# 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.