Automatic control
2. Analysis
Lesson 6
(absolute) Stability
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Stability
• Internal behavior
– The effect of all characteristic roots.
• External behavior
– The effect by cancellation of some transfer
function poles.
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Definition :
A system is internal (asymptotic) stable, if the zero-input
response decays to zero, as time approaches infinity, for
all possible initial conditions.
Asymptotic stable =>All the characteristic polynomial
roots are located in the LHP (left-half-plan)
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Definition :
A system is external (bounded-input, bounded-output)
stable, if the zero-state response is bounded, as time
approaches infinity, for all bounded inputs..
bounded-input, bounded-output stable =>All the poles
of transfer function are located in the LHP (left-halfplan)
Asymptotic stable => BIBO stable
BIBO stable=> Asymptotic stable
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System response
(i) First order system response
(ii) Second order system response
(iii) High order system response
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First order
dy
 a0 y  b0 r
dt
b0
1
Y (s) 
R( s) 
y ( 0)
s  a0
s  a0
let
r (t )  Au (t )
b0
b0
A
A
a0
a0
1
Y (s) 


y (0)
s
s  a0 s  a0
Ab0
Ab0  a0t
y (t ) 
u (t ) 
e  y ( 0) e  a 0 t
a0
a0
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Second order
d2y
dy
dr
 a1
 a0 y  b1  b0 r
2
dt
dt
dt
b1s  b0
sy(0)  ( a1  1) y (0)
Y (s)  2
R(s) 
s  a1s  a0
s 2  a1s  a0
Three cases :
(a) Two characteristic roots are real and distinct.
(b) Two characteristic roots are equal.
(c) Two characteristic roots are complex numbers.
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Two characteristic roots are real and distinct.
let
y(0)  y (0)  0
r (t )  u (t )
k3
k1
k2
Y ( s) 


s  s1 s  s2 s
y (t )  (k1e s1t  k 2 e s2t  k3 )u (t )
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Two characteristic roots are equal
let
y(0)  y (0)  0
r (t )  u (t )
k3
k1
k2
Y ( s) 


2
( s  s1 )
s  s1 s
y (t )  (k1e s1t  k 2tes1t  k3 )u (t )
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Two characteristic roots are complex numbers
let
y(0)  y (0)  0
r (t )  u (t )
k1
Y ( s) 
R( s)
2
2
(s   )  
 n2
Y ( s)  2
R( s)
2
s  2n s   n
  n
  n 1   2
Undamped natural frequency
Damping ratio
y (t )  1 
e  nt
1 2
sin(  n 1   2 t  cos 1  )
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Higher-order system
 8s 2  5
Y ( s)  4
s  9s 3  37 s 2  81s  52
k3 s  k 4
k1
k2
Y ( s) 

 2
s  1 s  4 s  4s  13
Dominant root
nondominant root
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Stability testing
Properties of the polynomial coefficients :
• Differing algebraic signs
7 s  5s  3s  2s  s  10
6
4
3
2
At least one RHP root
• Zero-valued coefficients
s 6  3s 5  2s 4  8s 2  3s  17
Has imaginary axis roots or RHP roots or both
• All of the same algebraic sign, non zero
8s 5  6s 4  3s 3  2s 2  7 s  10
No direct information
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Routh-Hurwitz testing
P( s )  an s n  an 1s n1    a1s  a0
an  2 an 1  an an 3
1 
an 1
sn
an
an  2
an  4  b
sn 1
an 1
an 3
an 5 
sn  2
b1
b2
sn 3
c1
c2
b3 
c3 

s
0
an  4 an 1  an an 5
b2 
an 1
an 3b1  an 1b2
c1 
b1
The number of RHP roots of P(s) is the number of
algebra sign changes in the elements of the left column of the array.
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Example 1
P ( s )  2 s 4  3s 3  5 s 2  2 s  6
s
4
2
5
s
3
s
2
2
18  0
6
3
s
1
3
15  4 11

3
3
 32
11
6
s0
6
Two roots in the RHP
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Example 2
P ( s )  s  2 s  3s  4 s  1
4
3
4
1
3 1
s3
2
4
s2
1
1
s1
2
s0
1
s
2
no root in the RHP
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Example 3
P ( s )  3s  6 s  2 s  4 s  5
4
3
2
4
3
2
s3
6
4
s2
A
0
5
B ( 1) 5
s
s1
s0
5
0
5
n
s
2
5 5
s
1
10
s
0
5
n 移位次數移
至0消失為止
Two roots in the RHP
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Example 4
P ( s )  s 5  2 s 4  8s 3  11s 2  16 s  12
s5
1
8
s4
2
11 12
s
3
s
2
s
1
s
0
16
factor
2.5 10
3
0
12
3s  12
2
0
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P ( s )  s 5  2 s 4  8s 3  11s 2  16 s  12
s5
1
8
s4
2
11 12
s
3
s
2
s
1
6
s
0
12
16
2.5 10
3
12
d
2
(3s  12)  6 s
ds
no root in the RHP
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Example 5
R(s) +
-
k
s
10
s 2  2  10
Y (s )
1
s2
Y ( s)
10k

R( s) s( s 2  s  10)( s  2)  10k
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