Ch.09 Design via Root Locus

of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics and Control 9.07 Design via Root Locus The system operating with a desirable transient response generated

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09 Design via Root Locus

HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Learning Outcome

After completing this chapter, the student will be able to

• Use the root locus to design cascade compensators to improve the steady-state error

• Use the root locus to design cascade compensators to improve the transient response

• Use the root locus to design cascade compensators to improve both the steady-state error and the transient response

• Use the root locus to design feedback compensators to improve the transient response

• Realize the designed compensators physically

§1.Introduction

- The root locus typically allows us to choose the proper loop

gain to meet a transient response specification

- Setting the gain at a particular value yields the transient

response dictated by the poles at that point on the root locus

Improving Transient Response

System Dynamics and Control 9.03 Design via Root Locus

Responses from poles at 𝐴 and 𝐵 Sample root locus, showing possible

design point via gain adjustment 𝐴 and

desired design point that cannot be met via

simple gain adjustment 𝐵

§1.Introduction Improving Steady-State Error

Compensators can be used independently to improve the steady-state error characteristics

Configurations

• Cascade compensation

• Feedback compensation

HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics and Control 9.04 Design via Root Locus

§1.Introduction

Compensators

- Ideal compensators

• pure integration for improving steady-state error

• pure differentiation for improving transient response

- Ideal compensators must be implemented with active networks,

which, in the case of electric networks, require the use of active

amplifiers and possible additional power source

- An advantage of ideal integral compensators is that

steady-state error is reduced to zero Electromechanical ideal

compensators, such as tachometers, are often used to improve

transient response, since they can be conveniently interfaced

with the plant

§2.Improving Steady-State Error via Cascade Compensation Ideal Integral Compensation (PI)

The system operating with a desirable transient response generated by the closed-loop poles at 𝐴

Add a pole at the origin to increase the open-loop poles at point 𝐴 is no longer

180 0 , and the root locus no longer goes through point 𝐴

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§2.Improving Steady-State Error via Cascade Compensation

The system operating with a desirable

transient response generated by the

closed-loop poles at 𝐴

Add a zero close to the pole at the origin, the angular contribution of the compensator zero and compensator pole cancel out, point 𝐴 is still on the root locus, and the system type has been increased

A compensator with a pole at the

origin and a zero close to the pole is

called an ideal integral compensator

- Ex.9.1 Effect of an Ideal Integral Compensator

Given the system, operating with 𝜁 = 0.174, show that the

addition of the ideal integral compensator reduces the steady-state error to zero for a step input without appreciably affecting transient response

The compensating network network is chosen with a pole at the origin to increase the system type and a zero at−0.1, close to the compensator pole,

so that the angular contribution of the compensator evaluated

at the original, dominant, 2nd-order poles is approximately zero

Thus, the original, dominant, 2nd-order closed-loop poles are still approximately on the new root locus

Solution

The dominant poles −0.694 ± 𝑗3.926 for a gain, 𝐾 = 164.6

The third pole −11.61

𝐾𝑝= lim

𝑠→0𝐾𝐺 𝑠 = 8.23 ⟹ 𝑒 ∞ = 1

1 + 𝐾𝑝=

1

1 + 8.23= 0.108 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Adding an ideal integral compensator with a zero at−0.1 The dominant poles −0.678 ± 𝑗3.837 for a gain, 𝐾 = 158.2 The third pole −11.55

The fourth pole −0.0902 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics and Control 9.10 Design via Root Locus

The step response of

• the ideal integral compensated system 𝑐(∞) → 1.000

• the uncompensated system 𝑐(∞) → 0.892

The compensated system reaches the uncompensated

system’s final value in about the same time The remaining time

is used to improve the steady-state error over that of the

uncompensated system

The ideal integral compensator that improved steady-state error was implemented with a proportional-plus-integral (PI) controller

𝐺𝑐𝑠 = 𝐾1+𝐾2

𝑠 =

𝐾1 𝑠 +𝐾1

𝐾2 𝑠 The value of the zero can be adjusted by varying𝐾2/𝐾1 In this implementation, the error and the integral of the error are fed forward to the plant,𝐺(𝑠)

(9.2)

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Lag Compensation

Type 1 uncompensated system

Root locus before lag compensation

Root locus after lag compensation

Type 1 compensated system

Compensate the system, whose root locus is shown, to

improve the steady-state error by a factor of10 if the system is operating with𝜁 = 0.174 Solution

The uncompensated system error (Ex.9.1) is0.108 with 𝐾𝑝= 8.23

A tenfold improvement means a steady-state error of 𝑒 ∞ = 0.108/10 = 0.0108

1 + 𝐾𝑝= 0.0108

⟹ 𝐾𝑝=1 − 𝑒 ∞

1 − 0.0108 0.0108 = 91.59 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

The improvement in𝐾𝑝from the uncompensated system to the

compensated system is the required ratio of the compensator

zero to the compensator pole, or

𝑧𝑐

𝑝𝑐=𝐾𝑝𝑁

𝐾𝑝 𝑂

=91.59

8.23 = 11.13 Arbitrarily selecting

𝑝𝑐= 0.01

to get

𝑧𝑐= 11.13𝑝𝑐= 11.13 × 0.01 ≈ 0.111

Compensated system

Sketch the root locus of the compensated system

The dominant poles −0.678 ± 𝑗3.836 for a gain, 𝐾 = 158.1 The third pole −11.55

The fourth pole −0.101 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics and Control 9.16 Design via Root Locus

All transient and steady-state results for both the uncompensated

and the compensated systems

The fourth pole of the compensated system cancels its zero

The three closed-loop poles of the compensated system≅ the

uncompensated system ⟹ the transient response of both

systems is approximately the same, as is the system gain

Step responses of uncompensated and lag compensated systems

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TryIt 9.1

Use the following MATLAB

and Control System Toolbox

statements to reproduce

Fig.9.13.

Gu=zpk([ ],

[-1 -2 -10],164.6);

Gc=zpk([-0.111],[-0.01],1);

Gce=Gu*Gc;

Tu=feedback(Gu,1);

Tc=feedback(Gce,1);

step(Tu)

hold

step(Tc)

§2.Improving Steady-State Error via Cascade Compensation Skill-Assessment Ex.9.1

Problem A unity feedback system with the forward TF

𝑠(𝑠 + 7)

is operating with a closed-loop step response that has 15% overshoot Do the following

a.Evaluate the steady-state error for a unit ramp input

b.Design a lag compensator to improve the steady-state error by a factor of20

c Evaluate the steady-state error for a unit ramp input

to your compensated system d.Evaluate how much improvement in steady-state error was realized

Solution a.Evaluate the steady-state error for a unit ramp input

The point on root locus

−3.5 + 𝑗5.8,𝐾 = 45.84

For uncompensated system

𝐾𝑣= lim

𝑠→0𝑠𝐺(𝑠)

=𝐾

7=

45.84

7 = 6.55 ቚ

𝑒

𝐾𝑣= 0.1527 b.Design a lag compensator to improve the

steady-state error by a factor of20

Compensator zero should be20 × further to the left

than the compensator pole Arbitrarily select

𝐺𝑐𝑠 = (𝑠 + 2)/(𝑠 + 0.01)

c Evaluate the steady-state error for a unit ramp input The point on root locus

−3.4 + 𝑗5.63,𝐾 = 44.64 For compensated system

𝐾𝑣= lim

𝑠→0𝑠𝐺(𝑠)

=𝐾 × 0.2

7 × 0.01= 127 5 ቚ

𝑒

𝐾𝑣= 0.0078 d.Evaluate how much improvement in steady-state error ȁ

𝑒𝑟𝑎𝑚𝑝𝑐𝑜𝑚𝑝𝑒𝑛𝑠𝑎𝑡𝑒𝑑 ȁ

𝑒𝑟𝑎𝑚𝑝𝑢𝑛𝑐𝑜𝑚𝑝𝑒𝑛𝑠𝑎𝑡𝑒𝑑=

0.1527 0.0078= 19.58

⟹ 19.58 times improvement System Dynamics and Control 9.22 Design via Root Locus

§3.Improving Transient Response via Cascade Compensation

Ideal Derivative Compensation (PD)

Ideal derivative, or PD controller 𝐺𝑐𝑠 = 𝑠 + 𝑧𝑐 (9.12)

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Summarizes the results obtained from the root locus of each of the design cases

Uncompensated system and ideal derivative compensation

solutions

- Ex.9.3 Ideal Derivative Compensator Design

Given the system, design an ideal derivative compensator to

yield a 16% overshoot, with a threefold reduction in settling time

𝜁 = − 𝑙𝑛 %𝑂𝑆/100

𝜋2+ 𝑙𝑛2%𝑂𝑆/100 = −

𝑙𝑛 16%

𝜋2+ 𝑙𝑛216% = 0.504 For the uncompensated system

• dominant, 2nd-order poles−1.205 ± 𝑗2.064

• settling time, 𝑇𝑠𝑢

𝑇𝑠𝑢= 4

𝜁𝜔𝑛=

4 1.205

= 3.320 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Solution

• the transient and steady-state error characteristics

For the compensated system

• settling time

• dominant, 2nd-order poles

𝜎 =4

𝑇𝑠= 4 1.107= 3.613

𝜔𝑑= −3.613 𝑡𝑎𝑛 120.260

= 6.193

𝑇𝑠=1

3× 𝑇𝑠

𝑢=1

3× 3.320 = 1.107

- real part

- imaginary part

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Design the location of the compensator zero

Using the open-loop poles and the desired dominant 2nd-order

pole,−3.613 ± 𝑗6.193

𝜃𝑧 𝑐− 𝜃𝑝 1− 𝜃𝑝 2− 𝜃𝑝 3= 1800

𝜃 𝑝 1 = 𝑎𝑛𝑔𝑙𝑒((−3.613 + 6.193 ∗ 𝑗) − 0) = 120.26 0

𝜃𝑝2= 𝑎𝑛𝑔𝑙𝑒((−3.613 + 6.193 ∗ 𝑗) − (−4)) = 86.42 0

𝜃𝑝3= 𝑎𝑛𝑔𝑙𝑒((−3.613 + 6.193 ∗ 𝑗) − (−6)) = 68.92 0

⟹ 𝜃𝑧 𝑐= 1800+ 𝜃𝑝 1+ 𝜃𝑝 2+ 𝜃𝑝 3= 455.60

= 95.60

The location of the compensator zero

6.193

−3.613 − (−𝜎)= 𝑡𝑎𝑛 95.6

0

⟹ 𝜎 = 3.006

Summarizes the results for both the uncompensated system and the compensated system

Uncompensated and compensated system step responses

Root locus for the compensated system

Run ch9p1 in Appendix B Learn how to use MATLAB to

• design a PD controller

• solve Ex.9.3

TryIt 9.2

Use MATLAB, the Control System Toolbox, and the following steps to use SISOTOOL

to perform the design of Ex.9.3

1 Type SISOTOOL in the MATLAB Command Window

2 Select Import in the File menu of the SISO Design for SISO Design Task Window

3 In the Data field for G, type zpk([ ], [0, −4, −6], 1) and hit ENTER on the keyboard Click OK

4 On the Edit menu choose SISO Tool Preferences … and select Zero/pole/gain : under the Options

tab Click OK

5 Right-click on the root locus white space and choose Design Requirements/New

6 Choose Percent overshoot and type in 16 Click OK

7 Right-click on the root locus white space and choose Design Requirements/New

8 Choose Settling time and click OK

9 Drag the settling time vertical line to the intersection of the root locus and 16% overshoot radial line

10.Read the settling time at the bottom of the window

11.Drag the settling time vertical line to a settling time that is 1/3 of the value found in Step 9

12.Click on a red zero icon in the menu bar Place the zero on the root locus real axis by clicking again

on the real axis

13.Left-click on the real-axis zero and drag it along the real axis until the root locus intersects the

settling time and percent overshoot lines

14.Drag a red square along the root locus until it is at the intersection of the root locus, settling time

line, and the percent overshoot line

15.Click the Compensator Editor tab of the Control and Estimation Tools Manager window to see the

resulting compensator, including the gain

- The ideal derivative compensator used to improve the transient response is implemented with a proportional-plus-derivative (PD) controller

𝐺𝑐𝑠 = 𝐾1+ 𝐾2𝑠 = 𝐾2 𝑠 +𝐾1

𝐾2

• 𝐾1/𝐾2is chosen to equal the negative of the compensator zero

• 𝐾2is chosen to contribute to the required loop-gain value System Dynamics and Control 9.36 Design via Root Locus

(9.17)

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- While the ideal derivative compensator can improve the

transient response of the system, it has two drawbacks

• First, it requires an active circuit to perform the differentiation

• Second, differentiation is a noisy process

The level of the noise is low, but the frequency of the noise is

high compared to the signal Differentiation of high

frequencies can lead to large unwanted signals or saturation

of amplifiers and other components

⟹ To overcome the disadvantages of ideal differentiation and still

retain the ability to improve the transient response: using the

passive networklead compensator

§3.Improving Transient Response via Cascade Compensation Lead Compensation

- An active ideal derivative compensator can be approximated with a passive lead compensator

- Advantages

• no additional power supplies are required, and

• noise due to differentiation is reduced

- Disadvantage

• the additional pole does not reduce the number of branches of the root locus that cross the imaginary axis into the right half-plane, or

• the addition of the single zero of the PD controller tends to reduce the number of branches of the root locus that cross into the right half-plane

Design three lead compensators for the system that will reduce

the settling time by a factor of2 while maintaining30% overshoot Compare the system characteristics between the three designs

Solution

Determine the characteristics of the uncompensated system operating at

30% overshoot

𝑂𝑆% = 30% ⟹ 𝜁 = 0.358 The uncompensated settling time

𝑇𝑠𝑢= 4

𝜁𝜔𝑛= 4 1.007= 3.972 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

The summarized characteristics of the uncompensated system

Find the design point

The settling time

The desired pole location

• real part −𝜁𝜔𝑛= −4/𝑇𝑠= −2.014

• imaginary part 𝜔𝑑= −2.014𝑡𝑎𝑛 110.980= 5.252

Designing the lead compensator

Arbitrarily assume a compensator zero at

− 5 on the real axis as a possible solution

𝜃𝑧𝑐− 𝜃𝑝𝑐− 𝜃1− 𝜃2− 𝜃3= 1800

𝜃 1 = 𝑎𝑛𝑔𝑙𝑒((−2.014 + 5.252 ∗ 𝑗) − 0) = 110.98 0

𝜃2= 𝑎𝑛𝑔𝑙𝑒((−2.014 + 5.252 ∗ 𝑗) − (−4)) = 69.29 0

𝜃 3 = 𝑎𝑛𝑔𝑙𝑒((−2.014 + 5.252 ∗ 𝑗) − (−6)) = 52.80 0

𝜃 𝑧 𝑐 = 𝑎𝑛𝑔𝑙𝑒((−2.014 + 5.252 ∗ 𝑗) − (−5)) = 60.38 0

⟹ 𝜃𝑝 𝑐= 𝜃𝑧 𝑐− 1800− 𝜃1− 𝜃2− 𝜃3

= −352.690= 7.310

𝑇𝑠=1

2× 𝑇𝑠

𝑢=1

2× 3.972 = 1.986

Find the design point

The location of the compensator pole 5.252

−2.014 − (−𝑝𝑐)= 𝑡𝑎𝑛7.310⟹ 𝑝𝑐= 42.96

Justify the estimates of percent overshoot and settling time

The third and fourth poles are at −43.8 and −5.134 Since

− 43.8 is more than 20 times the real part of the dominant pole, the effect of the third closed-loop pole is negligible Since the closed-loop pole at−5.134 is close to the zero at −5, we have pole-zero cancellation, and the 2nd-order approximation is valid System Dynamics and Control 9.42 Design via Root Locus

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Comparison of lead compensation designs

Uncompensated system and lead compensation responses

Run ch9p2 in Appendix B

Learn how to use MATLAB to

• design a lead compensator

• solve Ex.9.4

§3.Improving Transient Response via Cascade Compensation Skill-Assessment Ex.9.2

Problem A unity feedback system with the forward TF

𝑠(𝑠 + 7)

is operating with a closed-loop step response that has 15% overshoot Do the following

a.Evaluate the settling time b.Design a lead compensator to decrease the settling time by three times Choose thecompensator’s zero

to be at−10 System Dynamics and Control 9.46 Design via Root Locus

Solution a.Evaluate the settling time

𝜁 = − 𝑙𝑛 %𝑂𝑆/100

𝜋 2 +𝑙𝑛 2 %𝑂𝑆/100 o

= − 𝑙𝑛 15%

𝜋2+ 𝑙𝑛215%

= 0.517

The desired point

𝑝 = −3.5 ± 𝑗5.8

𝐾 = 45.84

Setting time for uncompensated system

𝑇𝑠𝑢= 4

ȁ𝑅𝑒ȁ=

4 3.5= 1.143

b.Design a lead compensator For the compensated system

• settling time

𝑇𝑠=1

3× 𝑇𝑠

𝑢=1

3× 1.143 = 0.381

• dominant, 2nd-order poles

𝜎 =4

𝑇𝑠=

4 0.381= 10.50

𝜃0= 𝑎𝑛𝑔𝑙𝑒((−3.5 + 5.8 ∗ 𝑗) − 0) = 121.110

𝜔𝑑= −10.499 𝑡𝑎𝑛 121.110= 17.40 other way, the real part of the design point must be three times larger than the uncompensatedpole’s real part3 −3.5 + 𝑗5.8 = −10.50 + 𝑗17.40

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Designing the lead compensator

Compensator zero at−10 on the real axis

𝜃𝑧𝑐− 𝜃𝑝𝑐− 𝜃1− 𝜃2= 1800

𝜃1= 𝑎𝑛𝑔𝑙𝑒 −10.50 + 17.40 ∗ 𝑗 − 0

= 121.11 0

𝜃 2 = 𝑎𝑛𝑔𝑙𝑒 −10.50 + 17.40 ∗ 𝑗 − −7

= 101.37 0

𝜃 𝑧 𝑐 = 𝑎𝑛𝑔𝑙𝑒 −10.50 + 17.40 ∗ 𝑗 − −10

= 91.65 0

⟹ 𝜃𝑝 𝑐= 𝜃𝑧 𝑐− 1800− 𝜃1− 𝜃2

= 91.650− 1800− 121.110− 101.370

= −310.840

= 49.20

dominant, 2nd-order poles

𝜎 = 10.50

𝜔𝑑= 17.40

𝜔𝑝 𝑐= 49.20

The location of the compensator pole 17.40

−10.50 − (−𝑝𝑐)= 𝑡𝑎𝑛 49.2

0

⟹ 𝑝𝑐= 25.52 with this pole, the gain𝐾 = 476.3 A higher-order closed-loop pole is found to be at−11.54 This pole may not be close enough to the closed-loop zero at

− 10 Thus, we should simulate the system to be sure the design requirements have been met

§4.Improving Steady-State Error and Transient Response

PID Controller Design

A PID controller has two zeros plus a pole at the origin

• One zero and the pole at the origin can be designed as the

ideal integral compensator

• The other zero can be designed as the ideal derivative

compensator

(9.21)

𝐺𝑐𝑠 = 𝐾1+𝐾2

𝑠 + 𝐾2𝑠

=

𝐾3 𝑠2+𝐾1

𝐾3𝑠 +𝐾2

𝐾3

𝑠

The design technique consists of the following steps 1.Evaluate the performance of the uncompensated system to determine how much improvement in transient response is required 2.Design the PD controller to meet the transient response specifications The design includes the zero location and the loop gain

3.Simulate the system to be sure all requirements have been met 4.Redesign if the simulation shows that requirements have not been met

5.Design the PI controller to yield the required steady-state error 6.Determine the gains,𝐾1,𝐾2, and𝐾3

7.Simulate the system to be sure all requirements have been met 8.Redesign if simulation shows that requirements have not been met HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics and Control 9.52 Design via Root Locus

Given the system, design a PID controller so that the system

can operate with a peak time that is 2/3 that of the

uncompensated system at 20% overshoot and with zero

steady-state error for a step input

Solution

Step 1

Evaluate the performance of the uncompensated system operating

at 20% overshoot

• the dominant poles

−5.415 ± 𝑗10.57, 𝐾 = 121.5

• The third pole

−8.169, 𝐾 = 121.5 The complete performance of the uncompensated system is shown in the first column of Table 9.5, where we

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The complete performance of the uncompensated system

Step 2

Design the PD controller to meet the transient response specifications

The imaginary part of the compensated dominant pole

𝜔𝑑=𝜋

𝑇𝑝

=2 𝜋

3× 0.297

= 15.87

𝜎 = 𝜔𝑑 𝑡𝑎𝑛117.130

= −8.13 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien System Dynamics and Control 9.56 Design via Root Locus

The compensatingzero’s location

𝜃𝑧 𝑐+ 𝜃−8− 𝜃−3− 𝜃−6− 𝜃−10= 1800

𝜃 −8 = 𝑎𝑛𝑔𝑙𝑒((−8.13 + 15.87 ∗ 𝑗) − (−8))

= 90.47 0

𝜃−3= 𝑎𝑛𝑔𝑙𝑒((−8.13 + 15.87 ∗ 𝑗) − (−3))

= 107.91 0

𝜃 −6 = 𝑎𝑛𝑔𝑙𝑒((−8.13 + 15.87 ∗ 𝑗) − (−6))

= 97.64 0

𝜃 −10 = 𝑎𝑛𝑔𝑙𝑒((−8.13 + 15.87 ∗ 𝑗) − (−10))

= 83.28 0

⟹ 𝜃𝑧𝑐= 1800− 𝜃−8+ 𝜃−3+ 𝜃−6+ 𝜃−10

= 18.370

15.87

−8.13 − (−𝑧𝑐)= 𝑡𝑎𝑛18.370→ 𝑧𝑐= 55.92 The PD controller

𝐺𝑃𝐷𝑠 = 𝑠 + 55.92 HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

The complete root locus for the PD-compensated system

Complete specifications for ideal derivative compensation

§4.Improving Steady-State Error and Transient Response Step 3&4

Simulate the system to be sure all requirements have been met

Step responses for uncompensated, PD-compensated, and PID-compensated systems

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