GATE EC 2001

Question 1
The voltage e_o in figure is
A
2V
B
\frac{4}{3}V
C
4V
D
8V
Network Theory   Basics of Network Analysis
Question 1 Explanation: 
Applying KCL,
\begin{aligned} \frac{e_{0}-12}{4}+\frac{e_{0}}{4}+\frac{e_{0}}{4}&=0 \\ \Rightarrow \quad 3 e_{0}&=12 \\ \therefore \quad e_{0}&=4 \mathrm{V} \end{aligned}
Question 2
If each branch of a Delta circuit has impedance \sqrt{3}Z, then each branch of the equivalent Wye circuit has impedance.
A
\frac{Z}{\sqrt{3}}
B
3Z
C
3\sqrt{3}Z
D
\frac{Z}{3}
Network Theory   Basics of Network Analysis
Question 2 Explanation: 
\begin{aligned} Z_{\Delta} &=3 Z_{Y} \\ \Rightarrow \quad \sqrt{3} Z_{\Delta} &=3 Z_{Y} \\ Z_{Y} &=\frac{Z_{\Delta}}{\sqrt{3}} \end{aligned}
Question 3
The transfer function of a system is given by H(s)=\frac{1}{s^2(s-2)}. The impulse response of the system is: (* denotes convolution, and U(t) is unit step function)
A
(t^2 * e^{-2t})U(t)
B
(t * e^{2t})U(t)
C
(te^{-2t})U(t)
D
(te^{-2}t)U(t)
Signals and Systems   Laplace Transform
Question 3 Explanation: 
Impulse response of system is L^{-1}[H(s)]
\frac{1}{s^{2}(s-2)}=\frac{1}{s^{2}} \times \frac{1}{s-2}=\left(t * e^{+2 t}\right) u(t)
Question 4
The admittance parameter Y_{12} in the 2-port network in figure is
A
0.2 mho
B
0.1 mho
C
-0.05 mho
D
0.05 mho
Network Theory   Two Port Networks
Question 4 Explanation: 
\left[\begin{array}{cc} y_{1}+y_{3} & -y_{3} \\ -y_{3} & y_{2}+y_{3} \end{array}\right]=\left[\begin{array}{cc} y_{11} & y_{12} \\ y_{21} & y_{22} \end{array}\right]
y_{12}=-y_{3}

y_{12}=-\frac{1}{20}=-0.05 \mathrm{mho}
Question 5
The region of convergence of the z-transform of a unit step function is
A
\mid z\mid \gt 1
B
\mid z\mid \lt 1
C
(real part of Z) \gt 0
D
(real part of Z) \lt 0
Signals and Systems   Z-Transform
Question 5 Explanation: 
\begin{aligned} h(n)&=u(n) \\ H(z)&=\sum_{n=0}^{\infty} 1 . z^{-n} \end{aligned}
For the convergence of H(z)
\sum_{n=0}^{n}\left(z^{-1}\right)^{n}\lt \infty
\therefore ROC is the range of values of z for which
\left|z^{-1}\right| \lt 1\text{ or } |z|\gt |1|
Question 6
The current gain of a BJT is
A
g_{m}r_{0}
B
\frac{g_{m}}{r_{o}}
C
g_{m}r_{\pi }
D
\frac{g_{m}}{r_{\pi }}
Analog Circuits   BJT Analysis
Question 6 Explanation: 
h_{f e}=g_{m} \cdot r_{\pi}
Question 7
MOSFET can be used as a
A
current controlled capacitor
B
voltage controlled capacitor
C
current controlled inductor
D
voltage controlled inductor
Electronic Devices   BJT and FET Basics
Question 7 Explanation: 
Voltage controlled capacitor.
Question 8
The effective channel length of a MOSFET in saturation decreases with increase in
A
gate voltage
B
drain voltage
C
source voltage
D
body voltage
Electronic Devices   BJT and FET Basics
Question 8 Explanation: 
At the edge of saturation i.e. when drain to source voltage reaches V_{D_{sat}} the inversion layer charge at the drain end becomes zero (ideally). The channel is said to be pinched off at the drain end. If the drain to source voltage V_{DS} is increased even further beyond the saturation edge so that V_{DS} \gt V_{D_{sat}}, an even larger portion of the channel becomes pinched off and effective channel length is reduced.
Question 9
The ideal OP-AMP has the following characteristics.
A
R_{i}=\infty , A=\infty , R_{0}=0
B
R_{i}=0, A=\infty , R_{0}=0
C
R_{i}=\infty , A=\infty , R_{0}=\infty
D
R_{i}=0 , A=\infty , R_{0}=\infty
Analog Circuits   Operational Amplifiers
Question 10
The 2's complement representation of -17 is
A
101110
B
101111
C
111110
D
110001
Digital Circuits   Number Systems
Question 10 Explanation: 
\begin{aligned} 17 &=010001 \\ -17 &=101111(2 \text { 's complement }) \end{aligned}
There are 10 questions to complete.
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