lecture slides chapter07

7/25/2019 Lecture Slides CHAPTER07

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ME464-Sys Dyn & Ctrl Spring-2013 Dr. Shaukat Ali


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The Root Locus Method

Chapter 7


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Introduction

The root locus concept

The root locus procedure

PID controllers

Outline


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The root locus is a powerful tool for design and analysis offeedback control systems. The transient performance of a feedback control system is directly

related to the location of the poles in the s-plane.

As we change one or more system parameters, the location ofpoles in s-plane changes.

Therefore, it is necessary to determine how the poles move in the

s-plane as the parameters are varied. Root locus (RL):

It is the path of the roots of the characteristic equationtraced out in the s-plane as a system parameter is changed.

For example: changes in the parameters of a controller

This is crucial because one can adjust the system response to getthe desired performance through proper selection of controllerparameters.

7.1: Introduction


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Root Locus Method:

A graphical method for sketching the locus of roots in the s-plane as a parameter is varied.

This method gives a qualitative information about thestability and performance of the system.

7.1: Introduction


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Root locus:

It is the path of the roots of the characteristic equationtraced out in the s-plane as a system parameter is changed.

Example:

7.2: The Root Locus Concept

0102 Ksssq

10

1

sssG

Kss

KsT

10

2


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Pole location as a function of gain



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Root locus:

As K varies the locationof the poles is changed.

Hence root locus showsthe changes in the

transient response as theparameter K is varied.



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Remarks:

For K25, the system has complex roots, and hence is: Underdamped

In underdamped case the real parts of the poles are alwaysthe same regardless of the value of the gain. This means that

the settling time remains the same. Note: for higher order systems, this analysis is not

practical.



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The performance of a control system is described by the

closed-loop transfer function:

The roots of the characteristic equation determine the modes

of response of the system. Example:


)(

)(

sD

sN

sR

sYsT

01 sKG

sKG

sKGsT

1


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Characteristic equation:

Where K is a variable parameter.

The characteristic roots (poles) of the system must satisfythis equation.

The characteristic equation is a function of


01 sKG

js

01 sKG


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Complex number algebra:

In polar form: Magnitude/absolute value

Angle/phase


0y0,xifNAN

0y0,xif

0y0,xif

0y0,xiftan

0y0,xiftan

0xiftan

2

2

x

y1

x

y1

x

y1

s

Imaginary

Real

planes

s22 yxs

x

y


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Complex function:

The function of a complex number is also a complex number

In case of control systems:

The magnitude


assF

j

n

j

i

m

i

ps

zssF

1

1

lengthspole

lengthszero

1

1

j

n

j

i

m

i

ps

zs

sF


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The phase or angle:

Example:

Find F(s) at the point s=-3+j4


n

j

j

m

i

i pszs

sF

11

anglespoleangleszero

2

1

ss

ssF


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KG(s) is a function of s and hence is a complex number

In polar form

Therefore:


1 sKG

01 jsKG

01 jsKGsKG

1sKG

,540,180)21(180360180 kksKG

,3,2,1,0 k


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Conclusion:

A pole of a system causes the angle of KG(s) (or simply G(s)since K is a scalar) to be an odd multiple of 180.

A pole of a system causes the absolute value of KG(s) to beequal to 1.



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An orderly process for locating the locus of roots as a

parameter varies. STEP 1: Locate the poles and zeros of P(s) in the s-plane

Write the characteristic equation as:

Then rearrange it in such away that the parameter of interest, K,appears as a multiplying factor:

Factors P(s) and write it in form of poles and zeros:

7.3: The Root Locus Procedure

01 sF

KsKP 0for01

01

1

1

j

n

j

i

m

i

ps

zs

K


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Locate the poles (-pi) and zeros (-zi) on the s-plane. By

convention, we use x to denote poles and o to denotezeros.

Rearrange the characteristic equation:

For K=0, the roots of the characteristic equation are thepoles of the P(s), since:

For K=, the roots of the characteristic equation are the

zeros of the P(s), since:


011

i

m

i

j

n

j

zsKps

01

i

m

i

zs

01

j

n

j

ps


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Property of RL:

The RL of the characteristic equation 1+KP(s)=0 begins at thepoles of P(s) and ends at the zeros of P(s) as K increases fromzero to infinity. (verified)

STEP 2: Locate the segments of the real axis that are the root loci.

Property of RL: The root locus on the real axis always lies in a section of the real

axis to the left of an odd number of poles and zeros.

This property can be verified by the angle criterion.

A branch of the RL is the locus of one root when K varies.

The number of braches of RL is equal to the number ofpoles of P(s).



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Property of RL:

The RL are symmetrical with respect to the horizontal realaxis because the complex roots appears as pairs of complexconjugate roots.

Complex conjugate roots:

Roots having same real parts but with imaginary parts equalin magnitude and opposite in sign.



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Example:

characteristic equation of a control system:


01)(1 2

4

1

2

1 1

ss

sKsF


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STEP 3: proceed the loci to the zeros at infinity along

the asymptotes. Asymptotes: it is the path that the root locus follows as the

parameter becomes very large and approaches infinity.

The number of asymptotes is equal to the number of poles

minus the number of zeros. Angle of the asymptote: the angle that the asymptote makes

with the real axis.

Real axis intercept: The asymptotes are centered at a point onthe real axis:


1,2,1,0180

12

mnk

mn

k

A

mn

zp

mn

zerospoles

A

m

i

i

n

j

j

11


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Example


242

111

sss

sKsF


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STEP 4: Determine the break away point on the real

axis (if any). The breakaway is a point on the real axis at which the root

locus leave the real axis.

To compute the break away point, we determine the

maximum of K. That is:

The roots will give the break away point.

The breakaway point can also be evaluated graphically.


01 sKP

01 sPdsddsdK sPK

1


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STEP 5: Determine the angle of departure of the

locus from a pole. Angle of locus departure from a pole is the difference

between the net angle due to all other poles and zeros and thecriterion angle.



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STEP 6: Determine where the locus crosses the

imaginary axis (if it does so), using the Routh-Hurwitzcriterion.

STEP 7: complete the sketch of the root loci.



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Example 7.4:

Plot the RL for a system with following characteristicequation as K varies for K>0.

STEP1:

Zeros: this system has no finite zerosPoles at:

As the number of poles n=4, we have 4 separate loci.

The RL are symmetrical with respect to the real axis


01 44444 jsjsssK

444441

jsjssssP

jj,, 44,4440


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STEP2:

A segment of the RL exists on the real axis between s=0 ands=-4.

STEP3:

Number of asymptotes=n-m=4-0=4

The angles of the asymptotes

The intercept/center of the asymptotes


1,2,1,018012

mnk

mn

kA

mn

zp

A

m

i

i

n

j

i

11


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STEP4:

The breakaway point is evaluated through:

Or graphically, the breakaway point is between 0 and -4.

STEP5:

Angle of departure of RL from a pole:

STEP6:

The imaginary axis values of the loci.

STEP7:

Sketch the complete RL


01 sPdsd

ds

dK


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PID (Proportional-Integral-Derivative) controller:

In industry, the most common controller that has so far beenimplemented is the simple proportional plus integral plusderivative (PID) controller.

The general form of the PID control (actuating) signal is:

7.6: Three-Term (PID) Controller

dt

tde

DIP KdtteKteKtu

Controller Process

Sensor

Measured output

_

+ sY sR

sE sU

sG c sG

p

1


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The transfer function of the PID Controller is:

If KP is the only non-zero term, the controller is said to be aproportional controller or P-control for short.

If KI is the only non-zero term, the controller is said to be anintegral controller or I-control for short.

If KP and KI are the only non-zero terms, the controller issaid to be aproportional plus integral controller or PI-

control for short.

If KP and KD are the only non-zero terms, the controller issaid to be aproportional plus derivative controller or PD-control for short.


sKK Ds

K

PsE

sU I


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Objective:

Choose a suitable controller Gc(s) to make the closed-loopsystem:

1) stable;

2) the output y(t) of the plant be close to the reference signal

r(t) (i.e. error ess should be small)


1

1

s


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Proportional control (P-control)

The controller transfer function is:

The system transfer function:

Stability: from R-H criterion, the system is stable if:

Steady-state error


Pc

KsG

PP

Ks

K

sT 1

01

0

PK

P

ssK

e

1

1


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Proportional control (P-control)

The unit step response is:

For quick response (i.e. less settling time), the KP must be as

large as possible, for a given time constant .


tK PK

P ety

1

1

planes

Imaginary

Real

PK

1


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Integral-control (I-control)

The problem with a P controller is that the steady-state error,ess, is not zero for a step input in case of type zero system.

Can we design a controller which makes ess = 0?

Yes. We need to introduce an integral term.

The output signal of an Integral controller has the form:

The controller transfer function is:

The system transfer function (2nd order system):


dtteKtuI

s

K

c

IsG

I

I

Kss

KsT

2


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Stability: from R-H criterion, the system is stable if:

Steady-state error in case of step input:

An integral controller eliminated the steady-state error of astep response .


0

0

IK

0

lim10

s

sssG

Ae


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Unit step response:

For a given time constant, KI can be adjusted to get thedesired transient response of the system

Example:


22

2

12 2 nn

n

IK

IK

sssssT

n

IK

n 212

,

10

1

s

sGp


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PI-Control

A proportional plus integral (PI) controller has the transferfunction:

The task is to tune the control parameters KP and KI toachieve a better control.

By combining the advantages of P and I controllers, we havemore freedom of choice and can achieve better performancesince there are two parameters to tune.


s

K

Pc

IKsG

1

1

s


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Summary

s

KsKP 2s

KsKP 3s

KsKP


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Summary

11

s

KsKP

11

ss

KsKP

11

1

ss

sK asKP


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Summary

11

21

ss

KsKP 11

21

sss

KsKP

111

21

sss

sK asKP


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Example 7.7: Automobile Velocity Control

A velocity control system for maintaining the relative velocitybetween the two vehicles.

The output Y(s) is the relative velocity between the twovehicles.

The input R(s) is the desired relative velocity between the two

vehicles. The transfer function of the process (automobile system) is:

Objective: To design a controller that can maintain a prescribed velocity

between the vehicles and maneuver the active vehicle.

7.7: Design Examples

82

1

ssp sG


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Example 7.7: Automobile Velocity Control continued



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Control Goal:

Maintain a prescribed velocity between the vehicles and maneuver theactive vehicle.

Variable to be controlled:

The relative velocity between the vehicles, denoted by y(t)

Design specifications:

DS1: zero steady state error to a step input.

DS2: steady state error due to a ramp input of less than 25% of theinput magnitude.

DS3: Percent overshoot less than 5% to a step input

DS4: settling time less than 1.5 seconds to a step input (using a 2%criterion to establish settling time)



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System configuration:

Obtain mathematical models:


Controller Process

_

+

sY sR sG

c sG

p

82

1

ssp sG


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