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سیستمهای کنترل خطی
1389پاییز
بسم ا... الرحمن الرحيم
دکتر حسين بلندي- دکتر سید مجید اسما عیل زاده
Recap.
• Optimal Systems,• Index of Performance,• Introduction to Ess.
2
Steady-State Tracking & System Types
3
In General:
• Unity feedback control:
G(s) C(s)+
-r(s) e y(s)
plant
Go.l.(s)+
-r(s) e
y(s)ol
ol
G
G
sr
sy
1)(
)(
T.F. get & open, loop cut i.e.
yto e from T.F. loop open the is )()(
)(.. sG
se
sylo
4
s.an cancelcan otherwise ,0 need ,0 If
0
0but
0, 0
)1()1)(1(
)1()1)(1(
:into factored be alwayscan
0
0
011
011
11
1
01
21..
..
bN
ba
bK
anpN
aaaN
asasasasas
bsbsb
sTsTsTs
sTsTsTKG
G
mN
N
N
NN
NN
nn
n
mm
pN
mbalo
lo
5
psloss
slos
ss
lo
lo
lo
KsGe
ssR
sG
ssRsseete
sRsG
sysRse
sRsG
sGsy
1
1
)(1
1step to
1)(:input stepFor
)(1
)()(lim)(
: trackingstate-steady
)()(1
1)()()(:error tracking
)()(1
)()(:loop-closed
0..
0..0
..
..
..
6
r, respect to with 0" type" called is system the,0 If
)K control alproportion with confused be not to p, small use here(
1
1step toThen
const.error position static called
)0()(lim denote
P
....0
N
Ke
GsGK
pss
lolos
p
7
01
1step to
)0(
,2 or type ,1 typecalled is system
larger or ,2or ,1 If
. zero-non input with step acan track system 0 type
..
pss
lop
ss
Ke
GK
N
e
8
higheror 1 for type 0
0 for type 1
1
then,)(:unitnot is step If
higheror 1 for type 0
0 for type 1
1
:input stepunit For
. zero input with step acan track higher or 1 typeof systemA
RKe
s
Rsr
Ke
e
pss
pss
ss
9
const errorvelocity static called
:denote
ramp to
:ramp unit is If
)(lim
1
)(
1lim
)(
1lim
)(1
1
lim
)(1
)(lim
1)(
)(
0
0
0
2
0
0
2
ssGK
KssG
ssGssGs
s
sG
ssre
ssr
sr
ols
v
vols
ols
ols
ols
ss
r(t)
t
10
signal. input ramp a track not can system 0 type
ramp to
system, 0 type For
vss
s
n
mm
sv
Ke
a
bs
asas
bsbsbsK
baN
1
0lim
lim
0,0,0
0
0
0
01
01
0
00
11
. error statesteady zero-non
withramp tracks system 1 type
finiteramp to
finite,
1: type For
ss
vss
n
mm
s
n
mm
sv
e
Ke
a
b
asas
bsbsb
asas
bsbsbsK
baaN
,01
0lim
lim
!0,0,0,1
1
0
121
01
0
01
01
0
010
12
. no withinput ramp a
track can system higher or 2 type
ramp to
higher, or 2 type For
factor a as s has still ones cancel
ss
vss
n
mm
s
n
mm
sv
e
Ke
sasas
bsbsb
asasasas
bsbsbsK
baa
N
01
lim
lim
0,0
,3,2
22
31
01
0
0
012
23
3
01
0
010
13
1type if :then
unit, not is ramp If
2type if
1type if
0type if
:input ramp unit For
RK
e
s
Rsr
b
a
Ke
vss
vss
1
)(
0
1
2
0
1
14
)(12
1)(
1)(:
2
3
tttr
ssr
input onaccelerati unit
ass
ssss
ss
KsGssGss
sGs
s
sG
ssre
e
1
)(
1lim
)(
1lim
)(1
1
lim)(1
)(lim
20220
3
00
acc to
r(t)
t0
15
ass
s
n
mm
sa
sa
Ke
a
bs
asas
bsbsbsK
as
ssGK
1acc to
0lim
lim
0 den.in offactor no system, 0 For type
constant.error on accelerati theis )(lim
0
02
0
01
012
0
0
0
16
sig. acc. tract tcan' system 1 or 0 type
acc to
bu i.e.
den. in of factor one i.e.
. in of factor one system, 1 type For
ass
s
nn
n
mm
sa
Ke
a
bs
asasas
bsbsbsK
aa
s
sGs
1
0lim
lim
0,0
)(1
1
0
0
011
1
012
0
10
17
error. s.s. finite withsig.acc tract can system 2 type
acc to
or
den. in of factors two or,
in of factors two :2 type
01
)0(0
lim
0,0,0
1,2
0
2
02
0
22
11
012
0
210
b
a
Ke
ba
b
sasas
bsbsbsK
aaa
s
Gs
N
ass
nn
n
mm
sa
18
error. s.s. no withsig.acc tract can syst. higher or 3 type
acc to
higher or 3 type
01
0
lim
0,0
3
0
33
11
012
0
0210
ass
nn
n
mm
sa
Ke
b
sasas
bsbsbsK
baaa
N
19
stabilize. to difficult are system higher or 2 type but
tracting. bettertyper larger like seems
:Caution
by A. multiplied be to needs then,
:rather acc, unit not If
3type if
2type if
0,1type if
acc to
:input acc. forsummary
ss
ass
e
ttA
b
a
Ke
)(12
1
0
1
2
0
2
20
r(t)=R·1(t)r(s)=R/s
r(t)=R·t·1(t)r(s)=R/s2
r(t)=R·1/2·t2
r(s)=R/s3
type 0(N=0 a0≠0)
Kp=b0/a0
ess=R/(1+Kp)
Kv=0
ess=∞
Ka=0
ess=∞
type 1(N=1 a0=0 a1≠0 b0≠0 )
Kp= ∞
ess=0
Kv=b0/a1
ess=R/Kv
Ka=0
ess=∞
type 2, N=2a0=a1=0
a2≠0,b0≠0
Kp= ∞
ess=0
Kv= ∞
ess=0
Kp=b0/a2
ess=R/Ka
type≥3, N ≥ 3a0=a1=a2=0
b0≠0
Kp= ∞
ess=0
Kv= ∞
ess=0
Ka= ∞
ess=0
sys.type
ref.input
21
Example of tank
ass
vss
ppss
avpp
plo
p
p
Ke
Ke
RKKe
KKRKGK
NRAs
RKsHsCsG
KsCRAs
RsH
K
1acc to
1ramp to
1
1
1
1step to
0,)0(
0 type0,1
)()()(
)(,1
)(
:control
..
H+
- C
22
ass
Ivss
pss
aIs
vp
Ip
IpIp
Ke
RKKe
Ke
KRKssGKK
Ns
RAss
RKsKsHsCsG
s
KsK
s
KKsC
sH
1
11
01
1
0,)(lim,
1
)1(
)()()()(
)(
),(
0
acc to
ramp to
step to
type den, in of factor one
but
same :control PI
23
the to loop the following from path the in #
T.F. loop open the in #i.e.
in # is w.r.t.type sys.
type. sys. is tracking statesteady toKey
es
s
sGs
sr
1
1
)(1
)(
+r(s)
Kps+KI
s+
-
r(s) e ωn2
s(s+2ξ ωn)
1
Ts+1
)(1
srs
w.r.t.2 type 2 :path in #count
e.g.
24
example
1type one is there
:default Take
specified. dist. or input No :Note
acc. ramp. step. to error statesteady &
constants error type, system find
,1
)5.0)(5.1(
)15.3()(
s
sss
sKsG
G(s)r(s) e(s) y(s)
25
KKe
e
e
K
KK
ssGK
K
vss
ss
ss
a
sv
p
2.4
11
0
0
2.45.05.1
15.3)(lim
0
ramp to
1 type for acc to
1 type for step to
1 type for
1 type for
26
121
12
1
512
15)(lim
20
0
2,
221
)5)(12(
)1(5)(
2
0
2
ass
sa
ss
ss
vp
Ke
sGsK
e
e
KKs
sss
ssG
acc to
typeramp to
step to
type
type ,#
:Example
27
The stability of linear feedback systems.
28
The concept of stability.
29
30
The stability of a cone.
31
Tacoma Narrows Bridge
As oscillation begins
At catastrophic failureon November 7, 1940.
32
Bounded Input - Bounded Output criterion
An unconstrained linear system is stable if the output response is bounded for all bounded inputs.
Note: By a bounded input, we mean an input variable that stays within upper and lower limits for all values of time
33
Example: BI-BO criterion
Q.If the step function is applied at the input of continuous system and the output remains below a certain level for all time, is the system stable?
34
Example: BI-BO criterion
Q.If the step function is applied at the input of continuous system and the output remains below a certain level for all time, is the system stable?
A.The system is not necessarily stable since the output must be bounded for every bounded input.
A bounded output to one specific bounded input does not ensure stability.
35
Example: BI-BO criterion
Q. If the step function is applied at the input of continuous system and the output is of the form y = t, is the system stable or unstable?
36
Example: BI-BO criterion
Q. If the step function is applied at the input of continuous system and the output is of the form y = t, is the system stable or unstable?
A. This system is unstable since a bounded input produced an unbounded output.
37
Characteristic root locations criterion
A system is stable if all the poles of the transfer function have negative real parts.
Stability in the s-plane.
38
Stable system
A necessary and sufficient condition for a feedback system to be stable is that all the poles of the transfer function have negative real parts.
This means that all the poles are in the left-hand s-plane.
39
• Determine in each case if the set of roots represents, stable, marginally stable, or unstable systems.
a) -1, -2 e) -2 + j, -2 - j, 2j, -2jb) -1, +1 f) 2,-1,-3c) -3,-2,0 g) -6,-4,7d) -1 + j, -1- j h) -2 + 3j, -2 -3j, -2
40
• Determine in each case if the set of roots represents, stable, marginally stable, or unstable systems.
a) -1, -2 e) -2 + j, -2 - j, 2j, -2jb) -1, +1 f) 2,-1,-3c) -3,-2,0 g) -6,-4,7d) -1 + j, -1- j h) -2 + 3j, -2 -3j, -2
41
Marginally stable
If there are any poles on the jω-axis (with none on the right hand side), then the system is considered marginally stable.
42
Unstable
If there are one or more poles in the right-hand s-plane or there are repeated roots on the jω-axis, the system is unstable.
43
Example: Poles and zeros
Q. A system has poles at -1 and -5 and zeros at 1 and -2. Is the system stable?
44
Example: Poles and zeros
Q. A system has poles at -1 and -5 and zeros at 1 and -2. Is the system stable?
A. The system is stable since the poles are roots of the system characteristic equation which have negative real parts.
The fact that the system has a zero with a positive real part does not affect its stability.
45
Example: Characteristic equation
Q. Determine if the system with the following characteristic equation is stable:
(s+1)(s+2)(s-3) = 0
46
Example: Characteristic equation
Q. Determine if the system with the following characteristic equation is stable:
(s+1)(s+2)(s-3) = 0
A. This characteristic equation has the roots
-1,-2, and 3 and therefore represent an unstable systems since there is a positive real root.
47
Example: integrator
Q. The differential equation of an integrator may be written as follows:
dy/dt = x
Determine if an integrator is stable.
48
Example: integrator
y(t) x(t)dt0
Y(s) 1
sX(s)
Y(s)X(s)
1s
s s1 0 0 j0
49
Example: integrator
A. The characteristic equation of his system is s = 0. Since the root does not have a negative real part, an integrator is not stable.
Since it has no roots with positive real parts, integrator is marginally stable
50
Stability in the s-plane.
NOT STABLESTABLE
The Routh-Hurwitz Stability Criterion.
51
52
Introduction
• Goal: Determining whether the system is stable or unstable from a characteristic equation in polynomial form without actually solving for the roots.
• Routh’s stability criterion is useful for determining the ranges of coefficients of polynomials for stability, especially when the coefficients are in symbolic (nonnumerical) form.
53
A necessary condition for Routh Stability
• A necessary condition for stability of the system is that all of the roots of its characteristic equation have negative real parts, which in turn requires that all the coefficients be positive.
A necessary (but not sufficient) condition for stability is that all the coefficients of the polynomial
characteristic equation be positive.
54
A necessary and sufficient condition for Stability
• Routh’s formulation requires the computation of a triangular array that is a function of the coefficients of the polynomial characteristic equation.
A system is stable if and only if all the elements of
the first column of the Routh array are positive
55
Characteristic Equation
• Consider an nth-order system whose the characteristic equation (which is also the denominator of the transfer function) is:
q(s) ansn an 1s
n 1 an 2sn 2 a1s a0
56
Determining the Routh array
• Consider the characteristic equation:
• First arrange the coefficients of the characteristic equation in two rows, beginning with the first and second coefficients and followed by the even-numbered and odd-numbered coefficients:
q(s) an sn an 1sn 1 an 2s
n 2 an 3sn 3 a1s a0
sn : an an 2 an 4 sn 1 : an 1 an 3 an 5
57
Routh array: method (cont’d)
• Then add subsequent rows to complete the Routh array:
sn : an an 2 an 4 sn 1 : an 1 an 3 an 5 sn 2 : bn 1 bn 3 bn 5 sn 3 : cn 1 cn 3 cn 5
s2 :
s1 :
s0 : hn 1
?!
58
Routh array: method (cont’d)
• Compute elements for the 3rd row:
sn : an an 2 an 4 sn 1 : an 1 an 3 an 5 sn 2 : bn 1 bn 3 bn 5 sn 3 : cn 1 cn 3 cn 5
s2 : * *
s1 : *
s0 : *
bn 1 1
an 1
an an 2
an 1 an 3
,
bn 3 1
an 1
an an 4
an 1 an 5
,
cn 1 1
bn 1
an 1 an 3
bn 1 bn 3
59
Routh-Hurwitz Criterion
• 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.
60
Three distinct cases of the first column array
1. No element in the first column is zero.
2. There is a zero in the first column and in this row.
3. There is a zero in the first column and the other zero
in this row.
61
Case1. No element in the first column is zero
012
2)( asasasq
0
00
1
1
2
0
1
2 a
b
aa
s
ss
01
02
11
2011 0
1)0(a
a
aa
aa
aaab
62
Second-order system
012
2)( asasasq
The requirement for a stable second-order system is that all the coefficients be positive or all coefficients be negative.
63
Third-order system
012
23
3)( asasasasq
s3
s2
s1
s0
a3
a2
b1
c1
a1
a0
0
0
64
Third-order system
2
30121 a
aaaab
01
011 a
b
abc
012
23
3)( asasasasq
For system to be stable, it is necessary and sufficient
that the coefficients a0 ,a1, a2, a3 be positive
and a2a1 > a0a3.
65
Unstable system
s3
s2
s1
s0
1
1 22
24
2
240
0
Two roots of q(s) lie in the right-hand s-plane.
q(s) s3 s2 2s 24
q(s) (s 1 j 7)(s 1 j 7)
66
Two changes in sign
67
Case 2: There is a zero in the first column and in this row.
Ksssssq 234)(
s4
s3
s2
s1
s0
1
1
c1
K
1
1
K0
0
K
0
00
0
KK
c1
The system is unstable for all values of gain K.
68
Case 3:A zero in the first column and the other zero in this row.
q(s) s3 2s2 4sKFor a stable system:
U(s) = 2s2 + 8 = 2 (s + j2)(s - j2)
q(s) (s 2)(s j2)(s j2)
For a marginal stability: K=8
0 < K < 8
s3 1 4
s2 2 K
s1 8 K
20
s0 K 0
69
Ex. A six-legged micro robot
It is equipped with sensor network that includes 150 sensors of 12 different types. The legs are instrumented so that the robot can determine the lay of the terrain, the surface texture, hardness, and color.
70
Micro robot
71
Micro robot
633244)( 2345 ssssssq
s5
s4
s3
s2
s1
1
1
20
210
4
24
60
630
3
63
0
00
)3)(3(21)3(216321)( 22 jsjssssU
The auxiliary polynomial:
72
The auxiliary polynomial
• The equation that immediately precedes the zero entry in the Routh array.
73
Micro robot
213
)( 232
ssss
sq
s3
s2
s1
s0
1
1 20
21
1
210
0
s 1 j 6
The robot is using flexible legs with high-gaincontrollers that may become unstable and oscillate.
74
Example :
Given the characteristic equation,
is the system described by this characteristic equation stable?
Answer:• One coefficient (-2) is negative.
• Therefore, the system does not satisfy the necessary condition for stability.
4s4ss2s3s4s)s(a 23456
75
Example :
Given the characteristic equation,
is the system described by this characteristic equation stable?
Answer:• All the coefficients are positive and nonzero.• Therefore, the system satisfies the necessary condition for
stability.• We should determine whether any of the coefficients of
the first column of the Routh array are negative.
44234)( 23456 sssssssa
76
Example (cont’d): Routh array
44234)( 23456 sssssssa
?:s
??:s
??:s
???:s
???:s
0424:s
4131:s
0
1
2
3
4
5
6
77
Example (cont’d): Routh array
?:
??:
??:
???:
4025:
0424:
4131:
0
1
2
3
4
5
6
s
s
s
s
s
s
s
44234)( 23456 sssssssa
2
5
4
10
4
4321
04
0
4
4141
44
16
4
4401
78
Example (cont’d): Routh array
?:
??:
??:
05122:
4025:
0424:
4131:
0
1
2
3
4
5
6
s
s
s
s
s
s
s
44234)( 23456 sssssssa
2)2(25
25204
5
12
5
2032
25
25444
79
Example (cont’d): Routh array
4:
01576:
43:
05122:
4025:
0424:
4131:
0
1
2
3
4
5
6
s
s
s
s
s
s
s
44234)( 23456 sssssssa
The elements of the 1st column are not all positive:
the system is unstable
80
Example : Stability versus Parameter Range
Consider a feedback system such as:
The stability properties of this system are a function of the proportional feedback gain K. Determine the range of K over which the system is stable.
81
Example (cont’d)
0)6)(1(
11
sss
sK
82
Example (cont’d)
• Expressing the characteristic equation in polynomial form, we obtain:
0)6(5 23 KsKss
0)6)(1(
11
sss
sK
83
Example (cont’d)
• The corresponding Routh array is:
• Therefore, the system is stable if and only if
q(s) s3 5s2 (K 6)sK
Ks
Ks
Ks
Ks
:
5)304(:
5:
61:
0
1
2
3
4K 30
5 0 and K 0
K 7.5 and K 0
K 7.5
84
Example (cont’d)
• Solving for the roots gives:
-5 and ±1.22j for K=7.5=> The system is unstable (or critically stable) for K=7.5
-4.06 and –0.47 ±1.7j for K=13-1.90 and –1.54 ±3.27j for K=25
=> The system is stable for both K=13 and K=25
KsKsssa )6(5)( 23
85
Example : Stability versus Two Parameter Range
Consider a Proportional-Integral (PI) control such as:
Find the range of the controller gains so that the PI feedback system is stable.
)K,K( 1
86
Example (cont’d)
• The characteristic equation for the system is given by:
021
11 1
)s)(s(s
KK
87
Example (cont’d)
• Expressing the characteristic equation in polynomial form, we obtain:
023 123 Ks)K(ss
021
11 1
)s)(s(s
KK
88
Example 4 (cont’d)
• The corresponding Routh array is:
• For stability, we must have:
KsKsssa )6(5)( 23
10
11
12
3
336
3
21
K:s
)KK(:s
K:s
K:s
23
0 11
KKandK
89
The Routh-Hurwitz criterion
90
Relative Stability
• Further characterization of the degree of stability of a stable closed loop system.
• Can be measured by the relative real part of each root or pair of roots.
91
Root r2 is relatively more stable than roots r1 and r1’.
92
Axis shift
464)( 23 ssssq
14)1(6)1(4)1( 2323 nnnnnn ssssss
sn3
sn2
sn1
sn0
1
10
1
1
10
0
)1)(1())((1)( 2 jsjsjsjsssU nnnn
93
EX. Flow graph diagram
11
sL
12 3 sL
L3 Ks 2
)3()3(1)(1 221121321
sKsssLLLLL
s2 2s (K 3) 0 K 3 for stability
94
Ex. Block diagram model.
T(s) KG1(s)G2(s)
1 KG1(s)G2(s)
(s) 1KG1(s)G2(s) 0
0)3(2)3)(1()( 2 KssKsss
1
1)(1
ssG
G2(s) 1
s 3
K 3 for stability
95
Tracked vehicle turning control.
Select K and a so that the system is stable and ess≤ 24%of a ramp command.
96
Characteristic equation:
1Gc (s)G(s) 0
1K(s a)
s(s1)(s 2)(s 5)0
s4 8s3 17s2 (K 10)sKa 0
97
Routh array
s4
s3
s2
s1
s0
1
6
b1
c1
Ka
17
K 10
Ka
Ka
b1 126 K
8c1
b1(K 10) 8Ka
b1
b1 126 K
8c1
b1(K 10) 8Ka
b1
98
Ramp response for a=0.6 and K=70
99
The steady-state error
E(s) 1
1Gc (s)G(s)R(s)
E(s) s(s1)(s 2)(s 5)
s(s1)(s 2)(s 5) K(s a)
A
s2
ess(t) limsE (s) 10A
aKfor s 0
100
The closed-loop T(s)
T(s) 70s 42
s4 8s3 17s2 80s 42s 7.08
s 0.58
s 0.17 3.2 j
s 0.17 3.2 j
101
The stable region
K 126
Ka 0
(K 10)(126 K) 64Ka 0
ess AKa
10
0.24
Ka 42
K 70 a 0.6
102
Absolute Stability
• A closed loop feedback system which could be characterized as either stable or not stable.
103
Stable System
• A stable system is a dynamic system with a bounded response to a bounded input.
104
Stability of S.S. Eqn.
xx A0)( xAI
0)det( AI
105
Example: Closed epidemic system
2
1
00
10
01
0
0
0
u
u
dt
dx
x
0
0
0
00
00
00
det)det(
AI
106
0)(
0)(
det)det( AI
]))([()det( 2 AI
107
0)]()([()det( 22 AI
