GATE ECE 2010 | State Space Analysis Question 18 | Control Systems

GATE ECE 2010

MCQ (Single Correct Answer)

-0.6

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

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

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

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

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

GATE ECE 2007

MCQ (Single Correct Answer)

-0.6

The state space representation of a separately excited DC servo motor dynamics is given as $$$\left[ {\matrix{ {{{d\omega } \over {dt}}} \cr {{{d{i_a}} \over {dt}}} \cr } } \right] = \left[ {\matrix{ { - 1} & 1 \cr { - 1} & { - 10} \cr } } \right]\left[ {\matrix{ \omega \cr {{i_a}} \cr } } \right] + \left[ {\matrix{ 0 \cr {10} \cr } } \right]u.$$$

Where 'ω' is the speed of the motor, 'i_a' is the armature current and u is the armature voltage. The transfer function $${{\omega \left( s \right)} \over {U\left( s \right)}}$$ of the motor is

$${{10} \over {{s^2} + 11s + 11}}$$

$${1 \over {{s^2} + 11s + 11}}$$

$${{10s + 10} \over {{s^2} + 11s + 11}}$$

$${1 \over {{s^2} + s + 11}}$$

GATE ECE 2007

MCQ (Single Correct Answer)

-0.6

Consider a linear system whose state space Representation is $$\mathop x\limits^ \bullet \left( t \right) = AX\left( t \right).$$
If the initial state vector of the system is $$x\left( 0 \right) = \left[ {\matrix{ 1 \cr { - 2} \cr } } \right],$$
then the system response is $$x\left( t \right) = \left[ {\matrix{ {{e^{ - 2t}}} \cr { - 2{e^{ - 2t}}} \cr } } \right].$$
If the initial state vector of the system changes to $$x\left( 0 \right) = \left[ {\matrix{ 1 \cr { - 1} \cr } } \right],$$
then the system response becomes $$x\left( t \right) = \left[ {\matrix{ {{e^{ - t}}} \cr { - {e^{ - t}}} \cr } } \right].$$

The system matrix a is

$$\left[ {\matrix{ 0 & 1 \cr { - 1} & 1 \cr } } \right]$$

$$\left[ {\matrix{ 1 & 1 \cr { - 1} & { - 2} \cr } } \right]$$

$$\left[ {\matrix{ 2 & 1 \cr { - 1} & { - 1} \cr } } \right]$$

$$\left[ {\matrix{ 0 & 1 \cr { - 2} & { - 3} \cr } } \right]$$

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