System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
Question
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Chapter 10, Problem 10.22P
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 10.

Expert Solution
Check Mark

Answer to Problem 10.22P

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

Case 1: For ζ=0.707, KI=200 and KP=36.

Case 2: For ζ=1, KI=100 and KP=36.

Case 3: For a root separation factor of 10, KI=1000 and KP=216.

Explanation of Solution

Given:

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

System Dynamics, Chapter 10, Problem 10.22P , additional homework tip  1

Where, the parameter values are as given:

I=c=4

Also, the performance specifications require the time constant of the system to be τ=0.2.

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

  3. 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 .
  4. Also, the corresponding time constant for the system would be τ=1ζωn.

  5. 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

    For roots s1ands2 ,

    s1s2=K

  6. 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=04s2+s(4+KP)+KI=0 I=c=4s2+s(4+KP)4+KI4=0

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

Undamped natural frequency:

ωn2=KI4ωn=KI4ωn=KI2rad/sec

Damping ratio:

2ζωn=(4+KP)4ζ=(4+KP)8ωnζ=(4+KP)8KI2ζ=(4+KP)4KI

Thus, the time constant for the system is:

τ=1ζωnτ=1(4+KP)4KIKI2 ζ=(4+KP)4KI,ωn=KI2rad/secτ=8(4+KP)

As the performance specifications require the time constant of the system to be τ=0.2.

Therefore, at τ=0.2

τ=8(4+KP)0.2=8(4+KP)4+KP=40KP=36

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

Since, ζ=(4+KP)4KI

0.707=(4+36)4KI KP=362.828KI=40KI=402.828=14.144KI=200.05200

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

Since, ζ=(4+KP)4KI

1=(4+36)4KI KP=36KI=10KI=100

Case 3. When the root separation factor is 10.

For this value of the root separation factor s1s2=10, the ratio of time constants for the roots would be:

τ1τ2=110

As given in the question system, performance requires time constant to be 0.2seconds or say the dominant time constant to be 0.2 seconds, then the secondary time constant would be 0.02 seconds. Therefore, the characteristic roots corresponding to these time constants would be:

s=10.2=5,s=10.02=50

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

s=5,s=50(s+5)(s+50)=0s2+55s+250=0

On comparing this equation with s2+s(4+KP)4+KI4=0, we get

KI4=250KI=1000

And

(4+KP)4=554+KP=220KP=216.

Conclusion:

The values of gains for the PI controller areas follows:

Case 1. For ζ=0.707, KI=200 and KP=36.

Case 2. For ζ=1, KI=100 and KP=36.

Case 3. For a root separation factor of 10, KI=1000 and KP=216.

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 10.22P

The response Ω(s) for the unit-step command response Ωr(s) for all the cases is shown in Figure 1. At the damping ratio ζ=0.707, there is a high overshoot in the response in comparison to the other cases where damping ratios are 1 and 1.739 respectively.

Explanation of Solution

Given:

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

System Dynamics, Chapter 10, Problem 10.22P , additional homework tip  2

Where, the parameter values are as given:

I=c=4

Also, the performance specifications require the time constant of the system to be τ=0.2.

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

Case 1. For ζ=0.707, KI=200 and KP=36.

Case 2. For ζ=1, KI=100 and KP=36.

Case 3. For a root separation factor of 10, KI=1000 and KP=216.

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)Ω(s)=sKP+KI4s2+s(4+KP)+KIΩr(s)s4s2+s(4+KP)+KITd(s) I=c=4

Case 1. When ζ=0.707, the gain values are KI=200 and KP=36:

Since, Ω(s)=sKP+KI4s2+s(4+KP)+KIΩr(s)s4s2+s(4+KP)+KITd(s)

Ω(s)=36s+2004s2+s(4+36)+200Ωr(s)s4s2+s(4+36)+200Td(s)KP=36,KI=200Ω(s)=36s+2004s2+40s+200Ωr(s)s4s2+40s+200Td(s)

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

Ω(s)=36s+2004s2+40s+200Ωr(s)s4s2+40s+200Td(s)Ω(s)=36s+2004s2+40s+2001ss4s2+40s+2000Ω(s)=36s+2004s2+40s+2001s

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

Ω(s)=36s+2004s2+40s+2001sΩ(s)=1s(s+1)s2+10s+50

On taking inverse Laplace transform of this, we have

Ω(s)=1s(s+1)s2+10s+50=1s(s+5)(s+5)2+(5)2+451(s+5)2+(5)2ω(t)=1e5tcos(5t)+0.8e5tsin(5t)

Case 2. When ζ=1, the gain values are KI=100 and KP=36:

Since, Ω(s)=sKP+KI4s2+s(4+KP)+KIΩr(s)s4s2+s(4+KP)+KITd(s)

Ω(s)=36s+1004s2+s(4+36)+100Ωr(s)s4s2+s(4+36)+100Td(s)KP=36,KI=100Ω(s)=36s+1004s2+40s+100Ωr(s)s4s2+40s+100Td(s)

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

Ω(s)=36s+1004s2+40s+100Ωr(s)s4s2+40s+100Td(s)Ω(s)=36s+1004s2+40s+1001ss4s2+40s+1000Ω(s)=36s+1004s2+40s+1001s

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

Ω(s)=36s+1004s2+40s+1001sΩ(s)=1s+4(s+5)21(s+5)

On taking inverse Laplace transform of this, we have

Ω(s)=1s+4(s+5)21(s+5)ω(t)=1+e5tte5t

Case 3. When the root separation factor is 10.

That is, the gain values are KI=1000 and KP=216:

Since, Ω(s)=sKP+KI4s2+s(4+KP)+KIΩr(s)s4s2+s(4+KP)+KITd(s)

Ω(s)=216s+10004s2+s(4+216)+1000Ωr(s)s4s2+s(4+216)+1000Td(s)KP=216,KI=1000Ω(s)=216s+10004s2+220s+1000Ωr(s)s4s2+220s+1000Td(s)

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

Ω(s)=216s+10004s2+220s+1000Ωr(s)s4s2+220s+1000Td(s)Ω(s)=216s+10004s2+220s+10001ss4s2+220s+10000

Ω(s)=216s+10004s2+220s+10001s

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

Ω(s)=216s+10004s2+220s+10001sΩ(s)=1s+445(s+5)4945(s+50)

On taking inverse Laplace transform of this, we have

Ω(s)=1s+445(s+5)4945(s+50)ω(t)=1+445e5t4945e50t

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

System Dynamics, Chapter 10, Problem 10.22P , additional homework tip  3

Figure 1

From Figure 1,it is observed that there is a high overshoot in the responseat the damping ratio ζ=0.707.

Conclusion:

Therefore, the response Ω(s) for the unit-step command response Ωr(s) for all the cases is shown in Figure 1. At the damping ratio ζ=0.707, there is a high overshoot in the response in comparison to the other cases where damping ratios are 1 and 1.739 respectively.

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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...
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