GATE ECE 2011 | State Space Analysis Question 16 | Control Systems

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

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

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

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

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

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 24 English

The transfer function of the system is

$${{s + 1} \over {{s^2} + 1}}$$

$${{s - 1} \over {{s^2} + 1}}$$

$${{s + 1} \over {{s^2} + s + 1}}$$

$${{s - 1} \over {{s^2} + s + 1}}$$

GATE ECE 2008

MCQ (Single Correct Answer)

-0.6

A signal flow graph of a system is given below. GATE ECE 2008 Control Systems - State Space Analysis Question 26 English

The set of equations that correspond to this signal flow graph is

$${d \over {dt}}\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] = \left[ {\matrix{ \beta & { - \gamma } & 0 \cr \gamma & \alpha & 0 \cr { - \alpha } & { - \beta } & 0 \cr } } \right]\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] + \left[ {\matrix{ 1 & 0 \cr 0 & 0 \cr 0 & 1 \cr } } \right]\left( {\matrix{ {{u_1}} \cr {{u_2}} \cr } } \right)$$

$${d \over {dt}}\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] = \left[ {\matrix{ 0 & \alpha & \gamma \cr 0 & { - \alpha } & { - \gamma } \cr 0 & \beta & { - \beta } \cr } } \right]\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] + \left[ {\matrix{ 0 & 0 \cr 0 & 1 \cr 1 & 0 \cr } } \right]\left( {\matrix{ {{u_1}} \cr {{u_2}} \cr } } \right)$$

$${d \over {dt}}\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] = \left[ {\matrix{ { - \alpha } & \beta & 0 \cr { - \beta } & { - \gamma } & 0 \cr \alpha & \gamma & 0 \cr } } \right]\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] + \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr 0 & 0 \cr } } \right]\left( {\matrix{ {{u_1}} \cr {{u_2}} \cr } } \right)$$

$${d \over {dt}}\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] = \left[ {\matrix{ { - \gamma } & 0 & \beta \cr \gamma & 0 & \alpha \cr { - \beta } & 0 & { - \alpha } \cr } } \right]\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] + \left[ {\matrix{ 0 & 1 \cr 0 & 0 \cr 1 & 0 \cr } } \right]\left( {\matrix{ {{u_1}} \cr {{u_2}} \cr } } \right)$$

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