GATE EC 2016 SET-3

Question 1
Consider a 2 x 2 square matrix

A=\begin{bmatrix} \sigma & x\\ \omega & \sigma \end{bmatrix}

where x is unknown. If the eigenvalues of the matrix A are (\sigma +j\omega ) \; and \; (\sigma -j\omega ), then x is equal to
A
+j\omega
B
-j\omega
C
+\omega
D
-\omega
Engineering Mathematics   Linear Algebra
Question 1 Explanation: 
\begin{aligned} A &=\left[\begin{array}{ll} \sigma & x \\ \omega & \sigma \end{array}\right] \\ \text { Trace } &=\text { sum of eigen values } \\ 2 \sigma &=\sigma+j \omega+\sigma-j \omega \\ |A| &=\text { product of eigens } \\ \sigma^{2}-x \omega &=(\sigma+j \omega)(\sigma-j \omega)=\sigma^{2}+\omega^{2} \end{aligned}
which is possible only when, x=-\omega
Question 2
For f(z)=\frac{sin(z)}{z^{2}}, the residue of the pole at z=0 is __________
A
0.5
B
1
C
2
D
3
Engineering Mathematics   Complex Analysis
Question 2 Explanation: 
Residue of \frac{\sin z}{z^{2}}
= coefficient of \frac{1}{2} in \left\{\frac{z-\frac{z^{3}}{3 !}+\frac{z^{5}}{5 !}---}{z^{2}}\right\}
= coefficient of \frac{1}{z} in \left\{\frac{1}{z}-\frac{z}{3 !}+\frac{z^{3}}{5 !}---\right\}
=1
Question 3
The probability of getting a "head" in a single toss of a biased coin is 0.3. The coin is tossed repeatedly till a "head" is obtained. If the tosses are independent, then the probability of getting "head" for the first time in the fifth toss is __________
A
0.01
B
0.04
C
0.07
D
0.1
Engineering Mathematics   Probability and Statistics
Question 3 Explanation: 
\begin{aligned} P(H)&=0.3 \\ P(T)&=0.7 \end{aligned}
since all tosses are independent
so, probability of getting head for the first time in 5^{\text {th }} toss is
\begin{aligned} &=P(T) P(T) P(T) P(T) P(H) \\ &=0.7 \times 0.7 \times 0.7 \times 0.7 \times 0.3=0.072 \end{aligned}
Question 4
The integral \int_{0}^{1}\frac{dx}{\sqrt{(1-x)}} is equal to __________
A
1
B
2
C
3
D
4
Engineering Mathematics   Calculus
Question 4 Explanation: 
\begin{aligned} \int_{0}^{1} \frac{1}{\sqrt{1-x}} d x &=-2 \int_{0}^{1} \frac{1}{2 \sqrt{1-x}} d x \\ &=-\left.2(\sqrt{1-x})\right|_{0} ^{1} \\ &=-2(0-1)=2 \end{aligned}
Question 5
Consider the first order initial value problem

y'=y+2x-x^{2},y(0)=1,(0\leq x\leq \infty )

with exact solution y(x)=x^{2}+e^{x}. For x = 0.1, the percentage difference between the exact solution and the solution obtained using a single iteration of the second-order Runge-Kutta method with step-size h = 0.1 is __________
A
0.06
B
0.012
C
0.6
D
0.12
Engineering Mathematics   Numerical Methods
Question 5 Explanation: 
\begin{aligned} \frac{d y}{d x} &=y+2 x-x^{2} \\ y(0) &=1 \quad \quad 0 \leq x \leq \infty \\ f(x, y) &=y+2 x-x^{2} \\ x_{0}=0 ; y_{0}=1 ; h &=0.1 \\ k_{1} &=h f\left(x_{0}, y_{0}\right) \\ &=0.1\left(1+2 \times 0-0^{2}\right)=0.1 \\ k_{3} &=h /\left(x_{0}+h_{1} y_{0}+k_{1}\right) \\ &=0.1\left(\left(y_{0}+k_{1}\right)+2\left(x_{0}+h\right)-\left(x_{0}+h\right)^{2}\right) \\ &\left.=0.1(11+0.1)+2(0.1)-(0.1)^{2}\right) \\ &=0.129\\ y_{1} &=y_{0}+\frac{1}{2}\left(k_{1}+k_{2}\right) \\ &=1+\frac{1}{2}(0.1+0.129) \\ &=1.1145 \\ \text { Exact solution } y(x) &=x^{2}+e^{x} \\ &=(0.1)^{2}+e^{0.1} \\ &=1.1152 \\ \text { Error } &=1.1152-1.1145=0.00069 \\ \% \text { error } &=0.06 \% \end{aligned}
Question 6
Consider the signal x(t)=cos(6\pi t)+sin(8\pi t), where t is in seconds. The Nyquist sampling rate (in samples/second) for the signal y(t)=x(2t+ 5) is
A
8
B
12
C
16
D
32
Signals and Systems   Sampling
Question 6 Explanation: 
\begin{aligned} X(t) & =\cos (6 \pi t)+\sin (8 \pi t) \\ Y(t) & =x(2 t+5) \\ Y(t) & =\cos [6 \pi(2 t+5)+\sin (8 \pi(2 t+5)] \\ & =\cos (12 \pi t+30 \pi)+\sin (16 \pi t+40 \pi) \\ f_{m_{1}} & =6 \mathrm{Hz}_{i} \qquad f_{m 2}=8 \mathrm{Hz}\\ \text{Nyquist }&\text{sampling rate},\\ f_{s} &=2 f_{\max } \\ &=16 \text { samples/second } \end{aligned}
Question 7
If the signal x(t)=\frac{sin(t)}{\pi t}*\frac{sin(t)}{\pi t} with * denoting the convolution operation, then x(t) is equal to
A
\frac{sin(t)}{\pi t}
B
\frac{sin(2t)}{2\pi t}
C
\frac{2sin(t)}{\pi t}
D
(\frac{sin(t)}{\pi t})^{2}
Signals and Systems   Fourier Transforms, Frequency Response and Correlation
Question 7 Explanation: 




\begin{aligned} x(t) &=x_{1}(t) * x_{1}(t) \\ X(\omega) &=X_{1}(\omega) \cdot X_{1}(\omega) \end{aligned}


\therefore \quad x(t)=\frac{\sin t}{\pi t}
Question 8
A discrete-time signal x[n]=\delta [n-3]+2\delta [n-5] has z-transform X(z). If X(z)=X(-z) is the z-transform of another signal y[n], then
A
y[n]=x[n]
B
y[n]=x[-n]
C
y[n]=-x[n]
D
y[n]=-x[-n]
Signals and Systems   Z-Transform
Question 8 Explanation: 
Given,
\begin{aligned} x[n] &=\delta[n-3]+2 \delta[n-5] \\ X(z) &=z^{3}+2 z^{5} \\ X(-z) &=(-z)^{-3}+2(-z)^{-5} \\ Y(z) &=X(-z)=-z^{-3}-2 z^{-5} \\ &=-\left[z^{-3}+2 z^{-5}\right] \\ y[n] &=-x[n] \end{aligned}
Question 9
the RLC circuit shown in the figure, the input voltage is given by

v_{i}(t)=2cos(200t)+4sin(500t)

The output voltage v_{0}(t) is
A
cos(200t) + 2 sin(500t)
B
2cos(200t) + 4 sin(500t)
C
sin(200t) + 2 cos(500t)
D
2sin(200t) + 4 cos(500t)
Network Theory   Sinusoidal Steady State Analysis
Question 9 Explanation: 


Where, V_{i}(t)=2 \cos (200 t)+4 \sin (500 t)
As different frequencies are operating, using superposition theorem, we get
for \omega=200 \mathrm{rad} / \mathrm{sec}
\begin{array}{l} x_{L}=\omega L=(200)(0.25)=50 \Omega \\ x_{C}=\frac{1}{\omega C}=\frac{1}{200 \times 100 \times 10^{-6}}=50 \Omega \end{array}


\begin{aligned} V_{0}(t) &=V_{i}(t) \\ X_{L} &=0.4 \times 500=200 \Omega \\ X_{C} &=\frac{1}{10 \times 10^{-6} \times 500}=200 \Omega \end{aligned}


\begin{aligned} \therefore \quad V_{0}(t)&=V_{i}(t) \\ \text { Therefore } \quad V_{0}(t)&=2 \cos (200 t)+4 \sin (500 t) \end{aligned}
Question 10
The I-V characteristics of three types of diodes at the room temperature, made of semiconductors X, Y and Z, are shown in the figure. Assume that the diodes are uniformly doped and identical in all respects except their materials. If E_{gX},E_{gY} \; and \; E_{gZ} are the band gaps of X, Y and Z, respectively, then
A
E_{gX} \gt E_{gY}\gt E_{gZ}
B
E_{gX}=E_{gY} = E_{gZ}
C
E_{gX} \lt E_{gY}\lt E_{gZ}
D
no relationship among these band gaps exists.
Electronic Devices   PN-Junction Diodes and Special Diodes
Question 10 Explanation: 
\begin{array}{l} X \rightarrow \text { Ge Diode } \ldots \ldots \ldots \ldots \ldots E_{G}=0.2 \mathrm{V} \\ Y \rightarrow \text { Si Diode } \ldots \ldots \ldots \ldots E_{G}=0.7 \mathrm{V} \\ Z \rightarrow \text { GaAs Diode }(\text { or }) \mathrm{LED} \ldots \ldots \ldots . E_{G}=1.3 \mathrm{V} \end{array}
There are 10 questions to complete.
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