Question 1 |
The output expression for the Karnaugh map shown below is
A+\bar{B} | |
A+\bar{C} | |
\bar{A}+\bar{C} | |
\bar{A}+C |
Question 1 Explanation:
F=A+\bar{C}
Question 2 |
The circuit shown below is an example of a
low pass filter. | |
band pass filter | |
high pass filter. | |
notch filter. |
Question 2 Explanation:
\frac{V_{out}}{V_{in}}=-\left [\frac{\frac{ R_2\cdot \frac{1}{j\omega C}}{R_2+\frac{1}{j\omega C}}}{R_1} \right ]
\frac{V_{out}}{V_{in}}=-\left [ \frac{R_2}{R_1(R_2j\omega C+1)} \right ]
So the system is a low pass filter.
Question 3 |
The following figure shows the connection of an ideal transformer with primary to secondary turns ratio of 1:100. The applied primary voltage is 100 V (rms), 50 Hz, AC. The rms value of the current I, in ampere, is __________.
100 | |
10 | |
20 | |
50 |
Question 3 Explanation:
The above circuit can be drown by transferring secondary circuit to primary side.
I=\frac{100V}{(8+10j-4j)\Omega }
\;\;=\frac{100V}{(8+6j)\Omega }
So the rms value of I will be 10 A.
I=\frac{100V}{(8+10j-4j)\Omega }
\;\;=\frac{100V}{(8+6j)\Omega }
So the rms value of I will be 10 A.
Question 4 |
Consider a causal LTI system characterized by differential equation \frac{dy(t)}{dt}+\frac{1}{6}y(t)=3x(t). The response of the system to the input x(t)=3e^{-\frac{t}{3}}u(t). where u(t) denotes the unit step function, is
9e^{-\frac{t}{3}}u(t) | |
9e^{-\frac{t}{6}}u(t) | |
9e^{-\frac{t}{3}}u(t)-6e^{-\frac{t}{6}}u(t) | |
54e^{-\frac{t}{6}}u(t)-54e^{-\frac{t}{3}}u(t) |
Question 4 Explanation:
The differential equation
\begin{aligned} \frac{dy(t)}{dt} &+\frac{1}{6}y(t)=3x(t) \\ \text{So, }sY(s)&+\frac{1}{6}Y(s) =3X(s) \\ Y(s) &=\frac{3X(s)}{\left ( s+\frac{1}{6} \right )} \\ X(s) &=\frac{9}{\left ( s+\frac{1}{3} \right )} \\ \text{So, } Y(s)&=\frac{9}{\left ( s+\frac{1}{3} \right )\left ( s+\frac{1}{6} \right )} \\ &=\frac{54}{\left ( s+\frac{1}{6} \right )} -\frac{54}{\left ( s+\frac{1}{3} \right )}\\ \text{So, }y(t) &= (54e^{-1/6t}-54e^{-1/3t})u(t) \end{aligned}
\begin{aligned} \frac{dy(t)}{dt} &+\frac{1}{6}y(t)=3x(t) \\ \text{So, }sY(s)&+\frac{1}{6}Y(s) =3X(s) \\ Y(s) &=\frac{3X(s)}{\left ( s+\frac{1}{6} \right )} \\ X(s) &=\frac{9}{\left ( s+\frac{1}{3} \right )} \\ \text{So, } Y(s)&=\frac{9}{\left ( s+\frac{1}{3} \right )\left ( s+\frac{1}{6} \right )} \\ &=\frac{54}{\left ( s+\frac{1}{6} \right )} -\frac{54}{\left ( s+\frac{1}{3} \right )}\\ \text{So, }y(t) &= (54e^{-1/6t}-54e^{-1/3t})u(t) \end{aligned}
Question 5 |
Suppose the maximum frequency in a band-limited signal x(t) is 5 kHz. Then, the maximum frequency in x(t)\cos (2000\pi t), in kHz, is ________.
5 | |
6 | |
7 | |
8 |
Question 5 Explanation:
Maximum possible frequency of x(t)(2000 \pi t)=f_1+f_2=5+1=6kHz
Question 6 |
Consider the function f(z)=z+z^* where z is a complex variable and z^* denotes its complex conjugate. Which one of the following is TRUE?
f(z) is both continuous and analytic | |
f(z) is continuous but not analytic | |
f(z) is not continuous but is analytic | |
f(z) is neither continuous nor analytic |
Question 6 Explanation:
\begin{aligned} f(z) &=z+z^* \\ f(z)&=2x \text{ is continuous (polynomial)} \\ u &=2x, v=0 \\ u_x &=2, u_y=0 \\ v_x&=0, v_y=0 \end{aligned}
C.R. equation not satisfied.
Therefore, no where analytic.
C.R. equation not satisfied.
Therefore, no where analytic.
Question 7 |
A 3 x 3 matrix P is such that, P^{3} = P. Then the eigenvalues of P are
1, 1, -1 | |
1, 0.5 + j0.866, 0.5 - j0.866 | |
1, -0.5 + j0.866, -0.5 - j0.866 | |
0, 1, -1 |
Question 7 Explanation:
By Calyey Hamilton theorem,
\lambda ^3=\lambda
\lambda =0, 1, -1
\lambda ^3=\lambda
\lambda =0, 1, -1
Question 8 |
The solution of the differential equation, for t \gt 0, y''(t)+2y'(t)+y(t)=0 with initial conditions y(0) = 0 and y'(0) = 1, is (u(t) denotes the unit step function),
te^{-t}u(t) | |
(e^{-t}-te^{-t})u(t) | |
(-e^{-t}+te^{-t})u(t) | |
e^{-t}u(t) |
Question 8 Explanation:
The differentail equation is
y'(t)+2y'(t)+y(t)=0
So, (s^2y(s)-sy(0)-y'(0))+2[sy(s)-y(0)]+y(s) =0
So, y(s)=\frac{sy(0)+y'(0)+2y(0)}{s^2+2s+1}
Given that y'(0)=1, y(0)=0
So, y(s)=\frac{1}{(s+1)^2}
So, y(t)=te^{-t}u(t)
y'(t)+2y'(t)+y(t)=0
So, (s^2y(s)-sy(0)-y'(0))+2[sy(s)-y(0)]+y(s) =0
So, y(s)=\frac{sy(0)+y'(0)+2y(0)}{s^2+2s+1}
Given that y'(0)=1, y(0)=0
So, y(s)=\frac{1}{(s+1)^2}
So, y(t)=te^{-t}u(t)
Question 9 |
The value of the line integral
\int_{c}(2xy^{2}dx+2x^{2}ydy+dz)
along a path joining the origin (0,0,0) and the point (1,1,1) is
\int_{c}(2xy^{2}dx+2x^{2}ydy+dz)
along a path joining the origin (0,0,0) and the point (1,1,1) is
0 | |
2 | |
4 | |
6 |
Question 9 Explanation:
\int _C\bar{F}\cdot \bar{dr}
where,
\bar{F}=xy^2\bar{i}+2x^2y\bar{j}+\bar{k}
\bigtriangledown \times \vec{F}=\vec{O}
(\vec{F} is irrotational \Rightarrow \vec{F} is conservative)
\begin{aligned} \vec{F}&=\bigtriangledown \phi \\ \phi _x&=2xy^2 \\ \phi _y&=2x^2y \\ \phi _z&= 1\\ \Rightarrow \; \phi &=x^2y^2+z+C \end{aligned}
where, \vec{F} is conservative
\int _C \bar{F}\bar{dr}=\int_{(0,0,0)}^{(1,1,1)}d\phi =[x^2y^2+z]_{(0,0,0)}^{(1,1,1,)}=2
where,
\bar{F}=xy^2\bar{i}+2x^2y\bar{j}+\bar{k}
\bigtriangledown \times \vec{F}=\vec{O}
(\vec{F} is irrotational \Rightarrow \vec{F} is conservative)
\begin{aligned} \vec{F}&=\bigtriangledown \phi \\ \phi _x&=2xy^2 \\ \phi _y&=2x^2y \\ \phi _z&= 1\\ \Rightarrow \; \phi &=x^2y^2+z+C \end{aligned}
where, \vec{F} is conservative
\int _C \bar{F}\bar{dr}=\int_{(0,0,0)}^{(1,1,1)}d\phi =[x^2y^2+z]_{(0,0,0)}^{(1,1,1,)}=2
Question 10 |
Let f(x) be a real, periodic function satisfying f(-x)=-f(x). The general form of its Fourier series representation would be
f(x)=a_{0}+\sum_{k=1}^{\infty }a_{k}cos(kx) | |
f(x)=\sum_{k=1}^{\infty }b_{k}sin(kx) | |
f(x)=a_{0}+\sum_{k=1}^{\infty }a_{2k}cos(kx) | |
f(x)=\sum_{k=0}^{\infty }a_{2k+1}sin(2k+1)x |
Question 10 Explanation:
Given that,
f(-x)=-f(x)
So, function is an odd function.
So, the fourier series will have sine term only. So,
f(x)=\sum_{k=1}^{\infty }b_x \sin (kx)
f(-x)=-f(x)
So, function is an odd function.
So, the fourier series will have sine term only. So,
f(x)=\sum_{k=1}^{\infty }b_x \sin (kx)
There are 10 questions to complete.