Lecture introduction to control systems chapter 4 system stability analysis (dr huynh thai hoang)

Routh’s stability criterion stability criterion Routh’s Routh’s criterion statement criterion statement æ The necessary and sufficient condition for a system to be stable isthat all the

Lecture Notes

Introduction to Control Systems

Instructor: Dr Huynh Thai Hoang Department of Automatic Control Faculty of Electrical & Electronics Engineering

Ho Chi Minh City University of Technology

Email: hthoang@hcmut.edu.vn

huynhthaihoang@yahoo.com Homepage: www4.hcmut.edu.vn/~hthoang/

Chapter 4

SYSTEM STABILITY ANALYSIS SYSTEM STABILITY ANALYSIS

æ Root locus method

Ø Root locus definition

Ø Rules for drawing root loci

Ø Stability analysis using root locus

æ Frequency response analysis

æ Frequency response analysis

Ø Bode criterion

Ø Nyquist’s stability criterionNyquist s stability criterion

Stability concept

BIBO stability

æ A system is defined to be BIBO stable if every bounded input to the

æ A system is defined to be BIBO stable if every bounded input to thesystem results in a bounded output over the time interval [t0,+∞) forall initial times t0

System

Stable system System at Instable system

stability boundary

Poles and zeros

n n

m m

m m

b s b

s b s

b U

s

Y s

− 1

1

1 1

0

)(

)

()

æ Consider a system described by the transfer function (TF):

n n

n n

a s

a s

a s

a s

A = + −1 + + −1 +

1 0

)

n n

n n

a s

a s

a s

a s

1

1 1 0

)(

)(

m m

m m

b s b

s b s

b s

B = + −1 + + −1 +

1 0

has m zeros denoted as z i =1 2 m

æ Poles: are the roots of the denominator of the transfer function, i.e.the roots of the equation A(s) = 0 Since A(s) is of order n, the

has m zeros denoted as z i, i =1,2,…m.

Pole

Pole – – zero plot zero plot

æ Pole zero plot is a graph which represents the position of poles and

æ Pole – zero plot is a graph which represents the position of poles andzeros in the complex s-plane

PoleZero

Stability analysis in the complex plane

æ The stability of a system depends on the location of its poles

æ If all the poles of the system lie in the left-half s-plane then the

i blsystem is stable

æ If any of the poles of the system lie in the right-half s-plane then thesystem is unstable

system is unstable

æ If some of the poles of the system lie in the imaginary axis and theothers lie in the left-half s-plane then the system is at the stabilityboundary

Characteristic equation

æ Ch t i ti ti i th ti A( ) 0

æ Characteristic equation: is the equation A(s) = 0

æ Characteristic polynomial: is the denominator A(s)

(

)()

()

(

t t

y

t u t

t

Cx

B Ax

x

Y fb (s)

0)

()(

1 G

Characteristic equation Characteristic equation

0)

()(

1+ G s H s = det(s I − A) = 0

Algebraic stability criteria

æ Th diti f li t t b t bl i th t ll

Necessary condition

æ The necessary condition for a linear system to be stable is that allthe coefficients of the characteristic equation of the system must bepositive

æ Example: Consider the systems which have the characteristicequations:

01

4

UnstableUnstable0

12

Routh’s stability criterion stability criterion

Rules for forming the Rules for forming the Routh Routh table table

æ Consider a linear system whose characteristic function is:

Rules for forming the Rules for forming the Routh Routh table table

01

1 1

a s

a s

a s

a0s + a1s + + a n−1s + a n 0

æ In order to analysis the system stability using Routh’s criterion, it isnecessary to form the Routh table according to the rules below:ecessa y to o t e out tab e acco d g to t e u es be ow:

Ø The Routh table has n+1 rows

Ø The 1st row consist of the even-index coefficients

Ø The 2nd row consist of the odd-index coefficients

Ø The element at row i th column j th (i ≥ 3) is calculated as:

1 , 1 1

−

= c i

α

Routh’s stability criterion stability criterion

R th

R th t bl t bl Routh

Routh table table

1 , 1 1

−

= c i

α

1 , 1

Routh’s stability criterion stability criterion

Routh’s Routh’s criterion statement criterion statement

æ The necessary and sufficient condition for a system to be stable isthat all the coefficients of the characteristic equation are positive and

Routh s Routh s criterion statement criterion statement

that all the coefficients of the characteristic equation are positive andall terms in the first column of the Routh table have positive signs

æ The number of sign changes in the first column of the Routh table isequal the number of roots lying in the right-half s-plane

Routh’s stability criterion stability criterion – – Example 1 Example 1

æ Analyze the stability of the system which have the following

æ Analyze the stability of the system which have the followingcharacteristic equation:

01

25

4 + s + s + s + =

s

æ Solution: Routh table

æ Conclusion: The system is stable because all the terms in the first

æ Conclusion: The system is stable because all the terms in the firstcolumn are positive

Routh’s stability criterion stability criterion – – Example 2 Example 2

æ Analyze the system described by the following block diagram:

æ Analyze the system described by the following block diagram:

) 5 )(

3 (

50 )

+ + +

=

s s

s s

s G

Y (s)

R (s)

) 5 )(

3 (s + s + s +

s

2

1 )

æ Solution: The characteristic equation of the system:

0 )

( ).

) 5 )(

3 (

+ +

+ +

+

s s

s s

s

⇔

0 50

) 2 )(

5 )(

3 ( s + s2 + s + s + + =

s

⇔

Routh’s stability criterion stability criterion – – Example 3 Example 3

æ Find the condition of K for the following system to be stable

æ Find the condition of K for the following system to be stable

) 2 )(

1 (

)

+ +

+

=

s s

s s

K s

G

æ Solution: The characteristic equation of the system is:q y

0)

2)(

1(

++

+

+

s s

s s

02

Routh’s stability criterion stability criterion – – Special case #1 Special case #1

æ If a first-column term in any row is zero, but the remaining terms inthat row are not zero or there is no remaining term, then the zeroterm is replaced by a very small positive numberp y y p ε and the rest rows

of the Routh table is calculated as the normal case

Routh’s stability criterion stability criterion – – Example 4 Example 4

æ Analyze the stability of the system whose characteristic equation is:

æ Analyze the stability of the system whose characteristic equation is:

æ Solution:

0 3

8 4

Routh’s stability criterion stability criterion – – Special case #2 Special case #2

æ If all the coefficients in any row are zero:

æ If all the coefficients in any row are zero:

Ø Forming an auxiliary polynomial with coefficients of the last rowabove the “all-zero-term row”, denote the auxiliary polynomialy p y

as A0(s).

Ø Replace the “all-zero-term row” by another row whose elementsare the coefficients of the derivative dA ( )/d

are the coefficients of the derivative dA0(s)/ds.

Ø Then continue to calculate the Routh table as the normal case

æ Note: The roots of A0(s) are also the roots of characteristicequation

Routh’s stability criterion stability criterion – – Example 5 Example 5

æ Analyze the stability of the system whose characteristic equation is:

æ Analyze the stability of the system whose characteristic equation is:

0 4 7

8 8

Routh’s stability criterion stability criterion – – Example 5 (cont’) Example 5 (cont’)

æ The auxiliary polynomial:

æ The auxiliary polynomial:

4 4

⇒

æ The roots of the auxiliary polynomial (are also the roots thecharacteristic equation):

0 4 4

Ø The characteristic equation has two roots lying in the imaginary

Ø The characteristic equation has two roots lying in the imaginary axis

Ø The number of roots lying in the left-half s-plane is 5 – 2 = 3

Hurwitz’s stability criterion

Rules for forming the Hurwitz matrix

æ Given a system whose characteristic equation is:

Rules for forming the Hurwitz matrix

01

1 1

a s

a s

a s

a0s + a1s + + a n−1s + a n = 0

æ In order to analysis the system stability using Hurwitz’s criterion, it

is necessary to form the Hurwitz matrix according to the rules belowy g

Ø The Hurwitz matrix is a square matrix of order n×n.

coefficients of the characteristic polynomial; the indexes increase

on the right and decrease on the left of the diagonal

on the right and decrease on the left of the diagonal

Ø The even row of the Hurwitz matrix consists of the even-indexedcoefficients of the characteristic polynomial; the indexes increase

on the right and decrease on the left of the diagonal

Hurwitz’s stability criterion

a

a a

a a

KK

00

00

0

5 3

1

6 4

2 0

a

MM

MM

M

K0

0

0 0 2 4

⎦

⎣ 0 K K K K a n

Hurwitz’s criterion statement

that all the determinants of the principal sub-matrices of the Hurwitz

Hurwitz’s stability criterion

Hurwitz’s stability criterion – – Example 1 Example 1

æ Analyze the stability of the system whose characteristic equation is:

æ Analyze the stability of the system whose characteristic equation is:

0 2 3

0 3 1

0 2 4

0

0 0

3 1

2 0

3 1

a a

a a

a a

1 3 4

2 4 3

1

Δ a a

The determinants:

10 2

1 3

4 3

1 2

0

3 1

Δ

a a

2 4

0

3 1

3 1

a a

a a

20 10

2 3

1

2 0

0

2 0

3 1

3 3

1

2 0

Δ

a a

a a

a

a a

æ Conclusion: The system is stable because all the determinants are

æ Conclusion: The system is stable because all the determinants arepositive

Hurwitz’s stability criterion

Hurwitz’s stability criterion – – Some corollaries Some corollaries

æ A 2 nd order system is stable if the coefficients of the characteristic

æ A 2 order system is stable if the coefficients of the characteristic polynomial satisfy the conditions:

2,0

,0

3 0 2

1a a a a

,0

3 0 2

1a a a a

i

a i

⎪

The root locus method

The concept of root locus (RL)

æ Example: Plot of all the roots of the following characteristic

æ Example: Plot of all the roots of the following characteristicequation when K changes from 0 → +∞

04

2 + s + K =

s

æ Definition: Root locus is the set of all the roots of the characteristic

Magnitude and phase condition of the root locus

æ In order to apply the rules for construction of the root locus first we

æ In order to apply the rules for construction of the root locus, first wehave to equivalently transform the characteristic equation tostandard form:

)

(s

N

0)

)

()

(0

s D

s

N K s

Denote:

where K is the changing parameter

)(

0 )

)

12

()(

G

⇔

Rules for construction of the root locus

characteristic equation = number of poles of G0(s) = n.

Ø For K = 0: the root locus begins at the poles of G0(s).

Ø As K goes to +∞ : m branches of the root locus end at m zeros of

Ø As K goes to +∞ : m branches of the root locus end at m zeros of

G0(s), the n−m remaining branches go to infinity approaching theasymptote defined by the rule 5 and rule 6

æ Rule 3: The root locus is symmetric with respect to the real axis

æ Rule 4: A point on the real axis belongs to the root locus if the total

Rules for construction of the root locus (cont’)

m i

p m

n

i i

æ Rule 7: : Breakaway / break-in points (or break points for short), if

any, are located in the real axis and are satisfied the equation:

Rules for construction of the root locus (cont’)

æ Rule 8: The intersections of the root locus with the imaginary axiscan be determined by using the Routh-Hurwitz criteria or bysubstituting s=jω into the characteristic equation

æ Rule 9: The departure angle of the root locus from a pole p j (of

æ Rule 9: The departure angle of the root locus from a pole p j (ofmultiplicity 1) is given by:

i j

i

i j

1 1

) arg(

) arg(

180

θ

The geometric form of the above formula is:g

θj = 1800 + (∑angle from the zero zi (i=1 m) to the pole p j )

− (∑ angle from the poles p( g p p i i (i=1 m, i( ≠j) to the pole p j) p p j j ))

The root locus method

The root locus method – – Example 1 Example 1

2(

)(

++

=

s s

s

K s

G

æ Solution:

æ The characteristic equation of the system:

0)

3)(

2(

++

+

s s

s

æ Zeros: none

The root locus method

The root locus method – – Example 1 (cont’) Example 1 (cont’)

æ The asymptotes:

æ The asymptotes:

)1(

0)(

3

)12

()

12

π α

π

π α

1)(

)1(

3

0

3

)(

)(

n

π α

α α

3

50

3

0)]

3()2(0[zero

3( 2 + +

The root locus method

The root locus method – – Example 1 (cont’) Example 1 (cont’)

æ The intersections of the root locus with the imaginary axis:

æ The intersections of the root locus with the imaginary axis:

Method 1: Using the Hurwitz’s criterion

65

1

j s

s

⇔

⎪

⎩s3 = − j 6

The root locus method

The root locus method – – Example 1 (cont’) Example 1 (cont’)

6

K

ω

The root locus method

The root locus method – – Example 1 (cont’) Example 1 (cont’)

The root locus method

The root locus method – – Example 2 Example 2

(

)

++

=

s s

s

s G

æ Solution:

æ The characteristic equation of the system:

0)

(

)208

(

++

+

s s

s

æ Poles: p1 = 0 p2,3 = − 4 ± j 2

The root locus method

The root locus method – – Example 2 (cont’) Example 2 (cont’)

π

æ The asymptotes:

)1(

0)(

3

)12

()

12

π α

π

π α

1)(

)1(

3

0

n

π α

α α

∑

∑

3

8 0

3

) 0 ( )]

2 4

( )

2 4

( 0 [ zero

æ The break points:

s s

0

dK

Th ⎨⎧s1 = −3.331 ((2 break points2 8 20) = 0

++

+

s s

1

s

(2 break points

accepted)

The root locus method

The root locus method – – Example 2 (cont’) Example 2 (cont’)

æ The intersections of the root locus with the imaginary axis:

æ The intersections of the root locus with the imaginary axis:

Substitute s=jω into the equation (2):

0)

(20)

(8)

The root locus method

The root locus method – – Example 2 (cont’) Example 2 (cont’)

æ The departure angle of the root locus from the pole p2

)]

()

The root locus method

The root locus method – – Example 2 (cont’) Example 2 (cont’)

The root locus method

The root locus method – – Example 3 Example 3

æ Sketch the root locus of the system shown below when K=0→+∞

æ Sketch the root locus of the system shown below when K=0→+∞.

)1

()

)208

)(

3(

)1

()

++

+

+

=

s s

s s

s

K s

(

)208

)(

3(

)1

(

++

+

++

s s

s s

s K

)208

)(

3(s + s + s +

s

æ Poles: p1 = 0 p2 = − 3 p3,4 = − 4 ± j 2

æ Zeros: z1 = − 1

The root locus method

The root locus method – – Example 3 (cont’) Example 3 (cont’)

ỉ The asymptotes:

ỉ The asymptotes:

) 1 (

0) (

3

) 1 2

( )

1 2

πα

ππ

1) (

) 1 (

3

1

4

) (

) (

n

πα

αα

3

10 1

4

) 1 ( )]

2 4

( ) 2 4

( ) 3 ( 0 [

− +

OA cực

ỉ The break points:

(1) ⇔

) 1 (

) 20 8

)(

3 ( + 2 + +

−

= s s s s

) 1 (

60 88

77 26

−

d dK

0

)1

(

( )

) 1

s

⇔ ⎨ = − 1+± (rejected) (rejected) = 0

The root locus method

The root locus method – – Example 3 (cont’) Example 3 (cont’)

æ The intersections of the root locus with the imaginary axis:

æ The intersections of the root locus with the imaginary axis:

(1) ⇔ s4 + 11s3 + 44s2 + ( 60 + K)s + K = 0 (2)

S b tit t j i t th ti (2)

Substitute s=jω into the equation (2):

0 )

60 ( 44

+

−

= +

−

0 )

60 ( 11

0 44

3

2 4

ωω

893 , 5

K

ω

⎧

0)

208

)(

3(

)1

314 ,

)(

3(s + s2 + s +

s

Then the intersections are: s = ± j5 , 893 Critical gain: K cr = 322

The root locus method

The root locus method – – Example 3 (cont’) Example 3 (cont’)

æ The departure angle of the root locus from the pole p3

)(

)906

,1164

,153(

3,146

The root locus method

The root locus method – – Example 3 (cont’) Example 3 (cont’)

The root locus method

The root locus method – – Example 4 (cont’) Example 4 (cont’)

æ Given the system below:

æ Given the system below:

)39

(

10)

(s = 2

G

)39

K

s

K K

The root locus method

The root locus method – – Example 4 (cont’) Example 4 (cont’)

S l i

æ Solution:

æ The characteristic equation of the system:

0)

()(

1+ G C s G s =

0

107

10

⇔

03

3)(

9(

++

+

s s

⇔

æ Poles: p = − 9 p2 = + j 3 p3 = − j 3

æ Zeros: z1 = 0

æ Poles: p1 = 9 p2 + j 3 p3 = j 3

The root locus method

The root locus method – – Example 4 (cont’) Example 4 (cont’)

æ The asymptotes:

0) (l

2

/ )

1 2

( )

1 2

2

/

1

α

m n

9)

0()]

3(

)3(

9[

3

)()]

()(

2

s s

The root locus has two break points at the same location 3

The root locus has two break points at the same location −3