Controllers and Compensators MCQ Quiz - Objective Question with Answer for Controllers and Compensators - Download Free PDF

Last updated on Apr 1, 2025

Latest Controllers and Compensators MCQ Objective Questions

Controllers and Compensators Question 1:

qImage6793221b98b98557594b6428

What will be the natural frequency (wn) for the above controller? 

  1. \(\rm \omega_n=\sqrt{\frac{1}{Jk_p}}\)
  2. \(\rm \omega_n=\sqrt{\frac{k_p}{J}}\)
  3. \(\rm \omega_n=\sqrt{Jk_p}\)
  4. \(\rm \omega_n=\sqrt{\frac{J}{k_p}}\)

Answer (Detailed Solution Below)

Option 2 : \(\rm \omega_n=\sqrt{\frac{k_p}{J}}\)

Controllers and Compensators Question 1 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:

 

Jd2θdt2+kpθ=0" id="MathJax-Element-1-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">Jd2θdt2+kpθ=0

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:

 

d2θdt2+ωn2θ=0" id="MathJax-Element-2-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">d2θdt2+ωn2θ=0

By comparing the coefficients, we get:

 

ωn2=kpJ" id="MathJax-Element-3-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">ωn2=kpJ

Therefore, the natural frequency is:

 

ωn=kpJ" id="MathJax-Element-4-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">ωn=kpJ

Correct Option Analysis:

The correct option is:

Option 2: ωn=kpJ" id="MathJax-Element-5-Frame" role="presentation" style="position: relative;" tabindex="0">ωn=kpJ

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: ωn=1Jkp" id="MathJax-Element-6-Frame" role="presentation" style="position: relative;" tabindex="0">ωn=1Jkp

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: ωn=Jkp" id="MathJax-Element-7-Frame" role="presentation" style="position: relative;" tabindex="0">ωn=Jkp

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: ωn=Jkp" id="MathJax-Element-8-Frame" role="presentation" style="position: relative;" tabindex="0">ωn=Jkp

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 2:

In a closed-loop system, what is the primary purpose of the integral controller (Ki)?

  1. To increase system bandwidth.
  2. To adjust the damping ratio. 
  3. To amplify the system gain.
  4. To reduce the steady state error.

Answer (Detailed Solution Below)

Option 4 : To reduce the steady state error.

Controllers and Compensators Question 2 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 3:

If a pole is added to a system, it causes ______

  1. Lag compensation
  2. Lead compensation
  3. Lead lag compensation
  4. None of these

Answer (Detailed Solution Below)

Option 1 : Lag compensation

Controllers and Compensators Question 3 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 4:

The transfer function G(s) of a PID controller is

  1. \({K_1} + {K_2}s + {K_3}{s^2}\)
  2. \({K_1} + \frac{{{K_2}}}{s} + {K_3}s\)
  3. \({K_1} + \frac{{{K_2}}}{s}\)
  4. \({K_1}s + {K_2}{s^2} + {K_3}{s^3}\)
  5. None of these

Answer (Detailed Solution Below)

Option 2 : \({K_1} + \frac{{{K_2}}}{s} + {K_3}s\)

Controllers and Compensators Question 4 Detailed Solution

Proportion + Integral + Derivative:

 

Electronics Janesh Assignment 4 9Q S shubham D3

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:

\(\frac{{U\left( s \right)}}{{E\left( s \right)}} = {K_p} + \frac{{{K_I}}}{s} + {K_D} \cdot s\)

26 June 1

Integral control:

Electronics Janesh Assignment 4 9Q S shubham D1

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:

Electronics Janesh Assignment 4 9Q S shubham D2

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.

Controllers and Compensators Question 5:

What is the full form of PID?

  1. Proportional Integral Derivative
  2. Proportional Integral Device
  3. Programmable Integral Device
  4. Programmable Integral Derivative
  5. None of these

Answer (Detailed Solution Below)

Option 1 : Proportional Integral Derivative

Controllers and Compensators Question 5 Detailed Solution

Proportion + Integral + Derivative (PID):

 

Electronics Janesh Assignment 4 9Q S shubham D3

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:

\(\frac{{U\left( s \right)}}{{E\left( s \right)}} = {K_p} + \frac{{{K_I}}}{s} + {K_D} \cdot s\)

26 June 1

Integral control:

Electronics Janesh Assignment 4 9Q S shubham D1

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:

Electronics Janesh Assignment 4 9Q S shubham D2

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 _____

  1. phase shifted
  2. Low pass filter
  3. High pass filter
  4. resistance

Answer (Detailed Solution Below)

Option 2 : Low pass filter

Controllers and Compensators Question 6 Detailed Solution

Download Solution PDF

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

Pole zero plot:

F1 U.B Madhu 2.12.19 D1

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:

 \({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)

ϕis 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:

F1 U.B Madhu 2.12.19 D16

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:

 \({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)

ϕis positive

Slow response of an over-damped system can be made faster with the help of ______ controller.

  1. PD
  2. P
  3. PI
  4. Remote

Answer (Detailed Solution Below)

Option 2 : P

Controllers and Compensators Question 7 Detailed Solution

Download Solution PDF

Explanation:

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:

F1 R.D. N.J 26.09.2019 D 7 2

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) = Ke(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?

  1. The proportional controller
  2. The proportional and integral controllers
  3. The integral controller
  4. The derivative controller

Answer (Detailed Solution Below)

Option 4 : The derivative controller

Controllers and Compensators Question 8 Detailed Solution

Download Solution PDF

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 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 -

F1 Savita Engineering 28-6-22 D20

  1. Lead compensator
  2. Lag compensator
  3. Lead-lag compensator
  4. None of the above

Answer (Detailed Solution Below)

Option 1 : Lead compensator

Controllers and Compensators Question 9 Detailed Solution

Download Solution PDF

Concept:

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:

F2 S.B 6.8.20 Pallavi D9

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

F2 S.B 6.8.20 Pallavi D2

F2 S.B 6.8.20 Pallavi D1

Lag

F2 S.B 6.8.20 Pallavi D4

F2 S.B 6.8.20 Pallavi D3

Lag-lead

F2 S.B 6.8.20 Pallavi D6

F2 S.B 6.8.20 Pallavi D5

Lead-lag

F2 S.B 6.8.20 Pallavi D8

F2 S.B 6.8.20 Pallavi D7

 

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

  1. 15° 
  2. 30° 
  3. 45° 
  4. 60° 

Answer (Detailed Solution Below)

Option 2 : 30° 

Controllers and Compensators Question 10 Detailed Solution

Download Solution PDF

Concept:

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

  1. improves both steady state and transient response
  2. improves steady state only
  3. improves transient only
  4. improves steady state and reduced speed of transient response

Answer (Detailed Solution Below)

Option 4 : improves steady state and reduced speed of transient response

Controllers and Compensators Question 11 Detailed Solution

Download Solution PDF

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:

F1 U.B Madhu 2.12.19 D16

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:

F1 U.B Madhu 2.12.19 D1

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.

  1. Fast
  2. Lead
  3. Lag
  4. Lag-Lead

Answer (Detailed Solution Below)

Option 4 : Lag-Lead

Controllers and Compensators Question 12 Detailed Solution

Download Solution PDF

Analysis:

Lag Compensator:

F1 T.S 15.7.20 Pallavi D5

\(\frac{{{V_0}\left( s \right)}}{{{V_1}\left( s \right)}} = \frac{{1 + ST}}{{1 + \beta ST}}\)

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:

F1 T.S 15.7.20 Pallavi D6

\(\frac{{{V_0}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{{\alpha \left( {1 + ST} \right)}}{{1 + \alpha ST}}\)

\(\alpha = \frac{{{R_2}}}{{{R_1} + {R_2}}}\;;\alpha < 1\)

In lead comp. steady-state error increased So, the steady-state response is decreased.

In lead comp. Transient response Improves.

Lag-lead Compensator:

F1 T.S 15.7.20 Pallavi D7

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

  1. Lag 
  2. Lead
  3. Lag-lead
  4. Zero

Answer (Detailed Solution Below)

Option 1 : Lag 

Controllers and Compensators Question 13 Detailed Solution

Download Solution PDF

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::

\({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)

ϕis negative

Pole zero plot:

F1 U.B Madhu 2.12.19 D1

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

26 June 1

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:

\({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)

ϕis positive

Pole zero plot:

F1 U.B Madhu 2.12.19 D16

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, ___________.

Gate EE 2016 paper 2 Images-Q21

Answer (Detailed Solution Below) 0.30 - 0.33

Controllers and Compensators Question 14 Detailed Solution

Download Solution PDF

Given 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?

  1. The integral term
  2. Both proportion-integral
  3. The proportional term
  4. The derivative term

Answer (Detailed Solution Below)

Option 4 : The derivative term

Controllers and Compensators Question 15 Detailed Solution

Download Solution PDF

Proportional + Derivate:

Electronics Janesh Assignment 4 9Q S shubham D2

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.

26 June 1

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
Get Free Access Now
Hot Links: teen patti joy 51 bonus teen patti joy official teen patti octro 3 patti rummy teen patti comfun card online