ContSys1 L7 Stability

8/12/2019 ContSys1 L7 Stability

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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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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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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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Thus

for a>0 we will have a decaying oscillatory

motion (asymptotically stable)for a


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Re(s)

Im(s)

UnstableStable

Marginally Stable


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Re(s)

UnstableStable

Marginally Stable Im(s)


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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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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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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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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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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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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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A remote position controller withproportional-plus-derivative (PD) control.

Find stability criteria.

+

-

qo

FsJs 21qi

DsK 1


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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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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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A remote position controller withproportional-plus-integral-plus-derivative

(PID) control. Find stability criteria.

+

-

qo

FsJs 21qi

s

R

DsK1


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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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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 )(

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ContSys1 L7 Stability