yeek02 oct 10

II. FEEDBACK CONTROL SYSTEMS

II.A Advantage of Feedback

• Use Step input here for illustration

• Idealized performance of control system:

Control SystemR(s) Y(s)

Output Y(s) = Desired performance R(s)

1 1

(Up to possibly a scaling constant)

• Process G(s) inside block diagram of overall system

• G(s) variations T(s) variations System output variations

II.A.1. Performance Sensitivity to Parameter Variations

• Process parameters change due to environment, aging, etc Process transfer function G(s) changes with time

G(s)

System transfer function T(s)

ProcessR(s) Y(s)

• System Sensitivity:

When

Def:

)()()()()()(

sTsTsTsGsGsG

GG

TTT

GS

(% change in T over % change in Gfor small incremental change)

TG

GTT

GS

Very small G

• Lower/higher(?) sensitivity is better system output less affected by variations in parameter

variations

• Generally,

- Open loop system:

1 TGSsGsT )()(

- Closed loop system:

GHS

GHGsT T

G

11

1)(

)(sG Y(s)R(s)

-+ )(sG

)(sH

Y(s)R(s)

- Normally |1+GH(s)|>>1 for s values of our interest (Later!) i.e., closed loop system much lower

sensitivity (to variations in G)1T

GS

• Example 4.1 Feedback Amplifier

1

0

TK

a

ina

aS

KTVKV

GainaKinV oV

(a) In open loop configuration

If Ka changes by 10%, T and hence Vo changes by 10% for same Vin

(b) In feedback configuration

)(

)(

)(

a

TK

a

a

ina

p

KS

KKT

VVKVRR

a

11

1

00

2

9991

1010

4

TK

a

aS

K.

2R

GainaK

inV oVpR

oV

-+

aK

oVinVWith

If Ka changes by 10%, T and Vo changesby for same Vin%01.0

999%10

• Note: Other kind of Sensitivity, e.g., for closed loop system:Sensitivity due to variations in H(s):

-+ )(sG

)(sH

Y(s)R(s)

11

GHGH

TH

HT

HH

TT

THS

This means:

Hence, important to have small variations in feedback loop forclosed loop system

For 1+GH>>1

1TGS

1THS

variation in G(s) has small effect on T(s)

variation in H(s) has strong effect on T(s)

• For depending on parameter i.e.,

TTT

TTS

• For depending on G(s) depending on

Sensitivity dueto variation in

),()( sTsT )( sT

GTG

T SSST

T

)( sT

s1)(s

Speed)(s

Position

aL • Example: Armature-controlled DC Motor

- Recall: Block diagram with (Reduced Model)

- Let Output=angular speed (s) (not angle position )

- is input voltage to DC motor- In Textbook,

sEkVa

2

Step input of E (volts) with potentiometer factor k2 (Eqn. 4.40 in textbook)

- Here, we just use as output of controller block- For simplicity, we also set Kb=0 in deriving the equations

Nature inducedFeedback

aV

aV

)(s SpeedKa

R(s)Desiredperformance

Va(s)

)()(

)()(

)( sTsRsR

sKKs d

a

11 1

1

1

1

)(,)(

,)( bma

a

bma

m

bma

a

KKbRRR

KKbRKK

KKbRJR

111

- Ka is open loop controller gain to be designed

- Open loop speed control:

- Transfer function:

where

Armature-controlledDC Motor

Kb=0

Kb=0 Kb=0Kb=0

Armature-controlledDC Motor

- Closed loop speed control

)()(

)()(

)( sTKKKs

RsRKKKs

KKs dtata

a

11

1

11

1

11

- Ka is closed loop controller gain to be designed

- Transfer function:

- Typical values for speed control motor: K1=2, Kt=1, Ka large (from 40100)

Va

Kb=0

(XX)

- Derivation of (XX):

- Td(s)=0,

- R(s)=0,

- Hence,

- (XX) obtained by

)())(1(

)()( sRsPGH

sPGs

_ +P(s)+

_R(s) (s)

Td(s)

H(s)

G(s)q(s)

)()()()()()()(

sTssPHsqsqsGs

d

)(

))(1()()( sT

sPGHsGs d

)())(1(

)()())(1(

)()( sTsPGH

sGsRsPGH

sPGs d

1)(,1)(,)(

sHbJs

sGRKKsPa

ma

0dT• Sensitivity analysis for Armature-controlled DC motor

)()(

)( sRsKKs aol

11

1

- Let disturbance torque

* Ka -- control gain that we set* K1 -- motor gain that varies (slightly) with each unit and time* T(s) changes with change in K1:

* Hence, 10% change in K1 10% change in T(s)

10% change in

- Open loop speed control T(s)

1

111

KK

TT

TKS

)()()( sRsTsol

Sensitivity with respect to motor gain


)()(

)( sRKKKs

KKsta

acl

11

1

1

T(s)

* T(s) change with change in K1:

* Typical values: K1=2, Kt=1, Ka large (from 40100)

* Change in K1 yield vey small change in )()()( sRsTscl

)1(1

11

11

ta

TK KKKs

sS

- Results consistent with previous sensitivity analysis on OL and CL system

Sensitivity with respect to motor gain

ta KKK 1 11 T

KSfrom 80 200

* Hence, upon knowing K1, one can let toachieve desirable steady state value

* With step input R(s)=1/s, output given by

* Output steady state value using FVT

10)(lim)( KKss a

ol

s

ol

1)(ol1

1KKa

OL speed control system1

• Specific case study for sensitivity analysis

- Open loop speed control

)1()(

)1()(

1

1

1

1

ssKKsR

sKKs aaol

1KKa

)(tol

t t

- For step input, desired steady state 1)(

)(ol

* For example, when motor gain K1=2, we can set gain Ka=0.5so that

* However, if K1 were to change value by 10%, i.e., , without our knowledge, so we still kept Ka=0.5 as before, then

* Any change in K1 will result in same percentage ofchange in steady state value of (becausein this case)

* Open loop control can attain desired steady state value only if K1 is known and unchanged!

2.21 K

1.1)2.2)(5.0()( 1 KKaol

1)2(5.0)( 1 KKaol

11T

KS

10% change in )(ol


)1()(

11

1

ta

acl

KKKssKKs

* With ,

* Steady state value using FVT

ssR 1)(

at

acl

s

cl

KKKKK

ss1

10 1

)(lim)(

* One can set Ka large so that to yield close-to-desirable 1)( cl

11 ta KKK

CL speed control system

1at

a

KKKKK

1

1

1

)(tcl

t t

* For example, when K1=2, Kt=1, Ka =50,

1990090101100

1 1

1

.)(at

acl

KKKKK

1001 ta KKK

quite insensitive to variations in K1!

199099.0111110

50)2.2(150)2.2()(

cl

)(cl

* In this case, if K1 were to change by 10%, i.e., 2.21 K

* Hence, if K1 stays constant forever, OL speed control is“better” in steady state value because always

* On the other hand, CL speed control yields onlybut the steady state value is insensitive to change in motorgain K1

* Life is always changing, CL speed control is better overall

1)(ol

1)( cl

Summary: 1st advantage of CL system over OL system:Lower sensitivity in system parameter variations (of G(s) in forward loop)

Process

Can we adjust for “good” transient response?

II.A.2. Improving Transient Response

• Open loop speed control (Td=0 still)

)()()(

11

1

sKK

sRs a

- With R(s) a step functions

sR 1)(

- Open loop speed response:

)()(

11

1

ssKKs aol

- Open loop speed response:)(

)(11

1

ssKKs aol

)()( 111t

aol eKKt

- Inverse Laplace Transform:

1- Time constant , hence transient response, fixed by process parameter and cannot be changed

- Typically, for speed control motor Time to reach steady state long

(roughly 40-50 sec) for

sec101

)(tol

• Closed Loop Control (Td=0 still)

- Inverse Laplace transform:

)()(

)( cl

ta

at

eKKK

KKtcl

11 1

1

- Closed loop speed response:

)()()(

ta

a

KKKsKK

sRs

11

1

1

- With R(s) a step function

)()(

ta

a

KKKssKKs

11

1

1

where is adjustable by ta KK and

Time constant for closed loop system

tacl KKK1

1

1

- With

much smaller than 1 response to final value much faster

11 1 ta KKK

- Specifically, withand

KKK acl

1

1

1

sec101 1001 ta KKK

Summary: 2nd advantage of CL system over OL system:Transient adjustable and faster with closed loop system

• Open loop speed control

- For step disturbance torquesDsTd )(

DRssEe ol

s

ol10

)(lim)(

- Steady state error due to disturbance

II.A.3. Ability to Reject Disturbance

)()(

)()(

)( sTsRsR

sKKs d

a

11 1

1

1

1

Governing transient and steady state performance

Undesirable Error E(s) if 0dT

)()(

11

1

ssDRsEol

)( bma

a

KKbRR

R

1

Recall

(By FVT)

• Closed loop speed control

11 ta KKK

)()(

)()(

)( sTKKKs

RsRKKKs

KKs dtata

a

11

1

11

1

11

- For step disturbance

Hence,

sDsTd )(

)()()(

)(lim

ol

ta

cl

s

cl eKKK

DRssEe1

1

0 1

Undesirable Error E(s) if 0dT

)()(

ta

cl

KKKssDRsE

11

1

1

(By FVT)

Governing transient and steady state performance

Summary: 3rd advantage of CL system over OL system:Effects of disturbance much smaller in CL system

CL Final value=0.99

OL Final value=1DC motor example

II.A.4. Disadvantages of Feedback

• Require feedback signal: more instrumentations (sensors, comparators, etc.) system more complex, cost, maintenance

• May have small steady state error: not reaching desired value

• Require more powerful actuators – from DC Motor example:motor input is 50 times more for CL control than OL control

• System may be unstable due to improper controller assignmentand system uncertainties, etc. require proper design

V a(t)

for

OL

and

CL

spee

d co

ntro

l

Time (sec)

Ka=0.5

Va(s)

Motor and arm G(s)Armature-current controlled

II.A.5. Design Example: Disk Drive Read System (CL Position Control)

• Output y(t) = head position = (t) from DC Motor• Zero back emf: Kb=0 • Parameters:

Disturbance

)()()(

)()()(

)()()( sDsGsGK

sGR(s)sGsGK

sGsGKsYaa

a

21

2

21

21

1

1

)(sDKsss

sR(s)Ksss

K

aa

a

50002000010201000

50002000010205000

2323

Response due to R(s) governing transient and steady state performance

Undesirable Error Due to D(s)

)(sR

• Close Loop Performance (using Full Model)

)(sD

)(sY

• Performance due to R(s) only- R(s)=1/s (Step), D(s)=0

* Ka=10 – too slow?* Ka=80 – too oscillatory?

(Textbook Fig. 4.36)

- Transient changeable by Ka

- R(s)=0 (No input), Disturbance D(s)=1/s (Step)

- Disturbance error for Ka=10 is 8 times the disturbance error for Ka=80

(Textbook Fig. 4.37)

• Error due to disturbance D(s) only

- Overall motor response with R(s)=1/s and D(s)=0.1/s• Performance with BOTH input and disturbance

Design issue “Best” value of Ka?

yeek02 oct 10

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