System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
Question
Book Icon
Chapter 10, Problem 10.31P
To determine

(a)

The values of gain KPandKI of the PI controller of first-order plant for the following cases:

Case 1. For ζ=0.707.

Case 2. For ζ=1.

Case 3. For a root separation factor of 5.

Expert Solution
Check Mark

Answer to Problem 10.31P

The values of gains for the PI controller are as follows:

Case 1. For ζ=0.707, KI=5 and KP=5.

Case 2. For ζ=1, KI=2.5 and KP=5.

Case 3. For a root separation factor of 5, KI=12.5 and KP=25.

Explanation of Solution

Given:

The proportional integral controller of first order plant is as shown below:

System Dynamics, Chapter 10, Problem 10.31P , additional homework tip  1

Where, the parameter values are as given:

I=10,c=5

Also, the performance specifications require the time constant of the system to be τ=2.

Concept Used:

  1. The transfer functions for the block diagram are as shown below:
  2. Ω(s)Ωr(s)=sKP+KIIs2+s(c+KP)+KIΩ(s)Td(s)=sIs2+s(c+KP)+KI

    For a second order system having a characteristic equation of the form s2+2ζωns+ωn2=0, the characteristic roots are of the form s=ζωns±ωn1ζ2ζ1 .

    Also, the corresponding time constant for the system would be τ=1ζωn.

  3. For a second order system, if the root separation factor is K then, we have following conclusions for the roots and their corresponding time constant, that is
  4. For roots s1ands2 ,

    s1s2=K

    The corresponding time constants of the system would be:

    τ1τ2=1Re(s1)1Re(s2)=Re(s2)Re(s1)=1K

Calculation:

From the block diagram as shown, the transfer functions are as:

Ω(s)Ωr(s)=sKP+KIIs2+s(c+KP)+KIΩ(s)Td(s)=sIs2+s(c+KP)+KI

Therefore, the characteristic equation for the system is:

Is2+s(c+KP)+KI=0

On putting the values of parameters in this expression of characteristic equation:

Is2+s(c+KP)+KI=010s2+s(5+KP)+KI=0 I=10,c=5s2+s(5+KP)10+KI10=0

On comparing this equation with form s2+2ζωns+ωn2=0, we get

Undamped natural frequency:

ωn2=KI10ωn=KI10rad/sec

Damping ratio:

2ζωn=(5+KP)10ζ=(5+KP)20ωnζ=(5+KP)20KI10ζ=(5+KP)210KI

Thus, the time constant for the system is:

τ=1ζωnτ=1(5+KP)210KIKI10 ζ=(5+KP)210KI,ωn=KI10rad/secτ=20(5+KP)

As the performance specifications require the time constant of the system to be τ=2.

Therefore, at τ=2

τ=20(5+KP)2=20(5+KP)5+KP=10KP=5

Case 1. When ζ=0.707, the gain value of KI is:

Since, ζ=(5+KP)210KI

0.707=(5+KP)210KI KP=51.41410KI=10KI=101.414=2.24KI=5.015

Case 2. When ζ=1, the gain value of KI is:

Since, ζ=(5+KP)210KI

1=(5+KP)210KI KP=5KI=10210KI=2.5

Case 3. When the root separation factor is 5.

For this value of the root separation factor s1s2=5, the ratio of time constants for the roots would be:

τ1τ2=15

As given in the question system, performance requires time constant to be 2seconds or say the dominant time constant to be 2 seconds then the secondary time constant would be 0.4 seconds. Therefore, the characteristic roots corresponding to these time constants would be:

s=12=0.5,s=10.4=2.5

Thus, the corresponding characteristic equation for the system will be:

s=0.5,s=2.5(s+0.5)(s+2.5)=0s2+3s+1.25=0

On comparing this equation with s2+s(5+KP)10+KI10=0, we get

KI10=1.25KI=12.5

And

(5+KP)10=35+KP=30KP=25.

Conclusion:

The values of gains for the PI controller areas follows:

Case 1. For ζ=0.707, KI=5 and KP=5.

Case 2. For ζ=1, KI=2.5 and KP=5.

Case 3. For a root separation factor of 5, KI=12.5 and KP=25.

To determine

(b)

To plot:

The response Ω(s) for the unit-step command response Ωr(s) for all the cases in sub-part (a)

Also, compare the responses obtained in all these cases.

Expert Solution
Check Mark

Answer to Problem 10.31P

The response Ω(s) for the unit-step command response Ωr(s) for all the cases in sub-part (a) is shown in Figure 1. At the damping ratio, ζ=0.707, there is a small overshoot encountered in the response while for the cases where damping ratios are 1 and 1.34 respectively shows the overdamping in the response. And the responses in the first and second case are sluggish, while the response for the third case is faster with the less settling time.

Explanation of Solution

Given:

The proportional integral controller of first order plant is as shown below:

System Dynamics, Chapter 10, Problem 10.31P , additional homework tip  2

Where, the parameter values are as given:

I=10,c=5

Also, the performance specifications require the time constant of the system to be τ=2.

The values of gains for the PI controller are as follows:

Case 1. For ζ=0.707, KI=5 and KP=5.

Case 2. For ζ=1, KI=2.5 and KP=5.

Case 3. For a root separation factor of 5, KI=12.5 and KP=25.

Concept Used:

  1. The transfer functions for the block diagram are as shown below:

Ω(s)Ωr(s)=sKP+KIIs2+s(c+KP)+KIΩ(s)Td(s)=sIs2+s(c+KP)+KI.

Calculation:

From the block diagram as shown, the transfer functions are as:

Ω(s)Ωr(s)=sKP+KIIs2+s(c+KP)+KIΩ(s)Td(s)=sIs2+s(c+KP)+KI

Therefore, the response Ω(s) for the system is:

Ω(s)=sKP+KIIs2+s(c+KP)+KIΩr(s)sIs2+s(c+KP)+KITd(s)

On putting the values of parameters in this expression of characteristic equation:

Ω(s)=sKP+KIIs2+s(c+KP)+KIΩr(s)sIs2+s(c+KP)+KITd(s)I=10,c=5Ω(s)=sKP+KI10s2+s(5+KP)+KIΩr(s)s10s2+s(5+KP)+KITd(s)

Case 1. When ζ=0.707, the gain values are KI=5 and KP=5:

Since, Ω(s)=sKP+KI10s2+s(5+KP)+KIΩr(s)s10s2+s(5+KP)+KITd(s)

Ω(s)=5s+510s2+s(5+5)+5Ωr(s)s10s2+s(5+5)+5Td(s)KP=5,KI=5Ω(s)=5s+510s2+10s+5Ωr(s)s10s2+10s+5Td(s)

Therefore, for unit-step command response Ωr(s) and zero disturbance torque Td(s), we have

Ω(s)=5s+510s2+10s+5Ωr(s)s10s2+10s+5Td(s)Ω(s)=5s+510s2+10s+51ss10s2+10s+50Ω(s)=5s+510s2+10s+51s

On simplifying this response of Ω(s) using partial fraction expansion, we get

Ω(s)=5s+510s2+10s+51s=s+12s2+2s+11sΩ(s)=1s(2s+1)2s2+2s+1

On taking inverse Laplace transform of this, we have

Ω(s)=1s(2s+1)2s2+2s+1=1s(s+12)s2+s+12=1s(s+12)(s+12)2+(12)2ω(t)=1e12tcos(12t)

Case 2. When ζ=1, the gain values are KI=2.5 and KP=5:

Since, Ω(s)=sKP+KI10s2+s(5+KP)+KIΩr(s)s10s2+s(5+KP)+KITd(s)

Ω(s)=5s+2.510s2+s(5+5)+2.5Ωr(s)s10s2+s(5+5)+2.5Td(s)KP=5,KI=2.5

Ω(s)=5s+2.510s2+10s+2.5Ωr(s)s10s2+10s+2.5Td(s)

Therefore, for unit-step command response Ωr(s) and zero disturbance torque Td(s), we have

Ω(s)=5s+2.510s2+10s+2.5Ωr(s)s10s2+10s+2.5Td(s)Ω(s)=5s+2.510s2+10s+2.51ss10s2+10s+2.50Ω(s)=5s+2.510s2+10s+2.51s

On simplifying this response of Ω(s) using partial fraction expansion, we get

Ω(s)=5s+2.510s2+10s+2.51sΩ(s)=1s10s+510s2+10s+2.5

On taking inverse Laplace transform of this, we have

Ω(s)=1s10s+510s2+10s+2.5=1s(s+12)s2+s+14=1s(s+12)(s+12)2Ω(s)=1s1(s+12)ω(t)=1e12t

Case 3. When the root separation factor is 5.

That is, the gain values are KI=12.5 and KP=25:

Since, Ω(s)=sKP+KI10s2+s(5+KP)+KIΩr(s)s10s2+s(5+KP)+KITd(s)

Ω(s)=25s+12.510s2+s(5+25)+12.5Ωr(s)s10s2+s(5+25)+12.5Td(s)KP=25,KI=12.5

Ω(s)=25s+12.510s2+30s+12.5Ωr(s)s10s2+30s+12.5Td(s)

Therefore, for unit-step command response Ωr(s) and zero disturbance torque Td(s), we have

Ω(s)=25s+12.510s2+30s+12.5Ωr(s)s10s2+30s+12.5Td(s)Ω(s)=25s+12.510s2+30s+12.51ss10s2+30s+12.50Ω(s)=25s+12.510s2+30s+12.51s

On simplifying this response of Ω(s) using partial fraction expansion, we get

Ω(s)=25s+12.510s2+30s+12.51sΩ(s)=1s10s+510s2+30s+12.5

On taking inverse Laplace transform of this, we have

Ω(s)=1s10s+510s2+30s+12.5=1s(s+12)s2+3s+1.25Ω(s)=1s(s+32)(s+32)2(1)2+1(s+32)2(1)2ω(t)=1e32tcosh(t)+e32tsinh(t)

On plotting the responses ω(t) for all the three cases, we have

System Dynamics, Chapter 10, Problem 10.31P , additional homework tip  3

Here, on comparing the responses for all the three cases, it is concluded that the response shifts from being underdamped to overdamped for the first two cases. Whereas, for the third case the response remains overdamped in nature however, its settling time is very less comparing the other two cases.

Conclusion:

At the damping ratio ζ=0.707, there is a small overshoot encountered in the response while for the cases where damping ratios are1 and 1.34 respectively shows the overdamping in the response. And the response in first and second case are sluggish while the response for the third case is faster with the less settling time.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 10 Solutions

System Dynamics

Ch. 10 - Prob. 10.15PCh. 10 - Prob. 10.16PCh. 10 - Prob. 10.17PCh. 10 - Prob. 10.19PCh. 10 - Prob. 10.20PCh. 10 - Prob. 10.21PCh. 10 - Prob. 10.22PCh. 10 - Prob. 10.23PCh. 10 - Prob. 10.24PCh. 10 - Consider the PI speed control system shown in...Ch. 10 - Prob. 10.26PCh. 10 - Prob. 10.27PCh. 10 - Prob. 10.28PCh. 10 - Prob. 10.29PCh. 10 - Prob. 10.30PCh. 10 - Prob. 10.31PCh. 10 - Prob. 10.32PCh. 10 - Prob. 10.33PCh. 10 - Prob. 10.34PCh. 10 - Prob. 10.35PCh. 10 - Prob. 10.36PCh. 10 - For the designs found in part (a) of Problem...Ch. 10 - Prob. 10.39PCh. 10 - Prob. 10.40PCh. 10 - Prob. 10.41PCh. 10 - Prob. 10.44PCh. 10 - Prob. 10.45PCh. 10 - Prob. 10.46PCh. 10 - For the system shown in Figure 10.7.1, / = c = 1....Ch. 10 - Prob. 10.48PCh. 10 - Prob. 10.51PCh. 10 - Prob. 10.52PCh. 10 - Prob. 10.53PCh. 10 - Prob. 10.54PCh. 10 - Prob. 10.56PCh. 10 - Prob. 10.57PCh. 10 - Prob. 10.58PCh. 10 - Prob. 10.59PCh. 10 - Prob. 10.60PCh. 10 - Prob. 10.61PCh. 10 - Prob. 10.62PCh. 10 - Prob. 10.63PCh. 10 - Prob. 10.64PCh. 10 - Prob. 10.65PCh. 10 - Consider Example 10.6.3. Modify the diagram in...Ch. 10 - Prob. 10.67PCh. 10 - 10.68 Consider Example 10.6.4. Modify the diagram...Ch. 10 - 10.69 Figure P10.7 shows a system for controlling...Ch. 10 - A speed control system using an...Ch. 10 - Prob. 10.72PCh. 10 - Prob. 10.73PCh. 10 - Prob. 10.74PCh. 10 - Consider Example 10.7.4. Use the diagram in Figure...Ch. 10 - Prob. 10.76PCh. 10 - Refer to Figure 10.3.9, which show s a speed...Ch. 10 - For the system in Problem 10.77 part (a), create a...
Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY