Stability

• BIBO stability:

Def: A system is BIBO-stable if any bounded input produces bounded output.

otherwise it’s not BIBO-stable.

))((

resp. impulse)( where

finite )(stable-BIBO :Thm

1

0

sH

th

dtth

L

stable- BIBO1:pole

1

1

)1)(1(

1

1

1)(

1,1seigenvalueschar.value

11

1

0

01

10

e.g.

cancelled pole/zero allafter

plane halfleft open in the

)( of poles"" all stable-BIBO

2

sss

s

s

ssH

xy

uxx

sH

stable.- BIBO

0both ,3,2:pole

)3)(2(

1

)3)(2)(1(

164

1)( :Example

0 part real

have TF of poles All stable-BIBO

23

ss

sss

ssss

ssH

stable BIBO 01:poles

1

2

0

1

1

1

30

1

2

0

1

1

2

100

01

10

001

1

302

0

1

1

200

010

001

302)()(

302

0

1

1

200

010

001

1

1

sss

s

s

s

s

s

s

BAsICsH

xy

uxx

Asymptotically Stable

A system is asymptotically stable if for any arbitrary initial conditions, all variables in the system converge to 0 as t→∞ when input=0.

All variables include y & its derivatives

all state variables

but y=Cx+Du=Cx

if x→0 then y→0

only need to check x→0

0

Criterion for A.S.

A.S. is system

0part real have all

4,2,1: of seigenvalue

01

1

0

0

7148

100

010

:Example

0part real have

seigenvalue all a.s. is systemA

A

xy

uxx

stable) BIBO is it (But

a.s. not is system

0part real have them of 2

: Aof seigenvalue

211

302

0

1

1

200

010

001

, , -

xy

uxx

case in this O.K. isbook the

as )(,0)0(when

)0(3)0(2

)0(302)()(but

A.S.not )0(0)(

)(

)(

)(

)0(

00

00

00

)0()(

200

010

001

0when

3

32

1

2

32

2

1

2

ttyx

xexe

xeetCxty

xtx

txe

txe

txe

x

e

e

e

xetx

xx

u

tt

tt

t

t

t

t

t

t

At

correctly in A.S.conclude wouldBook

as all

but

unstable as or If

when

:Consider

0)(

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

)0(22002)()(

)(,0)0()0(

)0(

00

00

00

)(:0

002

200

010

001

321

32

2

3

2

1

ty

ttyxxx

xeeetCxty

ttxxx

x

e

e

e

txu

xy

u

b

b

b

xx

ttt

t

t

t

If there is no pole/zero cancellation,

BIBO-stable A.S.

If system is C.C. & C.O. no pole/zero cancellation

BIBO-stable A.S.

Exact pole/zero cancellation only happens mathematically, not in real systems.

From now on, assume no p/z cancellation

BIBO stable A.S. all char. val<0

all eigenvalues<0 all poles<0

Thm:

If a system is A.S.

then it is BIBO-stable

But BIBO-stable A.S. (mathematically)

A polynomial

is said to be Hurwitz or stable

if all of its roots are in O.L.H.P

A system is stable if its char. polynomial is HurwitzA nxn matrix is called Hurwitz or stable

if its char. poly det(sI-A) is Hurwitzall eigenvalues<0

011

1)( asasasasF nn

nn

Routh-Hurwitz Method

From now on, when we say stability we mean A.S. / M.S. or unstable.

We assume no pole/zero cancellation,

A.S. BIBO stable

M.S./unstable not BIBO stable

Since stability is determined by denominator, so just work with d(s)

3

1

541

1

3212

5311

642

011

1

:

:

:

)(

n

n

nnnn

n

nnnnn

nnnn

nnnnn

nn

nn

s

a

aaaa

a

aaaas

aaas

aaaas

asasasasd

:table Routh

polynomialstic characteri the

called is d(s) T.F., c.l. of den. the be

Let

0

Routh Table

Repeat the process until s0 row

Stability criterion:

1) d(s) is A.S. iff 1st col have same sign

2) the # of sign changes in 1st col

= # of roots in right half plane

Note: if highest coeff in d(s) is 1,

A.S. 1st col >0

If all roots of d(s) are <0, d(s) is Hurwitz

Example:

unstable

RHP in roots 2 changes, sign 2

- :sign col first

65.2

065.2:

05.24

64:

64:

11:

64)(

0

1

2

3

23

s

s

s

s

ssssd ←has roots:3,2,-1

unstable

roots unstb 2

changes sign 2

10:

43.6:

107:

51:

1032:

10532)(

0

1

2

3

4

234

s

s

s

s

s

sssssd

(1x3-2x5)/1=-7

(1x10-2x0)/1=10

(-7x5-1x10)/-7

A.S.

change sign no 0,col 1st

1:

2:

11:

42:

131:

1432)(

0

1

2

3

4

234

s

s

s

s

s

sssssd

Remember this

sign same have coeff all iff

A.S.is system order 2nd

:system order 2nd

cs

bs

cas

cbsassd

:

:

:

)(

0

1

2

2

adbc

dcba

dsb

adbcs

dbs

cas

dcsbsassd

0all iff

A.S.is system order rd

:system order 3rd

,,,

3

:

:

:

:

)(

0

1

2

3

23

Remember this

A.S.

A.S.

A.S.Not 0coeff all

A.S.Not

A.S.Not

A.S.Not

A.S.Not

A.S.Not

A.S.

943)943(

943532

632632

235

1

13

2

3

15

23

23

23

23

23

23

2

2

2

2

sss

ssssss

sss

sss

ss

ss

s

ss

sse.g.

Routh Criteria

Regular case: (1) A.S. 1st col. all same sign(2)#sign changes in 1st col.

=#roots with Re(.)>0

Special case 1: one whole row=0Solution: 1) use prev. row to form aux. eq. A(s)=0

2) get

3) use coeff of in 0-row 4) continue

)(sAds

d

Example

)1(444)(

44:

0:

1414:

1616:

142814:

781:

47884)(

22

2

1

2

3

4

5

2345

sssA

s

s

s

s

s

s

ssssssd

:row prev.

←whole row=0

stable marginally

.originally col. 1st in 0 have did wesince A.S.Not But

R.H.P in roots no col. in change sign No

:atedifferenti

1

4:

8:

1414:

2

18

)(

0

1

2

2

s

s

s

s

ss

ds

sdA

1

0

3229.15.1

47884)(

).(

)1(444)(

)()()(

2345

22

-

j

j

ssssssd

sd

js

sssA

sAsd

A(s)

:roots has

Indeed

of roots are these&

:roots has

:example prev in :e.g.

original

of roots all are of root The :Fact

7

)474)(1()(

0

44

44

7

747

44

874

474478841

232

2

2

3

23

24

234

35

23

23452

sssssd

s-

s

ss-

sss

ss-

sss

ss-

sssssssss

0

11

44

44)(

)(121

00

121

121

122)(

1

2

3

3

244

3

4

5

2345

:s

:s

:s

ssds

sdA

sAsss

:s

:s

:s

ssssssd

:row From

e.g.

0)Re( withroots no

col. 1st in change sign No

:row From

1:

2:

2)(

1)(

0

1

22

s

s

sds

sdA

ssAs

unstable. is

roots. double are &

0))fence.(Re( the on roots areThey

at root double

at root double

of roots are

of roots But

)(

)()()1(12)(

).(

12)(

222224

24

sd

js

js

jsjsssssA

sd

sssA

e.g.

continue & 0by "0" replace :solution

0row wholebut

0col 1st in #an :2 case Special

3:

32:

3:

30:

21:

321:

322)(

0

1

2

2

3

4

234

s

-s

s

s

s

s

sssssd

replace

0)Re( have two these

:roots has :Verify

0)Re( i.e. RHP, in roots 2

col. 1st in changes sign 2

0 assume wesince

2928140570

902.009057.0

0)(

03

2

.j.

j

sd

Useful case: parameter in d(s)

How to use: 1) form table as usual

2) set 1st col. >0

3) solve for parameter range for A.S.

2’) set one in 1st col=0

3’) solve for parameter that leads to M.S. or leads to sustained oscillation

Example

s+3

s(s+2)(s+1) Kp

pp

p

p

plc

p

KsKss

sKssssd

sKsss

sKsG

K

3)2(3

)3()1)(2()(

)3()1)(2(

)3()(

23

..

char.poly

:Sol

stability for of range find:Q

+

0

03)2(3

03

0

3:3

3)2(3:

33:

21:

0

1

2

3

p

pp

p

p

pp

p

p

K

KK

K

Ks

KKs

Ks

Ks

col. 1st : A.S.For

:table Routh

=6

03

7)1(

03

42

0463

41)2(3

202

003

4)2(3)(

2

2

2

23

k

kk

kk

kk

kk

kk

skksssd

2)

1)

:need we: A.S.For

prod. outer two mid of prod 2)

0coeff all 1)

criteria? Routh order 3rd remember

e.g.

A.S.for

need weall over

also. and but

or

528.013

7

20

13

7

3

71

13

7

3

71

3

7)1( 2

k

kk

kk

kk

k

k>0.5

)137

(3

4

3

4

3

4

0)(43

)(,

13

7

4)2(3

22

1

k

kjs

sAkss

s

sdk

k

kk

:freq osci

noscillatio sustained to leads

:row From

0row And

M.S. is this At

:get we

set weIf

1

s(s2+2s+8)

+

-

K(s+z)

s+p