GATE EE 2012

Question 1
Two independent random variables X and Y are uniformly distributed in the interval [-1,1]. The probability that max [X,Y] is less than 1/2 is
A
3/4
B
9/16
C
1/4
D
2/3
Engineering Mathematics   Probability and Statistics
Question 1 Explanation: 
-1 \leq x\leq 1 \text{ and } -1 \leq y\leq 1 is the entire rectangle.The region in which maximum of {x,y} is less than 1/2 is shown below as shaded regioninside this rectangle.

\begin{aligned} P\left ( max{x,y} \lt \frac{1}{2} \right )&=\frac{\text{Area of shaded region}}{\text{Area of entire rectangle}} \\ &= \frac{\frac{3}{2} \times \frac{3}{2}}{2 \times 2}=\frac{9}{16} \end{aligned}
Question 2
If x = \sqrt{-1}, then the value of x^{x} is
A
e^{-\pi/2}
B
e^{\pi/2}
C
x
D
1
Engineering Mathematics   Complex Variables
Question 2 Explanation: 
\begin{aligned} x&=i, \text{ then in polar coordinates,} \\ x &=\cos \frac{\pi}{2} +i \sin \frac{\pi}{2}=e^{\frac{\pi}{2}i}\\ \text{Now, } x^x&=i^i =(e^{\frac{\pi}{2}i})^i=e^{i^2 \frac{\pi}{2}}=e^{-\frac{\pi}{2}} \end{aligned}
Question 3
Given f(z)=\frac{1}{z+1}-\frac{2}{z+3}. If C is a counter clockwise path in the z-plane such that |z+1|=1, the value of \frac{1}{2 \pi j}\oint_{C}f(z)dz is
A
-2
B
-1
C
1
D
2
Engineering Mathematics   Complex Variables
Question 3 Explanation: 
Given:
\begin{aligned} f(z)&=\frac{1}{z+1}-\frac{2}{z+3}\\ &=\frac{(Z+3)-2(Z+1)}{(Z+1)(Z+3)}\\ &=\frac{-Z+1}{(Z+1)(Z+3)} \end{aligned}
Poles are at -1 and -3 i.e. (-1,0) and (-3,0).
From figure below of |Z+1|=1,
we see that (-1,0) is inside the circle and (-3,0) is outside the circle.

Residue theorem says,
\frac{1}{2 \pi j}\oint _Cf(z)dz=Residue of those poles which are inside C.
So the required integral \frac{1}{2 \pi j}\oint _Cf(z)dz is given by the residue of function at poles (-1,0) ( which is inside the circle).
This residue is =\frac{-(-1)+1}{(-1+3)}=\frac{2}{2}=1
Question 4
In the circuit shown below, the current through the inductor is
A
\frac{2}{1+j}A
B
\frac{-1}{1+j}A
C
\frac{1}{1+j}A
D
0A
Electric Circuits   Basics
Question 4 Explanation: 


Apply KCL node at 'A'
so, current flowing through 1\Omega is (1-I_2)
Applying KVL in ABCD loop,
1\angle 0-1\angle 0+1(1-I_2)-jI_2=0
I_2=\frac{1}{1+j}
Question 5
The impedance looking into nodes 1 and 2 in the given circuit is
A
50 \Omega
B
100 \Omega
C
5 k\Omega
D
10.1 k\Omega
Electric Circuits   Network Theorems
Question 5 Explanation: 


To find thevenin impedance across node 1 and node 2. Connect a 1 V source and find the current through voltage source.
Then, Z_{Th}=\frac{1}{I_{Th}}
By applying KCL at node B and A
i_{AB}+99i_{b}=I_{Th}
i_b==i_A+i_{AB}
\Rightarrow \;\; i_b-i_A+99i_b=I_{Th}
\Rightarrow \;\; 100i_b-i_A=I_{Th}\;...(i)
By applying KVL in outer loop
10 \times 10^3i_b=1
i_b=10^{-4}A
and 10 \times 10^{3}i_b=-100i_A
i_A=-100i_b
\therefore From equation (i),
100i_b+100i_b=I_{Th}
\Rightarrow \; I_{Th}=200i_b
\;\;=200 \times 10^{-4}=0.02
\therefore \;\;Z_{Th}=\frac{1}{I_{Th}}=\frac{1}{0.02}=50\Omega
Question 6
A system with transfer function
G(s)=\frac{(s^{2}+9)(s+2)}{(s+1)(s+3)(s+4)}
is excited by sin(\omega t) . The steady-state output of the system is zero at
A
\omega = 1 rad/s
B
\omega = 2 rad/s
C
\omega = 3 rad/s
D
\omega = 4 rad/s
Control Systems   Frequency Response Analysis
Question 6 Explanation: 
Fpr sinusoidal excitation,
s=j\omega
\therefore \;\;G(j\omega )=\frac{(-\omega ^2+9)(j\omega +2)}{(j\omega +1)(j\omega +3)(j\omega +4)}
For zero steady state output
|G(j\omega )|=0
\;\;\;=\frac{(-\omega ^2+9)\sqrt{\omega ^2+4}}{(\omega ^2+1)(\sqrt{\omega ^2+9})(\sqrt{\omega ^2+16})}
For zero steady state output,
\Rightarrow \; \omega ^2=9
\Rightarrow \; \omega =3 \; rad/sec
Question 7
In the sum of products function f(X,Y,Z)=\sum(2,3,4,5), the prime implicants are
A
\bar{X}Y,X\bar{Y}
B
\bar{X}Y,X\bar{Y}\bar{Z},X\bar{Y}Z
C
\bar{X}Y\bar{Z},\bar{X}YZ,X\bar{Y}
D
\bar{X}Y\bar{Z},\bar{X}YZ,X\bar{Y}\bar{Z},X\bar{Y}Z
Digital Electronics   Boolean Algebra and Minimization
Question 7 Explanation: 
f(X,Y,Z)=\Sigma (2,3,4,5)

\therefore \;\; f(X,Y,Z)=X\bar{Y}+\bar{X}Y
Question 8
If x[n] = (1/3)^{|n|}-(1/2)^{n} u[n], then the region of convergence (ROC) of its Z-transform in the Z-plane will be
A
\frac{1}{3} \lt |z| \lt 3
B
\frac{1}{3} \lt |z| \lt 1/2
C
\frac{1}{2} \lt |z| \lt 3
D
\frac{1}{3} \lt |z|
Signals and Systems   Z-Transform
Question 8 Explanation: 
\begin{aligned} \text{Let, }x_1[n]&=\left ( \frac{1}{3} \right )^{|n|}\\ \text{and }x_2[n]&=\left ( \frac{1}{2} \right )^{n} u[n]\\ \Rightarrow \; x_1[n]&=\left ( \frac{1}{3} \right )^{n}u[n]+\left ( \frac{1}{3} \right )^{-n} u[-n-1]\\ \left ( \frac{1}{3} \right )^{n} u[n] & \overset{z}{\leftrightarrow} \frac{1}{1-\frac{1}{3}z^{-1}}; \; ROC: |z| \gt \frac{1}{3}\\ \left ( \frac{1}{3} \right )^{-n} u[-n-1] & \overset{z}{\leftrightarrow} \frac{-1}{1-\left (\frac{1}{3} \right )^{-1}z^{-1}}; \; ROC: |z| \lt 3\\ \text{and } x_2[n]=\left ( \frac{1}{2} \right )^{n} u[n] & \overset{z}{\leftrightarrow} \frac{1}{1-\frac{1}{2}z^{-1}}; \; ROC: |z| \gt \frac{1}{2}\\ \therefore \; & \text{ROC is }\frac{1}{2} \lt |z| \lt 3 \end{aligned}
Question 9
The bus admittance matrix of a three-bus three-line system is
Y=j\begin{bmatrix} -13 & 10 & 5\\ 10& -18 & 10\\ 5 & 10 & -13 \end{bmatrix}
If each transmission line between the two buses is represented by an equivalent \pi-network, the magnitude of the shunt susceptance of the line connecting bus 1 and 2 is
A
4
B
2
C
1
D
0
Power Systems   Load Flow Studies
Question 9 Explanation: 
y_{ik}= Series admittance of the line connecting buses I and k
\frac{y_{ik}'}{2}= Half line charging admittance in bus admittance matrix

\begin{aligned} Y_{11}&=\frac{y_{12}'}{2} +\frac{y_{31}'}{2} +y_{12}+y_{31}=-j13\\ y_{12} &=-y_{12}=j10 \\ y_{23} &=-y_{23}=j10 \\ y_{31} &=-y_{31}=j5 \\ \therefore \; Y_{12}'+y_{31}'&=2[-j13+j10+j5] \\ &= j4\\ &\text{similarly,} \\ y_{12}'+y_{23}' &=2[-j18+j10+j10]=j4 \\ y_{23}'+y_{31}' &=2[-j13+j10+j5]=j4 \\ \rightarrow \;\; y_{12}' &=j6-j4=j2 \end{aligned}
Question 10
The slip of an induction motor normally does not depend on
A
rotor speed
B
synchronous speed
C
shaft torque
D
core-loss component
Electrical Machines   Three Phase Induction Machines
Question 10 Explanation: 
\text{Slip}=\frac{N_s-N_r}{N_s}
From the above formula slip depends upon:
(1) Synchronous speed (N_s)
(2) Rotor speed (N_r)
as the shaft torque depends upon rotor speed therefore the slip also depends on shaft torque. And core-losses are independent of slip.
There are 10 questions to complete.