LINEAR CONTROL SYSTEMS Ali Karimpour Assistant Professor Ferdowsi University of Mashhad Lecture 10 Lecture 10 Stability analysis Topics to be covered include: v v Stability of linear control systems. u Bounded input.
Download ReportTranscript LINEAR CONTROL SYSTEMS Ali Karimpour Assistant Professor Ferdowsi University of Mashhad Lecture 10 Lecture 10 Stability analysis Topics to be covered include: v v Stability of linear control systems. u Bounded input.
LINEAR CONTROL
SYSTEMS
Ali Karimpour
Assistant Professor
Ferdowsi University of Mashhad
Lecture 10
Lecture 10
Stability analysis
Topics to be covered include:
v
v
Stability of linear control systems.
u
Bounded input bounded output stability (BIBO).
u
Zero input stability.
Stability of linear control systems through Routh
Hurwitz criterion.
2
Ali Karimpour Jun 2010
Lecture 10
Stability analysis
تجزیه تحلیل پایداری
The response of linear systems can always be decomposed as the zerostate response and zero-input response. We study
1. Input output stability of LTI system is called BIBO (bounded-input
bounded-output) stability ( the zero-state response )
2. Internal stability of LTI system is called Asymptotic stability ( the zero-input
response )
پاسخ سیستمهای خطی را می توان بصورت جمع پاسخ حالت صفر و پاسخ ورودی
.صفر بیان نمود
(ورودی کراندارBIBO پایداری ورودی خروجی سیستمهای خطی پایداری-1
) (پاسخ حالت صفر.خروجی کراندار) نامیده می شود
(پاسخ. پایداری داخلی سیستمهای خطی پایداری مجانبی نامیده می شود-2
3
) ورودی صفر
Ali Karimpour Jun 2010
Lecture 10
Input output stability of LTI system
LTI پایداری ورودی خروجی سیستمهای
Consider a SISO linear time-invariant system, then the output
can be described by
t
t
0
0
y(t ) g (t )u( )d g ( )u(t )d
(I )
where g(t)
y(t ) is
g (t the
)u( )dimpulse
g ( )u(t )dresponse of the system and system is
relaxed at t=0.
( خروجی را میLTI) در سیستم تک ورودی تک خروجی خطی غیر متغیر با زمان
توان بصورت
t
t
0
0
t
t
0
0
y(t ) g (t )u( )d g ( )u(t )d
(I )
. آرام استt=0 پاسخ ضربه بوده و سیستم درg(t) نمایش داد که
4
Ali Karimpour Jun 2010
Lecture 10
Input output stability of LTI system
LTI پایداری ورودی خروجی سیستمهای
Definition: A system is said to be BIBO stable (bounded-input
bounded-output) if every bounded input excited a bounded
output. This stability is defined for zero-state response and is
applicable only if the system is initially relaxed.
گویند اگر هر ورودی محدود خروجی محدودBIBO یک سیستم را پایدار:تعریف
این پایداری برای پاسخ حالت صفر تعریف شده و سیستم در ابتدا آرام.را تولید کند
.است
5
Ali Karimpour Jun 2010
Lecture 10
Input output stability of LTI system
LTI پایداری ورودی خروجی سیستمهای
Theorem: A SISO system described by (I) is BIBO stable if
and only if g(t) is absolutely integrable in [0,∞), or
0
g (t ) dt M
For some constant M.
گویند اگرBIBO ( را پایدارI) توصیف شده با معادالتSISO یک سیستم:قضیه
[ انتگرال پذیر باشد یا0,∞) در بازهg(t) و فقط اگر قدر مطلق
0
g (t ) dt M
. عدد ثابت می باشدM
6
Ali Karimpour Jun 2010
Lecture 10
Input output stability of LTI system
LTI پایداری ورودی خروجی سیستمهای
Proof: g(t) is absolutely integrable
system is BIBO
Let u(t ) be bounded so u(t ) um then
y (t )
0
g ( )u (t )d g ( ) u(t ) d um g ( ) d um M
0
0
So the output is bounded.
Now: System is BIBO stable
g(t) is absolutely integrable
If g(t) is not absolutely integrable, then there exists t1 such that:
Let us choose
t1
0
g ( ) d
if g ( ) 0
1
u(t1 )
1
t1
if g ( ) 0
7
y(t1 ) g ( )u(t1 )d g ( ) d So it is not BIBO
Ali Karimpour Jun 2010
0
t1
0
Lecture 10
Input output stability of LTI system
LTI پایداری ورودی خروجی سیستمهای
Theorem: A SISO system with proper rational transfer
function g(s) is BIBO stable if and only if every pole of g(s)
has negative real part.
گویندBIBO را پایدارg(s) با تابع انتقال مناسب گویایSISO یک سیستم:قضیه
. دارای قسمت حقیقی منفی باشدg(s) اگر و فقط اگر هر قطب
8
Ali Karimpour Jun 2010
Lecture 10
Internal stability
پایداری داخلی
The BIBO stability is defined for the zero-state response. Now we
study the stability of the zero-input response.
Definition: The zero-input response of x Ax is stable in the
sense of Lyapunov if every finite initial state x0 excites a
bounded response. In addition if the response approaches to
zero then it is asymptotically stable.
را به مفهوم لیاپانوف پایدار گویند اگرx Ax پاسخ ورودی صفر سیستم:تعریف
عالوه بر این اگر پاسخ به. پاسخ محدودی را بوجود آوردx0 هر حالت اولیه محدود
.صفر میل کند پایداری مجانبی حاصل می شود
9
Ali Karimpour Jun 2010
Lecture 10
Internal stability
پایداری داخلی
Theorem: The equation
x Ax is
asymptotically stable if and only if all
eigenvalues of A have negative real
parts.
پایدار مجانبی است اگر وx Ax معادله:قضیه
دارای قسمت حقیقیA فقط اگر تمام مقادیر ویژه
.منفی باشد
Relation between BIBO stability and
asymptotic stability?
10
Ali Karimpour Jun 2010
Lecture 10
R(s)
Example 1: Discuss the stability of the system .
s 1
s2
C (s )
1
s 1
BIBO stability:
C ( s)
1
G( s)
R( s ) s 2
There is no RHP root , so system is
BIBO stable.
Internal stability:
For internal stability we need
state-space model so we have:
1
x1 ( s )
s 1
s 1
x1 ( s )
R( s)
s2
x2 ( s )
1
x1 ( s) R( s)
x2 ( s )
s 1
3
x1 ( s )
R( s)
s2
R(s)
s 1
s2
x1
1
s 1
x2
C (s )
x2
C (s )
x2 x1 x2
x1 2 x1 r r
?
x2 x1 x2 r
x1 2 x1 3r
R(s)
3
s2
x1
+
+
1
s 1
11
11
Ali Karimpour Jun 2010
Example 1: Discuss the stability of the
system .
BIBO stability:
Lecture 10
R(s)
s 1
s2
C (s )
1
s 1
There is no RHP root , so system is BIBO stable.
Internal stability:
For internal stability we need
state-space model so we have:
x1 2 x1 3r
R(s)
x2 x1 x2 r
x1 2 0 x1 3
x 1 1 x 1 r
2
2
x1
c 0 1
x2
sI - A
3
s2
x1
+
+
1
s 1
x2
C (s )
12
s2 0
( s 1)(s 2)
1 s 1
1, 2
1
2
The system is not internally stable (neither asymptotic nor Lyapunov stable).
Very important note: If RHP poles and zeros between different part of system
12
omitted then the system is internally unstable although it may be BIBO stable.
Ali Karimpour Jun 2010
Lecture 10
مرور
Review
How can we check BIBO stability?
n( s )
G (s)
d (s)
Let d ( s) 0
If p1 , p 2 ,...,p n LHP
Find polesp1 , p 2 ,...,p n
System is BIBO stable
How can we check asymptotic stability?
Let sI A 0
If 1 , 2 ,...,n LHP
Find eigenvalues 1 , 2 ,...,n
System is asymptotically stable
For both kind of stability we need to compute the zero of some polynomial
13
Ali Karimpour Jun 2010
Lecture 10
Different regions in S plane
S نواحی مختلف در صفحه
LHP plane
Stable
RHP plane
Unstable
14
Ali Karimpour Jun 2010
Lecture 10
Stability and Polynomial Analysis
پایداری و تجزیه تحلیل چند جمله ای ها
Consider a polynomial of the following form:
The problem to be studied deals with the question of whether
that polynomial has any root in RHP or on the jw axis.
jw و یا روی محورRHP مساله این است که آیا چند جمله ای فوق ریشه ای در
.دارد و یا خیر
15
Ali Karimpour Jun 2010
Lecture 10
Some Polynomial Properties of Special Interest
چند خاصیت جالب چند جمله ای ها
Property 1: The coefficient an-1 satisfies
Property 2: The coefficient a0 satisfies
Property 3: If all roots of p(s) have negative real parts, it is necessary
that ai > 0, i {0, 1, …, n-1}.
Property 4: If any of the polynomial coefficients is nonpositive
(negative or zero), then, one or more of the roots have nonnegative
real plant.
16
Ali Karimpour Jun 2010
Lecture 10
Routh Hurwitz Algorithm
آلگوریتم روت هرویتز
The Routh Hurwitz algorithm is based on the following
numerical table.
sn
1
s n 1
s n2
an 1
an2
an 3
an4
an 5
2,1
2, 2
2,3
..............
..............
..............
s n 3
.
.
.
3,1
3, 2
3, 3
..............
s0
n,1
Routh’s table
17
Ali Karimpour Jun 2010
Lecture 10
Routh Hurwitz Algorithm
آلگوریتم روت هرویتز
2,1
2, 2
2,3
..............
..............
..............
s n 3
.
.
.
3,1
3, 2
3, 3
..............
s0
n,1
sn
1
s n 1
s n2
an 1
an2
an 3
an4
an 5
Routh’s table
a a a 1
2,1 n2 n1 n3
an 1
2, 2
an 4 an 1 an 51
an 1
2,3
an 6 an 1 an 71
an 1
18
Ali Karimpour Jun 2010
Lecture 10
Result
نتیجه
Consider a polynomial p(s) and its associated table.
Then the number of roots in RHP is equal to the number
of sign changes in the first column of the table.
تعداد ریشه های. و جدول متناظر آن را در نظر بگیریدp(s) چند جمله ای
. برابر با تعداد تغییر عالمت در ستون اول جدول استRHP واقع در
19
Ali Karimpour Jun 2010
Lecture 10
Routh Hurwitz Algorithm
آلگوریتم روت هرویتز
2,1
2, 2
2,3
..............
..............
..............
3,1
3, 2
3, 3
..............
sn
1
s n 1
s n2
an 1
s
n 3
an2
an 3
an4
an 5
.
.
.
s0
n,1
Routh’s table
Number of sign changes=number of roots in RHP
20
Ali Karimpour Jun 2010
Lecture 10
Example 1: Check the number of zeros in the RHP
. سیستم زیر را تعیین کنیدRHP تعداد صفر:1 مثال
p(s) 2s 4 s3 3s 2 5s 10 0
s4
2
3
10
s3
1
5
0
s2
7
45
7
10
10
0
0
0
0
0
s1
s0
Two roots in RHP
21
Ali Karimpour Jun 2010
Lecture 10
Routh Hurwitz special cases
حاالت خاص روت هرویتز
Routh Hurwitz special cases
1- The first element of a row is zero. (see example 2)
2- All elements of a row are zero. (see example 3)
22
Ali Karimpour Jun 2010
Lecture 10
Example 2: Check the number of zeros in the RHP
. سیستم زیر را تعیین کنیدRHP تعداد صفر:2 مثال
p(s) s 4 s3 2s 2 2s 3 0
s4
s3
s2
s1
s0
1
1
0
2 3 3
3
2
3
2
3
0
0
0
0
0
0
Two roots in RHP
for any
23
Ali Karimpour Jun 2010
Lecture 10
Example 3: Check the number of zeros in the RHP
. سیستم زیر را تعیین کنیدRHP تعداد صفر:3 مثال
p(s) s5 4s 4 8s3 8s 2 7s 4 0
s5
1
8
7
s4
4
8
4
s3
s2
s1
s1
s0
6
4
0
8
4
6
4
0
0
0
0
0
0
0
0
q( s ) 4s 2 4 0
q(s) 8s 0
No RHP roots + two roots on imaginary axis
24
Ali Karimpour Jun 2010
Example 4: Check the stability of following system
for different values of k
Lecture 10
. بررسی کنیدk پایداری سیستم زیر را بر حسب مقادیر:4 مثال
+
-
k
s(3s 1)
3
ks(3s 1)
s
2s 4
M ( s)
3
s(3s 1)
s
2s 4 ks(3s 1)
1 k 3
s 2s 4
s(3s 1)
s 3 2s 4
k
To check the stability we must check the RHP roots of
s3 3ks2 (2 k )s 4 0
s3
s
s
2
1
s0
1
2k
3k
3k (2 k ) 4
3k
4
4
0
3k 0
3k 6k 4
0
3k
2
We need k>0.528
for stability
0
25
Ali Karimpour Jun 2010
Lecture 10
Example 5: Check the BIBO and internal stability of the
following system.
. پایداری ورودی خروجی و پایداری داخلی سیستم زیر را تحقیق کنید5 مثال
r +
-
1
s 1
x2
s 1
s
x1
c
BIBO stability 1
c( s )
1
s
r ( s) 1 1 s 1
s
p 1
We have BIBO stability
Internal stability
x2 x2 r x1
x1 x2 x2
s 1
sI A
1
x1 1 0 x1 1
x 1 1 x 1 r
2
2
x1 x1 r
x2 x1 x2 r
0
s 2 1 0
s 1
1 1
2 1
We have not
26
Internal stability
Ali Karimpour Jun 2010
Example 6: The block Diagram of a control system is depicted in the Lecture 10
following figure. Find the region in K-α plane concluding the system stable.
ناحیه ای در.بلوک دیاگرام یک سیستم کنترل در شکل زیرنشان داده شده است: 6 مثال
. به دست آورید که سیستم پایدار باشدK-α صفحه
R(s) +
-
s
s
K ( s 2)
s2 1
Y(s)
27
Ali Karimpour Jun 2010
Lecture 10
Example 6: The block Diagram of a control system is depicted in the
following figure. Find the region in K-α plane concluding the system stable.
R(s) +
-
s
s
K ( s 2)
s2 1
C(s)
28
Ali Karimpour Jun 2010
Lecture 10
Exercises
1- Check the internal stability of following system.
2
x 0
1
y [1 2
3 0
3 2 x 1 u
4 1 0
0] x
1
2- a) Check the internal stability of following system.
3
b) Check the BIBO stability of following system.
6 4 6
x 1
2
2 x 3u
1 6 6 4
y [2 2 1]x
3- Are the real parts of all roots of following system less than -1.
p(s) s 4s 5s 6s 2s 4
5
4
3
2
29
Ali Karimpour Jun 2010
Lecture 10
Exercises (Cont.)
4- Check the internal stability of
following system versus k.
+
10
s( s 10)(s 5)
k
-
5- a)Check the BIBO stability of following system.
b) Check the internal stability of following system.
+
-
1
s 1
x2
s 1 x1
s ( s 10)
6- The eigenvalues of a system are -3,4,-5 and the poles of its transfer
function are -3 and -5.(Midterm spring 2008)
a) Check the BIBO stability of following system.
30
b) Check the internal stability of following system.
Ali Karimpour Jun 2010
Lecture 10
Exercises (Cont.)
7) The open-loop transfer function of a control system
with negative unit feedback is:
K ( s 2)
G( s) H ( s)
s(1 Ts)(1 2s)
Find the region in K-T plane concluding the system stable.
Answer:
Unstable
K
Unstable
Stable
0.33
0.66
Unstable
T
31
Ali Karimpour Jun 2010