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
There are 5 questions to complete.