System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 10, Problem 10.24P
To determine

The steady-state error due to a unit-ramp command and due to a unit-ramp disturbance of the PI controller used in first order plant for the given cases.

Expert Solution & Answer
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Answer to Problem 10.24P

The peak torque values for all the cases are:

Case 1. For ζ=0.707 ,

Error due to unit-ramp command: ess=0.02

Error due to unit-ramp disturbance: ess=0.005

Case 2. For ζ=1, Tm35Nm.

Error due to unit-ramp command: ess=0.04

Error due to unit-ramp disturbance: ess=0.01

Case 3. For a root separation factor of 10, Tm215Nm.

Error due to unit-ramp command: ess=0.004

Error due to unit-ramp disturbance: ess=0.001.

Explanation of Solution

Given:

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

System Dynamics, Chapter 10, Problem 10.24P

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

  3. The steady-state error of a system using final value theorem is:
  4. ess=lims0sE(s)

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)

And from the block diagram shown in figure, we have

E(s)=(Ωr(s)Ω(s))E(s)=(1sKP+KIIs2+s(c+KP)+KI)Ωr(s)+sIs2+s(c+KP)+KITd(s)Ω(s)=sKP+KIIs2+s(c+KP)+KIΩr(s)sIs2+s(c+KP)+KITd(s)

On keeping the values of the parameters such that I=c=4

E(s)=(1sKP+KIIs2+s(c+KP)+KI)Ωr(s)+sIs2+s(c+KP)+KITd(s)E(s)=(s(Is+c)Is2+s(c+KP)+KI)Ωr(s)+sIs2+s(c+KP)+KITd(s)E(s)=(s(4s+4)4s2+s(4+KP)+KI)Ωr(s)+s4s2+s(4+KP)+KITd(s)

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

Since,

E(s)=(s(4s+4)4s2+s(4+KP)+KI)Ωr(s)+s4s2+s(4+KP)+KITd(s)KI=200andKP=36E(s)=(s(4s+4)4s2+s(4+36)+200Ωr(s)+s4s2+s(4+36)+200Td(s))E(s)=(s(4s+4)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))

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

E(s)=(s(4s+4)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))E(s)=(s(4s+4)4s2+40s+2001s2+s4s2+40s+2000)E(s)=(4s+4)(4s2+40s+200)1s

E(s)=(s+1)(s2+10s+50)1s

Thus, the steady-state error for this design is:

ess=lims0sE(s)ess=lims0s(s+1)(s2+10s+50)1s E(s)=(s+1)(s2+10s+50)1sess=lims0(s+1)(s2+10s+50)=150ess=0.02

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

E(s)=(s(4s+4)4s2+40s+200Ωr(s)+s4s2+40s+200Td(s))E(s)=(s(4s+4)4s2+40s+2000+s4s2+40s+2001s2)E(s)=1(4s2+40s+200)1sE(s)=14(s2+10s+50)1s

Thus, the steady-state error for this design is:

ess=lims0sE(s)ess=lims0s14(s2+10s+50)1s E(s)=14(s2+10s+50)1sess=lims014(s2+10s+50)=1200ess=0.005

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

Since,

E(s)=(s(4s+4)4s2+s(4+KP)+KI)Ωr(s)+s4s2+s(4+KP)+KITd(s)KI=100andKP=36E(s)=(s(4s+4)4s2+s(4+36)+100Ωr(s)+s4s2+s(4+36)+100Td(s))E(s)=(s(4s+4)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))

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

E(s)=(s(4s+4)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))E(s)=(s(4s+4)4s2+40s+1001s2+s4s2+40s+1000)E(s)=(4s+4)(4s2+40s+100)1sE(s)=(s+1)(s2+10s+25)1s

Thus, the steady-state error for this design is:

ess=lims0sE(s)ess=lims0s(s+1)(s2+10s+25)1s E(s)=(s+1)(s2+10s+25)1sess=lims0(s+1)(s2+10s+25)=125ess=0.04

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

E(s)=(s(4s+4)4s2+40s+100Ωr(s)+s4s2+40s+100Td(s))E(s)=(s(4s+4)4s2+40s+1000+s4s2+40s+1001s2)

E(s)=1(4s2+40s+100)1sE(s)=14(s2+10s+25)1s

Thus, the steady-state error for this design is:

ess=lims0sE(s)ess=lims0s14(s2+10s+25)1s E(s)=14(s2+10s+25)1sess=lims014(s2+10s+25)=1100ess=0.01

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

Since,

E(s)=(s(4s+4)4s2+s(4+KP)+KI)Ωr(s)+s4s2+s(4+KP)+KITd(s)KI=1000andKP=216E(s)=(s(4s+4)4s2+s(4+216)+1000Ωr(s)+s4s2+s(4+216)+1000Td(s))E(s)=(s(4s+4)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))

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

E(s)=(s(4s+4)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))E(s)=(s(4s+4)4s2+220s+10001s2+s4s2+220s+10000)E(s)=(4s+4)(4s2+220s+1000)1sE(s)=(s+1)(s2+55s+250)1s

Thus, the steady-state error for this design is:

ess=lims0sE(s)ess=lims0s(s+1)(s2+55s+250)1s E(s)=(s+1)(s2+55s+250)1sess=lims0(s+1)(s2+55s+250)=1250ess=0.004

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

E(s)=(s(4s+4)4s2+220s+1000Ωr(s)+s4s2+220s+1000Td(s))E(s)=(s(4s+4)4s2+220s+10000+s4s2+220s+10001s2)E(s)=1(4s2+220s+1000)1sE(s)=14(s2+55s+250)1s

Thus, the steady-state error for this design is:

ess=lims0sE(s)ess=lims0s14(s2+55s+250)1s E(s)=14(s2+55s+250)1sess=lims0(s+1)4(s2+55s+250)=11000ess=0.001.

Conclusion:

The peak torque values for all the cases are:

Case 1. For ζ=0.707 ,

Error due to unit-ramp command: ess=0.02

Error due to unit-ramp disturbance: ess=0.005

Case 2. For ζ=1, Tm35Nm.

Error due to unit-ramp command: ess=0.04

Error due to unit-ramp disturbance: ess=0.01

Case 3. For a root separation factor of 10, Tm215Nm.

Error due to unit-ramp command: ess=0.004

Error due to unit-ramp disturbance: ess=0.001.

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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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