Part-b Pic Tkg Ss4 2011-Colour

2010 PIC Lecture 2010 – Prof. T.K.Ghoshal and Prof. Smita Sadhu 1

PROCESS INSTRUMENTATION AND CONTROL

Prepared and Compiled byProf. T.K.Ghoshal

& Prof. Smita Sadhu


Part-B


Proportional Control

Control Laws

&

Control Configurations


Proportional Control

�Simplest of all Control Laws

�Only a Proportional Controller– Whose output CO is proportional to its input e

which is the error = SP-PV.– CO = Kp e


Proportional Control-II

�The controller however is not perfect– Its output is limited, that is the output

saturates at some value.

– Often, the output is one sided, that is to be compatible to 4-20 mA standard.

em

-em

Kp.em

e

COsaturation

em

-em

Kp.em

e

COsaturation


Proportional Band

Kp Gp(s)+++

Kp+

Kp+

Kp+

Gp(s)Kp+

Gp(s)Kp+

Gp(s)Kp+

Gp(s)Kp+

Gp(s)Kp+

Gp(s)Kp+

Gp(s)Kp+

Gp(s)Kp+

e CO y=PVGp(s)Kp

+

v=SP

-

e CO y=PVGp(s)Kp

+

em

-em

Kp.em

e

COsaturation

Let Kp.em=100%

∴∴∴∴ em= 100%/Kp

∆Proportional band

Percentage change in controller input which would result in 100% change in controller

output


Proportional Band-II

Kp Gp(s)

-

+

vss is the input required for producing an output y=1

ess= 1/Kp

Percent steady state error≈≈≈≈ 100/Kp.

This is the Proportional band

yss=1

e-sT/(1+sττττ)

COss=1

ess

=1/Kp

Vss

=1+1/Kp

Let Kp.em=100%

∴∴∴∴ em= 100%/Kp

∆Proportional band

Higher gain=

lower PB

SS errors occur because the plant is Type Zero

Normalized Type 0 Plant


Proportional Band-III

Kp

Gp(s)

=e-sT/(1+sττττ)

ess=vss/(1+Kp) - Lss /(1+Kp)

For v=0,

ess= - Lss /(1+Kp) ≈≈≈≈ - Lss /Kp

ess ≈≈≈≈(100/ Kp)(- Lss /100)

ess % = PB% of (- Lss % )

Kp

Gp(s)

=e-sT/(1+sττττ)Kp

Gp(s)


Gp(s)


Gp(s)


Gp(s)


Gp(s)


Gp(s)


Gp(s)

=e-sT/(1+sττττ)

+

-

+

+yv e eKp

Load (L)

Kp

Gp(s)

=e-sT/(1+sττττ)

Example:

PB=5 % , L= - 30%

ess = 1.5%

Error in the presence of Steady loadNormally load is negative



Proportional Band-IV

�Significance of Proportional Band• PB % = 100/Kp %

• PB % = maximum % error before the controller saturates

• PB = % age ss error with 100% step input*

• PB = (negative) % age ss error with 100% load* • * requires 100% CO produces 100% PV

• Exercise:• % ss error with 40% input=0.4 PB

• % ss error with -67% load =0.67 PB

• % ss error with 30% input & -10% load = (0.3+0.1)PB = 0.4 PB.



Choosing Kp�Designers try to use as high Kp, that is as small

PB as possible

�The incentives are improved closed loop performance– Smaller steady state error .– Faster response.– Lower sensitivity to parameter variation.

�High Kp however, may lead to – oscillatory closed loop response, even instability– Higher actuator bandwidth requirement– Noisy output, controller saturation due to noise

�Choosing Kp is always a compromise!


Higher K => Faster Closed Loop response

0 0.5 1 1.5 2 2.5 30

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

0.6

0.65

0.7

K=1

K=1.5

K=2

( )1+

=s

KsG

p

*K=1, 90% of final

* 90%*

K=2,

90% of Final

SS Error=50%

SS Error=33%


Sensitivity to parameter variation

KGp(s)

=1/(1+sττττ)

v

-

+

e y

( )

( )sKGK

ssK

K

s

Ks

K

sG

p

cl 11

1

11

1

1

11

1

+=

++

=++

=

++

+=ττ

τ

τ

As K increases, effect of plant parameters

decreases

Plant parameters may vary


Close Loop BW increases with KBode Diagram

Frequency (rad/sec)Bode Diagram

Frequency (rad/sec)Bode Diagram

Frequency (rad/sec)

10-2

10-1

100

101

102

-10

0

10

20

Magnitu

de (

dB

)

System: untitled1

Frequency (rad/sec): 1.27

Magnitude (dB): -3.06

10-1

100

101

102

-10

0

10

20

Magnitu

de (

dB

)

System: untitled1



10-1

100

101

102

-10

0

10

20

Magnitu

de (

dB

)

System: untitled1



K=1

K=10

K=100

Higher BW =>

•Response speed ↑↑↑↑•Noise sensitivity↑↑↑↑


When gcf=pcf, PM=GM=0

10-1

100

101

Magnitu

de (

abs)

Bode Diagram

Frequency (rad/sec)

100

101

102

-180

-135

-90

-45

0

Phase (

deg)

pcf=wp

gcf=wg

Phase Margin

Gain

Margin

gcfincreaseswith gain

pcf does not

depend upon gain

PM decreases

with increase in

gain

GM decreases

with increase in

gain


Proportional Control: GCF

11

,

]1[1,

1

11

,

11

2

2

222

22

22

−=

≈

>>≈−=

+=

=+

==

+=

+=

−

pg

p

g

pppg

gp

g

p

g

p

Tj

p

pp

K

Kor

KforKKor

K

KMAt

K

j

eKGK

τω

τω

τω

τω

τωωω

τωωτ

ω

Exact relation

Approximate relation

Very handy but approximate

As Kp ↑, wg ↑, closed loop BW ↑,

and system becomes faster

When the plant has a gain K1, the term K1Kp should be used in place of Kp


Proportional Control: Gain Margin

)(log20

1

)(

1

1

;

10

22

22

GMGM

GMKM

GM

KMAt

dB

g

p

p

p

p

p

p

p

=

≈⇒+

==

+==

ω

ωτω

ω

τωωω


but the final expression of GM remains the same


Proportional Control: Phase Crossover Frequency

πτω

πω

πτωω

πτωω

ωω

ωτω

=

−+

=+

−=−−

=

−−=∠

−

−

−

−

p

p

pp

pp

p

PP

Tor

Tor

T

At

TGK

1tan

2,

tan,

tan

tan

1

1

1

1

T

T

T

ionapproximatFurther

T

TFor

p

p

p

p

p

p

p

2

2

,1

:

2

1

,1

πω

πω

ωτω

πτω

ω

ω

≈⇒

≈

<<

≈−

>>

πτω

πω

πτωω

πτωω

ωω

ωτω

=

−+

=+

−=−−

=

−−=∠

−

−

−

−

p

p

pp

pp

p

PP

Tor

Tor

T

At

TGK

1tan

2,

tan,

tan

tan

1

1

1

1

T

T

T


T

TFor

p

p

p

p

p

p

p

2

2

,1

:

2

1

,1

πω

πω

ωτω

πτω

ω

ω

≈⇒

≈

<<

≈−

>>

πτω

πω

πτωω

πτωω

ωω

ωτω

=

−+

=+

−=−−

=

−−=∠

−

−

−

−

p

p

pp

pp

p

PP

Tor

Tor

T

At

TGK

1tan

2,

tan,

tan

tan

1

1

1

1

T

T

T


T

TFor

p

p

p

p

p

p

p

2

2

,1

:

2

1

,1

πω

πω

ωτω

πτω

ω

ω

≈⇒

≈

<<

≈−

>>

πτω

πω

πτωω

πτωω

ωω

ωτω

=

−+

=+

−=−−

=

−−=∠

−

−

−

−

p

p

pp

pp

p

PP

Tor

Tor

T

At

TGK

1tan

2,

tan,

tan

tan

1

1

1

1

T

T

T


T

TFor

p

p

p

p

p

p

p

2

2

,1

:

2

1

,1

πω

πω

ωτω

πτω

ω

ω

≈⇒

≈

<<

≈−

>>

πτω

πω

πτωω

πτωω

ωω

ωτω

=

−+

=+

−=−−

=

−−=∠

−

−

−

−

p

p

pp

pp

p

PP

Tor

Tor

T

At

TGK

1tan

2,

tan,

tan

tan

1

1

1

1

T

T

T


T

TFor

p

p

p

p

p

p

p

2

2

,1

:

2

1

,1

πω

πω

ωτω

πτω

ω

ω

≈⇒

≈

<<

≈−

>>

Quadraticequation


Finding PCF

Finding ωp

�Two Methods:– Using quadratic equation Approximation

– Using exact equation and iterative method

�Problem:– Let τ = 20 s, T = 3 s, find ωp

– Quadratic approximation gives ωp=0.553

2

1 πτω

ω ≈−p

pT


Finding PCF-II

�Iterative Method

5.05.1

2≈≈≈

TTp

πω

553.0 :iteration3rd

553.0

)205566.0

1(tan

3

1

3

57.1

iteration2nd

5566.00333.03

57.1

)205.0

1(tan

1

2

iteration1st

1

1

≈

≈×

+≈

=+≈

×+≈

−

−

p

p

pTT

ω

ω

πωFirst value:

)1

(tan2

Using

1

τωπ

ωp

pT−+≈


Finding PCF-III

�Angle Check

( )π−≈−=−−=

×−×−=∠ −

14.348.166.1

20553.0tan3553.0 1

PPGK

�Angle Check

�Comments

– Truly PCF

– Both quadratic and iterative methods provide

good approximation of PCF


Critical (or ultimate) gain

�When the proportional gain is gradually changed from a low enough value, a time comes when the closed loop systems starts oscillating.

�At this condition – GCF equals PCF =>

– both the gain and phase margin become zero

– Oscillation occur at PCF

�The corresponding proportional gain is called the critical gain or the Ultimate Gain.

�The ultimate gain can be determined either experimentally or analytically.


Ultimate gain-II

�Critical gain is the value of KP such that ωg

is the same as ωp.PK̂

10-1

100

101

Magn

itude (

abs)

Bode Diagram

Frequency (rad/sec)

100

101

102

-180

-135

-90

-45

0

Phase

(deg)

ωωωωg=ωωωωp

PK̂


Ultimate Gain-III

TK

T

T

T

T

T

TT

TK

K

when

KGK

P

pP

p

P

pg

PPp

ττπ

τπ

τπ

τπ

τπ

τπτπ

τπ

τω

τω

ωωτω

5.1ˆ1

2

21

2

41

241

211ˆ

1

1

ˆ

,

1

2

2

2

2/1

2

2

22

22

22

22

22

22

≈⇒+≈

+≈

+=+=

+≈+=⇒

=+

=+

=

Can be used if system

parameters are known

Can be used only after

determining pcf



Finding Gain Margin

beforeasK

KGM

KKKWhen

KKGK

K

KGM

g

p

g

p

P

P

p

PgpPP

g

PPPp

P

P

≈

+

+==⇒

=+

==

=+

⇒+

=

=

ω

ω

τω

τω

τωωω

τωτω

22

22

22

2222

1

1ˆ

11

ˆ,,ˆ

111

ˆ definitionBy

GM = the factor by which gain may be enhanced without making the system unstable

You may convert GM to dB


Finding Phase Margin

τωω

π

τωπ

ωπ

πφ

ωτωφ

ωω

g

g

g

g

T

T

PM

T

g

1

2

)1

(tan2

tan

1

1

+−≈

+−−=

+=

−−=

−

=

− All angles and the margin in Radian

Don’t forget to convert to degree

Fairly accurate

when ωωωωg ττττ > 10


Examples-1

Problems:

(1) For a plant with K=1, τ = 10 s, T = 1 s, find the pcf of the system

(2) For the above plant with a proportional controller of gain KP=7.5, find the gcfand PM.


Solution to Example 1

sradTT

p /57.157.1

2:Prob(1) ≈≈≈

πω As T/ττττ is small,

the approximation is

fair

63.11063.1

157.1:2

;63.11057.1

157.1:1 :(1) Prob

≈×

+≈

≈×

+≈

p

p

Itn

Itn

ω

ω



sradKK pg /74.015.710

11)(

1

:Prob(2)

22 =−=−=τ

ω

rad

T

radTPM

g

g

g

g

965.01074.0

174.057.1

1

2 :approxeven with

deg26.55964.0)1

(tan2

1

=×

+−=

+−

==+−= −

τωω

πτω

ωπ


Examples- 2

Problem:

Let K=2, τ = 10 s, T = 1 s, and a proportional controller with gain KP=7.5/2=3.75 be used.

Find the pcf, gcf and PM of the system



sradTT

p /57.157.1

2≈≈≈

πω Same as Example 1

Approximation is good

63.1=pExact ω

sradKK pg /74.015.710

11

1 22 =−=−=τ

ω

deg3.551

2=+−=

τωω

π

g

gTPM

The combined effect of KKP is considered when the process has non-unity gain.


Example-3

Problem:

Let K=1, τ = 30 s, T = 10 s, find the pcf of the system

Design a proportional controller to provide a gain margin of 6dB. Find the gcf and PM of the controlled process.


Solution to Example-3

sradTT

p /157.057.1

2≈≈≈

πω

As T/ττττ is not small, the approximations are not

good

176.01030178.0

1157.0:2

;178.01030157.0

1157.0:1

=××

+≈

=××

+≈

p

p

Itn

Itn

ω

ω

37.530176.011ˆ 2222 =×+≈+= τω pPK

69.2995.1

37.5ˆˆ===⇒≈

GM

KKGM

K

K PP

P

P

995.110

6log20

20/6

10

==⇒

=

GM

dBGM



srad

K pg

/083.0169.230

1

11

2

2

=−=

−=τ

ω



( ) ( ) o4.6514.1/12/ =≈+−≈ radTPM gg τωωπ

( )deg2.6412.119.183.014.3

tan 1

==−−=−−= −

rad

TPM gg τωωπ


Proportional + Integral Control

(PI-Control)


PI Control� Proportional gain in a P-control cannot be increased

beyond a certain value due to PM and GM constraints

� Limited P control means– Steady state error

– Steady state error changing with load,

• requires resetting the set point to get constant PV

– Slow response

– Sensitivity to parameter variations

� Steady state error (offset) may be completely eliminated by reset (PI) PI control.

� The I-action obviates the need for frequent manual reset and hence called auto reset or simply reset action


Block representation of PI controller

+

COεεεε∫ dtK I

PK

++

+

CO∫ dtK I

PK

++

++

CO∫ dtK I

PK

++

+

I

PI

I

PI

PK

IP

K

KT

sTK

s

KKsK

dtKKsCO

=+=+=

+= ∫);

11()(

)( εεUnit?


Reset Time

�The Reset time indicates the time after which the

reset action becomes predominant

� It also provides an indication about the expected

time when the offset can be said to be practically

eliminated.

TimeReset

)1

()1

1()(

==

+=+=+=

I

PI

I

IP

I

PI

PK

K

KT

sT

sTK

sTK

s

KKsK


Reset Time-II

� The Reset time indicates the time after which the reset action becomes predominant

� At t =T1, the reset action output becomes equal to and just overtakes the proportional action output.

� T1=Ki/KP=Ti

� It also provides an indication about the expected time when the offset can be said to be practically eliminated.

Time �

e ↑a

aKP

aKit

T1Time �

CO ↑


Reset Time-III( )

( )

I

PP

I

PPP

P

PIPIPPP

IPP

I

IPI

I

I

IP

T

KK

T

KKKK

KTKTKKKs

TKKss

ssT

sTKssT

ssT

sR

sE

sRssT

sTKsE

)1/(),

)1/(1(

)1/(2

))1(/(21(1

2

/4)1()1(

/)1(

)1(

)1()1(

)1()(

]1

1)

1(1)[(

22

2

+−

+−

+−

+

+−±−=

−+±+−=

+++

+=

+++

+=

=+

++

τ

ττ

τ

ττ

ττ

ττ

� Consider a first order plant with just a time constant τ and a large reset time such that Ti~ τ .

� With large proportional gain, one of the closed loop pole is close to -1/Ti and the other is approx at Kp/ τ.

� The slower pole, with a time constant Ti is responsible for taking the steady state error to zero.

� The faster pole is the usual pole for proportional control.

This result is valid for only

slow reset


PI-Controller Frequency response

I

I

I

P

I

PK

TT

TKM

MTj

KjK

ωωπϕ

ω

ϕω

ω

1tan)(tan)2/(

)(

11

(say))1

1()(

11

2

−− −=+−=

+=

∠=+=

1/Ti Freq →

MKp


Why the steady state error should be zero with PI controlReductio Ad Absurdum Argument

� Proof Outline:– Assume the system to be linear and the closed loop system is

stable.

– Steady input and load are applied.

– Stability implies that in steady state all variables will remainconstant.

– If the error is not zero (say a constant) its integrated value would keep on changing like a ramp which is impossible as it contradicts the “steady state” assumption.

� Proof Limitation:– Not valid when the controller or actuator saturates, i.e. when the

output reaches maximum value, because the linearity assumption is violated.

– Consequence: expect some steady state error at large load even with PI control.


Why the steady state error should be zero with PI controlTF and Final Value Theorem

�Steady SP and Load

�Expression for error E(s)

+

-

+

+yv e eKk

Load (L)

PI Controller

Gp(s)

)())1((

)ˆˆ(

)()1

(1(

ˆˆ)(

)()()()1

1(ˆ

)()()(

sGsTKs

TLV

sGsT

sTKs

LVsE

sLsEsGsT

Ks

VsYsVsE

pIP

I

p

I

IP

p

I

P

++−

=+

+

−=

−+−=−=


Why the steady state error should be zero with PI controlTF and Final Value Theorem-II

�For any steady SP and any steady Load, the steady state error would be zero.

+

-

+

+yv e eKk

Load (L)

Kk(s) Gp(s)

0)ˆˆ( ifeven 0;

ehigher typor zero type;0)0(;)0(

)ˆˆ(

)())1((

)ˆˆ()(

0

00

≠−=⇒

≠−

=

++−

==

→

→→

LVe

GGK

TLVslt

sGsTKs

TLVsltssElte

ss

p

pP

I

s

pIP

I

ssss

Linearity

assumption is

implicit for

Laplace Transform

Final

value

theorem


Exercise

�Taking a standard plant transfer function

and a PI controller, use final value theorem to prove that the steady state error must be zero with constant set point and steady load.

Repeat the above with frequency domain analysis by letting jω→0.

( )τs

KesG

sT

p +=

−

1


PI control : points to ponder�Steady state error becomes zero whatever may

be the value of integral gain (even if close to zero)– What is the catch? Why should we at all use a higher

integral gain?

� If the load disturbance is such that even the highest possible output from the FCE cannot compensate the deviation – will PI control still assure zero offset?

– If not, what would be the behaviour when such load is suddenly withdrawn?


Freq Response: Plant with PI Controller

Freq →

M

1/τ ωg1/Ti

ωp

Unit –ve

slope

Unit –veslope

double –ve

slope

I

I

I

P

Tj

I

P

pK

TT

TT

T

KKM

Mj

Ke

TjK

jGjK

ωωτω

π

ωπ

ωτωϕ

ωωτ

ϕωτω

ωωω

1tan

1tan

2

)(tan2

)(tan

))(

11(

)(1

(say)1

)1

1(

)()(

11

11

22

−−

−−

−

−+−−=

+−−−=

++

=

∠=+

+

=

Igg

g

Igg

Pg

TTPM

T

KKM

g ωτωω

πϕπ

ωτωω

ωω1

tan1

tan2

|

1))(

11(

)(11,GCFat

11

22

−−= −+−=+=

=++

⇒==Extra

phase

lag


Freq Response: Plant with PI Controller-II

Freq →

M

1/τ ωg1/Ti

ωp

Unit –ve

slope

Unit –veslope

double –ve

slope

gg

Ig

gg

Pg

g

gg

gg

gg

T

KK

MM

M

g

ωωω

ωω

τω

ωωω

ωω

ωω

ωω

ˆ25.1Typically

))ˆ(

11(ˆ

wherefrom;1)(

ˆ

controlonly -Pfor GCF ˆ where

ˆ~ seed ay with iterativel used becan relation above The

|~

where

ofion approximatbetter a is~~ shown that bemay It

;~Let

2

2

~

≈

+≈

−=

==

=

≈

=

2

1tan

1tan

),( PCFat

11 πωτω

ω

πϕω

=+−

⇒−=

−−

Ipp

p

p

TT

*gω~

M~


Computing PCF; PI control

�Two methods:– Quadratic approx– Iterative

]}{

1tan

}{

1tan)[/1(

2}{

2}{ :rule iterative

equation quadratic:2

11

2

1tan

1tan

),( PCFat

)(

1

)(

1)1(

)0(

11

I

k

p

k

p

k

p

p

Ipp

p

Ipp

p

p

TT

T

T

TT

TT

ωτωπ

ω

πω

πωτω

ω

πωτω

ω

πϕω

−−+

−−

−+=

=

≈+−

=+−

⇒−=


Plant with PI Controller-Example

�Standard 1st order plant with delay , τ=1, T=0.1, Kp=8, Find pcf, GM, gcf, PM.

�Find the values of gcf, PM, pcf, GM, when an I-element with reset time Ti=0.3 s is added to the above P controller.



� Standard 1st order plant with delay , τ=1, T=0.1, Kp=8.

� pcf=16.31 rad/s;

� PM=51.7 deg,

� gcf=7.9 rad/s,

� GM=2.06=6.3 dB



� Standard 1st order plant with delay , τ=1, T=0.1, Kp=8. (pcf=16.31 rad/s; PM=51.7 deg, gcf=7.9 rad/s, GM=2.06=6.3 dB)

� Find the values of gcf, PM, pcf, GM, when an I-element with reset time Ti=0.3 s is added to the above P controller.

degree12261

tan1

tan2

PM

1.0017 freq at this ~

61.81.09 x 7.9))ˆ(

11(ˆ7.9;ˆ

11

2

.T

T

M

T

Igg

g

Ig

ggg

=−+−=

=

==+≈=

−−

ωτωω

π

ωωωω

PM nearly

halved

9% increase in GCF


Computing PCF; PI control-example

� Standard 1st order plant with delay ,Tau=1, T=0.1, pcf=16.31 rad/sKp=8, PM=51.7 deg, gcf=7.9 rad/s, Ti=0.3s

� First by quadratic equation

1.66 14.04, gives

011

22

11 2

=

=++−⇒=+−

p

I

pp

Ipp

pT

TT

T

ω

τω

πω

πωτω

ω

09.14}{,11.14}{,3.14}{

7.152

}{

)3()2()1(

)0(

===

==

ppp

pT

ωωω

πω

� Ignore the second value (why?)� By iteration


Effect of reset (PI) control

�Adding reset element to a P control system– Provides PI control configuration

• Which potentially eliminates steady state error (offset) in CL

– Results in substantial loss of phase margin• Which would need correction to maintain stability margin

– Results in minor increase in gcf• Minor decrease in rise time in CL

– Results in minor decrease in pcf– Results in higher overshoot in CL

�To restore stability margin, – Gcf must be reduced by reducing Kp, thereby

• Restoring Phase margin and reducing overshoot in CL• Reducing BW, consequently increasing rise time in CL• Increasing sensitivity


Effect of reset control (2)

Bode Diagram

Frequency (rad/sec)

10-1

100

101

Magnitu

de (

abs)

10-3

10-2

10-1

100

-180

-135

-90

-45

0

Phase (

deg)

wp~0.6 rad/s

wg~0.018 rad/s

Nominal system response with proportional gain

Additional phase lag for PI controller

Bode Diagram

Frequency (rad/sec)

10-3

10-2

10-1

100

-180

-135

-90

-45

0

Phase (

deg)

10-1

100

101

Magnitu

de (

abs)

Additional Phase lag

No change in gain at high frequency


Effect of reset control (3)

�Standard 1st order plant with delay

Tau=1, T=0.1, Kp=8, PM=51.7 deg, gcf=7.9 rad/s, Rise

time=0.095sec, SS error=11%

0.089.49.56PI with Ti=1/6

0.0868.721.9PI with Ti=1/4

0.0958.236.4PI with Ti=1/2

0.0957.951.7P only

Rise time (s)

gcf (rad/s)PM (deg)

Control


PI controller design trade off� Proportional gain should be high enough to ensure

– High enough gcf and low enough rise time– Acceptable sensitivity

� Proportional gain should be low enough to reduce the gcf so that the additional phase lag due to reset action can be accommodated

� Integral gain should be high enough (reset time should be small enough) to ensure quick reduction of offset (steady state error)– With adequate damping, the offset becomes negligible after a

period of 3Ti.

� Integral gain should be low enough (Ti large enough) to ensure adequate phase margin and damping ratio

� Typically Ti>3.3T


PI controller design approaches

Approaches

Empirical (e.g. Ziegler Nichols)

Analytical

Phase budget Gain reduction


Analytical design of PI controller

� Given PM, obtain appropriate values of Kp and Ti to ensure fast response

� Both the analytical methods (gain reduction and phase budget) start with designing a P controller with a specified gain margin

� In the gain reduction method, the proportional gain (equivalently the gcf) is reduced by a factor to enhance the PM. This additional phase is then allocated to compensate for the phase lag introduced by the reset.

� In the phase budget method, the phase lag introduced by the reset is budgeted. A proportional gain is chosen such that the PM is more than what is specified by the budgeted amount.

� Wait for the details……..


Steps for gain reduction method

�Obtain the frequency (ω1) at which the required PM is obtainable for P control.

�Choose GCF at ωg = 0.8ω1

�Determine the phase angle of the “plant” at ωgand the additional phase margin ∆φ.�Choose reset time Ti such that tan-1 (1/(ωg Ti ))= ∆φ�Determine the gain to satisfy the chosen GCF with PI control.

�Check GM.


Steps for gain reduction method

�Obtain the frequency (ω1) at which the required PM is obtainable for P control.�Choose GCF at ωg = 0.8ω1

�Determine the phase angle of the “plant” at ωg and the additional phase margin ∆φ.�Choose reset time Ti such that tan-1 (1/(ωg Ti ))= ∆φ�Determine the gain to satisfy the chosen GCF with PI control. �Check GM.

Bode Diagram

Frequency (rad/sec)

10-1

100

101M

agnitu

de (

abs)

10-3

10-2

10-1

100

-180

-135

-90

-45

0

Phase (

deg)

ωωωω1

Required PM


Steps for gain reduction method �Obtain the frequency (ω1) at which the required PM is obtainable for P control.

�Choose GCF at ωg = 0.8ω1�Determine the phase angle of the “plant” at ωg and the additional phase margin ∆φ.�Choose reset time Ti such that tan-1 (1/(ωg Ti ))= ∆φ�Determine the gain to satisfy the chosen GCF with PI control. �Check GM. Bode Diagram

Frequency (rad/sec)

10-1

100

101M

agnitu

de (

abs)

10-3

10-2

10-1

100

-180

-135

-90

-45

0

Phase (

deg)

ωωωω1

Required PM

ωωωωg=0.8ω=0.8ω=0.8ω=0.8ω1


Steps for gain reduction method �Obtain the frequency (ω1) at which the required PM is obtainable for P control.�Choose GCF at ωg = 0.8ω1

�Determine the phase angle of the “plant” at ωg and the additional phase margin ∆φ.�Choose reset time Ti such that tan-1 (1/(ωg Ti ))= ∆φ�Determine the gain to satisfy the chosen GCF with PI control. �Check GM.

Bode Diagram

Frequency (rad/sec)

10-1

100

101M

agnitu

de (

abs)

10-3

10-2

10-1

100

-180

-135

-90

-45

0

Phase (

deg)

ωωωω1

Required PM

ωωωωg=0.8ω=0.8ω=0.8ω=0.8ω1

Additional PM

∆φ∆φ∆φ∆φ

Phase angle of

plant at ωωωωg



�Determine the phase angle of the “plant” at ωg and the additional phase margin ∆φ.

�Choose reset time Ti such that tan-1 (1/(ωg Ti ))= ∆φ�Determine the gain to satisfy the chosen GCF with PI control.

�Check GM.Bode Diagram

Frequency (rad/sec)

10-1

100

101M

agnitu

de (

abs)

10-3

10-2

10-1

100

-180

-135

-90

-45

0

Phase (

deg)

ωωωω1

ωωωωg=0.8ω=0.8ω=0.8ω=0.8ω1

Additional PM

∆φ∆φ∆φ∆φ



�Determine the phase angle of the “plant” at ωg and the additional phase margin ∆φ.�Choose reset time Ti such that tan-1 (1/(ωg Ti ))= ∆φ

�Determine the gain to satisfy the chosen GCF with PI control. �Check GM.

))(

11(

)(1.

1

1))(

11(

)(1

1,GCFat

2

2

22

Ig

g

P

Igg

Pg

T

KK

T

KKM

ω

τω

ωτωω

+

+=⇒

=++

⇒==



�Determine the phase angle of the “plant” at ωg and the additional phase margin ∆φ.�Choose reset time Ti such that tan-1 (1/(ωg Ti ))= ∆φ

�Determine the gain to satisfy the chosen GCF with PI control.

�Check GM.

( )( ) )

)(

11(

)(1

where

1

22Ipp

Pp

p

T

KKM

MGM

ωτωω

ω

++

=

=


Gain reduction method: example

�Standard 1st order plant with delay , τ=1, T=0.1.

Required PM=45 °. Choose PI controller parameters.

�Find freq ω1 where PM=45 °, ω1=9.0. Choose ωg = 0.8ω1=7.0 r/s (say)

�Phase margin of the “plant” at 7.0 r/s =58°

�∆φ=13 °;1/(ωg Ti)=tan (13°)=0.23

�=> Ti=0.62s.

�Kp=6.9

�Check GM (exercise)


Steps for phase budget method

�Allocate a budget ∆φ for phase lag due to the integral action

�Obtain the frequency (ωg) at which the “plant”

phase angle is –π+ PM+ ∆φ

�Recheck phase

�Choose reset time Ti such that

tan-1 (1/(ωg Ti ))= ∆φ

�Determine the proportional gain to satisfy the

chosen GCF with PI control.

�Check GM


Steps for phase budget method

� Allocate a budget ∆φ for phase lag due to the integral action

� Obtain the frequency (ωg) at which the “plant” phase angle is –π+ PM+ ∆φ

� Recheck phase

� Choose reset time Ti such that

tan-1 (1/(ωg Ti ))= ∆φ

� Determine the proportional gain to satisfy the chosen GCF with PI control.

� Check GM

e.g. Let ∆φ =15 deg (say)


Steps for phase budget method� Allocate a budget ∆φ for phase lag due to the integral action

� Obtain the frequency (ωg) at which the “plant” phase angle is –π+

PM+ ∆φ� Recheck phase


tan-1 (1/(ωg Ti ))= ∆φ


� Check GM

Bode Diagram

Frequency (rad/sec)

10-1

100

101

Magnitu

de (

abs)

10-3

10-2

10-1

100

-180

-135

-90

-45

0

Phase (

deg)

ωωωωg

Phase angle

= –π+ PM+ ∆φ∆φ∆φ∆φ∆φ∆φ∆φ∆φ

PM




� Recheck phase� Choose reset time Ti such that

tan-1 (1/(ωg Ti ))= ∆φ


� Check GM

)(tan 1 τωωϕ ggT−−−=




� Recheck phase


tan-1 (1/(ωg Ti ))= ∆φ� Determine the proportional gain to satisfy the chosen GCF with PI control.

� Check GM

Ti =1/(ωg tan(∆φ) )



� Obtain the frequency (ωg) at which the “plant” phase angle is –π+ PM+ ∆φ� Recheck phase


tan-1 (1/(ωg Ti ))= ∆φ


� Check GM

))(

11(

)(1.

1

1))(

11(

)(1

1,GCFat

2

2

22

Ig

g

P

Igg

Pg

T

KK

T

KKM

ω

τω

ωτωω

+

+=⇒

=++

⇒==



� Obtain the frequency (ωg) at which the “plant” phase angle is –π+ PM+ ∆φ� Recheck phase


tan-1 (1/(ωg Ti ))= ∆φ� Determine the proportional gain to satisfy the chosen GCF with PI control.

� Check GM

( )( ) )

)(

11(

)(1

where

1

22Ipp

Pp

p

T

KKM

MGM

ωτωω

ω

++

=

=


Phase Budget method: example

Standard 1st order plant with delay , τ=1, T=0.1. Required

PM=45 °. Choose PI controller parameters.

Allocate ∆φ=13 ° (say), find ωg such that the phase angle is –π+ PM+ ∆φ= -122 °.

From iterative method, ωg = 7.0 r/s

Recheck phase

∆φ=13 °;1/(ωg Ti)=tan (13°)=0.23 => Ti=0.62s.

Kp=6.9

Check GM (exercise)


Proportional + Derivative Control

(PD-Control)


P-D Control

�Some shortcomings of P-only control may be reduced by introducing derivative action.

�The derivative action imparts phase lead

�PD control may be used to improve damping and/or increased response speed.– Keeping the same proportional gain, D-Control may

be used to improve phase margin, thereby increasing damping and reducing overshoots. (No Change of GCF or steady state error)

– D-Control allows increasing the proportional gain retaining the same phase margin, which improves speed of response (increasing GCF) and reducing sserror

– A mix of the above may also be used.


PD control-II

�Down side of D-Control

– Excessive increase of GCF

• may make the closed loop system noise-prone

• More actuator BW may be required

– GM may be reduced

– the controller output tends to be peaky and noisy

�PD control itself is not widely used except for

slow systems like thermal systems.

�The D-element is used in conjunction with PI,

making the popular PID controller.


PD controller

+

+

+

+

+

+

timederivative thecalled is where

)1(])/(1[)(

)(TF Controller

)()()(

D

DpDpp

Dp

T

sTKsKKKsE

sY

teKteKCOty

+=+==

+== &


P-D Controller Ramp Response

� Step response of PD Controller contains an impulse at t=0 and isdifficult to represent graphically– We would show the step response with roll off pole later

� The ramp response is more informative� At t=Td, the proportional action output overtakes the output from D-

action

t

input

at

t

CO

D-action output= Kd a

P-action output= Kp at

Total output= Kp at+ Kd a

t

CO




t

CO




t

CO




Td


P-D Controller Frequency Response

�Gain increases with frequency above ω= 1/TD

�Max phase lead

=90°

�Lead at corner freq

= 45°90°

45°

M

ω

1/TD

ω

φ

Unit positive

slope


P-D Controller with Roll-Off

� A roll-off pole is often used with P-D controller to reduce the effect of noise

� Does not noticeably affect GCF and phase lead� The time response in closed loop is only marginally

affected by addition of roll off pole

D

D

D

Dp

T

sT

sTK

sE

sY

εε

ε

1/-at is pole off roll The0.1 typicallynumber, small a is where

1

1

)(

)(off roll with TF Controller

+

+==


P-D Controller with Roll-Off-II

� Ratio ε=0.1� Max phase lead

reduced, occurs at 1/√ε from corner.

� Phase lead at first corner nearly at 45 deg,

� Phase lead before first corner unaltered.

� Max magnitude limited, Kp(1/ ε). Would limit noise

Freq Response

10-1

100

101

102

0

30

60

Ph

as

e (

de

g)

PD w Roll Off

Frequency (rad/sec)

101

102

Ma

gn

itu

de

(a

bs

)

1/TD

1/εTD


P-D Controller with Roll-Off-III

�Smoother response

�Step response peak=Kp/ε

0 0.2 0.4 0.6 0.8 10

5

10

15

20Ramp Response

Time (sec)

Am

plitu

de

0 0.2 0.4 0.6 0.8 1

10

20

30

40

50

60

70

80

90

100Step Response

Time (sec)

Am

plitu

de


Choosing P-D Control Parameters

�Considerations:– An important concern in PD Control is the GM

– With pure P-D control GM = 1/(Tdωg)– => Tdωg =1/GM– For a given GM, the obtainable phase lead at the

GCF is arctan (Tdωg) = arctan (1/GM)

ωg

GM

1/Td

1/τ

K1Kp

Frequency response of plant with PD controller


Choosing P-D Control Parameters-II

� Based on the considerations, the steps for choosing the parameters of P-D control are as follows:

1. Given PM and GM

2. Choose Tdωg=1/GM

3. Compute additional phase lead ϕd= tan-1 (1/GM)

4. Ignoring D-action determine ωg such that the phase ωg is [-π+PM - ϕd].

5. Obtain Td.

6. Find Kp to ensure the above GCF with PD control.



� Given PM and GM

� Choose Tdωg=1/GM� Compute additional phase lead ϕd= tan-1 (1/GM)

� Ignoring D-action determine ωg such that the phase ωg is [-π+PM - ϕd].

� Obtain Td.

� Find Kp to ensure the above GCF with PD control.

ωg

GM

1/Td

1/τ

K1KpAs the slope at ωg is a unit negative slope, we have,

ωg/(1/Td)=1/GM



� Given PM and GM

� Choose Tdωg=1/GM

� Compute additional phase lead ϕd= tan-1 (1/GM)� Ignoring D-action determine ωg such that the phase ωg is [-π+PM - ϕd].

� Obtain Td.


Bode Diagram

Frequency (rad/sec)

10-1

100

101

Magnitu

de (

abs)

10-3

10-2

10-1

100

-180

-135

-90

-45

0

Phase (

deg)

Uncontrolled plant

ωωωωg

90°

PD controller

45°

M

ω1/TD

ω

φ

Lead φd

ωg


Choosing P-D Control Parameters-II� Given PM and GM


� Compute additional phase lead ϕd= tan-1 (1/GM)

� Ignoring D-action determine ωg such that the phase ωg is [-π+PM -ϕd].

� Obtain Td.


Uncontrolled plant

90°

PD controller

45°

M

ω1/TD

ω

φ

Lead φd

ωg

Bode Diagram

Frequency (rad/sec)

10-1

100

101

Magnitu

de (

abs)

10-3

10-2

10-1

100

-180

-135

-90

-45

0

Phase (

deg)

Chosen ωg

-ππππ+PM-φφφφd



� Given PM and GM




� Obtain Td.� Find Kp to ensure the above GCF with PD control.

Td=1/(GM.ωg)



� Given PM and GM




� Obtain Td.


2

2

)(1

)(1.

1

Dg

g

pTK

Kω

τω

+

+=


Choosing P-D Control Parameters-III

� Example: Given τ=20, T=3, K= 1.5, desired PM=45 deg, desired GM=7dB

� Solution: PM= π/4, GM=7dB=2.24

� Td ωg =1/GM=1/2.24=0.447;

� ϕd= tan-1 (1/GM)=24 deg = 0.42 rad

� Angle at ωg = [-π+PM - ϕd]= -(3/4) π-0.42 = - 2.775 rad = -159 deg

� Solve for ωg using the following relation for phase angle,

� giving ωg = 0.44

� Td =.447/.44=1.016, say 1.0

� From the gain formula K1 Kp =8.1 => Kp =8.1/1.5 = 5.4

775.21

2−=+−−

τωω

π

g

gT


Performance comparison: P vs PD control

0000 5555 10101010 15151515 20202020 25252525 303030300000

0.50.50.50.5

1111

1.51.51.51.5

P-controlP-controlP-controlP-control

P-D controlP-D controlP-D controlP-D control

� Previous example, same plant and PM

� with only proportional control, the gain= 4.25 and GCF=0.315 are about

78% as obtained from PD

� Faster response, consistent with higher GCF, but higher overshoot in P-D. Marginally smaller SS error.

Pade artifacts(negative

suppressed)


Modified PD controller

�Derivative of PV, instead of error

+

+

e

+ CO

sKD

PK

PV

+

+


Modified P-D Control-II

� The modified controller gives same controller poles� But the closed loop response of modified D-Control does not have

the numerator differentiator� Result: Less overshoot

DCL

CLm

DP

PCLm

DP

DP

DPCL

DP

sTsK

sKsTK

K

SP

PVsK

PVdtdKPVSPKPV

sTK

sTK

SP

PVsK

PVSPsTKPV

+=

++==

−−=

++

+==⇒

−+=

1

1

)(

)()1(1

)(

)/()(:PD Modified

)1(1

)1()(

))(1(:PDOrdinary


PD & PDMod Time Response

� Same Parameters as before

� Noticeably Less overshoot, same as P

� Same settling time

� Marginally more rise time, but faster than P-only

0 5 10 15 20 25 30 35 400

0.5

1

1.5

Time----->

Ste

p R

esp

on

se

Mod P-D

PD


PD control exercise� Why is it necessary to place the zero of the PD controller

at a higher value compared to the prospective gcf?

� Why a roll off pole is recommended for a PD controller.

� Enumerate the advantages and disadvantages of PD control in comparison to P control.

� What is the typical value of the product ωgTD? Justify.

� What is the typical value of obtainable phase lead in a PD control?

� In which situations is PD control contra indicated?


Proportional, Integral + Derivative Control

(PID-Control)


PID Controller


PID Controller-II

�Most widely used

�Combines the good effects of PI and PD control

�SS error is zero

�Response is faster than P or PI controlled plant

Industrial PID

Controller


PID controller Transfer Function

� The TF of the PID Controller may be obtained as follows:

)III()1

1)(

1()(

)( when validis)II(

)II()1)(1

()(

)I()1

()1

1(

)(

)()(

2

LLL

LLL

LL

d

d

I

Ipk

IdIId

d

I

Ipk

I

IdIpd

I

p

dI

pk

sT

sT

sT

sTKsK

sTTTsTT

sTsT

sTKsK

sT

sTsTTKsT

sTK

sKs

KK

sE

sCOsK

ε+

++=

≈+⇒<<

++

≈

++=++=

++==Exact

No roll off

Product

form

approx

With

Roll off Pole


PID controller TF-II

)II()1)(1

()( LLLd

I

Ipk sT

sT

sTKsK +

+≈

)III()1

1)(

1()( LLL

d

d

I

Ipk

sT

sT

sT

sTKsK

ε+++

=

� The TF –II is convenient to apply for sketching asymptotic bode plot.

� This form was also popular in analog implementation, especially with roll-off pole, as shown in TF-III

� As the reset time is usually 5 to 8 times the derivative time, the error in approximation is not significant.

P-I action P-D action

Roll off


PID controller TF-III

� Bode Magnitude Plot� At low frequency, the reset action prevails and at higher

frequency the derivative action prevails.� In the mid-frequency, the behaviour is like proportional control

MKP

1/(εTd)ω

1/(Td)1/(TI)

KP/ ε


PID controller TF-IV

�Phase & Magnitude Plots

10-1

100

101

102

103

-90

-45

0

45

90

Ph

as

e (

de

g)

100

101

102

Ma

gn

itu

de

(a

bs

)Bode Diagram of PID Controller

Frequency (rad/sec)

Lag due to Reset

nullified by the D-action at this

frequency

At high frequency the controller

contributes zero

degree

Roll off pole

D-Zero


PID controller with the plant

�Bode Magnitude Plot

1/τ 1/Ti

1/Td

ωg

ω

M

1/τ 1/Ti

1/Td

ωg

ω

M τε s

eK

sT

sT

sT

sTKsGsK

sT

D

D

i

ippk ++

++=

−

1)

1

1)(

1()()(

Reset

term

Derivative

term

Plant


PID controller with the plant-II

1/τ 1/Ti

1/Td

ωgω

M

)tan(tan)tan2/()tan(

)}()({)(1)(1

)(1

)(

11

|)()(|

1111

2

1

2

2

2

DDi

pk

D

D

i

p

pk

TTTT

jGjK

K

T

T

TK

jGjKM

εωωωπωτωωωφ

ωτεω

ω

ω

ωω

−−−− −++−+−−=∠=

+×

+

+×+=

= K


PID controller with the plant-III

1/τ 1/Ti

1/Td

ωgω

M

)tantantan1

(tan

)tan(tan)tan2/()tan(

)}()({)(

1)(1)(1

)(1

)(

11

1GCFAt

1111

1111

2

1

2

2

2

DgDgigg

g

DgDgiggg

pkg

gDg

Dg

ig

p

TTTT

TTTT

jGjK

K

T

T

TK

M

εωωωωτω

π

εωωωπτωω

ωωωφ

τωεω

ω

ω

−−−−

−−−−

−++−+−=

−++−+−−=

∠=

=+

×+

+×+⇒

=


PID controller with the plant-IV

2

2

2

2

~

)ˆ(1

)ˆ(1)

)ˆ(

11(ˆ

wherefrom;)(1

ˆ

controlonly -Pfor GCF ˆ where

ˆ~ seed ay with iterativel used becan relation above The

|~

where

ofion approximatbetter a is~~ shown that bemay It ;~Let

Dg

Dg

Ig

gg

Pg

g

gg

gg

gg

T

T

T

KK

MM

M

g

εω

ω

ωωω

τω

ωωω

ωωωω

ωω

+

+×+≈

+=

==

=

≈

=

)tantan

1tan

1(tan

1

2

)( PCF,At

11

11

DPDP

iPP

P

P

TT

TTT

εωω

ωτωπ

ω

πωφ

−−

−−

−+

−+=

⇒−=

Using π/2T as the seed,the following relation may berecursively applied to get the PCF.Usually, 3 iterations give acceptably good result.If the roll off pole is present, do not ignore the epsilon term.


Qualitative effect of adding D to PI

derivative

mode on measurement


Analytical Tuning of PID controller

� Objective:

– Given plant parameters, GM (default=6dB) and PM (default=45°), determine PID gains.

� Background:

– φ 1= tan-1[ (ωgTd)] = Lead contribution of D element at GCF

– 1/GM ≅ ωgTd

– φ2= tan-11/(ωgTi)]= Lag contribution of I element at GCF

1/τ 1/Ti

ωg

1/TD

ω

M


Analytical Tuning of PID controller-II

�Background (contd.):

– Net lead obtainable at

GCF φ3= φ1- φ2

– From GM, the parameter ωgTd is determined and hence the total phase lead by the D-action.

– A portion of the lead needs to be allocated to the I-action.

1/τ 1/Ti

ωg

1/TD

ω

M


Analytical Tuning of PID controller-III

� Steps1. From GM, determine ωgTd ≅ 1/GM

2. Determine the obtainable phase lead given by the equation

φ 1= tan-1[ (ωgTd)]

3. Apportion an amount φ2 for nullifying the phase lag by the reset

action.

4. Compute available additional phase lead φ3= φ1- φ2

5. Determine ωg such that the lag of the plant is [-π+PM - ϕ3].

6. From ωg andωgTd determine Td

7. From φ2= tan-11/(ωgTi)] , determine Ti

8. Determine Kp to ensure the above GCF

Note: If Roll off pole is used, 10% of lead would be lost due to the roll off. This

should be accounted for in the design.


Analytical Tuning of PID controller-IV

� Example: Given τ =20, T=3, K1= 1.5, desired PM=45 deg, desired GM=7dB

1. Solution: PM= π/4, GM=7dB=2.24

2. Td ωg ≅ 1/GM=1/2.24=0.447; 3. φφφφ 1= tan-1[ (ωgTd)] =24º = 0.42 rad

4. Allocate φ2 =0.2 rad ≅ 12 º ; φ3= φ1- φ2 = 0.22 rad

5. Angle at ωg = [-π+PM - φ3]= -(3/4) π-0.42 = - 2.58 rad = -148 deg

6. Solve for ωg using the relation for phase angle, – giving ωg = 0.38

7. Td =.447/.38=1.18s, say 1.2s;

8. tan (φφφφ2 )=0.203 => TI= 1/(.447x.203)=11s9. From the gain formula K1 Kp =8.1 => Kp =8.1/1.5 = 5.4


Empirical Tuning

�Approaches:

– Ziegler Nichols

– Cohen-Coon

– Chien

– “Auto-Tuning” method

�Controller settings parameterized by

– Ultimate Gain & period of oscillation

– Plant parameters like T and τ, Combination of above


Empirical Tuning-II

�Plant parameters like T and τ, obtained from Process Reaction Curve,

– In open loop.

�The Ultimate Gain and the time period of oscillation are determined from

– closed loop experiment.

– Called Continuous cycling method

�“Auto-Tuning” method requires special on-off type controller


Process Reaction Curve

� Parameter Determination from Process Reaction Curve1. Wait until the process

reaches steady state.

2. Introduce a step change in the input.

3. Based on the output, obtain an approximate first order process with

a time constant τdelayed by T units from when the input step was introduced


Process Reaction Curve-II

� Advantages of Process Reaction Curve Method– Plant is not operated at the stability limit.

– There is less chance of saturating control loop components

– Determination of process dead time and process time lag often provides insight into the process not otherwise obtained with a closed-loop test

� Disadvantages – Since the test is conducted without controller feedback, any

significant change in process load may provide erroneous results

– If the process is noisy, it may be difficult to determine which part of the reaction curve has the steepest slope so as to construct the tangent line. In such cases, multiple trials may be necessary.

– The graphical method is considered valid only if 0.1< T/τ <0.5


Continuous Cycling Method

� A P- controller with independent bias control, PV transmitter and multi-channel recorder would be necessary.

� A portable field duty storage oscilloscope would also serve the purpose.


Continuous Cycling Method-II

� It may be necessary to measure the process gain by applying small step change in bias and measuring corresponding CO and PV in open loop

�Set the true plant under proportional control, with a very small gain.

�Adjust the bias to take the process at the operating point.

�Apply small impulsive input at the SP and note the PV and CO.

� Increase the gain until the loop starts oscillating.


Continuous Cycling Method-III� Note that linear oscillation is required and

that it should better be detected at the controller output.

� Linear oscillation would be sinusoidal and with more or less constant amplitude.

� Confirm that this indeed is the limiting stability condition by small changes of gain. A small negative change should make the oscillation decaying, a small positive change should make the oscillation growing

� Note this proportional gain as ultimate gain and the time period of oscillation as P. The oscillation generally occurs at the PCF.

� PCF (r/s)=2π/P.

0 10 20 30 40 50 60 70 80 90 100-0.5

0

0.5

1

1.5

2

2.5

0 10 20 30 40 50 60 70 80 90 100-1

-0.5

0

0.5

1

1.5

2

2.5

3

0 10 20 30 40 50 60 70 80 90 100-1

-0.5

0

0.5

1

1.5

2

2.5

3

Sustained

oscillation (almost)


Ziegler Nichols Controller Tuning Rules

� Most well known and widely used tuning rules.� Two variants

– Continuous cycling method

– Process Reaction Curve method

� The tuning rules for Continuous Cycling method is shown below

P/8P/20.6KuPID

--P/1.20.45KuPI

----0.5KuP

TdTiK1 KpControl

Law


Z-N Tuning-II� The Z-N tuning rules for Process Reaction Curve method

is shown below

� Removes the difficulties associated with cycling– Longer experimental time

– Cycling may not be allowed for all plants

� Method gives poorer results compared to continuous cycling.


Exercise

�Rewrite the above table in terms of T and

τ using the following:ωp=π/2T, P=2π/ωp ⇒ P= 4T

P/8P/20.6KuPID

--P/1.20.45KuPI

----0.5KuP

TdTiK1 KpControl Law

TK p

τ5.1ˆ =


Z-N Tuning-III

�For P-Control Z-N rule provides:

– GM = (Ku/KP)=2 ≅ 6dB

– PM varies mildly with β = τ /T

– For β = 20 to 6, PM varies from 46º to 56.3º only.

– Corresponding damping ratios are

approximately 0.46 to 0.56

– The transient responses to step input do

not differ greatly


Z-N Tuning-III

�For PI Control, Z-N rule provides:

– GM = (Ku/KP)=2.2 ≅ 6.8 dB


– For β = 20 to 6, PM varies moderately from 28º to 37.6º only.

– The corresponding closed loop damping ratios are misleading as the system is one order more than the P-control case and the TF has a zero.

– The transient responses to step input does not differ greatly compared to a P-controlled system with 45º phase margin.


Z-N Tuning-IV

�For PID Control, Z-N rule provides:– GM = (Ku/KP)=1/0.6=1.7 ≅ 4.6 dB


– For β = 20 to 5, PM varies moderately from 34.8ºto 44.1º only.

�The ordinary PID shows a fast but more peaky closed loop step response when compared to PI controlled system.– See example

�When the D-action is in the feed back loop, the peak is reduced.


Z-N Tuning-V

Example: Given τ =20, T=3, K1= 10;

Ku=11.1 ; P=11.4 s

P/8=1.4P/2=5.70.6Ku=6.2PID

--P/1.2=9.50.45Ku=5PI

----0.5Ku=5.6P

TdTiKpControl

Law


Correcting GCF and PCF values

1/τ 1/Ti

1/Td

ωg

ω

M

1112

21

2

1

1

1

12

2

1

1

1

1121

2

1

1

222

221

2

2

1

2

1

1

1

pppp

pp

p

p

p

p

gg

g

p

g

p

T

TT

T

TT

M

KM

KM

ωωφ

ωω

φωωπφφφφ

πωπφ

πω

τωπ

ωφ

ωωτω

τω

∆+=∆

+=

∆=+−+=∆=−

−−=−=

−−≈

−−−=

=

=+

=

+=


Example

�Given T=2 s, τ=7 s.

– Select Kp, Ti, TD using Z-N method.

– Calculate GCF, PCF, GM, PM.


Soln:

( )

srad

or

T

p

pp

p

p

/868.0

07

1

2,

1

2jG

PCF,At

2

=∴

=−−

+−−=−=∠

ω

ωπ

ω

τωπ

ωπω

� Given T=2 s, t=7 s, Select Kp, Ti, TD using Z-N method. Calculate GCF, PCF, GM, PM.


Soln: (contd)� Given T=2 s, τ=7 s, Select Kp, Ti, TD using Z-N method.

Calculate GCF, PCF, GM, PM.

( )( )

sPTsPT

sP

KK

KK

D

I

p

pp

p

p

p

905.08/62.32/

24.7/2

695.31577.66.0ˆ6.0

Hence,

1577.67868.01ˆ1

1

ˆ Also,

2

2

====

===×==

=×+=⇒=+

ωπ

τω


Soln: (contd2)� Given T=2 s, τ=7 s, Select Kp, Ti, TD using Z-N method. Calculate GCF, PCF, GM, PM.

( )

( )

( ) ( ) ( )

( ) ( )( ) 26.1905.05.01

62.35.0

11

75.01

695.3

11

1

1

5.01

71

695.3

1

1

Now,

2

221

2

221

2

2

1

11

1

1

=×+×

+×+

=

+++

==

=⇒=×+

=+

=

M

TT

KMGK

K

Dg

Igg

p

pp

g

g

g

p

g

ωωτω

ωω

τω

ωω



( ) ( ) ( )

( ) ( )( )

014.1

905.063.0162.363.0

11

763.01

695.3

11

1

1

63.026.15.0 Now,

2

2

222

2

22

2

1

2

22

2

12

=

×+×

+×+

=

+++

=

=

=×=×=

=

M

M

TT

K

MGK

M

Dg

Igg

p

pp

gg

g

ωωτω

ωω

ωω



( )

( )

srad

T

TT

T

srad

pp

Dp

Ipp

p

p

/046.12

356.0868.0

356.078.278.26658.0308.0163.057.1736.1

905.0868.0tan62.3868.0

1tan

7868.0

1tan57.12868.0

tan1

tan1

tan2

/868.0 Now,

1

11

111

1111

12

1

11

1

1

=+=

∆+=

=+−=+=∆−=+−+−−=

×+×

−×

+−×−=

+−+−−=

=

−−−

−−−

φωω

ππφφ

ωωτω

πωφ

ω



( )

( )

sradT

TT

T

pp

Dp

Ipp

p

/0461.12

123.004.1

123.0018.3018.3755.0259.0136.057.108.2

905.004.1tan62.304.1

1tan

704.1

1tan57.1204.1

tan1

tan1

tan2

2

22

111

1112

23

2

22

2

=+=∆

+=

=+−=+=∆−=+−+−−=

×+×

−×

+−×−=

+−+−−=

−−−

−−−

φωω

ππφφ

ωωτω

πωφ

( )

( )

sradT

TT

T

pp

Dp

Ipp

p

/12.12

0375.01.1

0375.0104.3104.3783.0246.0129.057.12.2

905.01.1tan62.31.1

1tan

71.1

1tan57.121.1

tan1

tan1

tan2

3

33

111

1113

34

3

33

3

=+=∆

+=

=+−=+=∆−=+−+−−=

×+×

−×

+−×−=

+−+−−=

−−−

−−−

φωω

ππφφ

ωωτω

πωφ



GM=3.21 dBPM=370


Z-N Tuning-VI

� Example: Comparative Responses

Rise time for PID

is expectedly

faster compared

to Z-N tuned P and PI controlled

systems

derivative action on

fwd path


Modified D in PID

0 5 10 15 20 25 30 35 400

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Time

Clo

se

d L

oo

p R

es

po

ns

eP

PI

PID-M

derivative action on

feedback path

Peak overshoot

reduced but so is

improvement in

rise time.


Cohen Coon Tuning Rule

�In order to provide closed loop responses with a damping ratio of 25%, Cohen and Coon suggested the design equation for an FOPTD model.

�Similar to the Ziegler and Nichols methods, this technique sometimes brings about oscillatory responses.


Cohen Coon Tuning Rule-II

� Mixed empirical and plant parameter� Results formatted for comparison with Z-N

� when τ>>T – For PID: 10% higher P- gain, 25% lower I-gain, 25% higher D-gain

– results in same gains as in ZN for P and PI

PID

--PI

----P

TdTiK1 KpControl

Law

uKT

)3

1(5.0τ

+

uKT

)11

1(45.0τ

+

2.1)

22.2

1.0(

P

T

T

++

ττ

uKT

)16

31(67.0

τ+

2)

6.0

2.0(25.1

P

T

T

++

ττ

8

74.0)

2.0(

P

T+ττ


Alternative Cohen Coon

�Mixed empirical and plant parameter

�r=τ / T


Chien Method

� Both the ZN and CC methods result in oscillatory closed loop response though with quarter amplitude damping

� In some applications, overshoots are not desirable and the above two methods are considered too “aggressive”.

� The method proposed by Chien guarantees either 0% overshoot or 20% overshoot.

� This method requires perform an open loop test to determine the dead time and time lag.

� The Chien method is valid only if T/τ < 0.11. Once you determine the process characteristics fall within this range, you can determine the controller settings as outlined in Table.


Chien Method-II

Kp is the process gain

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