Question 1 |
The maximum value attained by the function f(x) = x(x- 1)(x - 2) in the interval [1, 2] is _____.
0 | |
1 | |
2 | |
4 |
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.
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 ____.
1 | |
2 | |
3 | |
4 |
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
\frac{5}{s^{2}-4s+29} | |
\frac{5}{s^{2}+5} | |
\frac{s-2}{s^{2}-4s+29} | |
\frac{5}{s+5} |
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}
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)=3e^{-1}, is a solution of the differential equation \frac{d^{2}y}{dt^{2}}+2\frac{dy}{dt}+y=0. Then y(2) is
5e^{-1} | |
5e^{-2} | |
7e^{-1} | |
7e^{-2} |
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}
\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
\frac{24 \pi i}{13} | |
\frac{48 \pi i}{13} | |
\frac{24}{13} | |
\frac{12}{13} |
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}
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
\frac{1}{\sqrt{2}},-45^{\circ} | |
\frac{1}{\sqrt{2}},+45^{\circ} | |
\sqrt{2},-45^{\circ} | |
\sqrt{2},+45^{\circ} |
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
\sqrt{3} | |
1/\sqrt{3} | |
3 | |
3\sqrt{3} |
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
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
y(t) = x(t) cos(t)
This system is
linear and time-invariant | |
non-linear and time-invariant | |
linear and time-varying | |
non-linear and time-varying |
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.
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
\frac{1}{2e} | |
\frac{2}{e} | |
\frac{1}{e^{2}} | |
\frac{1}{2e^{2}} |
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}
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
8 | |
10 | |
12 | |
14 |
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.
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.
