Lecture Control system design: The stability of linear feedback systems - Nguyễn Công Phương
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Nguyễn Công Phương
CONTROL SYSTEM DESIGN
The Stability
of Linear Feedback Systems
Contents
I. Introduction
II. Mathematical Models of Systems
III. State Variable Models
IV. Feedback Control System Characteristics
V. The Performance of Feedback Control Systems
VI. The Stability of Linear Feedback Systems
VII. The Root Locus Method
VIII.Frequency Response Methods
IX. Stability in the Frequency Domain
X. The Design of Feedback Control Systems
XI. The Design of State Variable Feedback Systems
XII. Robust Control Systems
XIII.Digital Control Systems
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2
The Stability
of Linear Feedback Systems
1.
2.
3.
4.
The Concept of Stability
The Routh – Hurwitz Stability Criterion
The Stability of State Variable Systems
System Stability Using Control Design
Software
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3
The Concept of Stability (1)
• Stability is of the utmost importance.
• A close – loop feedback system that is unstable
is of little value.
• A stable system is a dynamic system with a
bounded response to a bounded input.
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4
The Concept of Stability (2)
/>
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5
The Concept of Stability (4)
1
Y ( s)
s
M
i 1
Ai
s i
Bk s Ck
y (t ) 1
2
2
2
k 1 s 2 k s ( k k )
N
M
Ae
i
i t
i 1
N
D e
k
k t
sin(k t k )
k 1
j
1
1
0
0
-1
-1
0
5
10
1
1
0
0
-1
-1
0
5
10
10
1
0
0
-1
0
5
10
0
5
10
0
10
0
5
10
-10
1
1
2
10
0.5
0.5
1
5
0
0
5
10
0
0
5
10
0
5
10
0
5
10
0
-1
5
0
10
1
0
-10
0
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5
10
0
0
5
6
10
The Stability
of Linear Feedback Systems
1.
2.
3.
4.
The Concept of Stability
The Routh – Hurwitz Stability Criterion
The Stability of State Variable Systems
System Stability Using Control Design
Software
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7
The Routh – Hurwitz Stability
Criterion (1)
q( s ) an s n an 1s n 1 an 2 s n 2 ... a1s a0 0
sn
an
an 2
an 4
s n 1
an 1
an 3
an 5
s n 2
bn 1
bn 3
bn 5
s n 3
cn 1
cn 3
cn 5
s
0
hn 1
1 an an 2 an 1an 2 an an 3
bn 1
,
an 1 an 1 an 3
an 1
bn 3
1 an an 4
,
an 1 an 1 an 5
cn 1
1 an 1 an 3
,
bn 1 bn 1 bn 3
The Routh – Hurwitz criterion states that
the number of roots of q(s) with positive real parts is equal to
the number of changes in sign of the first column of the Routh array
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The Routh – Hurwitz Stability
Criterion (2)
q( s ) an s n an 1s n 1 an 2 s n 2 ... a1s a0 0
sn
an
an 2
an 4
s n 1
an 1
an 3
an 5
s n 2
bn 1
bn 3
bn 5
s n 3
cn 1
cn 3
cn 5
s0
hn 1
1.
2.
3.
4.
No element in the 1st column is
zero.
There is a zero in the 1st column,
but some other elements of the row
containing the zero in the 1st
column are nonzero.
There is a zero in the 1st column,
and the other elements of the row
containing the zero are also zero.
Repeated roots of the characteristic
equation on the jω – axis.
The Routh – Hurwitz criterion states that
the number of roots of q(s) with positive real parts is equal to
the number of changes in sign of the first column of the Routh array
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9
The Routh – Hurwitz Stability
Criterion (3)
Ex. 1
q( s ) a2 s 2 a1s a0
sn
an
an 2
an 4
s2
a2
a0
s2
a2
a0
s n 1
an 1
an 3
an 5
s1
a1
0
s1
a1
0
s n 2
bn 1
bn 3
bn 5
s0
b1
0
s0
a0
0
s n 3
cn 1
cn 3
cn 5
s0
hn 1
1 an an 2
bn 1
an 1 an 1 an 3
1 a2
b1
a1 a1
a0
a0
0
The Routh – Hurwitz criterion states that
the number of roots of q(s) with positive real parts is equal to
the number of changes in sign of the first column of the Routh array
The system is stable if a2, a1 & a0 are all positive or all negative
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10
The Routh – Hurwitz Stability
Criterion (4)
Ex. 2
q( s ) s 3 s 2 2 s 50
sn
an
an 2
an 4
s3
1
2
s3
1
2
s n 1
an 1
an 3
an 5
s2
1
50
s2
1
50
s n 2
bn 1
bn 3
bn 5
s1
b1
b0
s1
48
0
s n 3
cn 1
cn 3
cn 5
s0
c1
c0
s0
50
0
s0
hn 1
b1
1 1 0
1 1 2
0,
48, b0
1 1 0
1 1 50
c1
1 1 50
1 1 0
50, c0
0
48 48 0
48 48 0
The Routh – Hurwitz criterion states that
the number of roots of q(s) with positive real parts is equal to
the number of changes in sign of the first column of the Routh array
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11
The Routh – Hurwitz Stability
Criterion (5)
Ex. 3
q( s ) a3s 3 a2 s 2 a1s a0
s3
a3
a1
s2
a2
a0
s1
b1
0
s0
c1
0
a3 0
a 0
2
b1 0
c1 0
b1
a3 0
a 0
2
a2 a1 a0a3
0
a2
a0 0
a2 a1 a0a3
, c1 a0
a2
a3 0
a 0
2
a2 a1 a0a3 0
a0 0
q( s ) s 3 2 s 2 6s 10, a2 a1 a0a3 2 6 10 1 2
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12
Ex. 4
The Routh – Hurwitz Stability
Criterion (6)
q( s ) s 5 2 s 4 2 s 3 4 s 2 11s 10
s5
1
2
11
s5
1
2
11
s4
2
4
10
s4
2
4
10
s3
0
6
0
s3
6
0
s2
c1
10
0
s2
c1
10
0
s1
d1
0
0
s1
d1
0
0
s0
10
s0
10
c1
4 12
12
6c 10
4 , d1 1
6 10
c1
Two sign changes two roots with positive real part the system is unstable
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13
Ex. 5
The Routh – Hurwitz Stability
Criterion (7)
q( s ) s 4 s 3 s 2 s K
s4
1
1
K
s4
1
1
K
s3
1
1
0
s3
1
1
0
s2
0
K
0
s2
K
0
s1
c1
0
0
s1
c1
0
0
s1
K
s1
K
c1
K
K
1
One sign change one root with positive real part
the system is unstable for all values of K
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14
Ex. 6
The Routh – Hurwitz Stability
Criterion (8)
3
2
3
q( s ) s 5 s 4 4 s 3 24 s 2 3s 63 ( s s s 21)( s 3)
s5
1
4
3
s5 s 4 4 s 3 24 s 2 3s 63 s 2 3
s4
1
24
63
s3
20
60
0
s 4 s 3 24 s 2 3s 63
s2
21
63
0
s4
1
s
0
0
s 3 s 2 s 21
3s 3
s5
3s 2
s 3 21s 2 3s 63
0
3s
s3
U ( s ) 21s 2 63 21( s 2 3)
21s 2
63
21s 2
63
0
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15
Ex. 6
The Routh – Hurwitz Stability
Criterion (9)
3
2
3
q( s ) s 5 s 4 4 s 3 24 s 2 3s 63 ( s s s 21)( s 3)
s5
1
4
3
s4
1
24
63
s
3
s
2
1
s
20
21
0
60
63
0
0
0
0
s3
1
1
s2
1
21
s1
20
0
s0
21
0
Two sign changes two roots with positive real part the system is unstable
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16
The Routh – Hurwitz Stability
Criterion (10)
Ex. 7
K (s a)
1
G( s)
s 1 s( s 2)( s 3)
T ( s)
R( s )
( )
1
s( s 2)( s 3)
G( s)
K (s a)
4
1 G ( s ) s 6s 3 11s 2 ( K 6) s Ka
q( s ) s 6s 11s ( K 6) s Ka
4
3
2
s4
1
11
Ka
s3
6
K 6
0
s2
b3
Ka
0
s1
c3
0
0
s1
Ka
b3
K (s a)
s 1
60 K
6 0
b3 ( K 6) 6 Ka
0
b
3
Ka 0
0 K 60
(60 K )( K 6)
a
36 K
b ( K 6) 6 Ka
60 K
, c3 3
6
b3
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17
Y ( s)
The Stability
of Linear Feedback Systems
1.
2.
3.
4.
The Concept of Stability
The Routh – Hurwitz Stability Criterion
The Stability of State Variable Systems
System Stability Using Control Design
Software
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18
Ex. 1
The Stability of State Variable
Systems (1)
3
1
x1 3x1 x2
x2 x2 Kx1 Ku
1
K
1
U ( s)
1/ s
L1 s 1 , L2 3s 1 , L3 Ks 2
1
N
L
n
n 1
Ln Lm
n ,m
nontouching
X 2 ( s)
1
X1 ( s)
1/ s
X2
Ln Lm Lp ...
n ,m , p
nontouching
1 ( L1 L2 L3 ) ( L1L2 ) 1 2s 1 ( K 3) s 2
q( s ) s 2 2 s ( K 3)
K 3 0 K 3
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19
Ex. 2
The Stability of State Variable
Systems (2)
a
dx
dt
0
1 0
u
0 x 0 1 1
u2
0 0
0
0 0 a
det( I A ) det 0 0
0 0
0
0
0
0
0
[ 2 ( ) ( 2 )]
q( ) [ 2 ( ) ( 2 )]
0
2
0
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20
The Stability
of Linear Feedback Systems
1.
2.
3.
4.
The Concept of Stability
The Routh – Hurwitz Stability Criterion
The Stability of State Variable Systems
System Stability Using Control Design
Software
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21
Ex. 1
System Stability Using Control
Design Software (1)
R( s )
( )
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10( s 2)
s 1
1
s( s 2)( s 3)
22
Y ( s)
Ex. 2
System Stability Using Control
Design Software (2)
q( s ) s 3 2 s 2 3s K
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Ex. 3
System Stability Using Control
Design Software (3)
Given a characteristic equation q(s) = s4 + 8s3 + 17s2 + (K + 10)s + aK = 0.
Find a & K such that the system is stable.
s4
1
17
aK
s3
8
K 10
0
s2
b3
b1
0
s1
c3
0
0
s1
d3
1
126 K
b3 [( K 10) 8 17]
8
8
1
b1 (1 0 8aK ) aK
8
c3
1
(126 K )( K 10)
[8b1 b3 ( K 10)]
8aK
b3
8
126 K
0
8
b3 0
(126 K )( K 10)
8aK 0
c3 0
8
d 0
3
aK 0
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d3
1
(b3 0 b1c3 ) b1 aK
c3
24
