# 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.