Classical Control - Lehigh University › ~eus204 › teaching › ME450_MRC › ...Classical Control – Prof. Eugenio Schuster – Lehigh University 18 • There is a need to worry

1

Classical Control

Topics covered:

Modeling. ODEs. Linearization. Laplace transform. Transfer functions. pBlock diagrams. Mason’s Rule. Time response specifications. Effects of zeros and poles. Stability via Routh-Hurwitz. Feedback: Disturbance rejection, Sensitivity, Steady-state tracking. PID controllers and Ziegler-Nichols tuning procedure. Actuator saturation and integrator wind-up

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

Actuator saturation and integrator wind up.

Root locus. Frequency response--Bode and Nyquist diagrams. Stability Margins.Design of dynamic compensators.

Classical Control

Text: Feedback Control of Dynamic Systems,4th Edition, G.F. Franklin, J.D. Powel and A. Emami-NaeiniPrentice Hall 2002.


2

What is control?

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

For any analysis we need a mathematical MODEL of the system

0.5

w

Block diagram for the cruise control plant:

)50(10

5 mph change in speed per each degree change of slope

Slope(degrees)


u +

-10 y

Control(degrees)

Output speed(mph)

)5.0(10 wuy −=

What is control?

w

Open-loop cruise control:

PLANT 10ru =

0.5

u +

-10 yol

Reference(mph)

wr

wrwuyol

5

)5.010

(10

)5.0(10

−=

−=

−=

1/10r

10


(mph)

rw

ryre

wyre

olol

olol

500[%]

5

=−

=

=−=%69.7,51,65

00,65==⇒==

=⇒==

olol

ol

eewrewr

mph

OK when:1- Plant is known exactly2- There is no disturbance

3

What is control?

w

Closed-loop cruise control:

PLANT( )clyru −= 20

0.5

u +

-10 ycl

wr

wuycl

2015

201200

)5.0(10

−=

−=

1/10r -

+


rw

ryre

wryre

clcl

clcl

2015

2011[%]

2015

2011

+=−

=

+=−=

%69.0655

2015

20111,65

%5.0%20110,65

=+=⇒==

==⇒==

cl

cl

ewr

ewr

What is control?

Feedback control can help:

• reference following (tracking)• disturbance rejectiondisturbance rejection• changing dynamic behavior

LARGE gain is essential but there is a STABILITY limit

“The issue of how to get the gain as large as possible to reduce the errors due to disturbances and uncertainties without making the system become unstable is what much of feedback


making the system become unstable is what much of feedback control design is all about”

First step in this design process: DYNAMIC MODEL

4

Dynamic Models

MECHANICAL SYSTEMS:

xbuxm &&& =

maF = Newton’s law

xvaxv

xbuxm

&&&

&

===

−=velocity

acceleration


mbsm

UV

muv

mbv

o

oeUueVv sto

sto +

=⎯⎯⎯⎯⎯ →⎯=+==

1,

& Transfer Function

sdtd→

Dynamic Models

MECHANICAL SYSTEMS: αIF = Newton’s law

2 sin Tlmgml +−= θθ&&

2

sin

mlI

Tlmgml c

=

==

=

+

θωαθω

θθ

&&&

& angular velocity

angular acceleration

moment of inertia


2sin2sin mlT

lg

mlT

lg cc =+⎯⎯ →⎯=+ ≈ θθθθ θθ &&&& Linearization

5

Dynamic Models

2mlT

lg c=+ θθ&&

Reduce to first order equations:

θθ&=

=

2

1

xx

212

21

mlTx

lgx

xx

c+−=

=

&

&

Tx ⎤⎡⎥⎤

⎢⎡⎤⎡ 0101 State Variable


uxlgx

mlTu

xx

x c ⎥⎦

⎤⎢⎣

⎡+

⎥⎥⎦⎢

⎢⎣−=⇒≡⎥⎦

⎤⎢⎣

⎡≡

10

0, 22

1 & State VariableRepresentation

General case:

JuHxyGuFxx+=+=&

Dynamic Models

ELECTRICAL SYSTEMS:

Kirchoff’s Current Law (KCL):

The algebraic sum of currents entering a node is zero at every instantevery instant

Kirchoff’s Voltage Law (KVL)

The algebraic sum of voltages around a loop is zero at every instant

iR iC iL)()()()( tGvtitRitv RRRR =⇔=

Resistors:

C it


iR

vR vC

iC

vL

L+ + +

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

C

t

CCC

C vdiCtv

dttdvCti +=⇔= ∫ ττ

∫ +=⇔=t

LLLL

L idvLti

dttdiLtv

0

)0()(1)()()( ττ

Capacitors:

Inductors:

6

Dynamic Models

ELECTRICAL SYSTEMS:

∞→−= AvvAvO )(

OP AMP:

+-

RIRO

A(vp-vn)

+

-

ip

in

iO vO

vn

vp

+

+

+ 0==

=

np

np

ii

vv

∞→AvvAv npO ),(


To work in the linear mode we need FEEDBACK!!!

Dynamic Models

ELECTRICAL SYSTEMS:

R C

R2

dv 11

KCL:

( ) ( ) ∫−=⇒∞=t

IOO dvCRvtvR

02)(10)( μμOC

+-

R1 Cv1 vO IO

O vCR

vCRdt

dv

12

11−=


Inverting integrator

Kv1 vO∫

∫CR 01

RCK 1−=

7

Dynamic ModelsELECTRO-MECHANICAL SYSTEMS: DC Motor

atiKT =

armature currenttorque

meKe θ&=

shaft velocityemf

+= TbJ θθ &&&


0=+++−

+−=

edtdiLiRv

TbJ

aaaa

mmm θθ

Obtain the State Variable Representation

Dynamic ModelsHEAT-FLOW:

( )TTR

q 1 21 −=

Temperature DifferenceHeat Flow

qC

T

R1

=&

Thermal resistanceThermal capacitance


( )IoI

I TTRRCT −⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

21

111&

8

Dynamic ModelsFLUID-FLOW:

outin wwm −=&

Mass rate Mass Conservation law

Outlet mass flowInlet mass flow

( )outin wwhhAm −=⇒= ρ1&&&


( )outinAρρ

A: area of the tankρ: density of fluidh: height of water

Linearization

( )uxfx ,=&Dynamic System:

( )oo uxf ,0 = Equilibrium( )oof

Denote oo uuuxxx −=−= δδ ,

( )uuxxfx oo δδδ ++= ,&Taylor Expansion

ff ∂∂


( ) uufx

xfuxfx

oooo uxuxoo δδδ

,,,

∂∂

+∂∂

+≈&

⇒∂∂

≡∂∂

≡oooo uxux u

fGxfF

,,, uGxFx δδδ +≈&

9

Linearization

ffff 1111⎥⎤

⎢⎡

∂∂

∂∂

⎥⎤

⎢⎡

∂∂

∂∂

LL

uGxFx δδδ +≈&

oo

oo

oo

oo

uxm

nn

m

ux

uxn

nn

n

ux

uf

uf

uu

ufG

xf

xf

xx

xfF

,1

1

,

,1

1

,,

⎥⎥⎥⎥⎥

⎦⎢⎢⎢⎢⎢

⎣ ∂∂

∂∂

∂∂=

∂∂

≡

⎥⎥⎥⎥⎥

⎦⎢⎢⎢⎢⎢

⎣ ∂∂

∂∂

∂∂=

∂∂

≡

L

MM

L

MM

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


Example: Pendulum with friction 0sin =++ θθθlm

xmkx

lgx

ox⎥⎥⎦

⎤

⎢⎢⎣

⎡−−= 1cos

10&

• Function f(t) of time– Piecewise continuous and exponential order

Laplace Transform

btKetf

10

Laplace transform examples• Step function – unit Heavyside Function

• After Oliver Heavyside (1850-1925)

0if1)()()(

>=−=−===∞+−∞−∞

−∞

− ∫∫ σωσeedtedtetusF

tjststst

⎩⎨⎧

≥<

=0for,10for,0

)(tt

tu

• Exponential function• After Oliver Exponential (1176 BC- 1066 BC)

0if)()(0000

>+−−

∫∫ σωσ sjsdtedtetusF

∫ ∫∞ ∞ ∞+−

+−−− >+

=+

−===0 0

)()( if1)( ασ

αα

ααα

ssedtedteesF

tstsstt


• Delta (impulse) function δ(t)

++0 0 0 αα ss

sdtetsF st allfor1)()(0

== ∫∞

−

−δ

Laplace Transform Pair TablesSignal Waveform Transform

impulse

step

ramp

)(tδ 1

s1

1

)(tu

)(tturamp

exponential

damped ramp

sine22 β

β

+ss

2s

α+s1

2)(

1

α+s

)(ttu

)(tue tα−

)(tutte α−

( ) )(sin tutβ


cosine

damped sine

damped cosine22)( βα

α

++

+

s

s

22)( βα

β

++s

22 β+s

s( ) )(cos tutβ

( ) )(sin tutte βα−

( ) )(cos tutte βα−

11

Laplace Transform Properties

{ } { } { } )()()()()()( 2121 sBFsAFtBtAtBftAf +=+=+ 21 fLfLLLinearity: (absolutely critical property)

( )Ft ⎫⎧ ( )ssFdf

t=

⎭⎬⎫

⎩⎨⎧∫0

)( ττLIntegration property:

)0()()( −−=⎭⎬⎫

⎩⎨⎧ fssF

dttdf

LDifferentiation property:

)0()0()(22)(2

−′−−−=⎪

⎪⎬⎫

⎪

⎪⎨⎧

fsfsFstfdL


)()()(2 ⎪⎭⎬

⎪⎩⎨ ff

dt

)0()0()0()()( )(21 −−−−′−−−=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧ −− mmm

m

mffsfssF

dttfd

LmsL


)()}({ αα +=− sFtfe tL

Translation properties:

s-domain translation:

{ } 0)()()( >=−− − asFeatuatf as forLt-domain translation:

Initial Value Property: )(lim)(lim0

ssFtfst ∞→+→

=

Fi l V l P t )(li)(li Ftf


Final Value Property: )(lim)(lim0

ssFtfst →∞→

=

If all poles of F(s) are in the LHP

12


)(1)}({asF

aatf =LTime Scaling:

Multiplication by time:ds

sdFttf )()}({ −=L

Convolution:

ds

)()(})()({0

sGsFdtgft

=−∫ τττL

Time product: λλπωσ

ωσdsGsF

jtgtf

j

j∫+

−−= )()(

21)}()({L


Laplace Transform

Exercise: Find the Laplace transform of the following waveform

[ ] )()2cos(2)2sin(22)( tutttf −+= ( )( )424)( 2 +

+=

ssssF ( )4+ss

Exercise: Find the Laplace transform of the following waveform

( )

[ ] ( )tudttedtuetf

dxxtuetft

t

tt

4040

04

5)(5)(

4sin5)()(−

−

−

+=

+= ∫ ( )( )

( )2

2

3

4020010)(

1648036)(

++

=

++++

=

sssF

ssssssF


( )40+sExercise: Find the Laplace transform of the following waveform

)2()(2)()( TtAuTtAutAutf −+−−= ( )seAsF

Ts 21)(−−

=

13

Laplace transformsLinearsystem

Differential Laplace Algebraicdom

ain)

Complex frequency domain(s domain)

equation

Classicaltechniques

transform L equation

Algebraictechniques

Tim

e do

mai

n (t

d


• The diagram commutes– Same answer whichever way you go

Responsesignal

Inverse Laplacetransform L-1

Responsetransform

Solving LTI ODE’s via Laplace Transform

( ) ( ) ( ) ( ) ubububyayay mmm

mn

nn

01

101

1 +++=+++−

−−

− LL

Initial Conditions: ( )( ) ( ) ( )( ) ( )0,,0,0,,0 11 uuyy mn KK −−Initial Conditions:

⎥⎦

⎤⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡−+− ∑∑∑∑∑

−

=

−−

=

−

=

−−−

=

−

=

−− ji

j

jiim

ii

ji

j

jiin

ii

n

j

jjnn susUsbsysYsasysYs1

0

)1(

0

1

0

)1(1

0

1

0

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

( ) ( ) ( ) ( )0,,0,0,,0 uuyy KK

Recall jk

j

jkkk

k

sfsFdt

tfd ∑−

=

−−−=⎭⎬⎫

⎩⎨⎧ 1

0

)1( )0()()( sL


011

1

1

0

)1(

0

1

0

)1(1

0

011

1

011

1

)0()0()()(

asasas

subsyasU

asasasbsbsbsbsY n

nn

ji

j

jim

ii

ji

j

jin

ii

nn

n

mm

mm

++++

−+

++++++++

= −−

−

=

−−

=

−

=

−−−

=−

−

−−

∑∑∑∑LL

L

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

14

Laplace Transform

Exercise: Find the Laplace transform V(s)

)(4)(6)( =+ tutvdt

tdv( ) 6

36

4)(+

−+

=sss

sV

3)0( −=−vdt ( ) 66 ++ sss

Exercise: Find the Laplace transform V(s)

( )( )( ) 12

3215)(

+−

+++=

sssssV5)(3)(4)( 22

2

=++ −etvdt

tdvdt

tvd t


( )( )( )2)0(',2)0( =−−=− vv

What about v(t)?

Transfer Functions

( ) ( ) ( ) ububyayay mmn

nn

01

101

1 ++=+++−

−−

− LL

Assume all Initial Conditions Zero:

)()()()()(

011

1

011

1n

nn

mm sU

sAsBsU

asasasbsbsbsY =++++

+++= −

−

−−

L

L

Assume all Initial Conditions Zero:

( ) ( ) )()( 01110111 sUbsbsbsYasasas mmnnn +++=++++ −−−− LLInputOutput


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

)()()(

21

21

011

1

011

1

n

m

nn

n

mm

pspspszszszsK

asasasbsbsb

sUsYsH

−−−−−−

=

+++++++

== −−

−−

L

L

L

L

15

Rational Functions

• We shall mostly be dealing with LTs which are rational functions – ratios of polynomials in s

)( 011

1m

mm

m bsbsbsbsF ++++=−

− L

• pi are the poles and zi are the zeros of the function• K is the scale factor or (sometimes) gain

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

)(

21

21

011

1

n

m

nn

nn

pspspszszszsK

asasasasF

−−−−−−

=

++++= −

−

L

L

L


• A proper rational function has n≥m• A strictly proper rational function has n>m• An improper rational function has n

16

Residues at multiple poles

rjrd1 ⎤⎡

−

Compute residues at poles of order r:

rr

rm

psk

psk

psk

pszszszsKsF

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

12

1

2

1

1

1

21

−++

−+

−=

−−−−

= LL

rjsFrpsjrds

dpsjr

k ii

j L1,)()(lim)!(1

=⎥⎦⎤

⎢⎣⎡ −

−→−=

( )31522

+

+

s

ss

)52()1( 23 ++ sss )52()1( 23 ⎥⎤

⎢⎡ ++ sssd 2)52()1(li1

232⎥⎤

⎢⎡ ++ sssd

Example:( ) ( )31

321

11

2

+−

++

+=

sss


3)1(

)52()1(lim 33−=

+++

−→ ssss

s1

)1()52()1(lim 31=⎥⎥⎦

⎤

⎢⎢⎣

⎡+

++−→ s

sssdsd

s2

)1()52()1(lim!2

1321

=⎥⎥⎦⎢

⎢⎣ +

++−→ s

sssdsd

s

( ) ( ) ( )( ) )(32

13

11

12

152 2

321

3

21 tutte

sssL

sssL t −+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+

−+

++

=⎟⎟⎠

⎞⎜⎜⎝

⎛++ −−−

Residues at complex poles

• Compute residues at the poles

• Bundle complex conjugate pole pairs into second-order terms if

)()(lim sFasas

−→

you want

• but you will need to be careful

• Inverse Laplace Transform is a sum of complex exponentialsTh ill b l

( )[ ]222 2))(( βααβαβα ++−=+−−− ssjsjs


• The answer will be real

17

Inverting Laplace Transforms in Practice

• We have a table of inverse LTs• Write F(s) as a partial fraction expansion

b sm + b sm−1 + + b s+ b

F(s) = bms + bm−1s +L+ b1s+ b0ans

n + an−1sn−1 +L+ a1s+ a0

=K (s− z1)(s− z2)L(s− zm)(s− p1)(s− p2)L(s− pn )

=α1

s− p1( )+

α2s− p2( )

+α31

(s− p3)+

α32s− p3( )2

+α33

s− p3( )3+ ...+

αqs− pq( )


• Now appeal to linearity to invert via the table– Surprise!– Nastiness: computing the partial fraction expansion is best

done by calculating the residues

Example 9-12

• Find the inverse LT of )52)(1(

)3(20)( 2 ++++=

sssssF

)(*221 kkksF ++=

21211)(

jsjsssF

+++

−++

+=

π45

255521)21)(1(

)3(20)()21(21

lim2

10152

2)3(20)()1(

1lim1

jej

jsjssssFjs

jsk

sss

ssFss

k

=−−=+−=+++

+=−+

+−→=

=−=++

+=+

−→=


)()4

52cos(21010

)(252510)( 45)21(

45)21(

tutee

tueeetf

tt

jtjjtjt

⎥⎦⎤

⎢⎣⎡ ++=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡++=

−−

−−−++−−

π

ππ

18

Not Strictly Proper Laplace Transforms

• Find the inverse LT of

– Convert to polynomial plus strictly proper rational function

348126)( 2

23

++

+++=

ssssssF

• Use polynomial division

35.0

15.02

3422)( 2

++

+++=

++

+++=

sss

sssssF


• Invert as normal

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

tdtf tt ⎥⎦⎤

⎢⎣⎡ +++= −−δδ

Block Diagrams

Series:1G 2G

1G

2G

+

+

Parallel:

21GGG =


21 GGG +=

19

Block DiagramsNegative Feedback:

G-

+)(sR )(sE )(sC BRER

−= Error signalReference input

H)(sB HCBGEC

== Output

Feedback signal

)1()1(

GHG

RCGRCGHGHCGRC

+=⇒=+⇒−=


)(

)1(1)1(GHR

EREGHHGERE+

=⇒=+⇒−=

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

Block DiagramsPositive Feedback:

G+

+)(sR )(sE )(sC BRER

+= Error signalReference input

H)(sB HCBGEC

== Output

Feedback signal

)1()1(

GHG

RCGRCGHGHCGRC

−=⇒=−⇒+=


)(

)1(1)1(GHR

EREGHHGERE−

=⇒=−⇒+=

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

20

Block Diagrams

Moving through a branching point:

G)(sR )(sCG)(sR )(sC G

G/1)(sB

G

)(sB

Moving through a summing point:

≡


G+

+)(sR )(sC

)(sB

G+

+

G

)(sR )(sC

)(sB

≡

Block Diagrams

1H

Example:

1G-

+)(sR )(sC2G 3G+

2H

-


212321

321

1 GGHGGHGGG++

)(sC)(sR

21

Mason’s Rule

6H

4H

1H+

+)(sU )(sY2H 3H+

5H

++

+

7H

++

Signal Flow Graph


)(sU )(sY1H 2H 3H

4H

6H

5H 7H

1 2 3 4

nodes branches

Mason’s RulePath: a sequence of connected branches in the direction of the signal flow without repetitionLoop: a closed path that returns to its starting nodeForward path: connects input and output

∑ ΔΔ== i iiG

sUsYsG 1)()()(

gains) loop (alltdeterminan system the

path forward ith the of gain

=

=Δ=

∑

i

-

G

1


path forward ith the touch not does that graph the of part the for of value

touch not do that loops three possible all of products gain

touch) not do that loops two possible all of products gain

g )p(

Δ=Δ+

−

+

∑∑∑

i

L

)(

(

22

Mason’s Rule

)(sU )(sY1H 2H 3H

4H

6H

1 2 3 4)(sU )(sY

5H 7H

6244321

1)()(

HHHHHHHHHHHHHHHHHHHHH

UsY −+

=


735156747362511)( HHHHHHHHHHHHHHsU +−−−−

Impulse Response

)()()(0

tudtu =−∫∞

ττδτDirac’s delta:

Integration is a limit of a sum g⇓

u(t) is represented as a sum of impulses

By superposition principle, we only need unit impulse response

( )τ−th Response at t to an impulse applied at τ


h )(ty)(tu

τττ dthuty )()()(0∫∞

−=

System Response:

23

Impulse Response

h )(ty)(tu

τττ dthuty )()()( ∫∞

−=

t-domain:

)()()()( thtyttu =⇒=δImpulse response

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

sUsHdthu =−∫∞

τττL

)(Y)(Us domain:

τττ dthuty )()()(0∫=

The system response is obtained by convolving the input with the impulse response of the system.

)()()()( thtyttu =⇒=δ


H

)()()( sUsHsY =

)(sY)(sUs-domain:

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

)()(1)()()( sHsYsUttu =⇒=⇒=δImpulse response

Time Response vs. PolesReal pole: teth

ssH σ

σ−=⇒

+= )(1)( Impulse

Response

1Ti C

00

<>

σσ Stable

Unstable


στ 1= Time Constant

24

Time Response vs. Poles

Real pole:

tethsH σσσ −=⇒= )()( Impulse eths

sH σσ

=⇒+

= )()(

στ 1= Time Constant

tetyss

sY σσ

σ −−=⇒+

= 1)(1)( Step R

Impulse Response


ss σ+ Response

Time Response vs. PolesComplex poles:

( )22

22

2

2)(

ω

ωζωω

=

++=

n

nn

n

sssH Impulse

Response

( ) ( )222 1 ζωζω −++ nns

::

ζωn Undamped natural frequency

Damping ratio

( )( )2

)( njsjs

sHωσωσ

ω−+++

=


( )( )

( ) 222

d

n

dd

s

jsjs

ωσω

ωσωσ

++=

+++

21, ζωωζωσ −== ndn

25


( ) ( ) ( )tethssH dtn

nn

n ωζ

ωζωζω

ω σ sin1

)(1

)(2222

2−

−=→

−++=

I l Impulse Response

0>σ Stable


0

26


( ) ( ) ( ) ( )⎥⎦⎤

⎢⎣

⎡+−=→

−++= − ttety

sssY d

dd

t

nn

n ωωσω

ζωζωω σ sincos1)(1

1)( 222

2

Step Response



Complex poles:

2

22

2

2)(

ω

ωζωω

++=

n

nn

n

sssH

( ) ( )222 1 ζωζω −++= nnn

s

( ) ( )nnn

s

s

ζωζωζ

ωζ

1:1

:0222

22

−++<

+=

CASES:

Undamped

Underdamped


( )( )[ ] ( )[ ]nn

n

ss

s

ωζζωζζζ

ωζ

11:1

:122

2

−−+−++>

+= Critically damped

Overdamped

27



Time Domain Specifications


28

Time Domain Specifications1- The rise time tr is the time it takes the system to reach the vicinity of its new set point2- The settling time ts is the time it takes the system transients to decaytransients to decay3- The overshoot Mp is the maximum amount the system overshoot its final value divided by its final value4- The peak time tp is the time it takes the system to reach the maximum overshoot point

Classical Control – Prof. Eugenio Schuster – Lehigh UniversityClassical Control – Prof. Eugenio Schuster – Lehigh University 55n

sp

nr

np

teM

tt

ζω

ωζωπ

ζ

ζπ 6.4

8.11

21

2

==

≅−

=

−−

Time Domain Specifications

Design specification are given in terms of

sppr tMtt ,,,

These specifications give the position of the poles

dn ωσζω ,, ⇒

Example: Find the pole positions that guarantee


Example: Find the pole positions that guarantee

sec3%,10sec,6.0 ≤

29

Effect of Zeros and Additional poles

Additional poles:1- can be neglected if they are sufficiently to the left of the dominant ones.2- can increase the rise time if the extra pole is within 2- can increase the rise time if the extra pole is within a 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 the time response.2- a zero in the LHP will increase the overshoot if the


zero 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 depress the overshoot and may cause the step response to start out in the wrong direction.

Stability

011

1

011

1

)()(

asasasbsbsbsb

sRsY

nn

n

mm

mm

++++++++

= −−

−−

L

L

)())(()( 21 mzszszsKsY −−−= L

Impulse response:

kkk

)())(()( 21 npspspsK

sR −−− L

)()()()()(

2

2

1

1

n

n

psk

psk

psk

sRsY

−++

−+

−= L


)()()()(1)(

2

2

1

1

n

n

psk

psk

psksYsR

−++

−+

−=⇒= L

tpn

tptp nekekekty +++= L21 21)(

30

Stability

We want:

tpn

tptp nekekekty +++= L21 21)(

nie ttpi K10 =∀⎯⎯→⎯ ∞→

Definition: A system is asymptotically stable (a.s.) if

ipi ∀< 0}Re{


0)(

)( 011

1

=

++++= −−

sa

asasassa nnn LCharacteristic polynomial:

Characteristic equation:

Stability

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

ia ∀> 0 iai ∀> 0

Use this as the first test!

If any ai

31

Routh’s Criterion

Necessary and sufficient conditionDo not have to find the roots pi!

Routh’s Array:

n

n

n

n

bbbsaaasaas

3321

2531

1421

L

L

−

− na Depends on whether

n is even or odd

761541321 aaabaaabaaab −−−


n

n

n

as

ddscccs

0

214

3213

M

−

−

MM

L

L

L

,,

,,

,,

1

31312

1

21211

1

31512

1

21311

1

7613

1

5412

1

3211

ccbbcd

ccbbcd

bbaabc

bbaabc

aaaab

aaaab

aaaab

−=

−=

−=

−=

===

Routh’s Criterion

How to remember this?

Routh’s Array:

L

L

L

L

4342413

3332312

2322211

131211

mmmsmmmsmmmsmmms

n

n

n

n

−

−

− ,

,

1,21,2

12,2

22,1

=

=

+−−

−

−

mmmm

am

am

jii

jj

jj


MMMM434241

3,1,1

1,11,1, ≥∀−=

−

+−− im

mmm

i

jiiji

32

Routh’s Criterion

The criterion:

• The system is asymptotically stable • The system is asymptotically stable if and only if all the elements in the first column of the Routh’s array are positive

• The number of roots with positive real parts is equal to the number of sign changes in the first column of the Routh


garray

Routh’s Criterion - Examples

Example 1: 0212 =++ asas

0322

13 =+++ asasasExample 2:

Example 3:

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

044234 23456 =++++++ ssssss


0)6(5 =+−++ kskssExample 4:

33

Routh’s Criterion - Examples

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

K-

+)(sR )(sY( )( )61

1+−

+sss

s


Routh’s Criterion

Special Case I: Zero in the first columnWe replace the zero with a small positive constant ε>0 and proceed as before. We then apply the p pp ystability criterion by taking the limit as ε→0

Example: 010842 234 =++++ ssss


34

Routh’s Criterion

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 , y qfrom the previous nonzero row:

L+++= −−+ 331

21

11 )(iii ssssa βββ

Where βi are the coefficients of the (i+1)th row in the array. We then replace the ith row by the coefficients of the derivative of the auxiliary polynomial


Example: 02010842 2345 =+++++ sssss

of the derivative of the auxiliary polynomial.

Properties of feedback

Disturbance Rejection:

wOpen loop

w

oKr +A+ y

Closed loop

wArKy o +=


cK-

+r +A+ y

wAK

rAK

AKycc

c

++

+=

11

1

35


Disturbance Rejection:

Choose control s.t. for w=0,y≈r

rwryA

Kc =+≈⇒>> 01

wryA

Ko +=⇒=1Open loop:

Closed loop:

F db k ll tt ti f di t b ith t


Feedback allows attenuation of disturbance without having access to it (without measuring it)!!!

IMPORTANT: High gain is dangerous for dynamic response!!!


Sensitivity to Gain Plant Changes

wOpen loop

w

oKr +A+ y

Closed loop

oo

o AKryT =⎟⎠⎞

⎜⎝⎛=


cK-

+r +A+ y

c

c

cc AK

AKryT

+=⎟

⎠⎞

⎜⎝⎛=

1

36


Sensitivity to Gain Plant Changes

Let the plant gain be

AT δδAA δ+

AA

TT

o

o δδ =

o

o

cc

c

TT

AA

AKAA

TT δδδδ

=

37

PID ControllerP Controller: Example (lecture06_a.m)

pK-

+)(sR )(sY12 ++ ss

A)(sE )(sU

)1()(1)(

)()(

2 AKssAK

sGKsGK

sRsY

p

p

p

p

+++=

+=

12 += pn AKω 11


Underdamped transient for large proportional gainSteady state error for small proportional gain

12 =n

pn

ζω0

121

21

⎯⎯ →⎯+

==⇒ ∞→pKpn AKω

ζ

PID ControllerPI Controller:

sKKsC Ip +=)(

-

+)(sR )(sY)(sG)(sE )(sU

1)(

,)()(1

)()()()(

sEsGsC

sGsCsRsY

=

+=

)()(,)()()(0

sEs

KKsUdeKteKtu Ipt

Ip ⎟⎠⎞

⎜⎝⎛ +=+= ∫ ττ

0)(1

1lim1

)(1

1lim)(lim1)(000

=⎟⎠⎞

⎜⎝⎛ ++

=⎟⎠⎞

⎜⎝⎛ ++

==⇒=→→→

sGs

KKssGs

KKsssEe

ssR

Ip

sIp

ssss

Step Reference:

.)()(1)( sGsCsR +

=


⎠⎝⎠⎝ ss

• 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!

Integral control achieves perfect steady state reference tracking!!!Note that this is valid even for Kp=0 as long as Ki≠0

38

PID ControllerPI Controller: Example (lecture06_b.m)

sKK Ip +

-

+)(sR )(sY12 ++ ss

A)(sE )(sU

( )AKsAKss

AKsK

sGKK

sGs

KK

sRsY

Ip

Ip

Ip

Ip

+++++

=⎟⎠⎞

⎜⎝⎛ ++

⎟⎠⎞

⎜⎝⎛ +

=)1()(1

)(

)()(

23


sp ⎟⎠⎜⎝

)(

DANGER: for large Ki the characteristic equation has roots in the RHP

0)1(23 =++++ AKsAKss IpAnalysis by Routh’s Criterion

PID ControllerPI Controller: Example (lecture06_b.m)

0)1(23 =++++ AKsAKss IpNecessary Conditions: 0,01 >>+ AKAK IpThis is satisfied because 0,0,0 >>> Ip KKA

AKs p3 11 +

Routh’s Conditions:

>−+ 01 AKAK Ip


AKsAKAKs

AKs

I

Ip

I

p

0

1

2

11−+

⇓Ip

AKK PI

1+

39

PID ControllerPD Controller:

sKKsC Dp +=)(-


1)(

,)()(1

)()()()(

sEsGsC

sGsCsRsY

=

+=

( ) )()(,)()()( sEsKKsUdt

tdeKteKtu DpDp +=+=

( ) )0(111

)(11lim)(lim1)(

00 GKssGsKKsssEe

ssR

pDpssss +

=++

==⇒=→→

Step Reference:

.)()(1)( sGsCsR +

=


PD controller fixes problems with stability and damping by adding “anticipative” action

∞→⇔= )0(0 GKe pss•Proportional gain is high•Plant has a pole at the origin

True when:

PID ControllerPD Controller: Example (lecture06_c.m)

-

+)(sR )(sY12 ++ ss

A)(sE )(sUsKKsC Dp +=)(

( )( )

( )( ) )1(1)(1

)()()(

2 AKsAKssKKA

sGsKKsGsKK

sRsY

pD

Dp

Dp

Dp

+++++

=++

+=

AK pn +=12ω AKAK ++ 11


The damping can be increased now independently of KpThe steady state error can be minimized by a large Kp

AKDn

pn

+=12ζω AKAKAKp

D

n

D

++

=+

=⇒12

12

1ω

ζ

40

PID ControllerPD Controller:

sKKsC Dp +=)(-


1)(

,)()(1

)()()()(

sEsGsC

sGsCsRsY

=

+=

( ) )()(,)()()( sEsKKsUdt

tdeKteKtu DpDp +=+=

NOTE: cannot apply pure differentiation.In practice,

.)()(1)( sGsCsR +

=


sKDis implemented as

1+ssK

D

D

τ

PID Controller

PID: Proportional – Integral – Derivative

⎟⎟⎠

⎞⎜⎜⎝

⎛++ sT

sTK Dp

11+)(sR )(sY)(sG)(sE )(sU⎟⎠

⎜⎝ sTI-

⎞⎛

⎥⎦

⎤⎢⎣

⎡++= ∫

U

dttdeTde

TteKtu D

d

Ip

1)(

)()(1)()(0

ττ DpDI

pI TKKT

KK == ,


⎟⎟⎠

⎞⎜⎜⎝

⎛++= sT

sTK

sEsU

DI

p11

)()(

PID Controller: Example (lecture06_d.m)

41

PID Controller: Ziegler-Nichols Tuning

• Empirical method (no proof that it works well but it works well for simple systems)• Only for stable plants• You do not need a model to apply the method• You do not need a model to apply the method

1)()(

+=

−

sKe

sUsY std

τ

Class of plants:


1)( +ssU τ


METHOD 1: Based on step response, tuning to decay ratio of 0.25.

T i T bl

dd

dI

dp

dp

tTtTKPID

tTt

KPD

tKP

50221:

3.0,9.0:

:

===

==

=

τ

τ

τTuning Table:


dDdId

p tTtTtKPID 5.0,2,2.1: ===

42


METHOD 2: Based on limit of stability, ultimate sensitivity method.

uK-


• Increase the constant gain Ku until the response


becomes purely oscillatory (no decay – marginally stable – pure imaginary poles)• Measure the period of oscillation Pu


METHOD 2: Based on limit of stability, ultimate sensitivity method.

T i T bl

8,

2,6.0:

2.1,45.0:

5.0:

uD

uIup

uIup

up

PTPTKKPID

PTKKPD

KKP

===

==

=

Tuning Table:


82

dud

u tPtK 4,2 == τ

The Tuning Tables are the same if you make:

43

PID Controller: Integrator Windup

Actuator Saturates: - valve (fully open)- aircraft rudder (fully deflected)

cu(O t t f th t ll )

u(Input of the plant)


(Output of the controller)

PID Controller: Integrator Windup

sKK Ip +

-

+)(sR )(sY)(sG)(sE )(sUc )(sU

What happens? - large step input in r- large e- large uc → u saturates

ll b ll


- eventually e becomes small- uc still large because the integrator is “charged”- u still at maximum- y overshoots for a long time

44

PID Controller: Anti-WindupPlant without Anti-Windup:

Plant with Anti-Windup:


PID Controller: Anti-WindupIn saturation, the plant behaves as:


For large Ka, this is a system with very low gain and very fast decay rate, i.e., the integration is turned off.

45

Steady State Tracking

The Unity Feedback Case

)(sC+)(sR )(sY)(sG)(sE )(sU)(sC-

)(sG

)()(11

)()(

sGsCsRsE

+=


11)(

)(1!

)(

+=

=

k

k

ssR

tkttrTest Inputs: k=0: step (position)

k=1: ramp (velocity)

k=2: parabola (acceleration)



)(sC+)(sR )(sY)(sG)(sE )(sUn

o sG )(

Type nSystem

-

11)(),()(1

1)(,)()()( +=+

== kn

on

o

ssRsR

ssGsEs

sGsGsC

Steady State Error:

ns


)0(lim1

)(lim1)(1

1lim)(lim)(lim00100

on

kn

sko

n

n

sk

nosst

ss Gss

ssGss

ss

sGsssEtee +=

+=

+===

−

→→+→→∞→

Steady State o

Final Value Theorem

46

Steady State TrackingThe Unity Feedback Case

likns −-

+)(sR )(sY)(sEn

o

ssG )(

Type nSystem

)0(lim

0o

nsss Gse

+=

→

Steady State Error:

Type (n) Step (k=0) Ramp (k=1) Parabola (k=2)Input (k)

111


Type 0

Type 1

Type 2

psoKsGsCG +

=+

=+

→1

1)()(lim1

1)0(1

1

0

vsoKsGssCG1

)()(lim1

)0(1

0

==→

asoKsGsCsG1

)()(lim1

)0(1

2

0

==→

∞ ∞

0

0 0

∞


2)()(lim

1)()(lim0)()(lim

20

0

==

==

==

→

→

nsGsCsK

nsGssCKnsGsCK

sv

spPosition Constant

Velocity Constant

Acceleration Constant2)()(lim0

==→

nsGsCsKsa

n: Degree of the poles of CG(s) at the origin (the number of integrators in the loop with unity gain feedback)

• Applying integral control to a plant with no zeros at the origin makes the system type ≥ I• All this is true ONLY for unity feedback systems


• All this is true ONLY for unity feedback systems• Since in Type I systems ess=0 for any CG(s), we say that the system type is a robust property.

47



+)(sR )(sY)(sE )(sU

)(sW

+

11)(

)(1!

)(

+=

=

k

k

ssW

tkttw

)()()()(1

)()()( sTssT

sGsCsG

sWsY

on==

+=

)(sC-

+)(sR )(sY)(sG)(sE )(sU +

Set r=0. Want Y(s)/W(s)=0.


k

n

osksstssss sssT

sssTssYtyye )(lim1)(lim)(lim)(lim

0100 →+→→∞→=====−

Steady State Error: e=r-y=-y Final Value Theorem

Steady State TrackingThe Unity Feedback Case

)(sC-

+)(sR )(sY)(sG)(sE )(sU +

)(sW

+

Steady State Output:

Type (n) Step (k=0) Ramp (k=1) Parabola (k=2)Disturbance (k)


Type 0

Type 1

Type 2

*

*

*

∞ ∞

0

0 0

∞

∞

48

Steady State TrackingExample:

)(sW

wKI to 1 type⇒≠ 0

sKK Ip +

-

+)(sR )(sY( )1+ss

Aτ

)(sE )(sU+

+


wKK IP to0type⇒=≠ 0,0

Root LocusController

)(sC+)(sR )(sY)(sG)(sE )(sU

Plant

-

)(1)()(

)()()(1)()(

)()(

sKLsGsC

sHsGsCsGsC

sRsY

+=

+=

)(sH

Sensor

⇒= )()( sKDsC


)(1)()()(1)( sKLsHsGsCsR ++

Writing the loop gain as KL(s) we are interested in tracking the closed-loop poles as “gain” K varies

49

Root LocusCharacteristic Equation:

0)(1 =+ sKL

The roots (zeros) of the characteristic equation are the ( ) qclosed-loop poles of the feedback system!!!

The closed-loop poles are a function of the “gain” K

Writing the loop gain as

nnmm

mm

asasasbsbsbs

sasbsL

++++++++

== −−

−

11

11

)()()( L


nn asasassa ++++ −11)( L

The closed loop poles are given indistinctly by the solution of:

KsLsKbsa

sasbKsKL 1)(,0)()(,0)()(1,0)(1 −==+=+=+

Root Locus

{ } { })()()(1 LKLsKL numdenrootszerosRL +=+=when K varies from 0 to ∞ (positive Root Locus) or

from 0 to -∞ (negative Root Locus)

⎪⎩

⎪⎨⎧

=∠

=⇔−=>o180)(

1)(1)(:0sL

KsL

KsLK

Phase condition

Magnitude condition


⎪⎩

⎪⎨⎧

=∠

−=⇔−=<o0)(

1)(1)(:0sL

KsL

KsLK

Phase condition

Magnitude condition

50

Root Locus by Characteristic Equation Solution

Example: K-

+)(sR )(sY( )( )110

1++ ss

)(sE )(sU

)10(11)()(

2 KssK

sRsY

+++=

Closed-loop poles: 0)10(110)(1 2 =+++⇔=+ KsssL

101s 0=K


24815.5 Ks −±−=

28145.5

5.52

4815.5

10,1

−±−=

−=

−±−=

−−=

Kis

s

Ks

s

0481048104810

−=

KKKK

Root Locus by Characteristic Equation Solution


We need a systematic approach to plot the closed-loop poles as function of the gain K → ROOT LOCUS

51

Root Locus by Phase Condition

Example: K-

+)(sR )(sY( )( )845

12 ++++

sssss)(sE )(sU

( )( )

( )( )( )isissss

ssssssL

222251

8451)( 2

−+++++

=

++++

=


iso 31+−=

belongs to the locus?


o

o43.108o87.36

o45

o70.78

o90


[ ] oooooo 18070.784587.3643.10890 −≈+++− iso 31+−=⇒ belongs to the locus!Note: Check code lecture09_a.m

52


iso 31+−=


We need a systematic approach to plot the closed-loop poles as function of the gain K → ROOT LOCUS

Root Locus{ } { })()()(1 LKLsKL numdenrootszerosRL +=+=

1

when K varies from 0 to ∞ (positive Root Locus) or from 0 to -∞ (negative Root Locus)

Basic Properties:

• Number of branches = number of open-loop poles • RL begins at open-loop poles

0)(0 =⇒= saK

0)()(1)(0)(1 =+⇔−=⇔=+ sKbsaK

sLsKL


• RL ends at open-loop zeros or asymptotes

• RL symmetrical about Re-axis

)(

⎩⎨⎧

>−∞→=

⇔=⇒∞=)0(

0)(0)(

mnssb

sLK

53

Root Locus

Rule 1: The n branches of the locus start at the poles of L(s)and m of these branches end on the zeros of L(s).n: order of the denominator of L(s)m: order of the numerator of L(s)m: order of the numerator of L(s)

Rule 2: The locus is on the real axis to the left of and odd number of poles and zeros.In other words, an interval on the real axis belongs to the root locus if the total number of poles and zeros to the right is odd.


This rule comes from the phase condition!!!

Root Locus

Rule 3: As K→∞, m of the closed-loop poles approach the open-loop zeros, and n-m of them approach n-m asymptotes with angles

1,,1,0,)12( −−=−

+= mnlmn

ll K πφ

and centered at

11011−− ∑∑ mnlab zerospolesα


1,,1,0,11 −−=−

=−

= ∑∑ mnlmnmn

K α

54

Root Locus

Rule 4: The locus crosses the jω axis (looses stability) where the Routh criterion shows a transition from roots in the left half-plane to roots in the right-half plane.

Example:

)54(5)( 2 ++

+=

sssssG


5,20 jsK ±==

Root Locus

Example:4673

1)( 234 +++++

=ssss

ssG


55

Root Locus

Design dangers revealed by the Root Locus:

• High relative degree: For n-m≥3 we have closed loop instability due to asymptotesinstability due to asymptotes.

• Nonminimum phase zeros: They attract closed loop poles into the RHP

46731)( 234 ++++

+=

ssssssG


11)( 2 ++

−=

ssssG

Note: Check code lecture09_b.m

Root Locus

Viete’s formula:

When the relative degree n-m≥2, the sum of the closed loop poles is constantpoles is constant

∑−= poles loop closed1a

mmmm bsbsbssbsL ++++== −−

11

11)()( L


nnnn asasassa

sL++++ −

−1

11)(

)(L

56

Phase and Magnitude of a Transfer Function

011

1

011

1)(asasas

bsbsbsbsG nn

n

mm

mm

++++++++

= −−

−−

L

L

)())(()( 21 mzszszsKsG −−−= L

The factors K, (s-zj) and (s-pk) are complex numbers:

)())(()(

21 npspspsKsG

−−− L

pk

zj

ipkk

izjj

pkerps

mjerzsφ

φ

==−

==−

L

K

1,)(

1,)(


Ki

kk

eKK φ=

pn

pp

zm

zzK

ipn

ipip

izm

izizi

erererererereKsG

φφφ

φφφφ

L

L

21

21

21

21)( =

Phase and Magnitude of a Transfer Function

( )zmzzK

pn

pp

zm

zzK

izzz

ipn

ipip

izm

izizi

errr

erererererereKsG

φφφ

φφφ

φφφφ

+++

=

LL

L

L

21

21

21

21

21)(

Now it is easy to give the phase and magnitude

( )( )

( ) ( )[ ]pnppzmzzK

pn

pp

K

ip

npp

zm

zz

ipn

ppmi

errrrrrK

errrerrreK

φφφφφφφ

φφφφ

+++−++++

+++

=

=

LL

L

L

L

L

L

2121

21

21

21

21

21


y g p gof the transfer function:

( ) ( )pnppzmzzKp

npp

zm

zz

sG

rrrrrrKsG

φφφφφφφ +++−++++=∠

=

LL

L

L

2121

21

21

)(

,)(

57

Phase and Magnitude of a Transfer FunctionExample: ( )( ) )20(51

)735.6()(+++

+=

sssssG

p1φ

p2φ

p3φ

z1φ

pr3zr1 pr2

pr1 ( )pppzppp

z

sG

rrrrsG

3211

321

1

)(

)(

φφφφ ++−=∠

=iss o 57 +−==


Root Locus- Magnitude and Phase Conditions

{ } { })()()(1 LKLsKL numdenrootszerosRL +=+=when K varies from 0 to ∞ (positive Root Locus) or

from 0 to -∞ (negative Root Locus)

⇔−=>K

sLK 1)(:0

( ) ( )[ ]pnppzmzzpKip

npp

zm

zz

pn

mp errr

rrrKpspspszszszsKsL φφφφφφφ +++−++++=

−−−−−−

= LLL

L

L

L2121

21

21

21

21

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

( ) ( ) oLLL

L

180)(

1)(

2121

21

21

=+++−++++=∠

==

pn

ppzm

zzK

pn

pp

zm

zz

p

psL

KrrrrrrKsL

φφφφφφφ


⇔−=<K

sLK 1)(:0

( ) ( )nm

( ) ( ) oLLL

L

0)(

1)(

2121

21

21

=+++−++++=∠

−==

pn

ppzm

zzK

pn

pp

zm

zz

p

psL

KrrrrrrKsL

φφφφφφφ

58

Root Locus

Selecting K for desired closed loop poles on Root Locus:

If so belongs to the root locus, it must satisfies the characteristic equation for some value of Kcharacteristic equation for some value of K

Then we can obtain K as

KsL o

1)( −=

1


)(1

)(1

o

o

sLK

sLK

=

−=

Root Locus

Example: ( )( )511)()(

++==

sssGsL

( ) ( ) 204242

54314351)(

143

2222 =++−=

++−++−=++==⇒+−=

iisssL

Kis ooo

o


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

Using MATLAB:

59

Root LocusExample: ( )( )51

1)()(++

==ss

sGsL

57

⇓

+−=o is

43 iso +−=

06.42)(

1==

osLK

57 iso +−=


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

Root Locus - Compensators

Example: ( )( )511)()(

++==

sssGsL

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

( )( )( )51110)()()(

+++==

ssssGsDsL( )( )51

1)()(++

==ss

sGsL


The zero attracts the locus!!!

60

Root Locus – Phase lead compensator

Pure derivative control is not normally practical because of the amplification of the noise due to the differentiation and must be approximated:

( )( ) zpzssGsDsL >+== ,1)()()(

zppszssD >

++

= ,)( Phase lead COMPENSATOR

When we study frequency response we will understand why we call “Phase Lead” to this compensator.


( )( ) psspsG +++ ,51)()()(

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

Root Locus – Phase lead compensatorExample:

Phase lead COMPENSATOR

( )( ) zpsspszssGsDsL <

++++

== ,51

1)()()(

o19.140o80.111o04.21

?1zφ


oooooo 03.9303.45304.2180.11119.1401801 ==+++=zφ 735.6−=⇒ z

( ) ( ) oLL 180)( 2121 =+++−++++=∠ pnppzmzzK psL φφφφφφφ

61

Root Locus – Phase lead compensatorExample:

Phase lead COMPENSATOR

( )( )511

20735.6)()()(

++++

==sss

ssGsDsL

COMPENSATOR

11757

=+−=

Kiso


Root Locus – Phase lead compensator

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

• The zero is placed in the neighborhood of the closed-loop natural frequency, as determined by rise-time or p q y, ysettling time requirements.• The poles is placed at a distance 5 to 20 times the value of the zero location. The pole is fast enough to avoid modifying the dominant pole behavior.

The exact position of the pole p is a compromise between:

• Noise suppression (we want a small value for p)


• Compensation effectiveness (we want large value for p)

62

Root Locus – Phase lag compensatorExample: ( )( )51

120735.6)()()(

++++

==sss

ssGsDsL

17356+s( )( )

2

00010735.6

511

20735.6lim)()(lim)(lim −

→→→×=

++++

===sss

ssGsDsLKsssp

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

( )( ) zpszssGsDsL

63

Root Locus – Phase lag compensator

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

• The ratio zero/pole is chosen based on the error constant specification.p• We pick z and p small to avoid affecting the dominant dynamic of the system (to avoid modifying the part of the locus representing the dominant dynamics)• Slow transient due to the small p is almost cancelled by an small z. The ratio zero/pole cannot be very big.

The exact position of z and p is a compromise between:


• Steady state error (we want a large value for z/p)• The transient response (we want the pole p placed far from the origin)

Root Locus - Compensators

Phase lead compensator: pzpszssD <

++

= ,)(

pzpszssD >

++

= ,)(Phase lag compensator:

We will see why we call “phase lead” and “phase lag” to these compensators when we study frequency response


64

Frequency Response

• We now know how to analyze and design systems via s-domain methods which yield dynamical information

• The responses are described by the exponential modesTh d d i d b h l f h L l– The modes are determined by the poles of the response Laplace Transform

• We next will look at describing cct performance via frequency response methods– This guides us in specifying the system pole and zero

i i


positions

Sinusoidal Steady-State Response

22)()cos()( ωωω+

=⇔=s

AsUtAtu

Consider a stable transfer function with a sinusoidal input:

• Where the natural modes lie

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

65


• Input

• Transform

φωφωφω cossinsincos)cos()( tAtAtAtu −=+=

2222 sincos)(ωφφ +−= AsAsU

• Response Transform

• Response Signal

2222 ωω ++ ss

N

N

psk

psk

psk

jsk

jsksUsGsY

−++

−+

−+

++

−== L

2

2

1

1*

)()()(ωω

forced response natural response


• Sinusoidal Steady State Response

44444 344444 21L44 344 21

responsenatural21

responseforced

* 21)( tpNtptptjtj Nekekekekkety +++++= − ωω

tjtjSS ekkety

ωω −+= *)(

∞→t

0


• Calculating the SSS response to• Residue calculation

[ ] [ ])()()(lim)()(limωω

ωωjsjs

sUsGjssYjsk→→

−=−=

)cos()( φω += tAtu

• Signal calculation

))(()(21

21)(

2sincos)(

))((sincos))((lim

ωφφ

ω

ωω

ωφωφωω

ωωφωφω

jGjj

js

ejGAejAG

jjAjG

jsjssAjssG

∠+

→

==

⎥⎦

⎤⎢⎣

⎡ −=⎥

⎦

⎤⎢⎣

⎡+−

−−=

kkLtyss ⎬⎫

⎨⎧

+= −)(*

1


( )Ktkeekeek

jsjsLty

tjKjtjKj

ss

∠+=

+=⎭⎬

⎩⎨ +

+−

−∠−∠

ω

ωωωω

cos2

)(

( ))(cos)()( ωφωω jGtjGAtyss ∠++=

66


• Response to• is ))(cos()(

)cos()(ωφωω

φωjGtAjGy

tAtu

ss ∠++=

+=

– Output frequency = input frequency– Output amplitude = input amplitude × |G(jω)|– Output phase = input phase + G(jω)

• The Frequency Response of the transfer function G(s) is given by its evaluation as a function of a complex variable at s=jω

∠


• We speak of the amplitude response and of the phase response– They cannot independently be varied

» Bode’s relations of analytic function theory

Frequency Response

• Find the steady state output for v1(t)=Acos(ωt+φ)

+_V1(s)sL

R V2(s)

+

• Compute the s-domain transfer function T(s)– Voltage divider

• Compute the frequency response

-

RsLRsT+

=)(

⎟⎠⎞

⎜⎝⎛−=∠

+= −

RLjT

LR

RjT ωωω

ω 122

tan)(,)(

)(


• Compute the steady state output

⎠⎝+ LR ω )(

( )[ ]RLtLR

ARtv SS /tancos)(

)( 1222

ωφωω

−−++

=

67

Bode Diagram

)

-5

0

Bode Diagrams• Log-log plot of mag(T), log-linear plot of arg(T) versus ω

(dB

))

Mag

nitu

de (d

B

-25

-20

-15

-10

0

Mag

nitu

de (

eg)


Frequency (rad/sec)

Phas

e (d

eg

104

105

106

-90

-45

Phas

e (d

e

Frequency Response

)cos()( φω += tAtu )(sG ))(cos()( ωφωω jGtAjGyss ∠++=

Stable Transfer Function

• After a transient, the output settles to a sinusoid with an amplitude magnified by and phase shifted by .

• Since all signals can be represented by sinusoids (Fourier i d t f ) th titi d

)( ωjG )( ωjG∠

)( jG )( jG∠


series and transform), the quantities and are extremely important.

• Bode developed methods for quickly finding and for a given and for using them in control design.

)( ωjG )( ωjG∠

)( ωjG )( ωjG∠)(sG

68

Frequency Response

• Find the steady state output for v1(t)=Acos(ωt+φ)

+_V1(s)sL

R V2(s)

+

• Compute the s-domain transfer function T(s)– Voltage divider

• Compute the frequency response

-

RsLRsT+

=)(

⎟⎠⎞

⎜⎝⎛−=∠

+= −

RLjT

LR

RjT ωωω

ω 122

tan)(,)(

)(


• Compute the steady state output

⎠⎝+ LR ω )(

( )[ ]RLtLR

ARtv SS /tancos)(

)( 1222

ωφωω

−−++

=

Bode Diagram

)

-5

0

Frequency Response - Bode Diagrams• Log-log plot of mag(T), log-linear plot of arg(T) versus ω

(dB

))

Mag

nitu

de (d

B

-25

-20

-15

-10

0

Mag

nitu

de (

eg)


Frequency (rad/sec)

Phas

e (d

eg

104

105

106

-90

-45

Phas

e (d

e

[ ] [ ] Hzffrad === ,2sec,/ πωω

69

Bode Diagrams

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

21

21

n

m

pspspszszszsKsG

−−−−−−

=L

L

The magnitude and phase of G(s) when s=jω is given by:

( ) ( )[ ]pnppzmzzKip

npp

zm

zz

errrrrrKsG φφφφφφφ +++−++++= LL

L

L2121

21

21)(

zzz

Nonlinear in the magnitudes


( ) ( )pnppzmzzKp

npp

zm

zz

jG

rrrrrrKjG

φφφφφφφω

ω

+++−++++=∠

=

LL

L

L

2121

21

21

)(

,)(

Linear in the phases

Bode Diagrams

)(log20)( ωω jGjG dB =

Why do we express in decibels?)( ωjG

zzz rrr

( ) ( )pnppzmzmz rrrrrrKsG log20log20log20log20log20log20log20)(log20 211 +++−++++= LLBy properties of the logarithm we can write:

?)()(21

21 =⇒= dBpn

ppm jGrrrrrrKjG ωω

L

L

The magnitude and phase of G(s) when s=jω is given by:


( ) ( )( ) ( )pnppzmzzK

dBp

ndBp

dBp

dBz

mdBz

dBz

dBdB

sG

rrrrrrKsG

φφφφφφφ +++−++++=∠

+++−++++=

LL

LL

2121

2121

)(

)(

Linear in the phases

Linear in the magnitudes (dB)

70

Bode DiagramsWhy do we use a logarithmic scale? Let’s have a look at our example:

222

1

1)(

)()(

⎟⎞

⎜⎛+

=+

=⇒+

=LLR

RjTRsL

RsTωω

ω

1 ⎟⎠

⎜⎝

+R

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+−=⎟

⎠⎞

⎜⎝⎛+−=

22

1log101log201log20)(RL

RLjT dB

ωωω

Expressing the magnitude in dB:

Asymptotic behavior:


( ) ωωωωω log20log20/log20/

log20)(: −=−=⎟⎠⎞

⎜⎝⎛−→∞→

dBdB L

RLRLR

jT

0)(:0 →→ dBjT ωω

LINEAR FUNCTION in logω!!! We plot as a function of logω.dBjG )( ω

Bode Diagrams

A cutoff frequency occurs when the gain is reduced from its maximum passband value by a factor :2/1

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

dB3log202log20log202

1log20 −≈−=⎟⎠⎞

⎜⎝⎛

MAXMAXMAX TTT

Bandwith: frequency range spanned by the gain passband

Let’s have a look at our example:

⎧ )(


⎩⎨⎧

====

⇒

⎟⎠⎞

⎜⎝⎛+

=2/1)(/

1)(0

1

1)(2 ωω

ωω

ωω

jTLRjT

RL

jT

This is a low-pass filter!!! Passband gain=1, Cutoff frequency=R/LThe Bandwith is R/L!

71

General Transfer Function (Bode Form)

( ) ( ) ( )q

nn

nmo

jjjjKjG⎥⎥⎦

⎤

⎢⎢⎣

⎡++⎟⎟

⎠

⎞⎜⎜⎝

⎛+= 121

2

ωωζ

ωωωτωω

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

Classes of terms:

1-

2-

( ) oKjG =ω( ) ( )mjjG ωω =

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.


3-

4-

( ) ( )jjG ωω =( ) ( )njjG 1+= ωτω

( )q

nn

jjjG⎥⎥⎦

⎤

⎢⎢⎣

⎡++⎟⎟

⎠

⎞⎜⎜⎝

⎛= 12

2

ωωζ

ωωω

General Transfer Function: DC gain

( ) oKjG =ω

( )⎧

= log20 odB KjωGMagnitude and Phase:

25

30

35

40

(dB

)

120

140

160

180

200

eg)

( )⎩⎨⎧

=∠000

o

o

KK

jG if

ifπ

ω( ) 10−=sG


10-1 100 101 1020

5

10

15

20

Frequency (rad/sec)

Mag

nitu

de

10-1 100 101 1020

20

40

60

80

100

Frequency (rad/sec)

Pha

se (d

e

72

General Transfer Function: Poles/zeros at origin

( ) ( )mjjG ωω =

( ) log20 ωmjωG dB ⋅=Magnitude and Phase:

0

10

20

e (d

B) -40

-20

0

deg)

( )2πω mjG =∠( )

ssGm 1,1 =−=

( ) 01 =dBG


10-1 100 101 102-40

-30

-20

-10

Frequency (rad/sec)

Mag

nitu

de

10-1 100 101 102-120

-100

-80

-60

Frequency (rad/sec)

Pha

se (d

decdBm 20⋅

General Transfer Function: Real poles/zeros

( ) ( )njjG 1+= ωτωMagnitude and Phase:

( ) ( )( ) ( )ωτω

τω1

22

tan

1log10−=∠

+⋅=

njG

njωG dB



ωτω

ω

τω

τω

log20)(

0)(

/1

/1

⋅+⋅⎯⎯ →⎯

⎯⎯ →⎯

>>

>

73

0

5

10


10/1,1 =−= τn

( ) 1GdB

30

-25

-20

-15

-10

-5

Mag

nitu

de (d

B)

dBjG dB = 0)0(

( )1

10+

= ssGdBn 3⋅ decdBn 20⋅


10-1 100 101 102 103-40

-35

-30

Frequency (rad/sec)

dBnG

dBnjG

dB

dB

∞=∞

⋅=

)sgn()(

3)/1( τ


-10

0

10 10/1,1 =−= τn

( ) 1G

o0)0( =∠ jG-70-60

-50

-40

-30

-20

Pha

se (d

eg)

( )1

10+

= ssG

o45⋅n


o

o

90)(

45)/1(

0)0(

⋅=∞∠

⋅=∠

∠

njG

njG

jG

τ

10-1 100 101 102 103-100

-90

-80

Frequency (rad/sec)

74

General Transfer Function: Complex poles/zeros

( )q

nn

jjjG⎥⎥⎦

⎤

⎢⎢⎣

⎡++⎟⎟

⎠

⎞⎜⎜⎝

⎛= 12

2

ωωζ

ωωω

⎤⎡ ⎞⎛⎞⎛22

Magnitude and Phase:

( )

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−

⋅=∠

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−⋅=

−22

1

22

2

2

/1/2tan

21log10

n

n

nndB

qjG

qjG

ωωωζωω

ωωζ

ωωω



y p

ωωω

ω

ωω

ωω

log402)(

0)(

⋅+⋅−⎯⎯ →⎯

⎯⎯ →⎯

>>

>

75

-20

0

20

General Transfer Function: Complex poles/zeros

05.0,1,1 ==−= ζωnq

( ) 1=sG

-140

-120

-100

-80

-60

-40

Pha

se (d

eg)

o

o

o

180)(

90)/1(

0)0(

⋅=∞∠

⋅=∠

=∠

qjG

qjG

jG

τo90⋅q

( )11.02 ++

=ss

sG


10-2 10-1 100 101 102-200

-180

-160

Frequency (rad/sec)

180)( ∞∠ qjG

Frequency Response

( ) ( ))50)(10(

5.02000++

+=

sssssGExample:


76

Frequency Response: Poles/Zeros in the RHP

• Same .

• The effect on is opposite than the stable case.

)( ωjG

)( ωjG∠

( )2

1−

=s

sGExample:

An unstable pole behaves like a stable zeroAn “unstable” zero behaves like a “stable” pole


This frequency response cannot be found experimentally but can be computed and used for control design.

First order LOW PASS

α⇓

+=

K

sKsT )(

αωω

+=

jKjT )(

αωω)(

22 +=

KjT

Gain and Phase:


( )αωωθαω

/tan)( 1−−∠=+

K

⎩⎨⎧

=∠018000

KK

Ko

77

First order LOW PASS

( )αωωθαω

ω

/tan)(

)(

1

22

−−∠=+

=

K

KjT

0)(,)0( =∞= TK

Tα

ωK

jT ⎯⎯ →⎯)(

αωα

ααα =⇒==

+= c

TKKjT2

)0(2

)(22

Cutoff frequency


ωω

αω

αω

αω

KjT

jT

⎯⎯ →⎯

→

>>

>

78

First Order HIGH PASS

α⇓

+=

Kj

sKssT )(

αωωω+

=jKjjT )(

αωω

ω)(22 +

=K

jT

Gain and Phase:


( )αωωθαω

/tan90)( 1−−+∠=+

oK

⎩⎨⎧

=∠018000

KK

Ko

First order HIGH PASS

( )αωωθαω

ωω

/tan90)(

)(

1

22

−−+∠=+

=

oK

KjT

KTT =∞= )(,0)0(

KjT ⎯⎯ →⎯

79

First order HIGH PASS

( )αωω�

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