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2009 Spring ME451 - GGZ Page 1Week 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
2009 Spring ME451 - GGZ Page 2Week 7-8: Stability
• 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
2009 Spring ME451 - GGZ Page 3Week 7-8: Stability
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?
2009 Spring ME451 - GGZ Page 4Week 7-8: Stability
• 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
2009 Spring ME451 - GGZ Page 5Week 7-8: Stability
• 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
2009 Spring ME451 - GGZ Page 6Week 7-8: Stability
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
2009 Spring ME451 - GGZ Page 7Week 7-8: 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
2009 Spring ME451 - GGZ Page 8Week 7-8: Stability
• 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
2009 Spring ME451 - GGZ Page 9Week 7-8: Stability
• 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
2009 Spring ME451 - GGZ Page 10Week 7-8: Stability
• 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
2009 Spring ME451 - GGZ Page 11Week 7-8: Stability
• (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
2009 Spring ME451 - GGZ Page 12Week 7-8: Stability
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
2009 Spring ME451 - GGZ Page 13Week 7-8: Stability
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
2009 Spring ME451 - GGZ Page 14Week 7-8: Stability
• 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
2009 Spring ME451 - GGZ Page 15Week 7-8: Stability
• 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
2009 Spring ME451 - GGZ Page 16Week 7-8: Stability
• 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
2009 Spring ME451 - GGZ Page 17Week 7-8: Stability
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
−
−
−−−−
−
−−−
2009 Spring ME451 - GGZ Page 18Week 7-8: Stability
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
2009 Spring ME451 - GGZ Page 19Week 7-8: Stability
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
−−
−−
−=
−=
2009 Spring ME451 - GGZ Page 20Week 7-8: Stability
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
2009 Spring ME451 - GGZ Page 21Week 7-8: Stability
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±
2009 Spring ME451 - GGZ Page 22Week 7-8: Stability
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
2009 Spring ME451 - GGZ Page 23Week 7-8: Stability
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)
Routh arrayRouth array
223)( 234++++= sssssQ
2
22
+s
2009 Spring ME451 - GGZ Page 24Week 7-8: Stability
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
2009 Spring ME451 - GGZ Page 25Week 7-8: Stability
• 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=⇒
2009 Spring ME451 - GGZ Page 26Week 7-8: Stability
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
2009 Spring ME451 - GGZ Page 27Week 7-8: Stability
1
0
11
11
0
1
2
3
s
s
s
s
Routh arrayRouth array
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
=+
2009 Spring ME451 - GGZ Page 28Week 7-8: Stability
1
0
11
00
121
121
0
1
2
3
4
5
s
s
s
s
s
s
Routh arrayRouth array
No sign changesNo sign changes
in the first columnin the first column
No root in OPEN(!) RHPNo root in OPEN(!) RHP
44
Derivative of auxiliary poly.Derivative of auxiliary poly.
44
22
Stability Stability –– RouthRouth--Hurwitz More Example 2Hurwitz More Example 2
222345 )1)(1(122)( ++=+++++= ssssssssQ
ssds
ssd44
)12( 3
24
+=++
sds
sd2
)1( 2
=+
2009 Spring ME451 - GGZ Page 29Week 7-8: Stability
1
10
00
101
0
41
2
3
4
−
−
−
s
s
s
s
s
ε
Routh arrayRouth array
One sign changesOne sign changes
in the first columnin the first column One root in OPEN(!) RHPOne root in OPEN(!) RHP
Derivative of auxiliary poly.Derivative of auxiliary poly.
44 00
Stability Stability –– RouthRouth--Hurwitz More Example 3Hurwitz More Example 3
)1)(1)(1(1)( 24+−+=−= sssssQ
3
4
4)1(
sds
sd=
−
ε
2009 Spring ME451 - GGZ Page 30Week 7-8: Stability
• 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
2009 Spring ME451 - GGZ Page 31Week 7-8: Stability
• (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)
2009 Spring ME451 - GGZ Page 32Week 7-8: Stability
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)
2009 Spring ME451 - GGZ Page 33Week 7-8: Stability
““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?
2009 Spring ME451 - GGZ Page 34Week 7-8: Stability
““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
2009 Spring ME451 - GGZ Page 35Week 7-8: Stability
• 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+=)(
2009 Spring ME451 - GGZ Page 36Week 7-8: Stability
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
2009 Spring ME451 - GGZ Page 37Week 7-8: Stability
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
Stability Stability –– RouthRouth--Hurwitz Control Example 1 (3)Hurwitz Control Example 1 (3)
s
KKsK I
P+=)(
• Characteristic equation
0254
2)(1
23=
+++
++
ssss
KsKIp
02)1(254 234=+++++
IPKsKsss
2009 Spring ME451 - GGZ Page 38Week 7-8: Stability
-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,
Stability Stability –– RouthRouth--Hurwitz Control Example 1 (4)Hurwitz Control Example 1 (4)
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
2009 Spring ME451 - GGZ Page 39Week 7-8: Stability
• Routh array
• If we select different KP, the range of KI changes.
Stability Stability –– RouthRouth--Hurwitz Control Example 1 (5)Hurwitz Control Example 1 (5)
?,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
2009 Spring ME451 - GGZ Page 40Week 7-8: Stability
• 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
Stability Stability –– RouthRouth--Hurwitz Control Example 1 (6)Hurwitz Control Example 1 (6)
?3 if happensWhat ==IP
KK
2063 2jss ±=⇔=+
)(sK254
223
+++ sss
(rad/sec) 2
(sec) 4.42
2≈
π
2009 Spring ME451 - GGZ Page 41Week 7-8: Stability
• Determine the range of K that stabilize the closed-loop system.
Stability Stability –– RouthRouth--Hurwitz Control Example 2 (1)Hurwitz Control Example 2 (1)
K
s
)3)(2(
1
++ sss
2009 Spring ME451 - GGZ Page 42Week 7-8: Stability
Stability Stability –– RouthRouth--Hurwitz Control Example 2 (2)Hurwitz Control Example 2 (2)
K
s
)3)(2(
1
++ sss
)3)(2(
11
)3)(2(
1
+++
++
ss
sssK
2009 Spring ME451 - GGZ Page 43Week 7-8: Stability
• Characteristic equation
Stability Stability –– RouthRouth--Hurwitz Control Example 2 (3)Hurwitz Control Example 2 (3)
0
)3)(2(
11
)3)(2(
1
1 =
+++
+++
ss
sssK
01)3)(2(
11 =
+++⋅+
sss
K
0)3)(2( =++++ Kssss
075 23=+++ Ksss
2009 Spring ME451 - GGZ Page 44Week 7-8: Stability
• Routh array of
• If K = 35, oscillation frequency is obtained by the auxiliary
equation
Stability Stability –– RouthRouth--Hurwitz Control Example 2 (4)Hurwitz Control Example 2 (4)
075 23=+++ Ksss
Ks
Ks
Ks
s
K
0
5
351
2
3
350
5
71
<<−
⇒±=⇔=+
7
2
2
:Period
7:Frequency70355
πjss
2009 Spring ME451 - GGZ Page 45Week 7-8: Stability
• 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
2009 Spring ME451 - GGZ Page 46Week 7-8: Stability
• 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
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