Controllers and Compensators MCQ Quiz - Objective Question with Answer for Controllers and Compensators - Download Free PDF
Last updated on Jun 10, 2025
Latest Controllers and Compensators MCQ Objective Questions
Controllers and Compensators Question 1:
The transfer function of a phase lead compensator is found to be the form (s+z1 ) / (s + p1 ) and that of a lag compensator to be of the form (s+z2 ) / (s + p2 ). Then which of the following conditions must be satisfied?
Answer (Detailed Solution Below)
Controllers and Compensators Question 1 Detailed Solution
Explanation:
Phase Lead and Lag Compensators
The transfer functions of phase lead and lag compensators are given as:
- Phase lead compensator: (s + z1) / (s + p1)
- Phase lag compensator: (s + z2) / (s + p2)
To determine the correct option, we analyze the conditions for these compensators:
Phase Lead Compensator:
- In a phase lead compensator, the purpose is to advance the phase of the system by introducing a zero z1 and a pole p1.
- The zero z1 is designed to be larger than the pole p1 (z1 > p1).
- This arrangement ensures that the phase contribution is positive in the frequency range of interest, which improves the transient response of the system.
Phase Lag Compensator:
- In a phase lag compensator, the purpose is to delay the phase of the system by introducing a zero z2 and a pole p2.
- The pole p2 is designed to be larger than the zero z2 (z2 < p2).
- This arrangement ensures that the phase contribution is negative in the frequency range of interest, which improves the steady-state accuracy of the system.
Correct Option Analysis:
The correct option is:
Option 1: z1 > p1 and z2 < p2
This option satisfies the conditions for both the phase lead and lag compensators. In the phase lead compensator, the zero is greater than the pole (z1 > p1), and in the phase lag compensator, the zero is smaller than the pole (z2 < p2). These arrangements ensure proper phase advancement and delay for their respective compensators.
Additional Information
To further understand the analysis, let’s evaluate the other options:
Option 2: z1 > p1 and z2 > p2
This option is incorrect as it assumes that the zero is greater than the pole for both the lead and lag compensators. While this is true for the phase lead compensator (z1 > p1), it contradicts the condition for the phase lag compensator, where the pole must be greater than the zero (z2 < p2).
Option 3: z1 < p1 and z2 < p2
This option is incorrect as it assumes that the zero is smaller than the pole for both the lead and lag compensators. While this is true for the phase lag compensator (z2 < p2), it contradicts the condition for the phase lead compensator, where the zero must be greater than the pole (z1 > p1).
Option 4: z1 < p1 and z2 > p2
This option is incorrect as it assumes that the zero is smaller than the pole for the lead compensator and larger for the lag compensator. This contradicts the correct conditions for both compensators. In the phase lead compensator, the zero must be greater than the pole (z1 > p1), and in the phase lag compensator, the pole must be greater than the zero (z2 < p2).
Conclusion:
Understanding the conditions for phase lead and lag compensators is essential for ensuring proper system performance. A phase lead compensator requires the zero to be greater than the pole, while a phase lag compensator requires the pole to be greater than the zero. The correct option, Option 1, satisfies these conditions, making it the appropriate choice.
```Controllers and Compensators Question 2:
What will be the natural frequency (wn) for the above controller?
Answer (Detailed Solution Below)
Controllers and Compensators Question 2 Detailed Solution
Explanation:
Natural Frequency (ωn)
Definition: Natural frequency (ωn) is a fundamental property of a mechanical system that describes the rate at which the system oscillates in the absence of any driving or damping force. It is dependent on the system's stiffness and mass.
Mathematical Derivation:
For a control system, the natural frequency can be derived from the standard second-order differential equation that describes the system's motion. This equation is:
Where:
- J is the moment of inertia.
- kp is the proportional gain or stiffness.
- θ is the angular displacement.
To find the natural frequency, we compare this equation to the standard form of the simple harmonic motion equation:
By comparing the coefficients, we get:
Therefore, the natural frequency is:
Correct Option Analysis:
The correct option is:
Option 2:
This option correctly represents the natural frequency for the given control system, as derived from the second-order differential equation of motion.
Additional Information
To further understand the analysis, let’s evaluate the other options:
Option 1:
This option incorrectly suggests that the natural frequency is the square root of the reciprocal of the product of moment of inertia (J) and stiffness (kp). The dimensions of this expression do not match those of frequency, making it an invalid option.
Option 3:
This option suggests that the natural frequency is the square root of the product of moment of inertia (J) and stiffness (kp). This expression is dimensionally incorrect for frequency and does not align with the standard form derived from the second-order differential equation.
Option 4:
This option suggests that the natural frequency is the square root of the ratio of moment of inertia (J) to stiffness (kp). This is dimensionally incorrect for frequency and does not align with the standard form derived from the second-order differential equation.
Conclusion:
Understanding the derivation and the correct form of the natural frequency is crucial for analyzing the behavior of mechanical systems. The natural frequency is a key parameter that determines how a system responds to external disturbances. By correctly identifying the natural frequency using the derived formula, we can better design and control mechanical systems for optimal performance.
Controllers and Compensators Question 3:
In a closed-loop system, what is the primary purpose of the integral controller (Ki)?
Answer (Detailed Solution Below)
Controllers and Compensators Question 3 Detailed Solution
Concept
In control systems, particularly in closed-loop systems, the purpose of various controllers is to improve the performance of the system. The Proportional-Integral-Derivative (PID) controller is one of the most commonly used types. Each component of a PID controller (Proportional, Integral, and Derivative) serves a unique function to enhance the system's performance.
An integral controller (Ki) is designed specifically to address the steady-state error. Steady-state error is the difference between the desired and actual output of a system when the system has reached a steady state. The integral action of the controller accumulates the error over time and adjusts the system's control input to eliminate this error, thus reducing the steady-state error to zero.
Additional Information
Let's look at the other options and understand why they are not correct:
1) To increase system bandwidth: Increasing system bandwidth is not the primary purpose of the integral controller. Bandwidth is more related to the speed of the system's response, which is generally influenced by the proportional (Kp) and derivative (Kd) components of the PID controller.
2) To adjust the damping ratio: The damping ratio is a measure of how oscillations in a system decay after a disturbance. This is primarily influenced by the proportional and derivative components of the PID controller, not the integral component.
3) To amplify the system gain: Amplifying system gain is related to the proportional controller (Kp), which adjusts the magnitude of the system's response to errors. The integral controller does not serve this purpose.
4) To reduce the steady state error: This is the correct answer. The integral controller continuously sums the error over time and adjusts the control input to eliminate the steady-state error, ensuring the actual output matches the desired output.
Controllers and Compensators Question 4:
If a pole is added to a system, it causes ______
Answer (Detailed Solution Below)
Controllers and Compensators Question 4 Detailed Solution
When we talk about adding a pole to a transfer function in a control system, we refer to introducing an additional term in the denominator of the transfer function. This affects the frequency response and stability characteristics of the system.
Definitions:
- Lag Compensation:
Lag compensation typically involves adding a pole and a zero such that the pole is closer to the origin than the zero (in the s-domain). This configuration slows down the system response and is commonly used to improve the steady-state performance (reduce steady-state error) while having a marginal impact on the transient response.
Lead Compensation: - Lead compensation, on the other hand, involves adding a zero and a pole where the zero is closer to the origin than the pole. This configuration speeds up the system response, enhances stability margins, and improves transient response by providing phase lead at higher frequencies.
Effects of Adding a Pole: - Adding a single pole to the system tends to reduce the overall system speed, increase lag, and potentially reduce the phase margin, which might adversely affect stability.
- Essentially, this means that the main result of adding a pole is akin to lag compensation, as it introduces a delay in the system's response to changes in input.
Controllers and Compensators Question 5:
The transfer function G(s) of a PID controller is
Answer (Detailed Solution Below)
Controllers and Compensators Question 5 Detailed Solution
Proportion + Integral + Derivative:
The PID controller produces on output, which is the combination of outputs of proportional, integral, and derivative controllers.
This is defined in terms of differential equations as:
\(u\left( t \right) = {K_p}\;e\left( t \right) + {K_I}\smallint e\left( t \right) \cdot dt + {K_D} \cdot \frac{{de\left( t \right)}}{{dt}}\)
Applying Laplace transform, we get:
\(U\left( s \right) = \left( {{K_p} + \frac{{{K_I}}}{s} + {K_D}s} \right)E\left( s \right)\)
Transfer function will be:
\(\frac{{U\left( s \right)}}{{E\left( s \right)}} = {K_p} + \frac{{{K_I}}}{s} + {K_D} \cdot s\)
Integral control:
It is the control mode where the controller Output is proportional to the integral of the error with respect to time.
Integral controller output = k × integral of error with time, i.e.
\({G_C}\left( s \right) = \frac{{U\left( s \right)}}{{E\left( s \right)}} = \frac{{{K_I}}}{s}\)
Proportional + Derivate:
The additive combination of proportional & Derivative control is known as P-D control.
Overall transfer function for a PD controller is given by:
\({G_C}\left( s \right) = \frac{{U\left( s \right)}}{{E\left( s \right)}} = {K_P} + s{K_D}\)
It is equivalent to a High-pass filter.
Top Controllers and Compensators MCQ Objective Questions
Lag compensator is a _____
Answer (Detailed Solution Below)
Controllers and Compensators Question 6 Detailed Solution
Download Solution PDFLag compensator:
Transfer function:
If it is in the form of
\(\frac{{1 + aTs}}{{1 + Ts}}\), then a < 1
If it is in the form of
\(\frac{{s + a}}{{s + b}}\), then a > b
Pole zero plot:
The pole is nearer to the origin.
Filter: It is a low pass filter (LPF)
Maximum phase lag frequency:
\({\omega _m} = \frac{1}{{T\sqrt a }}\)
Maximum phase lag:
\({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)
ϕm is negative
Lead compensator:
Transfer function:
If it is in the form of
\(\frac{{1 + aTs}}{{1 + Ts}}\), then a > 1
If it is in the form of
\(\frac{{s + a}}{{s + b}}\), then a < b
Pole zero plot:
The zero is nearer to the origin.
Filter: It is a high pass filter (HPF).
Maximum phase lead frequency:
\({\omega _m} = \frac{1}{{T\sqrt a }}\)
Maximum phase lead:
\({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)
ϕm is positive
Slow response of an over-damped system can be made faster with the help of ______ controller.
Answer (Detailed Solution Below)
Controllers and Compensators Question 7 Detailed Solution
Download Solution PDFExplanation:
The controller is a device that is used to alter or maintain the transient state & steady-state region performance parameter as per our requirement.
Proportional Controller-
The standard Proportional Controller as shown:
In space-form -
\({G_C}\left( s \right) = \frac{{U\left( s \right)}}{{E\left( s \right)}} = \frac{{{K_p}}}{s(s+1)}\)
In time-domain form -
p(t) = Kp e(t) + po
Where,
po = controller output with zero error
Kp = proportional gain constant.
Some effects of the proportional controller are as follows:
- The P-controller can stabilize a first-order system, can give a near-zero error, and improves the settling time by increasing the bandwidth.
- It also helps in reducing the steady-state error which makes the system more stable.
- The slow response of an over-damped system can be made faster with the help of the proportional controller. Hence option (2) is the correct answer.
Important Points
Effects of Proportional Integral (PI) controllers:
- Increases the type of the system by one
- Rise time and settling time increases and Bandwidth decreases
- The speed of response decreased i.e. transient response becomes slower
- Decreases the steady-state error and steady-state response is improved
- Decreases the stability
Effects of Proportional Derivative (PD) controllers:
- Decreases the type of the system by one
- Reduces the rise time and settling time
- Rise time and settling time decreases and Bandwidth increases
- The speed of response is increased i.e. transient response is improved
- Improves gain margin, phase margin, and resonant peak
- Increases the input noise
- Improves the stability
Which of the following controllers improves the transient response of a system?
Answer (Detailed Solution Below)
Controllers and Compensators Question 8 Detailed Solution
Download Solution PDFEffects of Proportional Derivative (PD) controllers:
- Decreases the type of the system by one
- Reduces the rise time and settling time
- Rise time and settling time decreases and Bandwidth increases
- The speed of response is increased i.e. the transient response is improved
- Improves gain margin, phase margin, and resonant peak
- Increases the input noise
- Improves the stability
Effects of Proportional Integral (PI) controllers:
- Increases the type of the system by one
- Rise time and settling time increases and Bandwidth decreases
- The speed of response decreased i.e. transient response becomes slower
- Decreases the steady-state error and steady-state response is improved
- Decreases the stability
Which of the following is true for the network shown below -
Answer (Detailed Solution Below)
Controllers and Compensators Question 9 Detailed Solution
Download Solution PDFConcept:
In general, the lead and lag compensator is represented by the below transfer function
\(\frac{{{V_0}\left( s \right)}}{{{V_i}\left( s \right)}} = k\frac{{s + a}}{{s + b}}\)
If a > b then that is lag compensator because pole comes first.
If a < b then that is the lead compensator since zero comes first.
Analysis:
Lead compensator:
1) When sinusoidal input applied to this it produces sinusoidal output with the phase lead input.
2) It speeds up the Transient response and increases the margin for stability.
A circuit diagram is as shown:
Response is:
\(\frac{{{V_0}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{{{R_2}\left( {1 + s{C_1}{R_1}} \right)}}{{{R_1} + {R_2} + s{C_1}{R_1}{R_2}}}\)
\(\frac{{{V_0}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{{1 + s\tau }}{{1 + \alpha s\tau }}\)
Lead constant \(\alpha = \frac{{{R_2}}}{{{R_1} + {R_2}}}\) < 1
Important Points
Compen- sator |
Pole zero plot |
Response |
Lead |
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Lag |
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Lag-lead |
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Lead-lag |
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The maximum phase shift that can be obtained by using a lead compensator with transfer function \(G_c (s)=\frac{4(1+0.15s)}{(1+0.05 s) }\) equal to
Answer (Detailed Solution Below)
Controllers and Compensators Question 10 Detailed Solution
Download Solution PDFConcept:
The standard T/F of the compensator is
\(\frac{{1 + aTs}}{{1 + Ts}}\)
Maximum phase lead
\( {ϕ _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\\ \)
Maximum phase lead frequency,
\({\omega _m} = \frac{1}{{T\sqrt a }}\)
Calculation:
The given transfer function is,
\(T/F = 4(\frac{{1 + 0.15s}}{{1 + 0.05s}})\)
By comparing both transfer functions,
aT = 0.15
T = 0.05
a = 3
Maximum phase lead
\(\begin{array}{l} {ϕ _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\\ = {\sin ^{ - 1}}\left( {\frac{{3 - 1}}{{3 + 1}}} \right)\\ = {\sin ^{ - 1}}\left( {\frac{1}{{2}}} \right) \end{array}\)
= sin-1 (0.5)
ϕm = 30°
The lag compensator
Answer (Detailed Solution Below)
Controllers and Compensators Question 11 Detailed Solution
Download Solution PDFLead compensator:
Transfer function:
If it is in the form of \(\frac{{1 + aTs}}{{1 + Ts}}\), then a > 1
If it is in the form of \(\frac{{s + a}}{{s + b}}\), then a < b
Maximum phase lead frequency: \({\omega _m} = \frac{1}{{T\sqrt a }}\)
Maximum phase lead: \({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)
ϕm is positive
Pole zero plot:
The zero is nearer to the origin.
Filter: It is a high pass filter (HPF)
Effect on the system:
- Rise time and settling time decreases and Bandwidth increases
- The transient response becomes faster
- The steady-state response is not affected
- Improves the stability
Lag compensator:
Transfer function:
If it is in the form of \(\frac{{1 + aTs}}{{1 + Ts}}\), then a < 1
If it is in the form of \(\frac{{s + a}}{{s + b}}\), then a > b
Maximum phase lag frequency: \({\omega _m} = \frac{1}{{T\sqrt a }}\)
Maximum phase lag: \({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)
ϕm is negative
Pole zero plot:
The pole is nearer to the origin.
Filter: It is a low pass filter (LPF)
Effect on the system:
- Rise time and settling time increases and Bandwidth decreases
- The transient response becomes slower
- The steady-state response is improved
- Stability decreases
If both fast response time and good steady-state accuracy are needed ______ compensators are used.
Answer (Detailed Solution Below)
Controllers and Compensators Question 12 Detailed Solution
Download Solution PDFAnalysis:
Lag Compensator:
\(\frac{{{V_0}\left( s \right)}}{{{V_1}\left( s \right)}} = \frac{{1 + ST}}{{1 + \beta ST}}\)
Where,
\(\beta = \frac{{{R_1} + {R_2}}}{{{R_2}}},\;\beta > 1\)
In Lag Compensator, the steady-state error is reduced So, steady-state Response is increased.
In lag compensator, Transient response decreases.
Lead Compensator:
\(\frac{{{V_0}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{{\alpha \left( {1 + ST} \right)}}{{1 + \alpha ST}}\)
\(\alpha = \frac{{{R_2}}}{{{R_1} + {R_2}}}\;;\alpha < 1\)
In lead comp. steady-state error increased So, the steady-state response is decreased.
In lead comp. Transient response Improves.
Lag-lead Compensator:
Lag lead Compensator is a combined form of a lead compensator and lag comp compensator.
\(\frac{{{V_0}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{{\left( {S + {Z_1}} \right)\left( {S + {Z_2}} \right)}}{{\left( {S + {P_1}} \right)\left( {S + {P_2}} \right)}}\)
\(\beta = \underbrace {\frac{{{Z_1}}}{{{P_1}}} > 1}_{Lag}, ~ \alpha = \underbrace {\frac{{{Z_2}}}{{{P_2}}} < 1}_{Lead}\)
So, lag-lead compensator improves both steady-state accuracy and fast response time.The compensator required to improve the steady state response of a system is
Answer (Detailed Solution Below)
Controllers and Compensators Question 13 Detailed Solution
Download Solution PDFLag compensator:
Transfer function:
If it is in the form of \(\frac{{1 + aTs}}{{1 + Ts}}\), then a < 1
If it is in the form of \(\frac{{s + a}}{{s + b}}\), then a > b
Maximum phase lag frequency:
\({\omega _m} = \frac{1}{{T\sqrt a }}\)
Maximum phase lag::
\({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)
ϕm is negative
Pole zero plot:
The pole is nearer to the origin.
Filter: It is a low pass filter (LPF)
Effect on the system:
- Rise time and settling time increases and Bandwidth decreases
- The transient response becomes slower
- The steady-state response is improved
- Stability decreases
Lead compensator:
Transfer function:
If it is in the form of \(\frac{{1 + aTs}}{{1 + Ts}}\), then a > 1
If it is in the form of \(\frac{{s + a}}{{s + b}}\), then a < b
Maximum phase lead frequency:
\({\omega _m} = \frac{1}{{T\sqrt a }}\)
Maximum phase lead:
\({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)
ϕm is positive
Pole zero plot:
The zero is nearer to the origin.
Filter: It is a high pass filter (HPF)
Effect on the system:
- Rise time and settling time decreases and Bandwidth increases
- The transient response becomes faster
- The steady-state response is not affected
- Improves the stability
For the network shown in the figure below, the frequency (in rad/s) at which the maximum phase lag occurs is, ___________.
Answer (Detailed Solution Below) 0.30 - 0.33
Controllers and Compensators Question 14 Detailed Solution
Download Solution PDFGiven circuit is a lag compensator and transfer function is given as
\(\frac{{{{\rm{V}}_0}\left( {\rm{s}} \right)}}{{{{\rm{V}}_{{\rm{in}}}}\left( {\rm{s}} \right)}} = \frac{{\left( {1 + \frac{1}{{\rm{s}}}} \right)}}{{10 + \frac{1}{{\rm{s}}}}} = \frac{{{\rm{s}} + 1}}{{10{\rm{s}} + 1}} = \frac{{\left( {1 + {\rm{sT}}} \right)}}{{\left( {1 + {\rm{\beta sT}}} \right)}}\)
On comparison we get, \(T = 1\)
\(\beta T = 10\; \Rightarrow \;\beta = 10\)
The frequency at which maximum lead occurs is \({{\rm{\omega }}_{\rm{m}}} = \frac{1}{{{\rm{T}}\sqrt {\rm{\beta }} }}\)
\(\therefore {{\rm{\omega }}_{\rm{m}}} = \frac{1}{{1\sqrt {10} }} = 0.316\frac{{{\rm{rad}}}}{{{\rm{sec}}}}\)Which of the following terms is responsible for noise measurement in the PID controller?
Answer (Detailed Solution Below)
Controllers and Compensators Question 15 Detailed Solution
Download Solution PDFProportional + Derivate:
The additive combination of proportional & Derivative control is known as P-D control.
The overall transfer function for a PD controller is given by:
\({G_C}\left( s \right) = \frac{{U\left( s \right)}}{{E\left( s \right)}} = {K_P} + s{K_D}\)
PD controller is nothing but a differentiator (or) a High Pass Filter.
The frequency of noise is very high. So this high pass filter will allow noise into the system which results in noise amplification.
Effects of Proportional Derivative (PD) controllers:
- Decreases the type of the system by one
- Reduces the rise time and settling time
- Rise time and settling time decreases and Bandwidth increases
- The speed of response is increased i.e. the transient response is improved
- Improves gain margin, phase margin, and resonant peak
- Increases the input noise
- Improves the stability
Effects of Proportional Integral (PI) controllers:
- Increases the type of the system by one
- Rise time and settling time increase and Bandwidth decreases
- The speed of response decreased i.e. transient response becomes slower
- Decreases the steady-state error and steady-state response is improved
- Decreases the stability