Chapter 6 Analysis of Feedback Control Systems Prof. Shi-Shang Jang Chemical Engineering Department...

Chapter 6Chapter 6Analysis of Feedback Analysis of Feedback Control SystemsControl Systems

Prof. Shi-Shang JangChemical Engineering DepartmentNational Tsing-Hua UniversityHsin Chu, TaiwanJune, 2015

6-1 Introduction6-1 IntroductionGiven a plant with a transfer function of

y(s)/m(s)=Gp(s), a controller is implemented with a transfer function of Gc(s), the system hence becomes a closed loop function.

It is very convenient to analyze the closed loop system in s-domain since only algebraic equations are involved.

In a closed loop system, there are two inputs for the system, the set point (servo) of the output and the disturbances (regulation) that affect the output.

6-1 Introduction 6-1 Introduction – Cont– Cont..

Variations in outlet temperature are sensed by the sensor-transmitter andsent to the controller causing the controller output signal to vary. This isin turn causes the control valve position and consequently the steam flowTo vary. The variations in steam flow cause the outlet temperature to vary, thus completing the loop.

Example: The Heat Exchanger Example: The Heat Exchanger System System

6-1.1 Closed-Loop Transfer Function6-1.1 Closed-Loop Transfer Function

(1)Set Point Change

0

1

setsp o

c

s v

o s s w

o

sp s v coseto s v c

E s K T C s

M s G s E s

W s G s M s

T s G s W s G s W s

C s H s T s

W s

K G s G s G sT s

T s H s G s G s G s

6-1.1 Closed-Loop Transfer Function - 6-1.1 Closed-Loop Transfer Function - ContinuedContinued

(2)Load Changes

0

( )

1

(3)General Case

% / %

/ % %

dimensionless

( )

1 1

(4)Closed

seto

o w

s v c

s v c

sp s v c setwo o

s v c s v c

T

T s G s

W s H s G s G s G s

TO C kg s COH s G s G s G s

C kg s CO TO

K G s G s G sG sT s W s T s

H s G s G s G s H s G s G s G s

Loop Transfer Functions

1f

i l

Z

Z

Z=OutputZi=Inputf=product in forward path from Zi to Zl=product of every transfer function infeedback loop


sFsGsG

sGsR

sGsG

sGsGsC

sHsGsG

sHsGsGsG

cc

c

w

sv

1

2

1

1

2

1

11

ExampleExample

Example – Example – Cont.Cont.

sWs

Ks

ss

ss

KsF

s

Ks

Laplacing

tUAtUAtWdt

dC

tUAcftUAtFTTcdt

dcV

ionLinearizat

tTtTUAtwdt

dTC

tTctftTtTUAtTctfdt

dTcV

c

w

cs

ssF

ss

M

psipp

ss

M

psipp

11

111


1100200

0100;

1

100

ln;

1

Dynamics Valve Control

TT

TT

vv

vv

Ks

KsG

rTransmitteSensor

wK

s

KsG


sGsWsGsF

s

K

ss

s

KsW

s

KsF

s

K

s

KsFs

sF

s

cc

wF

ss

F

11

1

11

11


sFsGsG

sGsR

sGsG

sGsGsC

sHsGsG

sHsGsGsG

cc

c

w

sv

1

2

1

1

2

1

11


1

2

1 2

1 1

1.652 1.183 1.0

0.2 1 8.34 1 0.502 1 0.75 1

3.34 0.524 1 1.0

8.34 1 0.502 1 0.75 1

1 1

v s

F

c

c c

G s G s G s H ss s s s

sG s G s H s

s s s

G s G s G sC s R s F s

G s G s G s G s

6.1.3 Steady State of Closed Loop 6.1.3 Steady State of Closed Loop SystemsSystems

OL

w

svcT

w

svc

wo

svc

wo

K

K

KKKK

K

GGGH

G

W

T

sGsGsGsH

sG

sW

sT

11

00001

0

0

0

1

OL

svcsp

svcT

svcsp

svc

svcsp

seto

o

svc

svcsp

seto

o

K

KKKK

KKKK

KKKK

GGGH

GGGK

T

T

sGsGsGsH

sGsGsGK

sT

sT

11

00001

000

0

0

1

ExampleExample

Example: New steady state of the Example: New steady state of the heating tankheating tank

1. Servo control with Kc=1;Tset=1

2 1.652 1.183 3.90860.7963

1 1 2 1.652 1.183 4.9086

1

150 1 150.7963

c v s

set c v s

set

K K KT

T K K K

T

T T

2. Regulation Control with F=1

3.34 -3.34-0.6804

1 1 2 1.652 1.183 4.9086

150 149.3196

F

c v s

KT

F K K K

T T

Example: Flow rate control Example: Flow rate control systemsystem

Example: Flow rate control Example: Flow rate control system – system – Cont.Cont.

1

1

11

1

11

1

;1

sKKKs

KKK

sF

sF

Set

sKKKss

sKKK

sF

sF

sKsG

sGsGK

sGsGK

sF

sF

KsHs

K

sM

sFsG

FCcvTv

cvTset

vI

IcvTvI

IcvTset

Icc

cvT

cvsp

set

Tv

vv

6.1.2 Characteristic Equation of 6.1.2 Characteristic Equation of the Loopthe Loop

11 0

1 2

1 2

1 2

1 2

1 ( ) 0

0

0

( )1 ( )

numerical terms( )

input terms

input terms

Unforced terms Forced terms

s v c

n nn n

n n

wo

s v c

on n

no

n

H s G s G s G s

a s a s a

a s r s r s r

G sT s W s

H s G s G s G s

T sa s r s r s r

bb bT t

s r s r s r

Example : Servo Problem for a First Order Example : Servo Problem for a First Order SystemSystem

11

1

11

11

1

11

11

11

1

PI)( ;1

1 : 2 Case

1'

'

11

1

11

1

1

; : 1 Case

0;11

sssKK

sKK

sK

ssK

sK

ssK

sK

sK

sK

sK

sR

sC

sKG

s

K

KKs

KKKK

KKs

KK

sK

K

sK

K

sR

sC

KG

sFs

KsG

IIc

Ic

I

Ic

I

Ic

Ic

Ic

Icc

c

c

c

c

c

c

c

cc

Proportional only

6.1.3 Steady State of Closed Loop 6.1.3 Steady State of Closed Loop SystemsSystems

Proposition 6-1 ： The closed loop dynamics of a first order system with a P-only controller behaves the same with another first order system, with a gain of KOL/(1+ KOL) and time constant of /(1+ KOL) for the servo problem.

6.1.3 Steady State of Closed Loop 6.1.3 Steady State of Closed Loop Systems– Systems– Cont.Cont.

Corollary 6-1 ： A P-only controller applied to a first order system can yield perfect control if Kc for a servo problem.

Proof: Consider (6-1), in case , we have K11, and 10. Corollary 6-2 ： Assume that a step change with a

magnitude of A at the set point Tset(s) is implemented to a P-only controller applied to a first order system, then the offset of the controller is A/(1+KOL) .

Proof: Let Tset=A/s , then To(s)=(A/s)[K1/ (1s+1)] , we have:

offset =

OL

OL

so

so

t K

AKAK

s

AKssTtT

11limlimlim 1

1

1

00

OLKAAKA

1

11


Kc 1 K1

1 1.67 0.667

2 1 0.8

5 5/11 10/11

5s+1

2

Transfer Fcn

simout

To Workspace

Scope

2

Gain

1

Constant

Add


Regulation problem (disturbance rejection) using a P-only control for a first order system:

01

10

)0(1

1

1

1

)(

0;1

;1

1

2

1

2

2

2

1

2

2

221

FKK

KOffset

KK

K

F

CsK

K

sK

sGsG

sG

sF

sC

sRs

KG

s

KG

c

c

cc


Corollary 6-3 ： A P-only controller applied to a first order system can yield perfect control if Kc for a regulation problem.

Corollary 6-4 ： Assume that a step change with a magnitude of A at the load F(s) is implemented to a P-only controller applied to a first order system, then the offset of the controller is .

01 1

2 FKK

K

c


Kc Offset

1 -0.124

2 -0.077

5 -0.036

time

Kc=1

Kc=2

Kc=5


Proposition 6-2 ： The offset of closed loop dynamics of a first order system with a PI controller results zero offset.

Proof:

10

0

11

1

c

c

IIc

Ic

KK

KK

R

C

sssKK

sKK

sR

sC

6-2 Stability of the Control 6-2 Stability of the Control LoopLoop

1 2

1 2 1

1 2

1 2

1 2

1 22 2

1

Let , , , be n roots of 1 0

input terms

= particular solution

At the cross over point:

other terms

'sin

n

n c

n

n

r tr t r tn

u

u

r r r G s G s

bb bC t

s r s r s r

b e b e b e

b s bC s

s

C t b t

6-2.1 Stability Criterion6-2.1 Stability CriterionProposition 6-3: For a feedback

control loop to be stable, all of the roots of its characteristic equation must be either negative real number or complex numbers with negative real parts.

Corollary 6-4: For a feedback control loop to be stable , all of the roots of its characteristic equation must fall on the left-hand half of the s plane, also known as the “left-hand plane” (LHP).

6-2.1 Stability 6-2.1 Stability Criterion Criterion - Example- Example

Kc p1 p2 p3

0 -3 -2 -1

0.23 -3.1 -1.75 -1.15

0.39 -3.16 -1.42* -1.42*

1.58 -3.45 -1.28-0.75j -1.28+0.75j

6.6 -4.11 -0.95-1.75j -0.95+1.75j

26.5 -5.1 -0.45-2.5j -0.45+2.5j

60* -6.0 0-3.32j* 0+3.32j*

100 -6.72 0.35-4j 0.35+4j

1

1

1 2 3P vG s G s G s H ss s s

11 2 3

11

1 2 3

C

C

KC s s s s

R s Ks s s

1 2 3

C

C

C s K

R s s s s K

Or

0321 CKsss

Characteristic equation can be derived:

6-2.1 Stability Criterion 6-2.1 Stability Criterion - - ExampleExample

6-2.1 Stability Criterion – 6-2.1 Stability Criterion – Cont.Cont.Consider the previous example, at Kc=0.39, the

system becomes oscillatory, at Kc=60, the system becomes unstable. At this point, we term the controller gain that makes the closed loop system unstable the ultimate gain (Ku=60) of the system, the frequency (u=3.32) that the system oscillates is termed the ultimate frequency.

Problem: Given the characteristic equation C(s,

Kc)=0, what is the ultimate gain and ultimate frequency of the system?

Answer: Direct substitution, Routh array, Root Locus

6-2.2 Method of Direct Substitution 6-2.2 Method of Direct Substitution – – Cont.Cont.Basic idea: Consider the characteristic

equation C(Kc,s)=0, in case of ultimate gain, i.e. Kc=Ku, then the root of the equation should be located at the imaginary axis, i.e. s=j. The problem of solving the ultimate condition, hence changed into C(Kc,s)=R(Kc, )+jI(Kc, )=0, i.e. R(Kc, )=0, I(Kc, )=0 (two unkowns, two equations)

Problem: many algebraic work invloved.

6-2.2 Method of Direct Substitution – 6-2.2 Method of Direct Substitution – Cont.Cont.

60;11

066 ;01111

0

6611

6116

6116321

23

23

23

23

uu

uuuu

uuuu

uuuu

uu

K

K

Kjjj

jjjK

sssKsssK

6-2.2 Method of Direct Substitution – 6-2.2 Method of Direct Substitution – Cont.Cont.

7.282186.0

2

8.23;/2186.0

0439008.01420

08.0143420900

08.013110130

013

016.0

110

1

130

501

%

/

13

016.0;

%

110

1;

/130

50

3.16 .

32

23

u

cuu

uucu

u

c

c

ccvs

vs

T

Ksrad

jK

jsLet

Ksss

Ksss

Ksss

sGsGsHsG

CO

skg

ssG

CssH

skg

C

ssG

Fig

6-2.3 Effect of Parameters6-2.3 Effect of Parameters

3 2

Let the temperature sensor/transmiter is re-calibrated (range reduced to

75-125 such that

2 %

10 150 2 0.016

1 1 030 110 1 3 1

900 420 43 1 1.6 0

20.2186 / ; 11.9;

0.

s v c c

c

u cu u

C

H ss C

G s H s G s G s Ks s s

s s s K

rad s K T

28.72186

Let the temperature sensor/transmiter is replaced to a faster one such that:

1 %

5 150 1 0.016

1 030 1 5 1 3 1

20.2345 / ; 18.7; 26.8

0.2186

s v c c

u cu u

H ss C

G s H s G s G s Ks s s

rad s K T

6-2.3 Effect of Parameters – 6-2.3 Effect of Parameters – Cont.Cont.

6-2.4 Effect of Dead Time – Pade 6-2.4 Effect of Dead Time – Pade ApproximationApproximation

...42

1

21

21

...62

1

330

220

00

0

330

220

00

ststst

st

st

ststste st

6-2.4 Effect of Dead Time – Pade 6-2.4 Effect of Dead Time – Pade Approximation Approximation – Cont.– Cont.

0

0

0

0020

0

0

1

1

21

12

01222

0121

211

01

11

10

0

tKK

t

t

KKst

KKt

st

sst

stKK

s

KeKsGK

s

KesG

cu

u

cc

c

stc

c

st

HomeworkHomework

Text , page 2206-1, 6-3, 6-5, 6-10, 6-12, 6-18, 6-

20, 6-25

ExampleExampleThe temperature of a tank is controlled by adjusting

the steam flow to the jacket of the tank. The temperature transmitter has a span of 100°F and is set between 100 and 200°F. The proportional controller is set at a proportional band of 20. The normal pressure to the valve is 8 psig and the valve is air to open type. The normal temperature of the tank is 170°F and the normal feed temperature is 65°F. The set point is moved from 170°F to 175°F. And the tank eventually comes to a steady state of 174.1°F. ◦ (i) What is the offset?◦ (ii) What is the pressure to the valve in the final steady state?◦ (iii) What is the process gain? (Assume the valve gain is unity)◦ (iv) What will be the offset if the proportional band was

changed to 10?

SolutionSolution

( )175 174.1 0.9

0.9( ) 12 0.108; 100 / 20 5;

1005 0.108 8 8.54

174.1 170( ) 7.59

5 0.10810 7.59

( ) 0.91 10 7.59 0.12

170 5 0.9 174.5

p

set

i F

ii e Kc

M psig

iii K

YivY

T F

Chapter 6 Analysis of Feedback Control Systems Prof. Shi-Shang Jang Chemical Engineering Department...

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