Classical Control

7/27/2019 Classical Control

1/192

Classical Control

Topics covered:

Modelin . ODEs. Linearization.

Laplace transform. Transfer functions.Block diagrams. Masons Rule.

Time res onse s ecifications.

Effects of zeros and poles.

Stability via Routh-Hurwitz. -

PID controllers and Ziegler-Nichols tuning procedure.

Actuator saturation and integrator wind-up.

Root locus.

Frequency response--Bode and Nyquist diagrams.

Classical Control Prof. Eugenio Schuster Lehigh UniversityClassical Control Prof. Eugenio Schuster Lehigh University 1

.

Design of dynamic compensators.


2/192

Classical Control

Text: Feedback Control of Dynamic Systems, , . . , . . . -

Prentice Hall 2002.



3/192

What is control?

Model Relation between gas pedal and speed:10 mph change in speed per each degree rotation of gas pedal

Disturbance Slope of road:5 mph change in speed per each degree change of slope

w

oc agram or e cru se con ro p an :

Slope(degrees)

0.5)5.0(10 wuy =

-

10

Control(degrees)

Output speed(mph)



4/192

What is control?

w

pen- oop cru se con ro :

r

0.5

PLANT

10

-

10 wr

ol

)5.0(10

.

=1/10 yolReference

(mph)wr 5=

r

wyr

wyre

ol

olol 5

==

%69.7,51,6500,65

=======

olol

ol

eewr

ewr

mph


rrol w en:

1- Plant is known exactly2- There is no disturbance


5/192

What is control?

w

ose - oop cru se con ro :

ru = 20

0.5wuycl )5.0(10 =

PLANTc

-

10wr

201

5

201

200=

1/10+yclr -

wryre clcl201201

+==

551

%5.0%201

10,65 ==== clewr

Classical Control Prof. Eugenio Schuster Lehigh UniversityClassical Control Prof. Eugenio Schuster Lehigh University 5rr

e clcl201201

[%] +=

=.

65201201, ==== clewr


6/192

What is control?

Feedback control can help:

disturbance rejection changing dynamic behavior

LARGE gain is essential but there is a STABILITY limit

the errors due to disturbances and uncertainties withoutmaking the system become unstable is what much of feedback

First step in this design process: DYNAMIC MODEL



7/192

Dynamic Models

MECHANICAL SYSTEMS: maF= Newtons law

xvxbuxm

&&&&

==

velocity

xva &&& == acceleration

mbsUmvmv o

oeUueVv sto

sto +==+ == ,& rans er unc ons

dt

d



8/192

Dynamic Models

MECHANICAL SYSTEMS: F= Newtons law

2

sin Tlmgml c=

+=

&

&&angular velocity

2

== &&& angular acceleration

&&&& 2sin2sin mllmllcc =+=+ Linearization



9/192

Dynamic Models

2ml

T

l

c=+ &&

=1x 21 xx =&Reduce to first order equations:

&=2x 212ml

Tx

l

gx c+=&

uxgxT

ux

x c

+

=

010

,1 & State Variable

lmx 2

=&


JuHxy +=


10/192

Dynamic Models

ELECTRICAL SYSTEMS:

Kirchoffs Current Law (KCL):

e a ge ra c sum o currents enter ng a no e s zero at

every instant

Kirchoffs Voltage Law (KVL)

The algebraic sum of voltages around a loop is zero atevery instant

Resistors:

iR iC iL+ + +

)()()()( GvtitRitv RRRR ==

Capacitors:

vR vC vL )0()(1)()()(0

CCCC

C vdiC

tvdt

dvCti +==

Inductors:

Classical Control Prof. Eugenio Schuster Lehigh UniversityClassical Control Prof. Eugenio Schuster Lehigh University 10+==

t

LLLL

L idv

L

ti

dt

tdiLtv

0

)0()(1

)()(

)(


11/192

Dynamic Models

ELECTRICAL SYSTEMS:

OP AMP:

+ip+= vvv npO ),(

RI

RO

i

iOvO

vp

+ ==

= np vv

- p- n

-n

vnnp



12/192

Dynamic Models

ELECTRICAL SYSTEMS:

R2KCL:

R1 Cv1

vO IOO v

CRv

CRdtdv

12

11 =

+-

( ) ( ) ==t

IOO dvCR

vtvR0

1

2 )(1

0)( OC

Kv1 vO

RCK

1=


Inverting integrator


13/192

Dynamic Models

ELECTRO-MECHANICAL SYSTEMS: DC Motor

armature currenttorque

me

at

Ke &=

=

shaft velocityemf

+= TbJ mmm &&&

0=+++ edt

diLiRv aaaa


Obtain the State Variable Representation


14/192

Dynamic Models

HEAT-FLOW:

1

Temperature DifferenceHeat Flow

R1

21

=

&C

Thermal resistanceThermal capacitance

( )IoI

I TTRRC

T

+=21

&



15/192

Dynamic Models

FLUID-FLOW:

&

Mass rate Mass Conservation aw

outin

Outlet mass flowIn et mass ow

( )outin wwA

hhAm ==

1&&&

A: area of the tank


h: height of water


16/192

Linearization

( )uxfx ,=&Dynamic System:

oouxf ,0 = Equilibrium

== oo ,

( )uuxxfx oo ++= ,&Taylor Expansion

ff &

ux oooo uxuxoo ,,,


oooo uxux ux ,,

, uxx +


17/192

Linearization

uGFx +&

mn uufG

xxfF

1

1

11

1

1

, =

=

MM

L

MM

L

oo

oo

oo

oo

uxm

nnux

uxn

nnux

u

f

u

f

x

f

x

f

,1

,

,1

,

LL

Example: Pendulum with friction 0sin =++ gk &&&

=

10&

Classical Control Prof. Eugenio Schuster Lehigh UniversityClassical Control Prof. Eugenio Schuster Lehigh University 17m

x

l ox

1cos


18/192

Laplace Transform

Functionf(t) of time

Piecewise continuous and exponential orderbtKetf


19/192

Laplace transform examples

After Oliver Heavyside (1850-1925)

=+====

sje

s

e

dtedtetusF

stst

Exponential function

After Oliver Exponential (1176 BC- 1066 BC)

+

+ >+

=+

===0 0 0

)( if1

)(

ss

edtedteesF

tstsstt

Delta (impulse) function (t)

sdtets st allfor1==


0


20/192

Laplace Transform Pair Tables

Si nal Waveform Transformimpulse

step

)(t 1

1)(tu

ramp

ex onential

21

s1

)(ttu

t

damped ramp

+s

2)(

1

+s)(tu

tte

sine

cosine

22 +s

22s

( ) )(sin tut

( ) )(cos tut

damped sine

22)(

++s( ) )(sin tutte


22)( ++s

( ) )(cos tutte


21/192

Laplace Transform Properties

sFsFtttftf +=+=+ LLL

Linearity: (absolutely critical property)

( )ssFdf

t = )( LIntegration property:

)0()()(

=

fssF

dt

tdfLDifferentiation property:

)0()0()(22

)(2=

fsfsFsdt

tfdL

)0()0()0()()( )(21 =

mmm

m

m

ffsfssFdt

tfdLmsL



22/192


= t

Translation properties:

-

{ } 0)()()( >= asFeatuatf as forL

t-domain translation:

Initial Value Property: )(lim)(lim0

ssFtfst +

=

Final Value Property: )(lim)(lim ssFtf =

If all poles ofF(s) are in the LHP



23/192


)()}({a

Fa

atf =LTime Scaling:

sdF )(

ds

=

Convolution: )()(})()({0

sGsFdtgft

= L

j+1

j j=

2



24/192

Laplace Transform

Exercise: Find the Laplace transform of the following waveform

)24 +s

( )42

+

=

ssExercise: Find the Laplace transform of the following waveform

( )dxxtuetft

t

0

4 4sin5)()( +=

( )( )2

3

164

8036)(

++

++=

sss

sssF

( )tudt

tetuetf t40

5)(5)(

+=

( )240

20010)(

+

+=

s

ssF

xerc se: Find the Laplace transform of the following waveform

22 TtuTtutut +=( )eA

sTs 21

=


s


25/192

Laplace transforms

Linearsystem

ain) Complex frequency domain

(s domain)

Differential

equation

Laplace

transform L

Algebraic

equationin(t

do

Classical

techni ues

Algebraic

techni uese

dom

Res onse Inverse La lace Res onse

Ti

signal transform L-1 transform


The diagram commutes

Same answer whichever way you go


26/192

Solving LTI ODEs via Laplace Transform

( ) ( ) ( ) ( ) ubububyayay mmm

m

n

n

n

01

101

1 +++=+++

LL

Initial Conditions:

( )

( ) ( )

( )

( ) ( )0,,0,0,,011 uuyy mn KK

kk 1

Recall j

j

jkk

ksfs

dt

=

=

0

)1( )0()(sL

111 imin

=

+

=

=

=

=

=

ji

j

jiim

i

ij

i

j

jiin

i

i

n

j

jjnnsusUsbsysYsasysYs

1

0

)1(

0

1

0

)1(1

0

1

0

)1( )0()()0()()0()(

011

1

0

)1(

00

)1(

0

011

1

011

1)0()0(

)()(asasas

subsyasU

asasas

bsbsbsbsY

n

n

n

j

j

ji

ii

j

j

ji

ii

n

n

n

m

m

m

m

++++

+++++

++++=

=

==

=

LL

L


For a given rational U(s) we get Y(s)=Q(s)/P(s)


27/192

Laplace Transform

Exercise: Find the Laplace transform V(s)

tdv

3)0( =

=+

v

uv

dt ( ) 66)(

+

+

=sss

sV

Exercise: Find the Laplace transform V(s)

( )( )( ) 12

321

5)(

+

+++=

sssssV

'

5)(3)(

4)( 2

2

==

=++ etvdt

tdv

dt

tvd t

,

What about v t?



28/192

Transfer Functions

( ) ( ) ( ) ububyayay mmn

n

n

01

101

1 ++=+++

LL

Assume all Initial Conditions Zero:

1m

011011 ssssasasas mn +++=++++

LL

InputOutput

)()(

)()(

1

011

1

011

m

n

n

n

m

bsbsbsY

sUsA

sUasasas

sY

+++

=++++

=

L

L

)())((

)(

21

011

1

m

nn

n

m

zszszs

asasassUs

++++==

L

L


)())(( 21 npspsps L


29/192

Rational Functions

We shall mostly be dealing with LTs which are rational

functions ratios of polynomials ins

)( 0111

011

nn

nn

mm

asasasa

bsbsbsb

sF ++++

++++

=

L

L

)())(( 21

21

n

m

pspspsK

=L

L

Kis the scale factor or (sometimes) gain

A strictly proper rational function has n>m


n mproper ra ona unc on as


30/192

Residues at simple poles

Functions o a comp ex varia e wit iso ate ,finite order poles have residues at the poles

)())(()( 2121 nm

kkkzszszsKsF

++

+

=

= L

L

L

2121 nn

sksksk )()()( 21 n

iipspsps

sps

= LL

)()(lim spsk ips

ii

= Residue at a simple pole:



31/192

Residues at multiple poles

rr

rm

s

k

s

k

s

k

s

zszszsKsF

)())(()(

22121

++

+

=

= LL

rjsFrpsjr

ds

jr

dpsjr

k ii

j L1,)()(lim)!(

1 =

=

31

522

+

+

s

ssExample:

31

3

21

1

1

2

+

+

+

+

=sss

3)52()1(

lim 3

23=++

sss 1)52()1(lim 3

23=

++

sssd 2)52()1(lim

!21

3

23

2

2=

++

sssd

31252 2 ss +


( ) ( ) ( )1111 323tutte

ssss+=

+

++

+=

+


32/192

Residues at complex poles

Compute residues at the poles )()(lim sasas

Bundle complex conjugate pole pairs into second-order terms if

you want

but you will need to be careful

( )[ ]222 2))(( ++=+ ssjsjs Inverse Laplace Transform is a sum of complex exponentials

The answer will be real



33/192

Inverting Laplace Transforms in Practice

We have a table of inverse LTs

Write F(s) as a partial fraction expansion

F(s) = bmsm

+ bm1sm1

+L+ b1s + b0ans

n + an1sn1 +L+ a1s + a0

= K 1 2 L m

(s p1)(s p2)L(s pn )

= 1 + 2 + 31 + 32 + 33 + ...+

q

Now appeal to linearity to invert via the table

s p1 s p2 s p3 s p3 s p3 s pq

Surprise! Nastiness: computing the partial fraction expansion is best


done by calculating the residues


34/192

Example 9-12

Find the inverse LT of)52)(1(

)(2 +++

=sss

ssF

*

21211)(221

jsjsssF ++++++=

5

10

1522

)3(20)()1(

1lim

1sss

ssFs

sk =

=++

+=+

=

4255521)21)(1(

)()21(21

lim2

ej

jsjss

ssFjs

jsk ==

+=++++

=++

=

)(252510)( 4)21(4)21( tueeetf jtjjtjt

++= ++


)()

4

52cos(21010 tutee tt

++=


35/192

Not Strictly Proper Laplace Transforms

Find the inverse LT of34

)(2 ++

=ss

ssssF

Convert to polynomial plus strictly proper rational function Use polynomial division 2

2)(+

++=s

ss

5.05.0

2

34

+++=

++

s

ss

Invert as normal

)(5.05.0)(2)()( 3 tueetdt

tdtf tt

+++=



36/192

Block Diagrams

Series:1G 2G

21GGG =

1G+Parallel:

2G+

21 GGG +=



37/192

Block Diagrams

+)(sR )(s )(sC

Reference input

-

)(sB GEC=

Output

= ee ac s gna

)1()1(

GHRGRCGHGHCGRC

+==+=

)1()1( GHREGHHGEE +==+=


Rule: Transfer Function=Forward Gain/(1+Loop Gain)


38/192

Block Diagrams

+)(sR )(s )(sC

Reference input

+

)(sB GEC=

Output

= ee ac s gna

)1()1(

GHRGRCGHGHCGRC

==+=

)1()1( GHREGHHGEE ==+=


Rule: Transfer Function=Forward Gain/(1-Loop Gain)


39/192

Block Diagrams

Moving through a branching point:

G

)(sR )(sC

G

)(sR )(sC

G/1

ss

Movin throu h a summin oint:

+)(sR )(sC +)(sR )(sC

+

)(sB

+

G


)(sB


40/192

Block Diagrams

Example:

+)(sR )(sC+

1

-

-

212321

321

1 GGHGGHGGG

++)(sC)(sR



41/192

Masons Rule

4

+)(sU )(sY+

6

++ +

+

+ +

7

4nodes branches

Signal Flow Graph

sU sY1H 2H 3

6

1 2 3 4

Classical Control Prof. Eugenio Schuster Lehigh UniversityClassical Control Prof. Eugenio Schuster Lehigh University 415H 7H

l


42/192

Masons Rule

a : a sequence o connec e ranc es n e rec on o esignal flow without repetitionLoop: a closed path that returns to its starting node

== GsY

sG1)(

is

pathforwardiththeofgain=iG

gains)loop(all

tdeterminansystemthe

=

=

-1

touchnotdothatloopsthreepossibleallofproductsgain

+

L

)(


pathforwardiththetouchnotdoesthatgraphtheofparttheforofvalue =i

M R l


43/192

Masons Rule

4H

6H

)(sU )(sY1H 2 3H1 2 3 4

5

73515674736251

6244321

1)(

)(

HHHHHHHHHHHHHHsU

sY

++

=


I l R


44/192

Impulse Response

)()()(0

tudtu =

Diracs delta:

Integration is a limit of a sum

By superposition principle, we only need unit impulse response

( )th Response at tto an impulse applied at

ytu


0=

I l R


45/192

Impulse Response

h )(ty)(tut-domain:

Impulse response

dthuty)()()( 0

=

)()()()( thtyttu ==

=

the impulse response of the system.

0

)(sY)(sUs-domain:

)()()( sUssY = )()(1)()()( ssYsUttu === Impulse response


The system response is obtained by multiplying the transferfunction and the Laplace transform of the input.

Ti R P l


46/192

Time Response vs. Poles

Real pole: eths

s

=+= )()(ImpulseResponse

0> Stable ns a e

1

= Time Constant


Time Response vs Poles


47/192


Real pole:

t

ethssH

=+= )()( ImpulseResponse

= Time Constant

tetyss

sY

=+

= 1)(1

)( StepResponse




48/192

Time Response vs. Poles2

Complex poles:

2

22 2)(

++= nnn

sssH mpu se

Response

( ) ( )222 1 ++=

nn

n

s

:

:

n Undamped natural frequency

Damping ratio

( )( ))(

dd

n

jsjssH

+++=

( ) 222

d

n

s

++=


21, == ndn



49/192


( )teths

sH dtnn

sin)(1

)(2222

2

=

++=

ImpulseResponse

0

0

Stable

Unstable




50/192


( )teth

ssH d

tnn

sin1

)(1

)(2222

2

=

++=

ImpulseResponse




51/192


( ) ( )( ) ( )

+=

++= ttety

sssY d

d

d

t

nn

n

sincos1)(1

1)(

222

2

Step




52/192


Complex poles:22 2

)(

++

=nn

n

sssH

( ) ( )222 1

++= nn

n

s

s :022

+=

:

Undam ed

( ) ( )nns 1:12

222 ++< Underdamped

( )[ ] ( )[ ]nnn

ss 11:1 22 +++>

=

Overdamped




53/192



Time Domain Specifications


54/192





55/192


1- The rise time tr is the time it takes the system toreach the vicinity of its new set point- s

transients to decay3- The overshoot Mp is the maximum amount the

sys em overs oo s na va ue v e y s navalue

4- The peak time t is the time it takes the system toreach the maximum overshoot point

8.1

n

r

n

p

1 2=


n

sp te

.==



56/192


Design specification are given in terms of

spr ,,,

These s ecifications ive the osition of the oles

dn ,,

Example: Find the pole positions that guarantee

sec3%,10sec,6.0


57/192

p

Additional poles:1- can be neglected if they are sufficiently to the left

.2- can increase the rise time if the extra pole is withina factor of 4 of the real part of the complex poles.

Zeros:1- a zero near a pole reduces the effect of that pole in

e me response.2- a zero in the LHP will increase the overshoot if thezero is within a factor of 4 of the real part of the

complex poles (due to differentiation).3- a zero in the RHP (nonminimum phase zero) will


response to start out in the wrong direction.

Stability


58/192

y

011

1

011

)()(

asasasbsbsbsb

sRsY

n

n

nmm

++++++++=

LL

)())(()( 21

21

n

m

pspsps

zszszsK

sR

s

=

L

L

)()()()( 2

2

1

1

n

n

pspspssR ++

+

= L

Impulse response:

)(1)( 21 nkkk

sYsR +++== L21 npspsps

tp

n

tptp nekekekty +++= L21 21)(


Stability


59/192

y

pnppnekekekty +++= L

2121)(

t

tK=

i< 0Re

)(011

++++=

asasassa n

n

n LCharacteristic polynomial:


=saarac er s c equa on:

Stability


60/192

y

Necessary condition for asymptotical stability (a.s.):

iai

> 0

Use this as the first test!

022 =+ ss

i ,

Example:

0)1)(2( =+ ss


Rouths Criterion


61/192

Necessary and sufficient condition

Do not have to find the roots !

Rouths Array:

n

n

aaas

aas1

421

L

L n

a Depends on whether

n is even or odd

n

n

cccs

bbbs

3213

3212

L,, 7613541

2321

1a

aaab

a

aaab

a

aaab

=

=

=

n

dds 214

M

L

L,,

31312121

1

315121

21311

cbbccbbc

bbaabc

bbaabc

=

=

==


nasMM

,,11 cc

Rouths Criterion


62/192

How to remember this?

Rouths Array:

Ln =

L

2

2322211

131211

mmms

n

n

,

,

12,2

,

=

am jj

jj

L

L

4342413

333231

mmmsn

31,11,1

1,21,2

= +

+

imm

mm

mjii

jii

1,1

,

mi


Rouths Criterion


63/192

The criterion:

The system is asymptotically stableif and only if all the elements in the first

u u rr y r v

The number of roots with positive realparts is equal to the number of signchanges in the first column of the Routh


Rouths Criterion - Examples


64/192

Example 1: 0212 =++ asas

032213 =+++ asasasExample 2:

Example 3: 044234 23456 =++++++ ssssss

0)6(5 23 =+++ kskssExample 4:


Rouths Criterion - Examples


65/192

Example: Determine the range of K over which thesystem is stable

+)(sR )(sY1+s

- 61 + sss


Rouths Criterion


66/192

Special Case I: Zero in the first columnWe replace the zero with a small positive constant>0 and proceed as before. We then apply the

stability criterion by taking the limit as 0

Example: 010842 234 =++++ ssss


Rouths Criterion


67/192

Special Case II: Entire row is zeroThis indicates that there are complex conjugate pairs.If the ith row is zero, we form an auxiliary equation

from the previous nonzero row:

L+++= + 331

21

11 )(iii ssssa

Where are the coefficients of the i+1 th row in thearray. We then replace the ith row by the coefficients

of the derivative of the auxiliary polynomial.

Example: 02010842 2345 =+++++ sssss


Properties of feedback


68/192

Disturbance Rejection:

r ++

w

o

wArKy o +=

+

wClosed loop

c-

wry c+

++

=1

1

1




69/192

Disturbance Rejection:

oose con ro s. . or w= ,yr

==1

O en loo :

rwryKc =+>> 01

o

Closed loop:

Feedback allows attenuation of disturbance withoutaving access to it wit out measuring it !!!


: g ga n s angerous or ynam c response



70/192

Sensitivity to Gain Plant Changes

r ++

w

o

oo AK

r

yT =

=

+

wClosed loop

c-

cc

K

AK

r

yT

+==

1




71/192

Sensitivity to Gain Plant Changes

TTo =Open loop:

o

o

cc

c

T

T

AAKAT

T =


72/192

P Controller:

pKsC =)(-

)(sG

.1)(

,)()(1)(

sE

sGsCsR

=

+=

)()(),()( ssUtetu pp == Step Reference:

)0(1)(1lim)(lim)(

00 GKssGKsssEe

ssR

ppss

ss +=

+===

= )0(0e pss

Plant has a pole at the origin

True when:


results in underdamped (or even unstable) transients.

PID Controller


73/192

P Controller: Example (lecture06_a.m)

+)(sR )(sY)(s )(sUp

- 12 ++ ss

11

)()(2

Kss

AK

sGK

sGK

sR

sY pp

+++

=

+

=

12 += pn AK0

11==

9 Underdamped transient for large proportional gain

12 =n

122 + ppn

K


9 Steady state error for small proportional gain

PID Controller


74/192

s

KKsC Ip +=)(

-

+)(sR )(sY)(sG)(s )(sU

,)()(1

)()(

)(

)(

sGsC

sGsC

sR

sY

+=

Kt

.

)()(1

1

)(

)(

sGsCsR

sE

+

=

,0

ss

seeu pIp

==

1111

Step Reference:

)(1m

)(1mm

000=

++

=

++

=== sG

s

KK

ssG

s

KK

ssses

sI

p

sI

p

ssss

It does not matter the value of the proportional gain

Plant does not need to have a pole at the origin. The controller has it!


Note that this is valid even forKp=0 as long asKi0

PID Controller


75/192

PI Controller: Example (lecture06_b.m)

I+)(sR )(sY)(s )(sU

sp

- 12 ++ ss

KsK

sGK

K

sY

Ip

+

+ )(

AKsAKsssG

s

KK

sR IpIp

++++=

++

= )1()(1

)( 23

DANGER: for largeKi the characteristic equation has roots in the RHP

23


p

Analysis by Rouths Criterion

PID Controller


76/192

PI Controller: Example (lecture06_b.m)

0123 =++++ KsKss

Necessary Conditions: 0,01 >>+ KK IpThis is satisfied because 0,0,0 >>> Ip KK

Rouths Conditions:

AKs p3 11 + >+ 01 AKAK Ip

AKAKs

s

Ip

I

1 1 +KK P

1+

o180)(

)(1)(:0

sLK

sL

KsLK

Phase condition

Magnitude condition

1

=

==

K


8145.5

=

Kis 0481


100/192


-as function of the gain K ROOT LOCUS

Root Locus by Phase Condition

Example: K+)(sR )(sY( )( )845 12 +++ + ssss s)(s )(sU


101/192

p- ( )( )845 +++ ssss

s 1+=( )

s

ssss

1

845 2

+=

+++

iso 31+=

belongs to the locus?




102/192

o43.108o87.36

o45o90

o70.78

oooooo 18070.784587.3643.10890 +++ is 31+= belon s to the locus!


Note: Check code lecture09_a.m



103/192

iso 31+=


-as function of the gain K ROOT LOCUS

Root Locus==

when Kvaries from 0 to (positive Root Locus) or


104/192

(p )

from 0 to - (negative Root Locus)

0)()(1

)(0)(1 =+==+ sKbsasLsKL

Basic Properties:

Number of branches = number of open-loop poles

RL begins at open-loop poles

0)(0 == saK

en s a open- oop zeros or asymp o es

>

===0

0)(0)(mns

sbsLK


RL symmetrical about Re-axis

Root Locus

Rule 1: The n branches of the locus start at the poles ofL(s)d f h b h d h f


105/192

and m of these branches end on the zeros ofL(s).

m: order of the numerator ofL(s)

Rule 2: The locus is on the real axis to the left of and oddnumber of poles and zeros.In other words an interval on the real axis belon s to theroot locus if the total number of poles and zeros to the rightis odd.This rule comes from the phase condition!!!


Root Locus

Rule 3: As K, m of the closed-loop poles approach thel d f th h t t


106/192

open-loop zeros, and n-m of them approach n-m asymptotes

1,,1,0,)12( =+=mnlll K

and centered at

1,,1,0,11 =

=

= mnlab Kzerospoles


Root Locus

Rule 4: The locus crosses the jaxis (looses stability) wherethe Routh criterion shows a transition from roots in the left


107/192

half-plane to roots in the right-half plane.

Example:

)54()(

2 ++=

sssssG

5,20 jsK ==


Root Locus

Example: 4673

1)( 234 ++++

+= ssss

ssG


108/192


Root Locus

Design dangers revealed by the Root Locus:


109/192

-

instability due to asymptotes.

4673)(

234 ++++=

sssssG

Nonminimum phase zeros: They attract closed loop polesinto the RHP

1

1)( 2 ++

= ss

s

sG


Note: Check code lecture09_b.m

Root Locus

Vietes formula:


110/192

- ,

poles is constant

= polesloopclosed1a

nn

nnmm

mm

asasas

bsbsbs

sa

sbsL

++++++++

==

11

1

11

1

)(

)()(

L

L


Phase and Magnitude of a Transfer Function

011

1

011)( asasas

bsbsbsbsG n

n

n mm ++++++++

=

L

L


111/192

)())(()(

21

21

n

m

pspsps

zszszsKsG

=L

L

e ac ors , s-zj an s-pk are comp ex num ers:

zjiz mjerzs == K1,)(

K

pkip

kk pkerps == L1,)(

e=zm

zz

Kiz

m

izizi ererer

L21 21


pn

ppip

n

ipip ererer L21 21

=

Phase and Magnitude of a Transfer Functionzzz

pn

pp

mK

ipipipmi

ererer

ererereKsG = L

L

21

21

21

21)(


112/192

n ererer 21( )

( )pnpp

zm

zz

K

ippp

izm

zzi

errr

errr

eK

+++

+++

= L

L

L

L

21

21

21

( ) ( )[ ]pnppzmzzKippp

zm

zz

errr

K +++++++= LL

L

L212121

Now it is easy to give the phase and magnitudeof the transfer function:

n21

pn

pp

z

m

zz

rrr

rrrKsG = L

L

21

21 ,)(


( ) ( )pnppzmzzKsG +++++++= LL 2121)(

Phase and Magnitude of a Transfer Function

Example:

( )( ) )20(51

.)(

+++=

ssssG


113/192

z

rsG 1=iss o 57 +==

ppp z

pr3zr1

pr2pr1 ( )pppzsG

rrr

3211

321

)( ++=


Root Locus- Magnitude and Phase Conditions

{ } { })()()(1 sL numdenrootszerosRL +=+=when Kvaries from 0 to (positive Root Locus) or


114/192

(p )

-

( ) ( )[ ]

p

n

ppz

m

zzpKi

ppp

z

m

zz

pmp errr

rrrKsss

zszszsKsL

+++++++

=

=LL

L

L

L

L21212121

)())(()(

1

nn

L 1)( 21 ==

ppp

zm

zz

p

rrrKsL

=> Ks:( ) ( ) oLL

L180)( 2121

21

=+++++++= pnppz

mzzK

n

psL

=< sLK 1)(:0 LL 1

)(21

21

== pnpp

zm

zz

p Krrr

rrr

KsL


oLL 0)( 2121 =+++++++=p

nppz

mzzpsL

Root Locus

Selecting Kfor desired closed loop poles on Root Locus:


115/192

o ,characteristic equation for some value ofK

Then we can obtain K as

sL o )( =

1=

1

)( o

K

s

=


o

Root Locus

Example:( )( )51

1)()(++

==ss

sGsL


116/192

( )( )51 ++ ss

54314351143 ++++=++==+= iissis

( ) ( ) 204242 2222 =++=

so

Using MATLAB:

sys=tf(1,poly([-1 -5]))so=-3+4i


[K,POLES]=rlocfind(sys,so)

Root Locus1

( )( )51 ++ ss57 +=o is

o =


117/192

06.421 ==Kos

57 is +=


When we use the absolute value formula we are assumingthat the point belongs to the Root Locus!

Root Locus - Compensators

Example: ( )( )511

)()( ++== sssGsL


118/192

Can we place the closed loop pole at so=-7+i5 only varying K?NO. We need a COMPENSATOR.

1==1==( )( )51 ++ ss( )( )51 ++ ss


The zero attracts the locus!!!

Root Locus Phase lead compensator

Pure derivative control is not normally practical because of theamplification of the noise due to the differentiation and must


119/192

zpzs

sD >+

= ,)(Phase lead

COMPENSATOR

When we study frequency response we will understand whywe call Phase Lead to this compensator.

( )( )zp

ssps

zssGsDsL >

++++

== ,51

1)()()(

How do we choose z and p to place the closed loop poleat so=-7+i5?


Root Locus Phase lead compensatorzs + 1

Phase lead( )( )ssps +++

,

51


120/192

COMPENSATOR

o19.140o80.111o04.21

?1z


oooooo 03.9303.45304.2180.11119.1401801 ==+++=z 735.6= z

LL 1802121 =+++++++= nmps


Example:

( )( )5120

.)()()(

+++==

ssssGsDs


121/192

COMPENSATOR

117=

+=so



Selecting z and p is a trial an error procedure. In general:

The zero is laced in the nei hborhood of the closed-


122/192

The zero is laced in the nei hborhood of the closedloop natural frequency, as determined by rise-time orsettling time requirements.

The poles is placed at a distance 5 to 20 times thevalue of the zero location. The pole is fast enough toavoid modifying the dominant pole behavior.

Noise suppression (we want a small value for p) Compensation effectiveness (we want large value forp)

Classical Control Prof. Eugenio Schuster Lehigh UniversityClassical Control Prof. Eugenio Schuster Lehigh University122

Root Locus Phase lag compensator

Example:

( )( )5120

.)()()(

+++==

ssssGsDs


123/192

( )( )

2

000

10735.651

1

20

735.6lim)()(lim)(lim

=+++

+===

sss

ssGsDsLK

sssp

What can we do to increase Kp? Suppose we want Kp=10.

( )( )zp

sss

s

ps

zssGsDsL

+

= ,)(Phase lag compensator:

We will see why we call phase lead and phase lag to


Frequency Response

We now know how to analyze and design systems via s-domainmethods which yield dynamical information e responses are escr e y e exponen a mo es


127/192

e responses are escr e y e exponen a mo es

The modes are determined by the poles of the response Laplace

Transform

We next will look at describing cct performance via frequency

This guides us in specifying the system pole and zeropositions


Sinusoidal Steady-State Response

ons er a s a e rans er unc on w a

sinusoidal input:

22


128/192

22 +s

The Laplace Transform of the response has poles Where the natural modes lie

These are in the open left half plane Re(s)


129/192

Transform2222

sincos)(

+

++

=ss

sU

esponse rans orm

N

s

k

s

k

s

k

s

k

s

ksUsGsY

++

+

+

++

== L21

*

)()()(

Response Signal

L* 21 tptptptjtj Nekekekekket +++++=

forced response natural response

Sinusoidal Steady State Response

44444 344444 2144 344 21

responsenaturalresponseforced

t


tjtjSS ekkety

+= *)( 0


Calculating the SSS response to Residue calculation

)cos()( += tAtu

mm sssss ==


130/192

2

sincos)(

sincos))((lim

mm

js

jsjs

j

AjGss

s

AjssG

sssss

=

+

=

Si nal calculation

))(()(2

1

2

1)( jGjj ejGAejAG +==

js

k

js

kLtyss

++

=

)(

*1

( )Ktk

eekeek tjKjtjKj

+=

+=

cos2


( ))(cos)()( jGtjGAtyss ++=


Response to is ))(cos()(

)cos()(

jGtAjGy

ttu

ss ++=

+=


131/192

Output frequency = input frequency

Out ut am litude = in ut am litude G Output phase = input phase + G(j)

T e Frequency Response o t e trans er unct on s s g venby its evaluation as a function of a complex variable ats=j We speak of the amplitude response and of the phase response

They cannot independently be varied

Bodes relations of analytic function theory


Frequency Response

Find the steady state output forv1(t)=Acos(t+)sL

+

V ( ) R V2(s)


132/192

_V1(s) RV2(s)

-

ompu e e s- oma n rans er unc on s

Voltage divider

RsL

RsT

+=)(

=

+=

RLjT

LR

RjT

122

tan)(,)(

)(

Compute the steady state output

( )[ ]RLtRtv SS /tancos)( 12 +=


R +

Bode Diagrams

Bode Diagram

0

Log-log plot of mag(T), log-linear plot of arg(T) versus

)

-

B


133/192

agn

itude(dB)

-15

-10

itude(dB

-25

-20

0

Mag

e(deg)

-45(deg)

Pha

-90

Phas


Frequency (rad/sec)

10 10 10

Frequency Response

)cos()( += ttu )(sG ))(cos()( jGtAjGyss ++=


134/192

Stable Transfer Function

After a transient, the output settles to a sinusoid with anamplitude magnified by and phase shifted by .)( jG )( jG

Since all signals can be represented by sinusoids (Fourierseries and transform), the quantities and are)( jG )( jG

.

Bode developed methods for quickly finding and)( jG )( G


.

Frequency Response

Find the steady state output forv1(t)=Acos(t+)sL

+

V (s) R V2(s)


135/192

_V1(s) R 2( )

-


Voltage divider

RsL

RsT

+=)(

=

+=

RLjT

LR

RjT

122

tan)(,)(

)(




R +

Frequency Response - Bode Diagrams

Bode Diagram

0


)

-

dB


136/192

agn

itude(dB)

-15

-10

itude(d

-25

-20

0

Mag

e(deg)

-45(deg)

Pha

-90

Phas


Frequency (rad/sec)

10 10 10

[ ] [ ] Hzffrad === ,2sec,/

Bode Diagrams

)())(()( 21

21

n

m

pspsps

zszszs

KsG

= LL


137/192

( ) ( )[ ]pn

ppzm

zzK

i

z

m

zz

rrr +++++++= LLL212121

The magnitude and phase ofG(s) when s=j is given by:

pn

p rrr L21

z

m

zz rrr=

L21

Nonlinear in the magnitudes

pn

ppzm

zzK

pn

p

jG

rrr

+++++++= LL

L

2121

21

)(

,


Linear in the phases

Bode Diagrams

)(log20)( jGjGdB

=

zzz


138/192

?)()( 21 ==dBppp

zm

zz

jGrrr

rrrKjG

L

L

( ) ( )pnppzmzmz rrrrrrKsG log20log20log20log20log20log20log20)(log20 211 +++++++= LL

By properties of the logarithm we can write:

Linear in the magnitudes (dB)

The magnitude and phase ofG(s) when s=j is given by:

zzzK

dB

p

ndB

p

dB

p

dB

z

mdB

z

dB

z

dBdB rrrrrrKsG +++++++= LL 2121)(


nm = LL 2121

Linear in the phases

Bode Diagrams

1)()( ===

RjT

RsT

1 +++ Ls


139/192

1

+R

Expressing the magnitude in dB:

+=

+=

22

1log101log201log20)(R

L

R

LjT

dB

Asymptotic behavior:

( )

log20log20/log20/log20)(: ==

dBdB L

R

LRLRjT

dB


LINEAR FUNCTION in log!!! We plot as a function of log.dB

jG )(

Bode Diagrams

A cutoff frequency occurs when the gain is reduced from its

Decade: Any frequency range whose end points have a 10:1 ratio

max mum pass an va ue y a ac or :21


140/192

1 =2 AXAXAX

Bandwith: frequency range spanned by the gain passband

Lets have a look at our example:

== 1)(01 jT

==

+2/1)(/

12 jTLR

R

L


s s a ow-pass ter ass an ga n=1, uto requency=R/L

The Bandwith is R/L!

General Transfer Function (Bode Form)q2

( ) ( ) ( ) nn

nm

o

jj

jjKjG +++=121

The magnitude (dB) (phase) is the sum of the magnitudes (dB)


141/192

g ( ) (p ) g ( )(phases) of each one of the terms. We learn how to plot each term,

we learn how to plot the whole magnitude and phase Bode Plot.

Classes of terms:

-

2-o

( ) ( )mjjG =-

4-

( ) ( )njjG 1+= q 2


nn

j

++

= 12

General Transfer Function: DC gain

( ) oKjG =

=Magnitude and Phase:

oB


142/192

( ) >

=

00 oK

jG

if

35

40

180

200

o10=sG

20

25

30

tude(dB)

100

120

140

se

(deg)

10

15Magni

40

60

80Pha


10-1

100

101

102

0

Frequency (rad/sec)10

-110

010

110

20

Frequency (rad/sec)

General Transfer Function: Poles/zeros at origin

( ) ( )

mjjG =

dB


143/192

( )

mjG =sGm 11 ==

10

20

-20

0s

01 =G

-10

0

nitude(dB)

-60

-40

ase(deg)

-30

-20Mag

-100

-80Ph

dec

dB

m 20


10-1

100

101

102

-40

Frequency (rad/sec)10

-110

010

110

2-120

Frequency (rad/sec)

General Transfer Function: Real poles/zeros

( ) ( )

n

jjG 1+=


144/192

( ) 22

1log10 += nj

G dB

( ) ( ) 1tan= njG

Asymptotic behavior:

log20)(

0)(

/1

/1

+

>>

>


145/192

-5

B)

( )

1+

=s

sG

dBn 3 dec

dBn 20

-20

-15

-

agnitude(

-30

-25M

dBnjG

dBjG

dB

dB

==

3)/1(

0)0(

10-1 100 101 102 103-40

-35

Fre uenc rad/sec

dBnGdB

= )sgn()(


General Transfer Function: Real poles/zeros

0

10 10/1,1 == n

-10( )

1=sG


146/192

-20( )

1+

=s

sG

-50

-40

-

hase(deg

o

o0)0( = jG-80

-70

-60

o

90)( =

=

njG

nj

10-1

100

101

102

103-100

-90


General Transfer Function: Complex poles/zerosq2

( ) nn

jj

jG ++=12

222

Magnitude and Phase:


147/192

( )

+

= 22

21log10dB qjG

1 /2 n

nn

22 /1 nAsymptotic behavior:

lo402

0)(

+

n >> n

General Transfer Function: Complex poles/zeros

0

20

05.0,1,1 === nqdB

q 2

-20 ( )1102

=sGdB


148/192

20

dB

)

( )11.02 ++ ss

dec

dBq 40

-60

-

Magnitude

-100

-80

( )dBqjGdBjG

dBdBn

dB

+==

3)(0)0(

10

-1

10

0

10

1

10

2

10

3-120

Frequency (rad/sec)

qdB

= sgn


2

2

112)()(

==== nrdBdBr

AX

dBqjGjG

General Transfer Function: Complex poles/zeros

0

20 05.0,1,1 === nq

-20 ( )1102

=sG


149/192

-40

( )11.02 ++ ss

-100

-80

-

hase(deg)

o0)0( = jGo

-160

-140

-120

o180)(901

==

qjGqj

10-2

10-1

100

101

102-200

-180


Frequency Response

( ) ( )5.02000 += ssGExample:


150/192


Frequency Response: Poles/Zeros in the RHP

Same .

)( jG.

An unstable pole behaves like a stable zero


151/192

An unstable pole behaves like a stable zero

An unstable zero behaves like a stable pole

1=Exam le:

2s

but can be computed and used for control design.


First order LOW PASS

+=ssT )(

KjT )(


152/192

+=

j

jT )(

Gain and Phase:

)(

22

+=jT

an=

> 00



153/192

0)(,)0( == TK

T

===

+= c

TKKjT

2

)0(

2)(

22Cutoff frequency

KjT

>)(

=== c


cB = Bandwith

First order LOW PASS

)(1

22

+=jT

an=


154/192

o45)1(tan)( 1 ==

= KK

o90)( >>


155/192

+=

j

jT )(

Gain and Phase:

)(

22

+=

o

jT

an+=

> 00



156/192

KTT == )(,0)0(

=

==

+= c

TKKjT

2

)(

2)(

22Cutoff frequency

KjT


=

First order HIGH PASS

)(

1

22

+

=

o

jT


157/192

o

oo 45)1(tan90)( 1 +=+= KK

K+ >

)(tan90)( 1o


First Order BANDPASS

+

+

==2

2

1

121 )()()(

ss

sTsTsT

= 21)(

KjK

jT


158/192

+

+=)(

jT

= 21)(

KK

jT

+

+

21

21

21

21)(

KKjT


159/192

2

12


160/192

2222

2

)(

2

)(oooo j

jjT

ss

ssT

++

=

++

=


:

:

o a ura requency

Damping Ratio

Second Order BANDPASS

2222 2)(2)( oooo j

jjTss

ssT

++=++=

K


161/192

K

=

KjT

oo

=


162/192

/ oK

2

)(

2

2)(

22

/)(

2MAXo

jTjT

jjT o

o

==+

==

!!s!frequenciecutofftheareofrootsThe

2= o

o

2

2

1

1 ++=oC 21

2

= CCo Center Frequency


2 = oC 12 CC

Second Order LOWPASS


163/192

2222

2

)(

2

)(oooo j

jT

ss

sT

++

=

++

=


:

:

o a ura requency

Damping Ratio

Second Order LOWPASS

2222 2)(2)( oooo jjTsssT ++=++=

)0()( TK

jT =


164/192

2)0()(

TjTo

>

o

o

==22

)0(/)(

TKjT oo ==

21)0()( === TjT


12

Second Order HIGHPASS


165/192

22

2222 2)(

2)(

oooo jjT

ss

ssT

++

=++

=


:

:

o a ura requency

Damping Ratio

Second Order HIGHPASS

22

2

22

2

2)(2)( oooo j

K

jTss

Ks

sT

++

=++=

2K


166/192

T 2

)()( == >>

TKjTo

oo

o

o

K

==2

)(

)(

==TK

jT o

)()( === oTjT


112

Frequency Response

)cos()( += ttu )(sG ))(cos()( jGtAjGyss ++=

Stable Transfer Function


167/192

)()()( jGjejGjG = BODE plots

{ } { })(Im)(Re)( jGjjGjG += NYQUIST plots


Frequency Response

{ } { } )()()(Im)(Re)( jGjejGjGjjGjG =+=

They are two way to represent the same information


168/192

{ })(Im jGj

)( jG

)( jG

{ })(Re jG


Frequency Response

Find the steady state output forv1(t)=Acos(t+)

sL+

_V1(s) RV2(s)

-


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Voltage divider

RsL

RsT

+=)(

=+= RLjTLR

RjT

122 tan)(,)()(




R +

Frequency Response - Bode Plots

Bode Diagram

0


)

gnitude(dB)

-15

-10

-

itude(dB

10)(

+=

ssG


170/192

a

=-25

-20

0

Mag

e(deg)

-45(deg)

Pha

-90

Phas


Frequency (rad/sec)

10 10 10

[ ] [ ] Hzfrad === ,2sec,/

Frequency Response Nyquist Plots

L

222)(

LRLjRLjRLjRjT

+

=

+=

+=

=

+=

RLR

22an,

)(

{ } { } 222222

2 RLjT

RjT

==


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{ } { } 222222 )(Im,)(Re

01:0 TT 1=T-

o90)(,0)(: jTjT 0)(

L

jjT2-

{ } == 0)(Re jT3-


{ } === ,00)(Im jT4-

Frequency Response - Nyquist Plots

)(Re)(Im jGjG vs.

10)(

+=

ssG


172/192

=


Nyquist Diagrams

General procedure for sketching Nyquist Diagrams:

Find G(j0)

Find G(j)


173/192

Fin * suc t at Re G(j*) =0; Im G(j*) is t eintersection with the imaginary axis.

=

intersection with the real axis.



( )21

1

)( += sssGExample:

( )( ) ( )2

2

2

2

2

22

12

1

1

1

1

1

1)(

+=

==

jjjjG


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2111 ++

= G 2:0 -

2- 01

)(:3

jjG

{ } == 0)(Re jG3-


4- { } === ,10)(Im jG 2)1(Re =jG


( )21

1

)( += sssGExample:


175/192


Nyquist Plots based on Bode Plots

( )2

1

)(

+

=ss

sG

dec20

dec

dB60


176/192

dec

o90o180

oo 90270 =


Nyquist Stability Criterion

)(sG-

+)(sU )(sY

When is this transfer function Stable?


177/192

NYQUIST: The closed loop is asymptotically stable if thenumber of counterclockwise encirclements of the point-

number of poles ofG(s) with positive real parts (unstablepoles)

Corollary: If the open-loop system G(s) is stable, then theclosed-loop system is also stable provided G(s) makes noencirclement of the point (-1+j0).



1335

1)(

234

++++

=ssss

sG

1332

1)(

234

++++

=ssss

sG


178/192



)(sG-

+)(sU )(sY

When is this transfer function Stable?

K


179/192

NYQUIST: The closed loop is asymptotically stable if thenumber of counterclockwise encirclements of the point-

number of poles ofG(s) with positive real parts (unstablepoles)


Neutral Stability

)(sKG-

+

21)( =sG

Root locus condition:


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o180)(,1)( == sGsKG

RL condition hold for s=jo,

Stability: At o180)( = jG


1)( >

+

+

= TsTs

sD


191/192


Frequency Response Phase Lag Compensators

1. Determine the open-loop gain Kthat will meet the PMrequirement without compensation.

. raw e o e p o o e uncompensa e sys em wcrossover frequency from step 1 and evaluate the low-frequency gain.

3 Determine to meet the low frequency gain error


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3. Determine to meet the low frequency gain errorrequirement.

4. Choose the corner frequency=1/T

(the zero of the

compensator) to be one decade below the new crossoverfrequency .

5. The other corner frequency (the pole of the compensator)

is then =1/ T.


6. Iterate on the design

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Classical Control