contsys1 l7 stability
TRANSCRIPT
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Mechanical Engineering Science 8
Dr. Daniil Yurchenko
Basic stability analysis
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Stability assessment
Lets consider a ball in the equilibrium
position
Unstable Marginally
Stable
Asymptotically
Stable
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Stability assessment
A system in which the transient part of theresponse decays with time is said to be
absolutely (asymptotically) stable If the transient part does not decay it is
said to be absolutely unstable
This stability manifests itself either as amonotonic increase in the error or as anoscillatory increase in the error.
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Stability assessment The first order system
If a>0 then and the solutionwill decay exponentially as(asymptotically stable)
If a
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Stability assessment The second order system
If we have TWO REAL roots,one of which is positive. Thus the systemis unstablesince when
0 bxxax 0)( 2 Axbass
2
42
2,1
baas
stAex
ba 42
x t
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Stability assessment
If we have TWO REALidentical roots -a/2. Thus the system isstable if a>0 and unstable otherwise
If then we have TWOCOMPLEX CONJUGATE roots with the
real part equal to -a/2 and imaginary partequal to
ba 42
ba 42
2/4 2ab
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Stability assessment
Thus
for a>0 we will have a decaying oscillatory
motion (asymptotically stable)for a
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Stability assessment
Re(s)
Im(s)
UnstableStable
Marginally Stable
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Stability assessment
Re(s)
UnstableStable
Marginally Stable Im(s)
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Stability assessment
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Stability assessment
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Stability assessment
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Stability assessment
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Stability assessment
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Stability assessment
We are able to solve analytically the firstand second order linear algebraic equations
(LAE). We can obtain analytical solution to the
third and sometimes fourth order LAEs.
What do we do if we have a higher orderequation?
Can we say something about stability?
0...... 11
10
nn
nnAsAsAsA
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Stability assessment
Luckily we can assess the system stabilitywithout solving the characteristic equation
for every system we encounter. We do thisthrough an examination of the coefficientsof the characteristic equation.
The criterion which we apply is known as
The Routh-Hurwitz Stability criterion
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Routh-Hurwitz Stability Criterion
First write the characteristic equation as apolynomial in s
For this polynomial we construct a matrix
containing the coefficients in the followingway
0...... 11
10
nnnn AsAsAsA
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Routh-Hurwitz Stability Criterion
nn
n
AA
A
AAA
AAA
AAAA
AAAA
...0000
0...0000
....................................
00...0
00...0
00...
00...
2
1
420
531
6420
7531
11 AR
Lets define a set
of determinants as
20
31
2AA
AAR
31
420
531
3
0 AA
AAA
AAA
R
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Routh-Hurwitz Stability Criterion
For a first order system 010 AsA
01A
For a system to be absolutely(asymptotically) stable it is requiredthat A0 >0 and a set of determinants
R1, R2,...Rn-1be positive.
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Routh-Hurwitz Stability Criterion
For a second order system
thus
For a third order system
0212
0 AsAsA
0,0 21 AA
011 AR 00
21
20
1
20
31
2 AAAA
A
AA
AAR
0322
1
3
0 AsAsAsA
011 AR 03021
20
31
2 AAAAAA
AA
R
0
0
0
0
0
330321
31
20
31
31
420
531
3 AAAAAA
AA
AA
AA
AA
AAA
AAA
R
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Routh-Hurwitz Stability Criterion
Consider
also since then
Thus
CLASS, DERIVE THE STABILITY
CONDITIONS FOR A 4thORDER SYSTEM
0,0,0,0 3021321 AAAAAAA
030212 AAAAR
0330213303213 AAAAAAAAAAAR
0,0 10 AA 0,0 32 AA
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Routh-Hurwitz example
A system has the characteristic equation:
Is it stable?
011072152240 2345 sssss
8544240107215272240
10152,152 21 RR
0542
3
3
2
4
1
5
0 AsAsAsAsAsA
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Routh-Hurwitz example Routh Hurwitz arrays
?
101520
172240
110152
3 R
98816 240*100152*152152*24010*72*152
152072240
10152
101520172240
110152
3R
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Routh-Hurwitz examples
If a system open-loop transfer function is
Then the characteristic equation is defined as
+
-
qoqi
)(
)()(0
sQ
sPsG
)()()(
)(
)(/)()(1
)(/)(
)()(1
)(
0
0
sHsPsQ
sP
sQsPsH
sQsP
sHsG
sGT
0)()(1 0 sHsG 0)(1 0 sG
0)()( sQsP
If H=1
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Routh-Hurwitz examples
A remote position controller withproportional-plus-derivative (PD) control.
Find stability criteria.
+
-
qo
FsJs 21qi
DsK 1
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Routh-Hurwitz examples
Open loop transfer function
Characteristic equation
FsJs
DsKDGO
2
1
011
112
2
2
FsJs
DsKFsJs
FsJs
DsKDGO
0)(2 KsKDFJs
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Routh-Hurwitz examples
Then for stability of a second order systemwe need all the coefficients to be positive
Remember we said that introducing
integral control may influence the systemstability
0)(,0,0 KDFKJ
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Routh-Hurwitz examples
A remote position controller withproportional-plus-integral-plus-derivative
(PID) control. Find stability criteria.
+
-
qo
FsJs 21qi
s
R
DsK1
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Routh-Hurwitz examples
Open loop transfer function
Characteristic equation FsJs
s
RDsK
DGO
2
1
0
11
11 2
2
2
FsJs
s
RDsKFsJs
FsJs
s
RDsK
DGO
0)( 23 KRKssKDFJs
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Routh-Hurwitz examples
Then
For stability we need all the coefficients tobe positive and
Thus we cannot take an arbitrary value of R
KRAKAKDFAJA 3210 ,),(,
03021 AAAA
JRKKKDF )(
JRKDF )(