1
GATE ECE 2014 Set 2
MCQ (Single Correct Answer)
+2
-0.6
An unforced linear time invariant (LTI) system is represented by $$$\left[ {\matrix{ {\mathop {{x_1}}\limits^ \bullet } \cr {\mathop {{x_2}}\limits^ \bullet } \cr } } \right] = \left[ {\matrix{ { - 1} & 0 \cr 0 & { - 2} \cr } } \right]\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr } } \right].$$$

If the initial conditions are x1(0)= 1 and x2(0)=-1, the solution of the state equation is

A
$${x_1}\left( t \right) = - 1,{x_2}\left( t \right) = 2$$
B
$${x_1}\left( t \right) = - {e^{ - t}},{x_2}\left( t \right) = 2{e^{ - t}}$$
C
$${x_1}\left( t \right) = {e^{ - t}},{x_2}\left( t \right) = - {e^{ - 2t}}$$
D
$${x_1}\left( t \right) = {e^{ - t}},{x_2}\left( t \right) = - 2{e^{ - t}}$$
2
GATE ECE 2014 Set 1
MCQ (Single Correct Answer)
+2
-0.6
Consider the state space model of a system, as given below GATE ECE 2014 Set 1 Control Systems - State Space Analysis Question 19 English

The system is

A
controllable and observable.
B
uncontrollable and observable.
C
uncontrollable and unobservable.
D
controllable and unobservable.
3
GATE ECE 2013
MCQ (Single Correct Answer)
+2
-0.6
The state diagram of a system is shown below. A system is shown below. A system is described by the state variable equations GATE ECE 2013 Control Systems - State Space Analysis Question 20 English

The state transition matrix eAt of the system shown in the figure above is

A
$$\left[ {\matrix{ {{e^{ - t}}} & 0 \cr {t{e^{ - t}}} & {{e^{ - t}}} \cr } } \right]$$ v
B
$$\left[ {\matrix{ {{e^{ - t}}} & 0 \cr { - t{e^{ - t}}} & {{e^{ - t}}} \cr } } \right]$$
C
$$\left[ {\matrix{ {{e^{ - t}}} & 0 \cr {{e^{ - t}}} & {{e^{ - t}}} \cr } } \right]$$
D
$$\left[ {\matrix{ {{e^{ - t}}} & { - t{e^{ - t}}} \cr 0 & {{e^{ - t}}} \cr } } \right]$$
4
GATE ECE 2013
MCQ (Single Correct Answer)
+2
-0.6
The state diagram of a system is shown below. A system is shown below. A system is described by the state variable equations GATE ECE 2013 Control Systems - State Space Analysis Question 21 English

The state-variable equations of the system shown in the figure above are

A
$$\eqalign{ & \mathop X\limits^ \bullet = \left[ {\matrix{ { - 1} & 0 \cr 1 & { - 1} \cr } } \right]X + \left[ {\matrix{ { - 1} \cr 1 \cr } } \right]u \cr & y = \left[ {\matrix{ 1 & { - 1} \cr } } \right]X + u \cr} $$
B
$$\eqalign{ & \mathop X\limits^ \bullet = \left[ {\matrix{ { - 1} & 0 \cr { - 1} & { - 1} \cr } } \right]X + \left[ {\matrix{ { - 1} \cr 1 \cr } } \right]u \cr & y = \left[ {\matrix{ { - 1} & { - 1} \cr } } \right]X + u \cr} $$
C
$$\eqalign{ & \mathop X\limits^ \bullet = \left[ {\matrix{ { - 1} & 0 \cr { - 1} & { - 1} \cr } } \right]X + \left[ {\matrix{ { - 1} \cr 1 \cr } } \right]u \cr & y = \left[ {\matrix{ { - 1} & { - 1} \cr } } \right]X - u \cr} $$
D
$$\eqalign{ & \mathop X\limits^ \bullet = \left[ {\matrix{ { - 1} & { - 1} \cr 0 & { - 1} \cr } } \right]X + \left[ {\matrix{ { - 1} \cr 1 \cr } } \right]u \cr & y = \left[ {\matrix{ 1 & { - 1} \cr } } \right]X - u \cr} $$
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