automatic control theory -...

Automatic Control Theory

Professor：Maode Yan

Department of Automation

School of Electronic and Control Engineering

E_mail: [email protected]

2

Chapter 6: The stability of linear feedback systems

重点掌握

Routh-Hurwitz稳定性判据

Main contents

1、Condition for a feedback system to be stable

2、Routh-hurwitz criterion

3

The concept of stability

A stable system is a dynamic system with a

bounded response to a bounded input.

A stable system is define as a system with a bounded

(limited) system response. That is, if the system is

subjected to a bounded input or disturbance and the

response is bounded in magnitude, the system is said

to be stable.

4 stable neutral unstable

Example

5

Condition for a feedback system to be stable

Consider the transfer function of a closed-loop system as

n

i

iss

sN

sD

sNsT

1

)(

)(

)(

)()(

Where , are roots of the characteristic equation

D(s) . Assume all roots are simple, then we have

nisi 2,1,

)()()( sRsTsC )(

)(

)(

1

sR

ss

sNn

i

i

n

i i

il

j rj

j

ss

A

ss

B

11

6

n

i

ts

i

l

j

ts

jirj eAeBtc

11

)(

Steady-state response Transient response

n

i

ts

it

ieA1

0lim

When is real number , let , we have isiis

0i 0lim

ts

it

ieA

0i i

ts

it

AeA i

lim

0i

ts

it

ieAlim


7

When is complex number, let , we have is iii js

tj

i

tj

iiiii eAeA)(

1

)(

)cos( tAe i

ti

0i 0)cos(lim

tAe i

t

t

i

0i

0i

)cos()cos(lim

tAtAe ii

t

t

i

)cos(lim tAe i

t

t

i

A necessary and sufficient condition for a feedback system to be stable is that all the poles of the system transfer function have negative real parts.


8

Routh-Hurwitz criterion

The characteristic equation of a feedback system

001

1

1

asasasa n

n

n

n

0naAssume: ,when n=5, Routh table is

5s 5a 3a 1a

4s

3s

2s

1s

0s

4a 2a 0a

1

4

5243 ba

aaaa

2

4

5041 ba

aaaa

0

1

1

4221 cb

abab

1

1

0121 dc

abbc

0a

0a

9


The Routh-Hurwitz criterion states that the number

of roots of the feedback system equation with

positive real parts is equal to the number of changes

in sign of the first column of the Routh array.

Several case:

Case 1. No element in the first column is zero.

10


Case 4. Repeated roots of the characteristic equation on the jw-axis.

Case 3. There is a zero in the first column, and the other elements of the row containing the zero are also zero.

Case 2. There is a zero in the first column, but some other elements of the row containing the zero in the first column are nonzero.

11


0asa...sasa)s(D n1n

1n

1

n

0

Hurwitz determinant of n-order system

n2n

1n

31

420

531

n

aa000

0a

0aa0

0aaa

0aaa

1 2 3 nTake the stage as a master determinant of order 1 ~ (n-1)-order determinant of Hurwitz

取各阶主子行列式作为1阶~（n-1）

阶赫尔维兹行列式

12


控制系统稳定的充分必要条件是：当a0>0时，各阶赫尔维茨行列式1、2、…、n均大于零。

nn

n

n

aa

a

aa

aaa

aaa

2

1

31

420

531

000

0

00

0

0

1 2 3 n

a0>0时, a1>0(全部系数数同号)

a0>0时, a1>0, a2>0(全部系数数同号)

0a.sasa)s(D 2

1

1

2

0

011 a

0asa)s(D 10

0a11

a0>0时

a0>0时

0aaaa

0a21

20

1

2

Second-order

system

First-order system

13


0asa.sasa)s(D 32

2

1

3

0

a0>0时, a1>0, a2>0, a3>0(全部系数数同号)

a0>0时

a1a2> a0 a3

n2n

1n

31

420

531

n

aa000

0a

0aa0

0aaa

0aaa

1 2 3

n0a11

0aaaaaa

aa3021

20

31

2

0a

aa0

0aa

0aa

22

31

20

31

3

Third-order

system

14


0asasa.sasa)s(D 43

2

2

3

1

4

0

0a11 0aaaaaa

aa3021

20

31

2

0aaaaaaa

aa0

aaa

0aa2

30

2

14321

31

420

31

3

0a

aaa0

0aa0

0aaa

00aa

34

420

31

420

31

4

a0>0时, a1>0, a2>0, a3>0 , a4>0 (全部系数数同号)

4

2

1

2

30321 aaaaaaa

When a0>0

nn

n

n

aa

a

aa

aaa

aaa

2

1

31

420

531

000

0

00

0

0

1 2 3

n

Fourth-order

system

15


First-order system a1>0(全部系数数同号)

a1>0, a2>0(全部系数数同号)

0asa...sasa)s(D n1n

1n

1

n

0

a1>0, a2>0, a3>0(全部系数数同号)

a1a2> a0 a3

a1>0, a2>0, a3>0 , a4>0(全部系数数同号)

4

2

1

2

30321 aaaaaaa

Sum up ：When a0>0

Third-order

system

Second-order

system

Fourth-order

system

16

Example 1 : Determine the stability of the closed-loop system that

has the following characteristic equations.

05432 234 ssss(1)

0433 ss(2)

01011422 2345 sssss(3)

Answer

(1)

5

6

51

42

531

0

1

2

3

4

s

s

s

s

s

0.2878 + 1.4161i

0.2878 - 1.4161i

-1.2878 + 0.8579i

-1.2878 - 0.8579i


17

0433 ss(2)

4

43

4

31

0

1

2

3

s

s

s

s


18

043)1)(43( 2343 sssssss

4

2

44

11

431

0

1

2

3

4

s

s

s

s

s


19

a=[1 2 2 4 11 10];

>> p=roots(a)

p =

0.8950 + 1.4561i

0.8950 - 1.4561i

-1.2407 + 1.0375i

-1.2407 - 1.0375i

-1.3087

01011422 2345 sssss(3)

10

107224

10124

660

1042

1121

0

2

1

2

3

4

5

s

s

s

s

s

s


20

0122 2345 sssss(4)

0852 23 sss(5)

5s 1 2 1

4s 1 2 1

3s 0 0 4 4

2s 1 1

1s 0 0 2 0

0s 1

-1.0000

0.0000 + 1.0000i

0.0000 - 1.0000i

-0.0000 + 1.0000i

-0.0000 - 1.0000i


21

Example 2 :Consider that the characteristic equation of a closed-loop

control system is

04)2(3 23 sKKss

Find the rang of K so that the system is stable.

Answer: 528.0K

Example 3:Determine the following characteristic equations,how many

roots are to the right of the line s=-1 in the s-plane.

0133 23 sss

01034 23 sss

(1)

(2)


22

Example 4: For the control system shown in following Fig, find the

value K, so that the steady-state error due to disturbance input

is –0.099.

)15.0)(12.0)(11.0(

10

sss

ssN

1)(

)(sR )(sC)(sE

K


23

Example 5:

4

2

1

2

30321 aaaaaaa

K)2s)(1ss(s

K

)s(R

)s(C2

0Ks2s3s3s)s(D 234

0K

22 213K332

0K9

14Stable range of K values

The coefficients are positive

When a0>0,


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