1
GATE ECE 1997
Subjective
+5
-0
Following fig. shows the block diagram representation of control system. The system in block A has an impulse response h(t ) = e−t u(t ). The system in block B has an impulse response h(t ) = e−2t u(t ). The block 'K' amplifies its inputs by a factor k. For the overall system with input x(t) and output y(t). GATE ECE 1997 Control Systems - Time Response Analysis Question 7 English (a) Find the transfer function $$\frac{y\left(s\right)}{x\left(s\right)}$$ when k =1.
(b) Find the impulse response when k = 0.
(c) Find the values of k for which the system becomes unstable.
2
GATE ECE 1997
MCQ (Single Correct Answer)
+2
-0.6
A certain linear time invariant system has the state and the output equations given below $$$\left[ {\matrix{ {\mathop {{x_1}}\limits^ \bullet } \cr {\mathop {{x_2}}\limits^ \bullet } \cr } } \right] = \left[ {\matrix{ 1 & { - 1} \cr 0 & 1 \cr } } \right]\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr } } \right] + \left[ {\matrix{ 0 \cr 1 \cr } } \right]u$$$ $$$y = \left[ {\matrix{ 1 & 1 \cr } } \right]\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr } } \right], if$$$ $${x_1}\left( 0 \right) =1 ,{x_2}\left( 0 \right) = - 1,$$ $$u\left( 0 \right) = 0,$$ then $${{dy} \over {dt}}{|_{t = 0}}$$ is
A
1
B
-1
C
0
D
None of the above
3
GATE ECE 1997
Subjective
+5
-0
GATE ECE 1997 Control Systems - State Space Analysis Question 8 English

For the circuit shown in the figure, choose state variables as $${x_{1,}}{x_{2,}}{x_3}$$ to be $${i_{L1}}\left( t \right),{v_{c2}}\left( t \right),{i_{L3}}\left( t \right)$$

Wriote the state equations

$$$\left[ {\matrix{ {\mathop {{x_1}}\limits^ \bullet } \cr {\mathop {{x_2}}\limits^ \bullet } \cr {\mathop {{x_3}}\limits^ \bullet } \cr } } \right] = A\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] + B\left[ {e\left( t \right)} \right]$$$
4
GATE ECE 1997
MCQ (Single Correct Answer)
+1
-0.3
A signed integer has been stored in a byte using the 2's complement format. We wish to store the same integer in a 16-bit word. We should
A
copy the original byte to the less significant byte of the word and fill the more significant with zeros
B
copy the original byte to the more significant byte of the word and fill the less significant with zeros
C
copy the original byte to the less significant byte of the word and make each bit of the more significant byte equal to the most significant bit of the original byte
D
copy the original byte to the less significant byte as well as the more significant byte of the word
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