Compensator Design Questions and Answers

Q: Given below are two statements :Statement I : The phase lag network allow to pass high frequencies and low frequencies are attenuated.Statement II : The steady state accuracy can be improved by using phase lead network.In the light of the above statements, choose the most appropriate answer from the options given below.

Both Statement I and Statement II are false.Statement I is false: a phase lag network behaves as a low-pass filter. It passes low frequencies and attenuates high frequencies, which is the reverse of what the statement says.Statement II is false: steady-state accuracy is improved by the lag network, which raises the low-frequency gain and so reduces the steady-state error. A phase lead network improves the transient response, increasing bandwidth and speed of response, not the steady-state accuracy. Since both statements are wrong, every option calling either one true is ruled out.

Q: A PI controller increases :

A PI controller increases the type and order of the system by 1.A PI controller has the transfer function\[ \begin{aligned} G_{c}(s) &= K_{p} + \frac{K_{i}}{s} \\ &= \frac{K_{p}s + K_{i}}{s} \end{aligned} \]It adds a pole at the origin and a zero. The extra pole at the origin raises the type of the system by 1, which is why integral action removes the steady state error to a step. The extra pole also raises the order of the characteristic equation by 1.Options naming only the type or only the order are each half of the effect, and the option saying both decrease has the effect backwards.

Q: Derivative control is desirable in a system in which :

Derivative control is desirable when the value of the controlled variable changes rapidly.Derivative action produces a controller output proportional to the rate of change of the error, \(K_{d}\,de/dt\). It anticipates where the error is heading, so it reacts strongly to fast changes, adds damping and reduces overshoot. This makes it useful in processes where the variable can change quickly.An output that settles below the set point is a steady-state offset, which needs integral action, not derivative action, since derivative does nothing to a constant error. Environmental influences and cost do not call for derivative control in particular.

Q: What is the purpose of the Smith predictor in control system design?

The Smith predictor is used to compensate for time delays in the system.A process with dead time, \(G(s)e^{-sT}\), is hard to control because the delay adds phase lag that grows with frequency and forces the controller gain down. O. J. M. Smith's scheme (1957) runs a model of the process inside the controller. The model's output without the delay is fed back in place of the real measurement, and the difference between the real output and the delayed model output corrects for model error. The controller can then be tuned as though the delay were not in the loop.Better transient response, less overshoot or improved frequency response can follow as consequences, but they are not the specific purpose. The predictor exists to deal with dead time.

Given below are two statements :

Statement I : The phase lag network allow to pass high frequencies and low frequencies are attenuated.
Statement II : The steady state accuracy can be improved by using phase lead network.

In the light of the above statements, choose the most appropriate answer from the options given below.

A.Both Statement I and Statement II are false.
B.Statement I is true but Statement II is false.
C.Statement I is false but Statement II is true.
D.Both Statement I and Statement II are true.

Exam Year(s)

RSMSSB JE - 2025, RSMSSB - 2025,

Answer: A

Both Statement I and Statement II are false.

Statement I is false: a phase lag network behaves as a low-pass filter. It passes low frequencies and attenuates high frequencies, which is the reverse of what the statement says.

Statement II is false: steady-state accuracy is improved by the lag network, which raises the low-frequency gain and so reduces the steady-state error. A phase lead network improves the transient response, increasing bandwidth and speed of response, not the steady-state accuracy. Since both statements are wrong, every option calling either one true is ruled out.

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A PI controller increases :

A.Type of system by 1
B.Order of system by 1
C.Type and order of system decreases by 1
D.Type and order of system increases by 1

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

A PI controller increases the type and order of the system by 1.

A PI controller has the transfer function

\[ \begin{aligned} G_{c}(s) &= K_{p} + \frac{K_{i}}{s} \\ &= \frac{K_{p}s + K_{i}}{s} \end{aligned} \]

It adds a pole at the origin and a zero. The extra pole at the origin raises the type of the system by 1, which is why integral action removes the steady state error to a step. The extra pole also raises the order of the characteristic equation by 1.

Options naming only the type or only the order are each half of the effect, and the option saying both decrease has the effect backwards.

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Derivative control is desirable in a system in which :

A.Output goes to steady state below the set level
B.Value of the controlled variable changes rapidly
C.Environmental variables play higher role
D.Cost effective design is necessary

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

Derivative control is desirable when the value of the controlled variable changes rapidly.

Derivative action produces a controller output proportional to the rate of change of the error, \(K_{d}\,de/dt\). It anticipates where the error is heading, so it reacts strongly to fast changes, adds damping and reduces overshoot. This makes it useful in processes where the variable can change quickly.

An output that settles below the set point is a steady-state offset, which needs integral action, not derivative action, since derivative does nothing to a constant error. Environmental influences and cost do not call for derivative control in particular.

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What is the purpose of the Smith predictor in control system design?

A.To compensate for time delays in the system
B.To enhance the transient response of the system
C.To reduce overshoot in the step response
D.To improve the frequency response of the system

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The Smith predictor is used to compensate for time delays in the system.

A process with dead time, \(G(s)e^{-sT}\), is hard to control because the delay adds phase lag that grows with frequency and forces the controller gain down. O. J. M. Smith's scheme (1957) runs a model of the process inside the controller. The model's output without the delay is fed back in place of the real measurement, and the difference between the real output and the delayed model output corrects for model error. The controller can then be tuned as though the delay were not in the loop.

Better transient response, less overshoot or improved frequency response can follow as consequences, but they are not the specific purpose. The predictor exists to deal with dead time.

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Following are the controllers :

1. On-off
2. PID
3. PI
The correct order of the decrease in error output is

A.1, 2, 3
B.1, 3, 2
C.3, 2, 1
D.3, 1, 2

Exam Year(s)

UGC NET - 2012,

Answer: B

The error is largest with the on-off controller, smaller with the PI controller, and smallest with the PID controller, so the order of decreasing error is 1, 3, 2 (option B). On-off control causes a cycling about the set point, the integral action in PI removes the steady-state offset, and the derivative action in PID also improves the response.

Options A, C and D place PID or PI ahead of the on-off controller, which would mean the more refined controller has the larger error.

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Which control system has hysteresis property ?

A.ON-OFF controller
B.Proportional controller
C.Integral controller
D.P-I-D controller

Exam Year(s)

UGC NET - 2012,

Answer: A

An ON-OFF controller has hysteresis: the output switches on when the measured value crosses one threshold and off at a different threshold, and the gap between them, the differential gap, prevents the switch from chattering around the set point, so option A is correct. Proportional, integral and P-I-D controllers produce an output that varies smoothly with the error, its integral or its derivative, and they are not characterised by a switching gap.

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The problem, with PI controller is its response to transients, is

A.very fast
B.having a dead zone
C.non-linear
D.sluggish

Exam Year(s)

UGC NET - 2012,

Answer: D

The problem with a PI controller is that its response to transients is sluggish, option (D). The integral action builds up slowly, and it adds a pole at the origin and a phase lag, so the controller is slow to react to a sudden disturbance, although it removes the steady state error. A PI controller is not very fast, option (A), since the lag slows it. It has no dead zone, option (B), which is a nonlinearity of some actuators. It is a linear controller, so option (C) is not right. The derivative action of a PID controller is what is added to speed up the response to transients.

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