A Presentation for Major Assignment“Bounded input bounded output system”

Submitted to:- Submitted by:- Mr. Somesh Chaturvedi MAYUR PANCHOLI(Dept. of Electrical engineering) K11038 B.TECH(Mechanical) 6th semester

CAREER POINT UNIVERSITY, KOTA

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

cancelled pole/zero allafter plane half leftopen in the )( of poles"" all stable-BIBO sH

Asymptotically StableA system is asymptotically stable if for any

arbitrary initial conditions, all variables in the system converge to 0 as t→∞ when input=0.

0part real have seigenvalue all a.s. is systemA

Thm: If a system is A.S. then it is BIBO-stable

But BIBO-stable does not guarantee A.S.in general

If there is no pole/zero cancellation, BIBO stable Asymp Stable

1

1 1 0 1 01

n n m

n mn n m

d d d d dy a y a y a y b u b u b udt dt dt dt dt

1, ( )

,x Ax Bu

H s D C sI A By Cx Du

1 01

1 1 0

( )( )( )

mm

n nn

b s b s bY sH sU s s a s a s a

Characteristic polynomialsThree types of models:

Assume no p/z cancellation

System characteristic polynomial is:1

1 1 0( ) det( ) n nnd s sI A s a s a s a

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 Hurwitz

A nxn matrix is called Hurwitz or stableif its char. poly det(sI-A) is Hurwitz, orif all eigenvalues have real parts<0

11 1 0( ) n n

n nd s a s a s a s a

Routh-Hurwitz MethodFrom now on, when we say stability we mean

A.S. / M.S. or unstable.

We assume no pole/zero cancellation,A.S. BIBO stableM.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

aaaaa

aaaaas

aaas

aaaas

asasasasd

:table Routh

polynomialstic characteri thecalled is d(s) T.F., c.l. of den. the be

Let

0

Routh Table

Repeat the process until s0 row

Routh-Hurwitz stability criteria:1) d(s) is A.S. iff 1st col have same sign2) 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, i.e., stable

Example:

unstableRHP 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

sss

ssssd ←has roots:3,2,-1

unstableroots unstb 2

changes sign 2

10:43.6:

107:51:

1032:

10532)(

0

1

2

3

4

234

sssss

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

sssss

sssssd

sign same thehave coeff all iff A.S. is systemorder 2nd

:::

)(

:systemorder 2nd

0

1

2

2

csbs

cas

cbsassd

Remember this

3 2

3

2

1

0

3rd order system:

( )

::

:

:

3rd order system is A.S., , , all same sign

iff

d s as bs cs d

s a cs b d

bc adsb

s d

a b c dbc ad

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

235113

23

15

23

23

23

23

23

23

2

2

2

2

sss

ssssss

sss

sssss

sss

ssss

e.g.

Routh CriteriaRegular 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

)(sAdsd

Example

)1(444)(

44: :row prev.

0: 4 4: 6 6:

4 8 4:781:

47884)(

22

2

1

2

3

4

5

2345

sssA

s

sssss

ssssssd

←whole row=0

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

R.H.Pin roots no col. 1in changesign No

4:8:

44:

8)( :atedifferenti

0

1

2

sss

sds

sdA

2 2

5 4

Fact: The roots of are all roots of original ( ) ( ) ( )e.g.: in prev example:

( ) 4 4 4( 1) has roots: &these are roots of ( ).

Indeed ( ) 4 8

A(s)d s A s

A s s ss j

d s

d s s s

3 28 7 4has roots: 1.5 1.3229 0 1

s s sj

j-

7

)474)(1()(

044

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

01144

44)()(121

00121121

122)(

1

2

3

3

244

3

4

5

2345

:s:s:s

ssds

sdAsAsss

:s:s:s

ssssssd

:row From

e.g.

0)Re( withroots nocol. 1st in change sign No

:row From

1:2:

2)(1)(

0

1

22

ss

sds

sdAssAs

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

jsjs

jsjsssssA

sdsssA

e.g.

continue & 0by "0" replace :solution0row wholebut 0col 1st in #an :2 case Special

3:

32:

3:30:21:

321:

322)(

0

1

2

2

3

4

234

s

-s

ssss

sssssd

Replace by

0)Re( have two these

:roots has :Verify

0)Re( i.e. RHP, in roots 2col. 1st in changes sign 2

0 assume wesince

2928140570902.009057.0

0)(

032

.j.j

sd

Useful case: parameter(s) in d(s)How to use: 1) form table as usual

2) set 1st col. >03) solve for parameter

range for A.S.2’) set one in 1st col=03’) 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

sKssssK

sG

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

030

3:3

3)2(3:

33:21:

0

1

2

3

p

pp

p

p

pp

p

p

K

KK

K

Ks

KKs

KsKs

col. 1st : A.S.For

:table Routh

037)1(

0342

0463

41)2(3202

003

4)2(3)(

2

2

2

23

k

kk

kk

kkkkkk

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

528.0137

20

137

371

137

371

37)1( 2

k

kk

kk

kk

k

)137(3

434

34

0)(43

)(,

137

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

Q: find region of stability in K- plane.

2

2

3 2

( ) ( 2)( )( 1) ( ) ( 2)

( ) ( 1) ( ) ( 2)

( ( 2) 1) 2:

01( 2) 1 0

22 0 0

1 2( ( 2) 1) 2

s K sH ss s s K s

d s s s s K s

s Ks K s KRouthCriteria

K

K K

K

K K K K

2

K