Lecture "Fundamentals of control systems - Chapter 4: System stability analysis system stability analysis" presentation of content: Stability concept, algebraic stability criteria, root locus method, frequency response analysis.
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Nội dung Text: Lecture Fundamentals of control systems: Chapter 4 - TS. Huỳnh Thái Hoàng
- Lecture Notes
Introduction of Control Systems
Instructor: Assoc. Prof. 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/
6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 1
- Chapter 4
SYSTEM STABILITY ANALYSIS
6 December 2013 © H. T. Hoàng - ÐHBK TPHCM 2
- Content
Stability concept
Algebraic stability criteria
Necessary y condition
Routh’s criterion
Hurwitz’s criterion
Root locus method
Root locus definition
Rules
R l for f drawing
d i root locil i
Stability analysis using root locus
Frequency response analysis
Bode criterion
Nyquist
Nyquist’ss stability criterion
6 December 2013 © H. T. Hoàng - ÐHBK TPHCM 3
- Stability concept
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 4
- BIBO stability
A system is defined to be BIBO stable if every bounded input
to the system results in a bounded output over the time
interval [t0,+∞) for all initial times t0.
u(t) y(t)
System
y(t) y(t) y(t)
Stable system System at Unstable
stability boundary system
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 5
- Poles and zeros
C
Consider
id a system
t d
described
ib d by
b the
th transfer
t f function
f ti (TF):
(TF)
Y ( s ) b0 s m b1s m 1 bm 1s bm
G(s)
U ( s ) a0 s n a1s n 1 an 1s an
Denote: A( s ) a0 s n a1s n1 an1s an ((TF’s denominator))
B ( s ) b0 s m b1s m1 bm1s bm (TF’ numerator)
Poles:
P l are theh roots off the
h denominator
d i off the
h transfer
f
function, i.e. the roots of the equation A(s) = 0. Since A(s) is
of order n,, the system
y has n ppoles denoted as pi , i =1,2,…n.
, ,
Zeros: are the roots of the numerator of the transfer function,
i.e. the roots of the equation B(s) = 0. Since B(s) is of order m,
the system has m zeros denoted as zi, i =1,2,…m.
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 6
- Pole – zero plot
Pole – zero plot is a graph which represents the position of
poles and zeros in the complex s-plane.
Pole
Zero
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 7
- Stability analysis in the complex plane
The stability of a system depends on the location of its poles.
If all the poles of the system lie in the left-half s-plane then the
system
t i stable.
is t bl
If any of the poles of the system lie in the right-half s-plane
then the system is unstable.
unstable
If some of the poles of the system lie in the imaginary axis
and the others lie in the left
left-half
half ss-plane
plane then the system is at
the stability boundary.
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 8
- Characteristic equation
Characteristic
Ch t i ti equation:
ti i the
is th equation
ti A(s)
A( ) = 0
Characteristic polynomial: is the denominator A(s)
Note:
Feedback systems Systems described
R(s) Y(s) by state equations
+_ G(s)
x (t ) Ax(t ) Bu (t )
H(s) y (t ) Cx(t )
Characteristic equation Characteristic equation
1 G(s) H (s) 0 det sI A 0
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 9
- Algebraic stability criteria
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 10
- Necessary condition
The necessary condition
Th diti f a linear
for li system
t t be
to b stable
t bl is
i
that all the coefficients of the characteristic equation of the
system must be positive.
Example: Consider the systems which have the characteristic
equations:
s 3 3s 2 2 s 1 0 Unstable
s 4 2 s 2 5s 3 0 Unstable
s 4 4 s 3 5s 2 2 s 1 0 Cannot conclude about the stability
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 11
- Routh’s stability criterion
Rules for forming the Routh table
Consider a linear system whose characteristic function is:
a0 s n a1s n1 an1s an 0
To analyze the system stability using Routh’s criterion, it is
necessary to form the Routh table according to the rules:
The Routh table has n+1 rows.
The 1st row consists of the even-indexed coefficients.
The 2nd row consists of the odd-indexed coefficients.
The element at row ith column jth (i 3) is calculated as:
cij ci 2, j 1 i .ci 1, j 1
ci 2,1
with
ith i
ci 1,1
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 12
- Routh’s stability criterion
Routh table
cij ci 2, j 1 i .ci 1, j 1
ci 2,1
i
ci 1,1
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 13
- Routh’s stability criterion
Routh’s
Routh s criterion statement
The necessary and sufficient condition for a system to be
stable is that all the coefficients of the characteristic equation
are positive and all terms in the first column of the Routh table
have positive signs.
The number of sign changes in the first column of the Routh
table is equal the number of roots lying in the right-half s-
plane.
plane
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 14
- Routh’s stability criterion – Example 1
Analyze the stability of the system which have the following
characteristic equation: s 4 4 s 3 5s 2 2 s 1 0
Solution: Routh table
Conclusion: The system is stable because all the terms in the
first column are positive.
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 15
- Routh’s stability criterion – Example 2
Analyze the system described by the following block diagram:
R(s) Y(s) 50
+_
_ G(s)
( ) G ( s)
s ( s 3)( s 2 s 5)
1
H (s)
H(s)
( ) s2
Solution: The characteristic equation of the system:
1 G ( s ).H ( s ) 0
50 1
1 . 0
s ( s 3)( s s 5) ( s 2)
2
s ( s 3)( s 2 s 5)( s 2) 50 0
s 5 6 s 4 16 s 3 31s 2 30 s 50 0
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 16
- Routh’s stability criterion – Example 2 (cont’)
Routh table
Conclusion: The system is unstable because the terms in the
g their signs
first column change g two times. The characteristic
equation has two roots with positive real parts.
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 17
- Routh’s stability criterion – Example 3
Find the condition of K for the following system to be stable.
stable
R(s)
( ) Y(s)
( )
+ G( )
G(s) K
G(s)
s ( s 2 s 1)( s 2)
Solution: The characteristic equation
q of the system
y is:
1 G(s) 0
K
1 0
s ( s s 1)( s 2)
2
s 4 3s 3 3s 2 2 s K 0
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 18
- Routh’s stability criterion – Example 3 (cont’)
Routh table
The necessary & sufficient condition for the system to be stable:
9
2 K 0 14
7 0K
K 0 9
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 19
- Routh’s stability criterion – Special case #1
If a first-column term in any row is zero, but the remaining
terms in that row are not zero or there is no remaining term,
then the zero term is replaced
p byy a veryy small p positive
number and the rest rows of the Routh table is calculated as
the normal case.
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 20
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