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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
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