System Modelling and Feedback - Control Systems - Electrical Engineering Previous Year Questions

1

2007 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2007

The system shown in figure below can be reduced to the form

with

A

X = c₀s + c₁, Y = 1/(s² + a₀s + a₁), Z = b₀s + b₁

B

X = 1, Y = (c₀s + c₁)/(s² + a₀s + a₁), Z = b₀s + b₁

C

X = c₀s + c₁, Y = (b₀s + b₁)/(s² + a₀s + a₁), Z = 1

D

X = c₀s + c₁, Y = 1/(s² + a₀s + a₁), Z = b₀s + b₀

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2

2007 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2007

Consider the feedback control system shown below which is subjected to a unit step input. The system is stable and has the following parameters k_p = 4, k_I = 10, ω = 500 and ζ = 0.7.

The steady state value of z is

A

1

B

0.25

C

0.1

D

0

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3

2008 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2008

The transfer functions of two compensators are given below:
C₁ = 10(s+1)/(s+10), C₂ = s+10/[10(s+1)]
Which one of the following statements is correct?

A

C₁ is a lead compensator and C₂ is a lag compensator

B

C₁ is a lag compensator and C₂ is a lead compensator

C

Both C₁ and C₂ are lead compensators

D

Both C₁ and C₂ are lag compensators

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4

2008 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2008

The transfer function G(s) of this system will be

A

s/((s+2))

B

(s+1)/(s(s-2))

C

s/(s(s-2))

D

1/(s(s+2))

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5

2008 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2008

A unity feedback is provided to the above system G(s) to make it a closed loop system as shown in figure.
For a unit step input r(t), the steady state error in the output will be

A

0

B

1

C

2

D

∞

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6

2010 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2010

As shown in the figure, a negative feedback system has an amplifier of gain 100 with ±10% tolerance in the forward path, and an attenuator of value 9/100 in the feedback path. The overall system gain is approximately:

A

10±1%

B

10±2%

C

10±5%

D

10±10%

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7

2010 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2010

If the electrical circuit of figure (b) is an equivalent of the coupled tank system of figure (a), then

A

A, B are resistances and C, D capacitances

B

A, C are resistances and B, D capacitances

C

A, B are capacitances and C, D resistances

D

A, C are capacitances and B, D resistances

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8

2011 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2011

An open loop system represented by the transfer function \(G(s) = \frac{(s-1)}{(s+2)(s+3)}\) is

A

stable and of the minimum phase type

B

stable and of the non-minimum phase type

C

unstable and of the minimum phase type

D

unstable and of the non-minimum phase type

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9

2011 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2011

A two-loop position control system is shown below. The gain \(k\) of the Tacho-generator influences mainly the

A

peak overshoot

B

natural frequency of oscillation

C

phase shift of the closed loop transfer function at very low frequencies (\(\omega \to 0\))

D

phase shift of the closed loop transfer function at very high frequencies (\(\omega \to \infty\))

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10

2013 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2013 [Session 1]

Assuming zero initial condition, the response y(t) of the system given below to a unit step input u(t) is

A

u(t)

B

t u(t)

C

\(\frac{t^2}{2} u(t)\)

D

\(e^{-t} u(t)\)

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11

2013 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2013 [Session 1]

The open-loop transfer function of a dc motor is given as \(\frac{\omega(s)}{V_a(s)} = \frac{10}{1+10s}\). When connected in feedback as shown below, the approximate value of K_a that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is

A

1

B

5

C

10

D

100

Check answer

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12

2013 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2013 [Session 1]

The signal flow graph for a system is given below. The transfer function \( \frac{Y(s)}{U(s)} \) for this system is

A

\(\frac{s+1}{5s^2+6s+2} \)

B

\(\frac{s+1}{s^2+6s+2} \)

C

\(\frac{s+1}{s^2+4s+2} \)

D

\(\frac{1}{5s^2+6s+2} \)

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13

2013 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2013 [Session 2]

In the feedback network shown below, if the feedback factor k is increased, then the

A

input impedance increases and output impedance decreases.

B

input impedance increases and output impedance also increases.

C

input impedance decreases and output impedance also decreases.

D

input impedance decreases and output impedance increases.

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14

2013 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2013 [Session 2]

The open-loop transfer function of a dc motor is given as \( \frac{\omega(s)}{V_a(s)} = \frac{10}{1+10s} \). When connected in feedback as shown below, the approximate value of \( K_a \) that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is

A

1

B

5

C

10

D

100

Check answer

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15

2013 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2013 [Session 3]

Assuming zero initial condition, the response \(y(t)\) of the system given below to a unit step input \(u(t)\) is

A

\(u(t)\)

B

\(t u(t)\)

C

\(\frac{t^2}{2} u(t)\)

D

\(e^{-t} u(t)\)

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16

2013 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2013 [Session 3]

The open-loop transfer function of a dc motor is given as \(\frac{\omega(s)}{V_a(s)}=\frac{10}{1+10s}\) . When connected in feedback as shown below, the approximate value of \(K_a\) that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is

A

1

B

5

C

10

D

100

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17

2013 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2013 [Session 3]

The signal flow graph for a system is given below. The transfer function \(\frac{Y(s)}{U(s)}\) for this system is

A

\(\frac{s+1}{5s^2+6s+2}\)

B

\(\frac{s+1}{s^2+6s+2}\)

C

\(\frac{s+1}{s^2+4s+2}\)

D

\(\frac{1}{5s^2+6s+2}\)

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18

2013 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2013 [Session 4]

The open-loop transfer function of a dc motor is given as \(\frac{\omega(s)}{V_a(s)} = \frac{10}{1+10s}\). When connected in feedback as shown below, the approximate value of \(K_a\) that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is

A

1

B

5

C

10

D

100

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19

2014 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2014 [Session 3]

The signal flow graph of a system is shown below. U(s) is the input and C(s) is the output.
Assuming, \( h_1 = b_1 \) and \( h_0 = b_0 - b_1 a_1 \), the input-output transfer function, \( G(s) = \frac{C(s)}{U(s)} \) of the system is given by

A

\(G(s) = \frac{b_0 s + b_1}{s^2 + a_0 s + a_1} \)

B

\(G(s) = \frac{a_1 s + a_0}{s^2 + b_1 s + b_0} \)

C

\(G(s) = \frac{b_1 s + b_0}{s^2 + a_1 s + a_0} \)

D

\(G(s) = \frac{a_0 s + a_1}{s^2 + b_0 s + b_1} \)

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20

2016 · Electrical Engineering · Control Systems · System Modelling and Feedback

Electrical Engineering (EE) 2016 [Session 2]

A second-order real system has the following properties:
a) the damping ratio \(\zeta = 0.5\) and undamped natural frequency \(\omega_n = 10\) rad/s,
b) the steady state value of the output, to a unit step input, is 1.02.
The transfer function of the system is

A

\(\frac{1.02}{s^2+5s+100}\)

B

\(\frac{102}{s^2+10s+100}\)

C

\(\frac{100}{s^2+10s+100}\)

D

\(\frac{102}{s^2+5s+100}\)

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