1
GATE EE 2008
MCQ (Single Correct Answer)
+1
-0.3
The equivalent circuits of a diode, during forward and reverse biased conditions are shown in figure.
(a) GATE EE 2008 Analog Electronics - Diode Circuits and Applications Question 12 English 1
(b) GATE EE 2008 Analog Electronics - Diode Circuits and Applications Question 12 English 2

If such diodes are used in the clipper circuit of figure given above, the output voltage (V0) of the circuit will be

A
GATE EE 2008 Analog Electronics - Diode Circuits and Applications Question 12 English Option 1
B
GATE EE 2008 Analog Electronics - Diode Circuits and Applications Question 12 English Option 2
C
GATE EE 2008 Analog Electronics - Diode Circuits and Applications Question 12 English Option 3
D
GATE EE 2008 Analog Electronics - Diode Circuits and Applications Question 12 English Option 4
2
GATE EE 2008
MCQ (Single Correct Answer)
+2
-0.6
The transfer function of a linear time invariant system is given as $$G\left( s \right) = {1 \over {{s^2} + 3s + 2}}$$

The steady state value of the output of the system for a unit impulse input applied at time instant $$t=1$$ will be

A
$$0$$
B
$$0.5$$
C
$$1$$
D
$$2$$
3
GATE EE 2008
MCQ (Single Correct Answer)
+2
-0.6
The state space equation of a system is described by $$\mathop X\limits^ \bullet = AX + BU,\,\,Y = Cx$$ where $$X$$ is state vector, $$U$$ is input, $$Y$$ is output and $$$A = \left( {\matrix{ 0 & 1 \cr 0 & { - 2} \cr } } \right)\,\,B = \left( {\matrix{ 0 \cr 1 \cr } } \right)\,\,C = \left[ {\matrix{ 1 & 0 \cr } } \right]$$$

A unity feedback is provided to the above system $$G(s)$$ to make it a closed loop system as shown in figure.

GATE EE 2008 Control Systems - State Variable Analysis Question 20 English

For a unit step input $$r(t),$$ the steady state error in the input will be

A
$$0$$
B
$$1$$
C
$$2$$
D
$$\infty $$
4
GATE EE 2008
MCQ (Single Correct Answer)
+2
-0.6
The state space equation of a system is described by $$\mathop X\limits^ \bullet = AX + BU,\,\,Y = Cx$$ where $$X$$ is state vector, $$U$$ is input, $$Y$$ is output and $$$A = \left( {\matrix{ 0 & 1 \cr 0 & { - 2} \cr } } \right)\,\,B = \left( {\matrix{ 0 \cr 1 \cr } } \right)\,\,C = \left[ {\matrix{ 1 & 0 \cr } } \right]$$$

The transfer function $$G(s)$$ of this system will be

A
$${s \over {\left( {s + 2} \right)}}$$
B
$${{s + 1} \over {s\left( {s - 2} \right)}}$$
C
$${s \over {\left( {s - 2} \right)}}$$
D
$${1 \over {s\left( {s + 2} \right)}}$$
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