b_{n}sin n\omega t$ of the periodic signal shown below will contain the following nonzero terms
 A $a_{0} \; and \; b_{n},n=1,3,5,...\infty$ B $a_{0} \; and \; a_{n},n=1,2,3,...\infty$ C $a_{0},a_{n} \; and \; b_{n},n=1,3,5,...\infty$ D $a_{0} \; and \; a_{n},n=1,3,5,...\infty$
Signals and Systems   Fourier Series
Question 4 Explanation:
Let, $x(t)$= Even and Hws

Fourier series expansion of $x(t)$ contains cos terms with odd harmonics.

NOw, $f(t)=1+x(t)$
Fourier series of $f(t)$ contains dc and cos terms with odd harmonics.
 Question 5
A 4-point starter is used to start and control the speed of a
 A dc shunt motor with armature resistance control B dc shunt motor with field weakening control C dc series motor D dc compound motor
Electrical Machines   DC Machines
Question 5 Explanation:
4-point starter is used to control the speed of shunt motor in field weakening region, i.e. above rates speeds.
In field weakening region field current will reduce in 3-point starter holding coil unable to hold the plunger in ON position.
 Question 6
A three phase, salient pole synchronous motor is connected to an infinite bus. It is operated at no load a normal excitation. The field excitation of the motor is first reduced to zero and then increased in reverse direction gradually. Then the armature current
 A Increases continuously B First increases and then decreases steeply C First decreases and then increases steeply D Remains constant
Electrical Machines   Synchronous Machines
Question 6 Explanation:
As field current reduces, the flux will start reducing, to keep this flux constant, the armature draws high current from bus.
Ar zero field current the motor acts as synchronous reluctance motor in this case the magnetizing current fully taken from bus. Ie we increase field current in reverse direction to keep flux constant, motor draws more current. During this process the load angle increases at one point the reversed field force dominates reluctance torque and rotor slip one pole pitch anf align to opposite pole. The instant align to opposite pole the flux will be very high, to reduce this flux current drops steeply t synchronous motor value. Here torque is reluctance + synchronous motor torque.
 Question 7
A nuclear power station of 500 MW capacity is located at 300 km away from a load center. Select the most suitable power evacuation transmission configuration among the following options
 A A B B C C D D
Power Systems   Switch Gear and Protection
Question 7 Explanation:
For transmission of bulk power of very long distance high voltage (400 kV) is used.
To increase reliability, double circuit is used.
 Question 8
The frequency response of a linear system G($j \omega$) is provided in the tubular form below

Gain Margin and phase margin are
 A 6 dB and 30$^{\circ}$ B 6 dB and -30$^{\circ}$ C -6 dB and 30$^{\circ}$ D -6 dB and -30$^{\circ}$
Control Systems   Frequency Response Analysis
Question 8 Explanation:
At gain crossover frequency $(\omega_{gc})$, magnitude of $G(j\omega)$ is 1.
$|G(j\omega_{gc})|=1$
Phase of $G(j\omega )=\angle G(j\omega_{gc})=-150^{\circ}$
Phase margin$=180^{\circ}+\angle G(j\omega_{gc})$
$\;\;\; =180^{\circ}-150^{\circ}=30^{\circ}$
At phase cross frequency $(\omega_{pc})$, phase of $G(j\omega )$ is $-180^{\circ}$,
$\angle G(j\omega_{pc})=-180^{\circ}$
M=magnitude of $G(j\omega )$ at $\omega _{pc}$
$=|G(j\omega_{pc})|=0.5$
Gain margin
$=20log\frac{1}{M}$
$=20 log \frac{1}{0.5}=6 dB$
 Question 9
The steady state error of a unity feedback linear system for a unit step input is 0.1. The steady state error of the same system, for a pulse input r(t) having a magnitude of 10 and a duration of one second, as shown in the figure is
 A 0 B 0.1 C 1 D 10
Control Systems   Time Response Analysis
Question 9 Explanation:
Let the system is represnted as

$\frac{Y(s)}{X(s)}=\frac{G(s)}{1+G(s)H(s)}$
$H(s)=1 (unity \; feedback)$
$Error=E(s)=\frac{X(s)}{1+G(s)}$
Steady state error for $X(s)=1/s, \; e_{ss}=0.1$
$e_{ss}=\lim_{s \to 0}sE(s)$
$\;\; =\lim_{s \to 0}\frac{sX(s)}{1+G(s)}$
$\Rightarrow \; 0.1=\lim_{s \to 0}\frac{s\cdot \frac{1}{s}}{1+G(s)}$
$\Rightarrow \; 0.1=\lim_{s \to 0}\frac{1}{1+G(s)}$
When input,

$X(t)=10[u(t)-u(t-1)]$
$X(s)=10\left ( \frac{1}{s}-\frac{e^{-s}}{s} \right )$
$\;\;=\frac{10}{s}(1-e^{-s})$
$e_{ss}=\lim_{s \to 0}sE(s)$
$\;\;=\lim_{s \to 0}\frac{sX(s)}{1+G(s)}$
$\;\;=\lim_{s \to 0}\frac{s \times \frac{10}{s}[1-e^{-s}]}{1+G(s)}$
$e_{ss}=\lim_{s \to 0}\frac{10(1-e^{-s})}{1+G(s)}=0$
 Question 10
Consider the following statement

(1) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the current coil.
(2) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the voltage coil circuit.
 A (1) is true but (2) is false B (1) is false but (2) is true C both (1) and (2) are true D both (1) and (2) are false
Electrical and Electronic Measurements   Measurement of Energy and Power
Question 10 Explanation:
The current coil carries a current of $I +I_P$ and produces a filed corresponding to this current. The compensating coil is connected in series with the pressure coil circuit and is made as nearly as possible identical and coincident with the current coil. It is so connected that it opposes the field of the current coil. The compensating coil carries a current $I_P$ and produces a field corresponding to this current. This field acts as in opposition to the current coil field. Thus the resultant field is due to current $I$ only. Hence the error caused by the pressure coil current flowing in the current coil is neutralized.
There are 10 questions to complete.