GATE ECE 2014 Set 2 | State Space Analysis Question 9 | Control Systems

GATE ECE 2014 Set 2

MCQ (Single Correct Answer)

-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

$${x_1}\left( t \right) = - 1,{x_2}\left( t \right) = 2$$

$${x_1}\left( t \right) = - {e^{ - t}},{x_2}\left( t \right) = 2{e^{ - t}}$$

$${x_1}\left( t \right) = {e^{ - t}},{x_2}\left( t \right) = - {e^{ - 2t}}$$

$${x_1}\left( t \right) = {e^{ - t}},{x_2}\left( t \right) = - 2{e^{ - t}}$$

GATE ECE 2014 Set 2

MCQ (Single Correct Answer)

-0.6

Consider the state space system expressed by the signal flow diagram shown in the figure. GATE ECE 2014 Set 2 Control Systems - State Space Analysis Question 16 English

GATE ECE 2014 Set 2 Control Systems - State Space Analysis Question 16 English

The corresponding system is

always controllable

always observable

always stable

always unstable

GATE ECE 2014 Set 1

MCQ (Single Correct Answer)

-0.6

Consider the state space model of a system, as given below GATE ECE 2014 Set 1 Control Systems - State Space Analysis Question 18 English

The system is

controllable and observable.

uncontrollable and observable.

uncontrollable and unobservable.

controllable and unobservable.

GATE ECE 2013

MCQ (Single Correct Answer)

-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

GATE ECE 2013 Control Systems - State Space Analysis Question 20 English

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

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

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

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

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