ees42042 fundamental of control systems stability...

EES42042 Fundamental of Control Systems

Stability Criterion – Routh Hurwitz

DR. Ir. Wahidin Wahab M.Sc.Ir. Aries Subiantoro M.Sc.

2

StabilityA system is stable if for a finite input the output is similarly finiteA system which is stable must have ALL its poles in the left half of the s-plane

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Figure 6.1Closed-loop polesand response:a. stable system;b. unstable system

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Figure 6.2Common causeof problems in finding closed-loop poles:a. original system;b. equivalent system

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Figure 6.3Equivalent closed-looptransfer function

6Routh-Hurwitz Stability Criterion

This is a means of detecting unstable poles from the denominator polynomial of a t.f. without actually calculating the roots.Write the denominator polynomial in the followingform and equate to zero - This is the characteristic equation.

Note that i.e. remove any zero root

a s a s a s a s aa

n n nn n

n

0 11

22

1 00

+ + + + + =≠

− −−.....

7Routh-Hurwitz Stability Criterion

If any of the coefficients is zero or negative in the presence of at least one positive coefficient thereare imaginary roots or roots in the right half planei.e. unstable roots

8Routh-Hurwitz Stability Criterion

if all coefficients are + ve form the Routh Array

s a a a a

s a a a a

s b b b b

s c c c c

s e e

s f

s g

n

n

n

n

0 2 4 61

1 3 5 71

1 2 3 42

1 2 3 4

21 2

11

01

.....

.....

.....

.......

−

−

−

9Routh-Hurwitz Stability Criterion

b a a a aa

b a a a aa

b a a a aa

11 2 0 3

1

21 4 0 5

1

31 6 0 7

1

=−

=−

=−

10Routh-Hurwitz Stability Criterion

c b a a bb

c b a a bb

c b a a bb

11 3 1 2

1

21 5 1 3

1

31 7 1 4

1

=−

=−

=−

11Routh-Hurwitz Stability Criterion

This process is continued until the nth row is completedThe number of roots of the characteristic lying in the right half of the s - plane (unstable roots) is equal to the numbe rof sign changes in the first column of the Routharray.

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Table 6.1Initial layout for Routh table

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Table 6.2Completed Routh table

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Figure 6.4a. Feedback system forExample 6.1;b. equivalentclosed-loop system

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Table 6.3Completed Routh table for Example 6.1

16

Example 2Determine if the following polynomial has roots in the right half of the s - plane

First two rows of Routh array formed from coefficientss s s s

s

s

4 3 2

4

3

2 3 4 5 0

1 3 5

2 4

+ + + + =

17

Example 2Form next row

s

s

s

4

3

2

1 3 5

2 4

1 5

1 2 3 1 42

=× − ×

5 2 5 1 02

=× − ×

18

Example 2Form next row

s

s

s

s

4

3

2

1

1 3 5

2 4

1 5

6−

− =× − ×6 1 4 2 5

1

19

Example 2Form next row

s

s

s

s

s

4

3

2

1

0

1 3 5

2 4

1 5

6

5

−

5 6 5 1 06

=− × − ×

−

Note two sign changes therefore two roots in RHP

20

Example 3

Apply Routh's criterion to the following polynomialto determine the condition for the existence of stable roots

a s a s a s a03

12

2 3 0+ + + =

21

Example 3

30

1

30211

312

203

322

13

0

ArrayRouth 0

as

aaaaas

aas

aas

asasasa

−

→=+++

22

Example 3

assuming all coefficients are positive the conditionfor stable roots is thata a a a1 2 0 3>

23

Routh Array - Special CasesCase of a zero in the 1st column

For example

Routh Arrays s s

s

ss

3 2

3

2

2 2 0

1 1

2 20

+ + + =

This presents a problem

when we come to obtain the

4th row - divide by zero

24

Routh Array - Special CasesCase of a zero in the 1st column

Define a small + ve number and evaluate whole arrayRouth Array

ε

ε

s

ss

3

2

1 1

2 2

1 2

Note no sign change indicating roots on imaginary axis

25

Routh Array - Special CasesCase of a zero in the 1st column

2

23

2031

arrayRouth 23

polynomial heConsider t

0

1

2

3

3

s

s

ss

ss

ε

ε

−−

≈

−

+−

Two sign changes therefore two

RHP roots

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Table 6.4Completed Routh table for Example 6.2

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Table 6.5Determining signs in first column of a Routh table with zero as first element in a row

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Table 6.6Routh table for Example 6.3

29

Routh Array - Special CasesCase of a row of zeros– roots of equal magnitude but opposite signs or

two conjugate imaginary rootss s s s s

s

s

s

5 4 3 2

5

4

3

2 24 48 25 50 0

1 24 25

2 48 50

0 0

+ + + − − =

−

−

Auxiliary eqn.

2 48 50 04 2s s+ − =

30

Routh Array - Special CasesCase of a row of zeros

P s s s

P s s sP s

( )

( )( )

= + − =

′ = +′

2 48 50 0

8 96

4 2

3

replace row with coeficients of

31

Routh Array - Special CasesCase of a row of zeros

s s s s s

s

s

s

s

s

s

5 4 3 2

5

4

3

2

1

0

2 24 48 25 50 0

1 24 25

2 48 50

8 96

24 50

112 7 0

50

+ + + − − =

−

−

−

−

.One sign change one root with +ve

real part

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Table 6.7Routh table for Example 6.4

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Figure 6.5Root positionsto generate evenpolynomials:A , B, C,or any combination

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Table 6.8Routh table for Example 6.5

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Table 6.9 Summary of pole locations for Example 6.5

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Figure 6.8Feedbackcontrol systemfor Example 6.8

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Table 6.13Routh table for Example 6.8

Control Systems Engineering, Fourth Edition by Norman S. NiseCopyright © 2004 by John Wiley & Sons. All rights reserved.

Table 6.14 Summary of pole locations for Example 6.5

39

Use of Routh TestRouth test only tells us whether or not a system is stabledoes not give the DEGREE of stabilityneed to have closed loop characteristic equation– would be more convenient to work from open

loop t.f.