GATE ECE 2013 | State Space Analysis Question 14 | Control Systems

GATE ECE 2013

MCQ (Single Correct Answer)

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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

GATE ECE 2013 Control Systems - State Space Analysis Question 21 English

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

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

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

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

GATE ECE 2012

MCQ (Single Correct Answer)

-0.6

The state variable description of an LTI system is given by GATE ECE 2012 Control Systems - State Space Analysis Question 22 English

where y is the output and u is input. The system is controllable for

$${a_1} \ne 0,{a_2} = 0,{a_3} \ne 0$$

$${a_1} = 0,{a_2} \ne 0,{a_3} \ne 0$$

$${a_1} = 0,{a_2} \ne 0,{a_3} = 0$$

$${a_1} \ne 0,{a_2} \ne 0,{a_3} = 0$$

GATE ECE 2011

MCQ (Single Correct Answer)

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The block diagram of a system with one input u and two outputs y₁ and y₂ is given below GATE ECE 2011 Control Systems - State Space Analysis Question 23 English

GATE ECE 2011 Control Systems - State Space Analysis Question 23 English

A state space model of the above system in terms of the state vector $$\underline x $$ and the output vector $$\underline y = {\left[ {\matrix{ {{y_1}} & {{y_2}} \cr } } \right]^\tau }$$ is

$$\mathop {\underline x }\limits^ \bullet = \left[ 2 \right]\underline x + \left[ 1 \right]u;\underline y = \left[ {\matrix{ 1 & 2 \cr } } \right]x$$

$$\mathop {\underline x }\limits^ \bullet = \left[ { - 2} \right]\underline x + \left[ 1 \right]u;\underline y = \left[ {\matrix{ 1 \cr 2 \cr } } \right]x$$

$$\mathop {\underline x }\limits^ \bullet = \left[ {\matrix{ { - 2} & 0 \cr 0 & { - 2} \cr } } \right]\underline x + \left[ {\matrix{ 1 \cr 1 \cr } } \right]u;\underline y = \left[ {\matrix{ 1 & 2 \cr } } \right]x$$

$$\mathop {\underline x }\limits^ \bullet = \left[ {\matrix{ 2 & 0 \cr 0 & 2 \cr } } \right]\underline x + \left[ {\matrix{ 1 \cr 1 \cr } } \right]u;\underline y = \left[ {\matrix{ 1 \cr 2 \cr } } \right]x$$

GATE ECE 2010

MCQ (Single Correct Answer)

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The signal flow graph of a system is shown below. GATE ECE 2010 Control Systems - State Space Analysis Question 25 English

The state variable representation of the system can be

$$\mathop x\limits^ \bullet = \left[ {\matrix{ 1 & 1 \cr { - 1} & 0 \cr } } \right]x + \left[ {\matrix{ 0 \cr 2 \cr } } \right]u$$
$$y = \left[ {\matrix{ 0 & {0.5} \cr } } \right]x$$

$$\eqalign{ & \mathop x\limits^ \bullet = \left[ {\matrix{ { - 1} & 1 \cr { - 1} & 0 \cr } } \right]x + \left[ {\matrix{ 0 \cr 2 \cr } } \right]u \cr & y = \left[ {\matrix{ 0 & {0.5} \cr } } \right]x \cr} $$

$$\eqalign{ & \mathop x\limits^ \bullet = \left[ {\matrix{ 1 & 1 \cr { - 1} & 0 \cr } } \right]x + \left[ {\matrix{ 0 \cr 2 \cr } } \right]u \cr & y = \left[ {\matrix{ {0.5} & {0.5} \cr } } \right]x \cr} $$

$$\eqalign{ & \mathop x\limits^ \bullet = \left[ {\matrix{ { - 1} & 1 \cr { - 1} & 0 \cr } } \right]x + \left[ {\matrix{ 0 \cr 2 \cr } } \right]u \cr & y = \left[ {\matrix{ {0.5} & {0.5} \cr } } \right]x \cr} $$

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