University of Virginia PID Controllers Jack Stankovic University of Virginia Spring 2015.

University of Virginia University of Virginia

PID Controllers

Jack Stankovic

University of VirginiaSpring 2015

University of Virginia University of Virginia 2/44

Outline• I Control: Integral Control

• PI Control: Proportional-Integral Control

• D Control: Derivative Control

• PD Control: Proportional-Derivative Control

• PID Control: Proportional-Integral-Derivative Control

• Summary


P Control• What is wrong with P control?

– Non-zero steady-state error

• Why?– When (current, instantaneous) error becomes zero

then there is no longer a control signal


I Control• What is I control?

– The controller output is proportional to the integral of all past errors

k

jI

I

jeKu

keKkuku

1

)()0(

)()1()(1)(

)(

z

zK

zE

zU I

Integral Controller

E(z) U(z)


Integral Control

KI G(z)

R(z) E(z) T(z)+

-

+ +U(z)

D(z)

+ +

N(z)

Y(z)z/

z-1

Adds Pole


PI Control

)()( -- response transient theIncrease -- Control P keKku PP

)()1()( --error state-steady Zero-- Control I keKkuku III

1

)(

)(

)(

)1()()()1()(

Control PI

z

KzKK

zE

zU

keKkeKKkuku

PIP

PIP


D Control• D Control: the control output is proportional to the

rate of change of the error

– D control is able to make an adjustment prior to the appearance of even larger errors.

– D control is never used alone, because of its zero output when the error remains constant.

)]1()([)( kekeKku DD z

zK

zE

zU DD )1(

)(

)(


PD Control

))1()(()()( kekeKkeKku DP

z

KzKK

z

zKK

zE

zU DDPDP

)()1(

)(

)(

control. Ian in does 1at pole thelike response,transient

system' down the slownot does which 0,zat pole a has controller The

z


PID Control

1

0

)]1()([)()()(k

iDIP kekeKieKkeKku

)1(

)2()(

)(

)( 2

zz

KzKKzKKK

zE

zU DDPDIP


I Control• Mostly used with P control

• Here, by itself to demonstrate its main effects

– Zero steady state error

– Slow response


Introducing I Control• What is I control?

– The controller output is proportional to the integral of all past errors

k

jI

I

jeKu

keKkuku

1

)()0(

)()1()(1)(

)(

z

zK

zE

zU I

Integral Controller

E(z) U(z)


I Control

KI G(z)

R(z) E(z) T(z)+

-

+ +U(z)

D(z)

+ +

N(z)

Y(z)z/

z-1


Steady-State Error with I Control• Start with Example 9.1: the

IBM Lotus Domino Server

43.0

47.0)(

z

zG

Recall

)()()(1

)()(

)(

)()(

zHzGzK

zKzG

zR

zYZFR

Here H(z) = 1


Example (continued)

.0 : iserror state-steady The

1

47.0

47.0

)]43.0/(47.0)][1/([1

)]43.0/(47.0)][1/([

)(

)()(

ss

I

I

I

I

R

e

K

K

zzzK

zzzK

zR

zYzF

Multiply top and bottom by (z-1) then set z = 1


Why is the steady state error zero

ssssR

ssssss

Rssss

ryF

yre

Fre

/)1(

)]1(1[

:iserror state-steady The


General Case for G(z)• The steady-state error of a system with I control is

0, as long as the close-loop system is stable.

0)]1(1[

1)1(

)()1(

)(

)()]1/([1

)()]1/([

)(

)()(

Rssss

R

I

I

I

IR

Fre

F

zzGKz

zzGK

zGzzK

zGzzK

zR

zYzF


The steady-state error due to disturbance

)1( iserror state-steady The Dssss Fde

0)1(

)()1(

)()1(

)()]1/([1

)(

)(

)()(

D

IID

F

zzGKz

zGz

zGzzK

zG

zD

zYzF

0)1( Dssss Fde

stable. is system loop-closed theas long as ,)( model systemany for and

of any valuefor zero is edisturbancconstant a error to state-steady theSo

zG

K I


• Disturbance Rejection in the IBM Lotus Domino Server

43.0

47.0)(

z

zG

0 5 10 15 20 250

5

10

r(k)

0 5 10 15 20 250

5

10

15

20

d(k

)0 5 10 15 20 25

-20

-10

0

10

20

u(k)

0 5 10 15 20 250

5

10

15

20

y(k)

0 5 10 15 20 25

-10

-5

0

5

10

e(k

)

Example 9.3


Transient Response with I Control• I Control eliminates the steady-state error, but it slows the

system down– The reason is the integrator adds an open-loop pole at 1,

which generates a closed-loop pole that is usually close to 1.

• Example 9.2: Closed-loop poles of the IBM Lotus Domino Server

43.0

47.0)(

z

zG


Example 9.2 (continued)

• Observe the root locus– The largest closed-loop pole is always closer to the unit

circle than the open-loop pole 0.43.

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1Root Locus

Real Axis

Imagin

ary

Axis

timesettlinglonger means This

)log(/4

rks

overshootlarger means also This

/

rM p

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1Root Locus

Real Axis

Imag

inar

y A

xis

Herethere isonlyP control

Here there isI control


PI Control• Common controller

• Fast response by P control

• Accurate response by I control

• Good combination

• Another example of Pole Placement Design– Previously we did pole placement for P controller


PI Control Design by Pole Placement• Slightly different than for P control (see p 305 in

text/handout)• Design Goals:

• Assumption: G(z) is a first-order system– A higher-order system is approximated by a

first-order system (chapter 3)

*

*

exceednot doesovershoot Maximum (4)

exceednot does timeSettling (3)

minimized iserror state-Steady (2)

stable is system loop-closed The (1)

P

s

M

k


PI Control Design by Pole Placement

• Example 9.5: Consider the IBM Lotus Domino server

• Determine transfer function of system and SASO requirements

%10 and 10k and 43.0

47.0)( **

s

PMzzG

Stable – poles within unit circleAccuracy – zero steady state error

Note: 2 poles (one from G(z) and one from I control)


PI Control Design by Pole Placement• Compute the desired closed loop poles

• Construct and expand the desired characteristic polynomial

jrez (

6.0,6.0let ve,conservati be To

70.0)1.0ln/(lnlog

log

67.0

*

1044 *

r

rM

r

eer

P

ks

36.0cos2 222 zzrzrz

) )( jrez

/rM p

)log(/4 rks

jre

sincos je j


Construct the modeled characteristic polynomial

)](])[[()1(

)(])[()(

zGKzKKz

zGKzKKzF

PIP

PIPR


Example 9.5 (continued)• Expand the modeled characteristic polynomial

PIP

PIP

PIP

PIPR KzKKz

KzKK

zGzKzKK

zGzKzKKzF

47.043.0]43.1)(47.0[

47.0)(47.0

)()}1/(]){[(1

)()}1/(]){[()(

2

PIP KzKKzzz 47.043.0]43.1)(47.0[36.0 22

Set the desired equal to the modeled


Example 9.5 (continued)• Solve for

36.047.043.0

143.1)(47.0

P

IP

K

KK

76.0

15.0

I

P

K

K

36.0

07.043.0...)(

equation above into substitute verify To

2

zz

zzFR


Example 9.5 (continued)• Then check

• Poles are

( .5 + .33) and (.5 - .33) so the system is stable

• = 1 so there is no steady state error)(zFR


Example 9.5 (continued)

0 5 10 15 20 250

5

10

r(k)

0 5 10 15 20 250

10

20

d(k)

0 5 10 15 20 25

-10

0

10

up(k

)

0 5 10 15 20 25-20

0

20

ui(k

)

0 5 10 15 20 25-20

0

20

u(k)

0 5 10 15 20 250

10

20

y(k)

10 timesettling

10%overshoot

P control leads to quicker response

I control leads to 0 steady-state error


PI Control Design Using Root Locus• The new issue:

– The root locus allows only one parameter to be varied– A PI controller has two parameters:

• The P control gain, and the I control gain

• Solution to this issue:– Determine possible locations of the PI controller’s

zero, relative to other poles and zeros– For each relative location of the zero, draw the root

locus– For the most promising relative locations, try a few

possible exact locations– Simulate to verify the result


Summary • I control adjust the control input based on the sum

of the control errors– Eliminate steady-state error– Increase the settling time

• D control adjust the control input based on the change in control error– Decrease settling time– Sensitive to noise

• P, I and D can be used in combination– PI control, PD control, PID control


Summary (continued)• Pole placement design

– Find the values of control parameters based on a specification of desired closed-loop properties.

• Root locus design– Observe how closed-loop poles change as controller

parameters are adjusted


Relationship to WSN and RTS

• Consider PRR problem in WSN

• Consider Miss Ratio Problem in RTS


Extra Slides


Example 9.4• Moving-average filter plus I control

– A moving average slows down the system responses– An I control also slows down the system response– So the combination leads to undesirable slow behavior

• Example: IBM Lotus Domino server + I control + Moving-average filer

cz

c

z

zK

zzG I

1 and

1 and

43.0

47.0)(

-1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5Root Locus

Real Axis

Imag

inar

y A

xis

circleunit the toclose polet (2)Dominan

gain ofregion d(1)Shrinke

:nsObservatio


Moving-average filter vs. I controller

• An I controller works like a moving-average filter:– More response to sustained change in the output

than a short transient disturbance

• An I controller drives the steady-state error to 0, but the moving-average filter does not.


PI Control


PI Control Design by Pole Placement

• Approaches:– Step 1: compute the desired closed-loop poles

– Step 2,3,4: find the P control gain and I control gain

– Step 5: Verify the result• Check that the closed-loop poles lie within the unit circle• Simulate transient response to assess if the design goals are

met

rker sks log4

*4 rMM

rP

P

*log

log

)](])[[()1(

)(])[()( ofr denominato theis )cos2( 22

zGKzKKz

zGKzKKzFrzrz

PIP

PIPR


PI Control

)()( -- response transient theIncrease -- Control P keKku PP

)()1()( --error state-steady Zero-- Control I keKkuku III

1

)(

)(

)(

)1()()()1()(

Control PI

z

KzKK

zE

zU

keKkeKKkuku

PIP

PIP


Steady-state Error with PI Control• PI has a zero steady-state error, in response to a

step change in the reference input– It also holds for the disturbance input

)]1(1[ Rssss Fre

)](])[[()1(

)(])[(

)()}1/(]){[(1

)()}1/(]){[(

)(

)()(

zGKzKKz

zGKzKK

zGzKzKK

zGzKzKK

zU

zYzF

PIP

PIP

PIP

PIPR

0)]1(1[

1)1(

Rssss

R

Fre

F


PI Control Design Using Root Locus• Example 9.6: PI control using root locus

43.0

47.0)(

z

zG

good.not circle,unit the toclose always is poledominant the(b),In


overshoot actualhigh of because good,Not

4gain overall and 0.2 zero Choose

Example 9.6 (continue)


9.0,6.0 IP KK


P control leads to quicker response

I control leads to zero steady-

state error


CHR Controller Design Method

)()()( :Assumption

input. control

in the change step a test,bump a with

system, target theof response step obtain the:Point Starting

kbukaynky


CHR Controller Design Method

axis time with theline tangent theofon intersecti -the-

react tostartsit after valuefinal its of 0.63reach tosystem by the needed time-the-

line tangent theof slop -the-

L

T

R


CHR Controller Design Method• Example 9.7:

design. controller PI for theAsk

response stepunit Given the



59at happens10)15(63.0

4

7.19/15

:have wefigure, theFrom

TTLk

L

R

No simulation is needed to verify it. Why?- Only one option in the table


D Control


D Control• A Real CS Example:

– An IBM Lotus Domino server is used for healthy consulting. (MaxUsr, RIS)

– Bird flu happens in this area. More and more people request for the service.

– More and more hardware is added to the server. – So the reference point keeps increasing.– To deal with the increasing reference point, do

we have better choices than P/I control? – How about setting the control output

proportional to the rate of error change?


D Control• D Control: the control output is proportional to the

rate of change of the error

– D control is able to make an adjustment prior to the appearance of even larger errors.

– D control is never used alone, because of its zero output when the error remains constant.

– The steady-state gain of a D control is 0.

)]1()([)( kekeKku DD z

zK

zE

zU DD )1(

)(

)(


PD Control


PD Control

))1()(()()( kekeKkeKku DP

z

KzKK

z

zKK

zE

zU DDPDP

)()1(

)(

)(

control. Ian in does 1at pole thelike respense,transient

system' down the slownot does which 0,zat pole a has controller The

z


PD Control• PD controllers are not appropriate for first-order

systems because pole placement is quite limited

• PD controllers can be used to reduce the overshoot for a system that exhibits a significant amount of oscillation with P control

• Example: consider a second-order system

49.03.1

1)(

2

zzzG


Example (continue)

control Ponly gConsiderin

control PD gConsiderin


Example (continue)

It is real good

It sounds good

09.0

09.0

18.0

5.0

D

P

DP

DP

D

K

K

KKKK

K


Example (continue)

09.067.03.1

09.018.0

)]49.03.1/(1}[/]){[(1

)]49.03.1/(1}[/]){[()(

23

2

2

zzz

z

zzzKzKK

zzzKzKKzF

DDP

DDPR

control D with Compare

control PI with Compare

error. state-steady theeliminate NOTcan control PD that means This

31.0)1( RF


Example (continu

e)


PID Control


PID Control

1

0

)]1()([)()()(k

iDIP kekeKieKkeKku

)1(

)2()(

)(

)( 2

zz

KzKKzKKK

zE

zU DDPDIP


PID Control• PI controllers are preferred over PID controller

– D control is sensitive to the stochastic variations– A low-pass filter can be applied to smooth the system

output. In that case, the D control only responds to large changes

– But the filter slows down the system response

• PID Control Design by Pole placement– Compute the dominant poles based on the design

goals– Compute the desired characteristic polynomial– Compute the modeled characteristic polynomial– Solve for the gains of the P, I and D control by

coefficient matching– Verify the result


PID control design by pole placement

• Example 9.8:– Consider the IBM Lotus Domino Server

%10 and 10k and 43.0

47.0)( **

s

PMzzG

6.0,6.0let ve,conservati be To

70.0)1.0ln/(lnlog

log

67.0

*

1044 *

r

rM

r

eer

P

ks

11.0063.070.0 :have We

3.0 Choose23

3

zzz

p



DDPDIP

DDPDIP

R

KzKKzKKKz

KzKKzKKK

zR

zYzF

47.0))2(47.043.0()43.1)(47.0(

))2()((47.0

)(

)()(

23

2

DDPDIP KzKKzKKKz

zzz

47.0))2(47.043.0()43.1)(47.0(

11.0063.070.023

23

23.0

01.1

31.0

7.043.1)(47.0

063.0)2(47.043.0

11.047.0

D

I

P

DIP

DP

D

K

K

K

KKK

KK

K


23.0

01.1

31.0

D

I

P

K

K

K


University of Virginia PID Controllers Jack Stankovic University of Virginia Spring 2015.

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Transcript of University of Virginia PID Controllers Jack Stankovic University of Virginia Spring 2015.