GATE EE 2016 SET 2


Question 1
The output expression for the Karnaugh map shown below is
A
A+\bar{B}
B
A+\bar{C}
C
\bar{A}+\bar{C}
D
\bar{A}+C
Digital Electronics   Boolean Algebra and Minimization
Question 1 Explanation: 


F=A+\bar{C}
Question 2
The circuit shown below is an example of a
A
low pass filter.
B
band pass filter
C
high pass filter.
D
notch filter.
Analog Electronics   Operational Amplifiers
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 __________.
A
100
B
10
C
20
D
50
Electric Circuits   Magnetically Coupled Circuits, Network Topology and Filters
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.
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
A
9e^{-\frac{t}{3}}u(t)
B
9e^{-\frac{t}{6}}u(t)
C
9e^{-\frac{t}{3}}u(t)-6e^{-\frac{t}{6}}u(t)
D
54e^{-\frac{t}{6}}u(t)-54e^{-\frac{t}{3}}u(t)
Signals and Systems   Linear Time Invariant Systems
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}
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 ________.
A
5
B
6
C
7
D
8
Signals and Systems   Fourier Transform
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.