# GATE EE 2016 SET 1

 Question 1
The maximum value attained by the function $f(x) = x(x- 1)(x - 2)$ in the interval [1, 2] is _____.
 A 0 B 1 C 2 D 4
Engineering Mathematics   Calculus
Question 1 Explanation:
\begin{aligned} f(x) &=x^3-3x^2+2x \\ f'(x)&=3x^2-6x+2\\ f'(x)&=0 \;\text{for stationary point} \end{aligned}
stationary points are $1+\frac{1}{\sqrt{3}}$
only $1+\frac{1}{\sqrt{3}}$ lies in [1,2]
\begin{aligned} f(1) &=0 \\ f(2)&=0 \\ f\left ( 1+\frac{1}{\sqrt{3}} \right )&=-\frac{2}{3\sqrt{3}} \end{aligned}
Maximum value is 0.
 Question 2
Consider a 3 x 3 matrix with every element being equal to 1. Its only non-zero eigenvalue is ____.
 A 1 B 2 C 3 D 4
Engineering Mathematics   Linear Algebra
Question 2 Explanation:
$A=\begin{bmatrix} 1 & 1 &1 \\ 1 & 1 &1 \\ 1& 1 & 1 \end{bmatrix}$ Eigen value are 0,0,3
 Question 3
The Laplace Transform of $f(t)=e^{2t}sin(5t)u(t)$ is
 A $\frac{5}{s^{2}-4s+29}$ B $\frac{5}{s^{2}+5}$ C $\frac{s-2}{s^{2}-4s+29}$ D $\frac{5}{s+5}$
Signals and Systems   Laplace Transform
Question 3 Explanation:
Laplace transform of $\sin 5t u(t)\rightarrow \frac{5}{s^2+25}$
$e^{2t}\sin 5t u(t)\rightarrow \frac{5}{(s-2)^2+25}=\frac{5}{s^2-4s+29}$
 Question 4
A function y(t), such that y(0)=1 and y(1)=3$e^{-1}$, is a solution of the differential equation $\frac{d^{2}y}{dt^{2}}+2\frac{dy}{dt}+y=0$. Then y(2) is
 A $5e^{-1}$ B $5e^{-2}$ C $7e^{-1}$ D $7e^{-2}$
Engineering Mathematics   Calculus
Question 4 Explanation:
Auxiliary equation,
\begin{aligned} m^2+2m+1&=0 \\ m&= -1,-1\\ y&=(c_1+c_2t)e^{-t} \\ y(0) &=1 \\ c_1&=1 \\ y&= (1+c_2t)e^{-t}\\ y(1)&=3e^{-1} \\ \Rightarrow \; (1+c_2)e^{-3} &=3e^{-3} \\ c_2&=2 \\ y&=(1+2t)e^{-t} \\ y(2)&=5e^{-2} \end{aligned}
 Question 5
The value of the integral $\oint_{c}\frac{2z+5}{(z-\frac{1}{2})(z^{2}-4z+5)}dz$ over the contour |z|=1, taken in the anti-clockwise direction, would be
 A $\frac{24 \pi i}{13}$ B $\frac{48 \pi i}{13}$ C $\frac{24}{13}$ D $\frac{12}{13}$
Engineering Mathematics   Calculus
Question 5 Explanation:
Singlarilies, $Z=\frac{1}{2}, 2\pm i$
Only, $Z=\frac{1}{2}$ lies inside C
By residue theorem,
$\oint _C=2 \pi i(R)=\frac{48 \pi i}{13}$
Residue at $\frac{1}{2}$
$=R_{1/2}=\lim_{Z \to 1/2}\left [ \left ( Z-\frac{1}{2} \right )\cdot \frac{2Z+5}{\left ( Z-\frac{1}{2} \right )(Z^2+4Z+5)} \right ]=\frac{24}{13}$
 Question 6
The transfer function of a system is $\frac{Y(s)}{R(s)}=\frac{s}{s+2}$. The steady state output y(t) is $A \cos (2t+\varphi)$ for the input cos(2t). The values of A and $\varphi$, respectively are
 A $\frac{1}{\sqrt{2}},-45^{\circ}$ B $\frac{1}{\sqrt{2}},+45^{\circ}$ C $\sqrt{2},-45^{\circ}$ D $\sqrt{2},+45^{\circ}$
Signals and Systems   Laplace Transform
Question 6 Explanation:
\begin{aligned} \frac{Y(s)}{R(s)}&=\frac{s}{s+2} \\ y(t)&=A \cos (2t+\phi ), \\ r(t)&=\cos 2t \\ \because \;H(s) &=\frac{s}{(s+2)} \\ H(j\omega )&=\frac{j\omega }{j\omega +2} \\ |H(j\omega )| &=\frac{\omega }{\sqrt{\omega ^2+4 }} \\ \angle H(j\omega ) &=90^{\circ} -\tan ^{-1}\left ( \frac{\omega }{2} \right ) \\ \because \; \omega &= 2 \text{ (as given)}\\ |H(j\omega )| &=\frac{2}{\sqrt{4+4}}=\frac{1}{\sqrt{2}} \\ |H(j\omega )| &=90^{\circ} -\tan ^{-1}(1)=45^{\circ} \\ \because \; \text{hence, }A &=1 \times |H(j\omega )|_{\omega =2}\\ &=1 \times \frac{1}{\sqrt{2}}=0.707\\ \phi &= 45^{\circ} \end{aligned}
 Question 7
The phase cross-over frequency of the transfer function $G(s)=\frac{100}{(s+3)^{3}}$ in rad/s is
 A $\sqrt{3}$ B $1/\sqrt{3}$ C 3 D $3\sqrt{3}$
Control Systems   Frequency Response Analysis
Question 7 Explanation:
$G(s)=\frac{100}{(s+1)^3}$
$G(j\omega )=\frac{100}{(1+j\omega )^3}$
$\;\;=\frac{100}{1+(j\omega )^3+3(j\omega )^2+3j\omega }$
$\;\;=\frac{100}{(1-3\omega^2 )+j(3\omega-\omega^3)}$
$\;\;=\frac{100[(1-3\omega^2 )+j\omega(3-\omega^2)^2]}{[ (1-3\omega^2 )+\omega^2(3-\omega^2)^2]}$
For phase corssover frequency $\omega_{ph} Img[G(j \omega )]=0;$
Hence, $\omega (3-\omega ^2)=0$
$\omega =0; \pm \sqrt{3}$
Therefore, $\omega _{ph}=\sqrt{3}$ rad/sec
 Question 8
Consider a continuous-time system with input x(t) and output y(t) given by

y(t) = x(t) cos(t)

This system is
 A linear and time-invariant B non-linear and time-invariant C linear and time-varying D non-linear and time-varying
Signals and Systems   Linear Time Invariant Systems
Question 8 Explanation:
\begin{aligned} y(t)&=x(t)\cos (t)\\ &\text{To check linearity,}\\ y_1(t)&=x_1(t)\cos (t)\\ &[y_1(t) \text{ is output for }x_1(t)]\\ y_2(t)&=x_2(t) \cos (t)\\ &[y_2(t) \text{ is output for }x_2(t)]\\ \text{so, the}& \text{ output for }(x_1(t)+ x_2(t)) \text{ will be}\\ y(t)&=[x_1(t)+ x_2(t)]\cos (t)\\ &=y_1(t)+y_2(t) \end{aligned}
So, the system is linear, to check time invariance.
The delayed output,
$y(t-t_0)=x(t-t_0)\cos (t-t_0)$
The output for delayed input,
$y(t, t_0)=x(t-t_0)\cos (t)$
Since, $y(t-t_0)\neq y(t,t_0)$
System is time varying.
 Question 9
The value of $\int_{-\infty }^{+\infty }e^{-t}\delta (2t-2)dt, \; where \; \delta (t)$ is the Dirac delta function, is
 A $\frac{1}{2e}$ B $\frac{2}{e}$ C $\frac{1}{e^{2}}$ D $\frac{1}{2e^{2}}$
Signals and Systems   Introduction of C.T. and D.T. Signals
Question 9 Explanation:
To find the value of $\int_{-\infty }^{\infty }e^{-t}\delta (2t-2)dt$
Since, $\delta (2t-2)=\frac{1}{2}\delta (1t-1)$ above integral can be written as
$\int_{-\infty }^{\infty }e^{-t}\frac{1}{2}\delta (t-1)dt=\frac{1}{2}e^{-1}=\frac{1}{2e}$
 Question 10
A temperature in the range of -40$^{\circ}$C to 55$^{\circ}$C is to be measured with a resolution of 0.1$^{\circ}$C. The minimum number of ADC bits required to get a matching dynamic range of the temperature sensor is
 A 8 B 10 C 12 D 14
Digital Electronics   A-D and D-A Converters
Question 10 Explanation:
Temperature range of $-40^{\circ}C \; to \; 55^{\circ}C$
So. Total range in $95^{\circ}C$
Since, resolution is $0.1^{\circ}C$
So, number of steps will be 950
To have 950 steps, we need at least 10 bits.
There are 10 questions to complete.