stability – a simple example...– routh-hurwitz stability criterion to determine stability...

2009 Spring ME451 - GGZ Week 7-8: Stability

• We want the mass to stay at x = 0, but wind gave some

initial speed (f(t) = 0). What will happen?

• How to characterize different behaviors with TF?

MM MM

KK

MM

BB

MM

BB

KK

Stability Stability –– A Simple ExampleA Simple Example

2

1

)(

)(

ssF

sx=

)(tf

)(tx

)(tf

)(tx

KssF

sx

+=

2

1

)(

)(

)(tf

)(txBsssF

sx

+=

2

1

)(

)(

KBsssF

sx

++=

2

1

)(

)(

)(tx

)(tf


• The most basic and important specification in control analysis and synthesis!

• Unstable systems have to be stabilized by feedback.

• Unstable closed-loop systems are useless.

– What happens if a system is unstable?

• may hit mechanical/electrical “stops” (saturation)

• may break down or burn out

Stability Stability –– ImportanceImportance


Tacoma Narrows Bridge (July 1-Nov.7, 1940)

2008…

WindWind--induced vibrationinduced vibration Collapsed!Collapsed!

StabilityStability––WhatWhat WillWill Happen to Unstable Systems?Happen to Unstable Systems?


• BIBO (Bounded-Input-Bounded-Output) stability : Any

bounded input generates a bounded output.

• Asymptotic stability :

Any ICs generates y(t) converging to zero.

BIBO stable BIBO stable

systemsystem

uu((tt)) Zero ICsZero ICs

Asymptotic stable Asymptotic stable

systemsystemuu((tt)=0)=0

yy((tt))

Stability Stability –– DefinitionDefinition

yy((tt))

Given any ICsGiven any ICs


• Zero : roots of n(s)

• Pole : roots of d(s)

• Characteristic polynomial : d(s)

• Characteristic equation : d(s) = 0

Ex.Ex.)(

)()(

sd

snsG =

Stability Stability –– Some TerminologiesSome Terminologies

Given the following transfer function

)1)(2(

)1)(1()(

2++

+−=

ss

sssG

1))( of Zeros( ±=sG

jsG ±−= ,2))( of Poles(

22)( 23+++= ssssd

022 23=+++ sss


For a system represented by a transfer For a system represented by a transfer

function function GG((ss),),

system is BIBO stablesystem is BIBO stable

system is asymptotically stablesystem is asymptotically stable

All the poles of All the poles of GG((ss)) are in the open left are in the open left

half of the complex plane.half of the complex plane.

Stability Stability –– ““ss”” Domain StabilityDomain Stability


Asymptotical Asymptotical

Stability: Stability:

((UU((ss)=)=0)0)

BIBO Stability: BIBO Stability:

((yy(0)=(0)=0)0)

Example:Example:

Bounded if ReBounded if Re(α)>(α)>00

Stability Stability –– ““IdeaIdea”” of Stability Conditionof Stability Condition

0)0( ),()()( yytutyty ==+α&

)()()0()( sUtYyssY =+− α

))0()((1

)( ysUs

sY ++

=α

0)Re( if 0)0(1

)]([)(0

11>→=

+==

−−− αα

αyey

sLsYLty

t

{ } ∫∫ −=−===−−−

tt

dtuedtugsUsGLsYLty00

11 )()()()()()]([)( τττττατ

∫∫ ⋅≤−≤−−

tt

udedtuety0

max

0

)()( τττ ατατ


• For a general system (nonlinear etc.), BIBO stability

condition and asymptotic stability condition are different.

• For linear time-invariant (LTI) systems (to which we can

use Laplace transform and we can obtain a transfer

function), the conditions happen to be the same.

• In this course, we are interested in only LTI systems, we

use simply “stable” to mean both BIBO and asymptotic

stability.

Stability Stability –– Remarks on Stability DefinitionRemarks on Stability Definition


• Marginally stable if

– G(s) has no pole in the open RHP (Right Half Plane), &

– G(s) has at least one simple pole on jω-axis, &

– G(s) has no multiple poles on jω-axis.

• Unstable if a system is neither stable nor marginally stable.

Marginally stableMarginally stable NOT marginally stableNOT marginally stable

Stability Stability –– ““Remarks on Stability Definition Remarks on Stability Definition (cont(cont’’d)d)

)1)(4(

1)(

2++

=sss

sG)1()4(

1)(

22++

=sss

sG

)1()4(

1)(

22++

=sss

sG)1(

1)(

−=

sssG


• Repeated poles

• Does marginal stability imply BIBO stability? No

– TF:

– Pick

– Output

Stability Stability –– ExamplesExamples

tts

ssUsGsYL sin

)1(

2)()()(

22

1=

+==

−

tts

sL ω

ω

ωsin

)(

2222

1=

+

− tts

sL ω

ω

ωcos

)(222

22

1=

+

−−

)1(

2)(

2+

=s

ssG

)1(

1)( sin)(

2+

=→=s

sUttuL


• (BIBO, asymptotically) stable if

Re(ssii) < 0 for all i.

• marginally stable if

– Re(ssii) ≤ 0 for all i, and

– simple root for Re(ssii) = 0

• unstable if

it is neither stable nor marginally stable.

Let Let ssii be be polespoles of of GG. Then, . Then, GG is is

……

Stability Stability –– SummarySummary


KK

MM

BB

MM

BB

KK

Poles= Poles=

stable?stable?

Poles= Poles=

stable?stable?

Poles= Poles=

stable?stable?

Poles= Poles=

stable?stable?

MM MM2

1

)(

)(

ssF

sx=

)(tf

)(tx

)(tf

)(tx

KssF

sx

+=

2

1

)(

)(

)(tf

)(tx

BsssF

sx

+=

2

1

)(

)(

KBsssF

sx

++=

2

1

)(

)(

)(tx

)(tf

Stability Stability –– Example RevisitedExample Revisited


StableStable

??

??

??

??

??

??

Stability Stability –– More ExamplesMore Examples

)(sG

)1)(1(

)2(52

+++

+

sss

s

)1)(1(

)2(52

+++

+−

sss

s

)3)(2(

52

+− ss

)1)(1(

22

2

+−+

+

sss

s

22 )1)(2(

5

++ ss

)1)(1(

12

+− ss

marginally stablemarginally stable unstableunstable


• Stability for LTI systems

– (BIBO and asymptotically) stable, marginally stable,

unstable

– Stability for G(s) is determined by poles of G.

• Next

– Routh-Hurwitz stability criterion to determine stability

without explicitly computing the poles of a system.

Stability Stability –– SummarySummary


• This is for LTI systems with a polynomial denominator

(without sin, cos, exponential etc.)

• It determines if all the roots of a polynomial

– lie in the open LHP (left half-plane),

– or equivalently, have negative real parts.

• It also determines the number of roots of a polynomial in

the open RHP (right half-plane).

• It does NOT explicitly compute the roots.

• No proof is provided in any control textbook.

Stability Stability –– Routh Hurwitz CriterionRouth Hurwitz Criterion


• Consider a polynomial

• Assume

– If this assumption does not hold, Q can be factored as

where

– The following method applies to the polynomial

Stability Stability –– Polynomial and an AssumptionPolynomial and an Assumption

01

1

1)( asasasasQ

n

n

n

n++++=

−

−L

00

≠a

0ˆ0

≠a

4444444 34444444 21L

)(ˆ

01

1

1 )ˆˆˆˆ()(

sQ

mn

mn

mn

mn

masasasassQ ++++=

−−

−−

−

−

)(ˆ sQ


From the given From the given

polynomialpolynomial

Stability Stability –– Routh ArrayRouth Array

1

0

1

1

21

2

321

3

321

2

7531

1

642

ms

ls

kks

cccs

bbbs

aaaas

aaaas

n

n

nnnn

n

nnnn

n

MMM

L

L

−

−

−−−−

−

−−−


Stability Stability –– Routh Array (3Routh Array (3rdrd row calculation)row calculation)

1

0

1

1

21

2

321

3

321

2

7531

1

642

ms

ls

kks

cccs

bbbs

aaaas

aaaas

n

n

nnnn

n

nnnn

n

MMM

L

L

−

−

−−−−

−

−−−

M

1

514

2

1

312

1

−

−−−

−

−−−

−=

−=

n

nnnn

n

nnnn

a

aaaab

a

aaaab


Stability Stability –– Routh Array (4Routh Array (4thth row calculation)row calculation)

1

0

1

1

21

2

321

3

321

2

7531

1

642

ms

ls

kks

cccs

bbbs

aaaas

aaaas

n

n

nnnn

n

nnnn

n

MMM

L

L

−

−

−−−−

−

−−−

M

1

3115

2

1

2113

1

b

babac

b

babac

nn

nn

−−

−−

−=

−=


1

0

1

1

21

2

321

3

321

2

7531

1

642

ms

ls

kks

cccs

bbbs

aaaas

aaaas

n

n

nnnn

n

nnnn

n

MMM

L

L

−

−

−−−−

−

−−−

The number of roots The number of roots

in the open right halfin the open right half--plane plane

is equal to is equal to

the number of sign changesthe number of sign changes

in the in the first columnfirst column of Routh array.of Routh array.

Stability Stability –– RouthRouth--Hurwitz CriterionHurwitz Criterion


Routh arrayRouth array

Two sign changesTwo sign changes

in the first columnin the first columnTwo roots in RHPTwo roots in RHP

Stability Stability –– RouthRouth--Hurwitz Example 1 Hurwitz Example 1 (case 1)(case 1)

)4)(2(82)( 223+−+=+++= sssssssQ

6

0)6(80

1

2

1

823

8

6

81

21

−

−−×

−

−

s

s

s

s

861 →−→ 2

15

2

1 j±


Routh arrayRouth arrayIf 0 appears in the first column of a If 0 appears in the first column of a

nonzero row in Routh array, replace it nonzero row in Routh array, replace it

with a small positive number. In this with a small positive number. In this

case, Q has some roots in RHP.case, Q has some roots in RHP.

Two sign changesTwo sign changes

in the first columnin the first columnTwo roots Two roots

in RHPin RHP

Stability Stability –– RouthRouth--Hurwitz Example 2 Hurwitz Example 2 (case 2)(case 2)

1011422)( 2345+++++= ssssssQ

10

6

10

60

1042

1121

0

1

1242

3

4

5

s

s

s

s

s

s

ε

ε −

ε

6124

0

→−

→

<

43421 ε

εε


2

0

21

21

231

0

1

2

3

4

s

s

s

s

sIf zero row appears in Routh array, Q If zero row appears in Routh array, Q

has roots either on the imaginary axis has roots either on the imaginary axis

or in RHP.or in RHP.

No sign changes No sign changes

in the first columnin the first columnNo roots No roots

in RHPin RHP

ButBut some some

roots are on roots are on

imagimag. axis.. axis.Take derivativeTake derivative of an of an auxiliary polynomialauxiliary polynomial

(which is a factor of (which is a factor of QQ((ss))))

Stability Stability –– RouthRouth--Hurwitz Example 3 Hurwitz Example 3 (Case 3)(Case 3)


223)( 234++++= sssssQ

2

22

+s


Routh arrayRouth arrayNo sign changes No sign changes

in the first columnin the first column

Find the range of K Find the range of K s.ts.t. . Q(sQ(s) has all roots in the left ) has all roots in the left

half plane. (Here, K is a design parameter.)half plane. (Here, K is a design parameter.)

Stability Stability –– RouthRouth--Hurwitz Example 4Hurwitz Example 4

4

43

21

0

3

4)2(31

2

3

s

s

Ks

Ks

K

KK −+

+

4)2(3)( 23++++= sKKsssQ

>−+

>

04)2(3

03

KK

K

3

211+−>K


• 1st order polynomial

• 2nd order polynomial

• Higher order polynomial

Necessary ConditionsNecessary Conditions

Stability Stability –– Simple & Useful Criteria for StabilitySimple & Useful Criteria for Stability

01)( asasQ +=

sign same thehave and LHPin are roots All01

aa⇔

01

2

2)( asasasQ ++=

sign same thehave and , LHPin are roots All012

aaa⇔

01

1

1)( asasasasQn

n

n

n++++=

−

−L

sign same thehave ),1,0( LHPin are roots All nkak

L=⇒


All roots in open LHP?All roots in open LHP?

Yes / NoYes / No

Yes / NoYes / No

Yes / NoYes / No

Yes / NoYes / No

Yes / NoYes / No

Stability Stability –– RouthRouth--Hurwitz Example 5Hurwitz Example 5

53 +s

)(sQ

1052 2−−− ss

1895723 2−− ss

)1)(1( 22++−+ ssss

3105 23−++ sss


1

0

11

11

0

1

2

3

s

s

s

s


No sign changesNo sign changes

in the first columnin the first column No root in OPEN(!) RHPNo root in OPEN(!) RHP

22

Derivative of auxiliary poly.Derivative of auxiliary poly.

(Auxiliary poly. is a factor of (Auxiliary poly. is a factor of Q(sQ(s).)).)

Stability Stability –– RouthRouth--Hurwitz More Example 1Hurwitz More Example 1

)1)(1(1)( 223++=+++= ssssssQ

sds

sd2

)1( 2

=+


1

0

11

00

121

121

0

1

2

3

4

5

s

s

s

s

s

s


No sign changesNo sign changes

in the first columnin the first column

No root in OPEN(!) RHPNo root in OPEN(!) RHP

44


44

22


222345 )1)(1(122)( ++=+++++= ssssssssQ

ssds

ssd44

)12( 3

24

+=++

sds

sd2

)1( 2

=+


1

10

00

101

0

41

2

3

4

−

−

−

s

s

s

s

s

ε


One sign changesOne sign changes

in the first columnin the first column One root in OPEN(!) RHPOne root in OPEN(!) RHP


44 00


)1)(1)(1(1)( 24+−+=−= sssssQ

3

4

4)1(

sds

sd=

−

ε


• Routh-Hurwitz stability criterion

– Routh array

– Routh-Hurwitz criterion is applicable to only polynomials (so, it is not possible to deal with exponential, sin, cos

etc.).

• Next,

– Routh-Hurwitz criterion in control examples

Stability Stability –– RouthRouth--Hurwitz SummaryHurwitz Summary


• (BIBO, asymptotically) stable if

Re(ssii) < 0 for all i.

• marginally stable if

– Re(ssii) ≤ 0 for all i, and

– simple root for Re(ssii) = 0

• unstable if

it is neither stable nor marginally stable.

Let Let ssii be be polespoles of of GG. .

Then, Then, GG is is ……

Stability Stability –– Summary (Review)Summary (Review)


1

0

1

1

21

2

321

3

321

2

7531

1

642

ms

ls

kks

cccs

bbbs

aaaas

aaaas

n

n

nnnn

n

nnnn

n

MMM

L

L

−

−

−−−−

−

−−−

The number of roots The number of roots

in the open right halfin the open right half--plane plane

is equal to is equal to

the number of sign changesthe number of sign changes

in the in the first columnfirst column of Routh array.of Routh array.

Stability Stability –– RouthRouth--Hurwitz Criterion (Review)Hurwitz Criterion (Review)


““most undergraduate students are exposed to the most undergraduate students are exposed to the

RouthRouth––Hurwitz criterion in their first introductory Hurwitz criterion in their first introductory

controls course. This exposure, however, is at the controls course. This exposure, however, is at the

purely algorithmic level in the sense that no attempt purely algorithmic level in the sense that no attempt

is made whatsoever to explain why or how such an is made whatsoever to explain why or how such an

algorithm works.algorithm works.””

An Elementary Derivation of the Routh–Hurwitz CriterionMing-Tzu Ho, Aniruddha Datta, and S. P. Bhattacharyya

IEEE Transactions on Automatic Controlvol. 43, no. 3, 1998, pp. 405-409.

StabilityStability––WhyWhyNoNoProofProofofofRouthRouth--HurwitzHurwitzCriterion?Criterion?


““The principal reason for this is that the classical The principal reason for this is that the classical

proof of the Routhproof of the Routh--Hurwitz criterion relies on the Hurwitz criterion relies on the

notion of Cauchy indexes and Sturmnotion of Cauchy indexes and Sturm’’s theorem, s theorem,

both of which are beyond the scope of both of which are beyond the scope of

undergraduate students.undergraduate students.””

““RouthRouth--Hurwitz criterion has become one of the few Hurwitz criterion has become one of the few

results in control theory that most control engineers results in control theory that most control engineers

are compelled to accept on faith.are compelled to accept on faith.””

StabilityStability–– ““WhyWhy””ContinuesContinues


• Design K(s) that stabilizes the closed-loop system for the

following cases.

– K(s) = K (constant)

–

(PI (Proportional and Integral) controller)

Stability Stability –– RouthRouth--Hurwitz Control Example 1Hurwitz Control Example 1

254

223

+++ sss)(sK

s

KKsK I

P+=)(


K(s) = K

• Characteristic equation

• Routh array

Stability Stability –– RouthRouth--Hurwitz Control Example 1 (2)Hurwitz Control Example 1 (2)

0254

21

23=

++++

sss

K

0)1(254 23=++++ Ksss

0)1(2

0

)1(24

51

0

2

91

2

3

>+

>

+

−

Ks

s

Ks

s

K

91 <<− K


02

9

2

)1(24

251

0

9

16)1)(9(21

2

92

3

4

>

<

+

−

−+−

−

II

PK

KKK

I

K

P

I

KKs

Ks

Ks

Ks

Ks

P

IPP

P

• Routh array


s

KKsK I

P+=)(


0254

2)(1

23=

+++

++

ssss

KsKIp

02)1(254 234=+++++

IPKsKsss


-1 0 1 2 3 4 5 6 7 8 90

0.5

1

1.5

2

2.5

3

3.5

• From Routh array,


IP

I

PKK

s

KKsK , of Range ,)( +=

08)9)(1(

0

9

>−−+

>

<

IpP

I

P

KKK

K

K

PK

)89(0

91-

if stable

be willsystem The

2

8

1PPI

P

KKK

K

−+<<

<<

)89(2

81

PPIKKK −+=

IK


• Routh array

• If we select different KP, the range of KI changes.


?,3Select ,)(IP

I

PKK

s

KKsK =+=

302

23

84

251

0

3

8241

2

3

4

<<

−

II

K

I

I

KKs

s

Ks

s

Ks

I

063 :Poly Aux. ,3 If 2=+= sK

I


• Auxiliary equation

• Oscillation frequency

• Period0 2 4 6 8 10 12 14 16 18 20

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Unit step responseUnit step response


?3 if happensWhat ==IP

KK

2063 2jss ±=⇔=+

)(sK254

223

+++ sss

(rad/sec) 2

(sec) 4.42

2≈

π


• Determine the range of K that stabilize the closed-loop system.


K

s

)3)(2(

1

++ sss



K

s

)3)(2(

1

++ sss

)3)(2(

11

)3)(2(

1

+++

++

ss

sssK




0

)3)(2(

11

)3)(2(

1

1 =

+++

+++

ss

sssK

01)3)(2(

11 =

+++⋅+

sss

K

0)3)(2( =++++ Kssss

075 23=+++ Ksss


• Routh array of

• If K = 35, oscillation frequency is obtained by the auxiliary

equation


075 23=+++ Ksss

Ks

Ks

Ks

s

K

0

5

351

2

3

350

5

71

<<−

⇒±=⇔=+

7

2

2

:Period

7:Frequency70355

πjss


• Step 1: Write the closed loop transfer function

• Step 2: Obtain the closed loop system characteristic

equation

• Step 3: Generate Routh Array

• Step 4: Let the 1st column of Routh Array greater than zero

to find constrain equations

• Step 5: Solve these constrain equation for control

parameters

Stability Stability –– RouthRouth--Hurwitz Hurwitz SynthesisSynthesis


• Control examples for Routh-Hurwitz criterion

– P controller gain range for stability

– PI controller gain range for stability

– Oscillation frequency

– Characteristic equation

• Next

– Root Locus

Stability Stability –– RouthRouth--Hurwitz Summary 2Hurwitz Summary 2

stability – a simple example...– routh-hurwitz stability criterion to determine stability...

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