control systems - chibum lee · pdf file08.03.2014 · chibum lee -seoultech control...

Control Systems

Stability

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Control SystemsChibum Lee -Seoultech

Outline

Stablity

Routh Hurwitz stability test.


Stability

The most important problem in control systems

Stability!!!

sta·bil·i·ty

• 1. The state or quality of being stable, especially:

a. Resistance to change, deterioration, or displacement.

b. Constancy of character or purpose; steadfastness.

c. Reliability; dependability.

• 2. The ability of an object, such as a ship or aircraft, to maintain

equilibrium or resume its original, upright position after displacement, as

by the sea or strong winds.


Stability

Control systems has many definitions on stability

• Stable in the sense of Lyapunov, Asymptotic stability….

• “A stable system is a dynamic system with a bounded response to a

bounded input.” - BIBO stability (in text)

Poles and stability

• The poles of the closed-loop transfer function of a given system are

located in the RHP the system becomes unstable.

• Tests of stability: Routh-Hurwitz Criteria, Root-Locus, Nyquist Stability

Criteria all test to check poles of the transfer function in the RHP


Stability

Maxwell 1868

Real rootswith + real part

Real rootswith - real part

Complex rootswith - real part

Complex rootswith + real part


Stability

Recall: transfer functions

For impulse input

r

k kkk

kkkkkkq

j j

j

rrrqn

m

m

m

m

n

m

n

n

n

n

m

m

m

m

ss

s

ps

sssspspsa

bsbsbsb

pspsps

zszszsK

asasasa

bsbsbsb

sR

sY

122

2

1

222

111

2

1

01

1

1

21

21

01

1

1

01

1

1

2

1)(

)2()2)(()(

)())((

)())((

)(

)(

r

k

kk

t

kkk

t

k

q

j

tp

j teteety kjkjj

1

22

1

1sin1cos)(


Stability

Poles and stability


Stability of a Control System

“A control system is stable if and only if all

closed loop poles lie in the left half s plane.”

• Closed loop

G(s))(sR

H(s)

)(sY)(sE

-+

01

1

1

01

1

1

)()(1

)(

)(

)(

asasasa

bsbsbsb

sHsG

sG

sR

sY

n

n

n

n

m

m

m

m


Routh-Hurwitz Stability Criterion

a necessary and sufficient criterion for the

stability of a LTI control system.

• Routh in 1876, Hurwitz in 1895

• determines if there are any poles in the RHP.

Characteristic equation

• The denominator of the closed loop transfer function

where the coefficient are real

• Check any root of Δ 𝑠 lies in RHP of the s-plane

0)( 01

1

1

asasasas n

n

n

n



Rewrite

• All coefficients of Δ(𝑠) must have the same sign if all the

roots are in the LHP

• All coefficients of Δ(𝑠) must be nonzero

1st Stability Criterion

• Note) it is necessary condition

0)1()(

)()(

)())(()(

321

3

12421321

2

13221

1

21

21

n

n

n

n

nnnn

n

nnn

n

nn

n

n

nn

ppppaspppppppppa

sppppppaspppasa

pspspsas

0)( 01

1

1

asasasas n

n

n

n

82)4)(2()( 232 sssssss



2nd Criterion

• 1st step: build an array formulation based on ordering

the coefficients

531

42

nnn

nnn

aaa

aaa

1n

n

s

s

0)( 01

1

1

asasasas n

n

n

n



• 2nd step: form the array

1

531

531

531

42

n

nnn

nnn

nnn

nnn

h

ccc

bbb

aaa

aaa

0

3

2

1

s

s

s

s

s

n

n

n

n

31

31

1

1

51

4

1

3

31

2

11

3211

1

1,

1

nn

nn

n

n

nn

nn

n

n

nn

nn

nn

nnnnn

bb

aa

bc

aa

aa

ab

aa

aa

aa

aaaab



The number of roots with positive real parts is

equal to the number of sign changes in that first

column. <necessary & sufficient cond.>

• 1st column

• 4 distinct cases of the 1st column

1

1

1

1

n

n

n

n

n

h

c

b

a

a



Case1: No element in the 1st column is zero

• Ex.

• Ex.

01

2

2)( asasas

0

1

02

a

a

aa

0

1

2

s

s

s

01

2

2

3

3)( asasasas

0

2

3012

02

13

0

a

a

aaaa

aa

aa

0

1

2

3

s

s

s

s

3021 aaaa



• Ex.

2 sign changes 2 roots with positive real parts

242)( 23 ssss

42

022

241

21

0

1

2

3

s

s

s

s

3,71 32,1 pjp



• Ex

2 sign changes 2 roots with positive real parts

5432)( 234 sssss

5

0 6

512

432042

531

0

1

2

3

4

s

s

s

s

s

4161.12878.0,8579.02878.1 4,32,1 jpjp



Case2: A zero in the 1st column, but some other

elements of the row containing the zero in the 1st

column are non zero

• Ex. 23)( 3 sss

2

02

3

20

31

0

1

2

3

s

s

s

s

2 sign changes2 roots with positive real parts

2,1 32,1 pp



Case3: A zero in the 1st column, and the other elements

of the row containing the zero are also zero

• form an auxiliary polynomial with the coefficients of the last row

and use the derivative of the polynomial in the next row.

• Ex. 842)( 23 ssss

8

04

82

0

1

2

s

s

s

2,2 3,21 jpp

Auxiliary polynomial U(s)

82)( 2 ssU

ss

sU4

)(

00

82

41

1

2

3

s

s

s

0 sign change no positive root


50

07.112

5024

968

50482

0

1

2

3

4

-s

s

s

s

s


• Ex.

502548242)( 2345 ssssss

00

50482

25241

3

4

5

s

s

s

Auxiliary polynomial U(s)

50482)( 24 sssU

sss

sU968

)( 3

1 sign change 1 positive root

5,1,1,2 5,4321 jpppp



Case4: Repeated roots of the characteristic equation

on the 𝑗𝜔-axis Unstable

• Ex. 122)( 2345 ssssss

1

0

11

0

121

121

0

1

2

3

4

5

s

s

s

s

s

s

)double(),double(,15,4,3,2,1 jjp 00

11

044

121

1

2

3

4

s

s

s

s

00

121

121

3

4

5

s

s

s



Nth order system


Determine the stability range of a parameter

• Ex

Application to Control Systems

Y(s)

Kssss

K

sR

sY

)2)(1()(

)(2

Ksssss 233)( 234

007

92

03

7

3

29023

31

1

3032

3

4

Ks

Ks

s

Ks

K

9

140

07

92&0

K

KK


Application to Control Systems

• Ex

Y(s)

Kas

K

KaKKs

KaK

s

Ks

Kas

060

36)6)(60(6

60

66

111

0

1

2

3

4

KasKssss )6(116)( 234

)()3)(2)(1(

)(

)(1

)(

)(

)(

asKssss

asK

sG

sG

sR

sY

K

KKa

K

Ka

K

36

)6)(60(

60

0

6

control systems - chibum lee · pdf file08.03.2014 · chibum lee -seoultech control...

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