Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 1
1
Chapter4 Time Domain Analysis of Control System
Ø Routh stability criterion
ØSteady state errors
ØTransient response of the first-order system
Ø Transient response of the second-order system
Ø Time domain performance specifications
Ø The relationship between the performance specifications and system parameters
Ø Transient response of higher-order systems
2
Definition of characteristic equation
)()()(
011
1
011
1 sGasasasabsbsbsb
sRsC
nn
nn
mm
mm =
++++++++
= −−
−−
LL
The characteristic equation of the system is defined as
0011
1 =++++ −− asasasa n
nn
n L
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 2
3
Transfer function:
2, 1j k k k n n kp j jσ ω ζ ω ω ζ− − ± =− ± − poles of the closed loop
∏ ∏
∏
= =
=
−++++
+== q
j
r
kkkkkj
m
ii
jsjsps
zsk
sRsCs
1 1
1
)]()][(([)(
)(
)()()(G
ωσωσ
zi : zeros of the closed loop
4
0
2 21 1
( )( 1 ) ( 1 )
q rj k k
j kjk n n k n n k
aa sC ss s p s j s j
α β
ζ ω ω ζ ζω ω ζ= =
+= + +
+ + + − + − −
∑ ∑
)1sin(
)1sin1cos()(
1 1
20
2
1 1
20
k
q
j
r
ikk
tk
tpj
kkk
q
j
r
kkkk
ttpj
teDeaa
tCtBeeaatC
kkj
kkj
ϕζω
ζωζω
ωζ
ωζ
+−++=
−+−++=
∑ ∑
∑ ∑
= =
−−
= =
−−
sjsjsps
zsksC q
j
r
kkkkkj
m
ii 1
)]()][(([)(
)()(
1 1
1 ⋅−++++
+=
∏ ∏
∏
= =
=
ωσωσ
Step response:
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 3
5
Consider that the characteristic equation of a LTI system
0)( 011
1 =++++= −− asasasasF n
nn
n L
Where all the coefficients are real numbers.
In order that there be no roots of the above equation with positivereal parts, it is necessary but not sufficient that
1. All the coefficients of the polynomial have the same sign.2. None of the coefficients vanishes.
4.1 Routh stability criterion
6
0000
000
00000
0000
000
60
61
662
1615133
61
061
1
5041
1
30214
5315
64206
FEFas
FE
CaEDsC
ACaC
ACaEC
ADBCsA
aADA
aaAaCA
BaAas
aa
aaaBa
aaaaAa
aaaas
aaasaaaas
×−
=−
=×−×
=×−
=−
=×−×
=−
=−
=×−
=−
=−
The Routh tabulation
0)( 12
21
10 =+++++= −−−
nnnnn asasasasasF L
for n = 6
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 4
7
The roots of the polynomial are all in the left half of the s-planeif all the elements of the first column of the Routh Array areof the same sign.
The necessary and sufficient condition for the stability of a system
Rout-Hurwitz criterion
If there are changes of signs in the elements of the first column, the number of sign changes indicates the number of roots with positive real parts.
8
Example: The characteristic equation of a system is 050104 23 =+++ sss
Determine the stability of the system using Routh criterion.
Solution:
05005.2
504101
0
1
2
3
ssss
−
The system has two roots located in the right half of the s-plane.
(2). Routh array is
(1). Check the necessary condition
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 5
9
The first element in any one row of the Routh Array is zero butthe other elements are not.
We can replace the zero element in the Routh tabulation by anarbitrary small positive number ε and then proceed with the Routhtest.
Special case 1
10
The characteristic equation of a system is
Determine the stability of the system using Routh criterion.
Example:01255 2345 =+++++ sssss
001
0015
15
011501)(0151251
0
21
2
3
4
5
s
s
s
sss
−−−
−
εεε
εε
ε
Solution: Routh array is
There are two sign changes in the first column of the tabulation, thesystem has two roots located in the right half of the s-plane.
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 6
11
The elements in one row of the Routh Array are all zero.
Special case 2
12
Example:
0161623 =+++ sss
The characteristic equation of the system is:
The Routh array is :
The auxiliary equation is:01600
161161
0
1
2
3
ssss
016)( 2 =+= ssA
01602
161161
0
1
2
3
ssss
The sign of the elements in the first column does not change, the system is stable .
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 7
13
Example: ( 1)( 1)(2 1)
K ss Ts s
++ +
( )R s ( )C s+
−
Determine the value of K and T to make the closed loop be stable.
Solution: the characteristic equation:
0)1()2(2 23 =+++++ KsKsTTs
0
02
2)1)(2(2
12
0
1
2
3
KsT
TKKTsKTs
KTs
+−++
++Routh array:
14
,02,0 >> TK
02)1)(2( >−++ TKKT
02 >+T
The condition for the stability is:
0>T
220
−+<<
TTK
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 8
15
K
T0
2
2
4
4 6
6
8
22
TKT
+<
−
220
−+<<
TTK
Stable
16
Example:
02108 23 =+++ sss
The characteristic equation of the system is
Determine the stability of the system. Analyze how many rootslie between the imaginary axis and the line .1−=s
Solution:
02075.928
101
0
1
2
3
ssss
the system is stable.
11 −= ssLet
The characteristic equation becomes:
0135 121
31 =−−+ sss
Routh array:
Im
Re01−
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 9
17
0135 121
31 =−−+ sss
Routh array for the above equation:
0108.21531
01
11
21
31
−−
−−
ssss
there is one root on the right side of the line s = -1.
Im
Re01−
18
4.6 Steady state errors
)()( ∞−∞=∞ rce )(
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 10
19
( )R s ( )C s( )G s
( )B s
( )E s+
−( )H s
)()(1)()(
sHsGsRsE
+=
0 0
1l i m ( ) l i m ( ) l i m ( )1 ( ) ( )s s t s s
e e t s E s s R sG s H s→ ∞ → →
= = =+
20
∏
∏
+=
=
+
+=
++++++
= q
vjj
m
iir
q
mr
pss
zsK
pspspsszszszsKsG
1
1
21
21
)(
)(
)())(()())(()(
νν L
L
ν = 0, type 0 system ν = 1, type 1 systemν = 2, type 2 system
the types of the control systems
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 11
21
a. position error constant
For a step input )()( tuRtr ⋅=
)()(1lim
)()(1/lim
)()(1)(lim)(lim)(lim
00
00
sHsGR
sHsGsRs
sHsGsRsssEtee
ss
sstss
+=
+=
+===
→→
→→∞→
00
lim1 ( ) ( ) 1 lim ( ) ( )ss s
s
R ReG s H s G s H s→
→
= =+ +
0lim ( ) ( ) Ps
G s H s K→
=
22
b. velocity error constant
For a ramp input: Rttr =)(
)()(lim
)()(1/lim
)()(1)(lim)(lim)(lim
0
2
0
00
sHssGR
sHsGsRs
sHsGsRsssEtee
ss
sstss
→→
→→∞→
=+
=
+===
0lim ( ) ( )v s
K sG s H s→
=v
ss KRe =
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 12
23
c. Acceleration error constant
for acceleration input 21( )2
r t Rt=
)()(lim
)()(1/lim
)()(1)(lim)(lim)(lim
20
3
0
00
sHsGsR
sHsGsRs
sHsGsRsssEtee
ss
sstss
→→
→→∞→
=+
=
+===
)()(lim 2
0sHsGsK
sa →=
ass K
Re =
24
Steady state error
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 13
25
Example:
+
−
( )R s ( )C s10( 5)s s +
Determine the steady state errors, when the input is
)(tu t 2
21 t, and .
Solution:
∞=+
==→→ )5(
10lim)()(lim00 ss
sHsGKssp
2)5(
10lim)()(lim00
=+
==→→ ss
ssHssGKssv
0)5(
10lim)()(lim 2
0
2
0=
+==
→→ ssssHsGsK
ssa
26
211
==v
ssv Ke
01
1=
+=
Pssp K
e
∞==a
ssa Ke 1
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 14
27
4.1 The transient response of the first-order system
4.1.1 The math model
TssRsC
+=
11
)()( ( )R s +
−( )C s
Ts1
28
4.1.2 the step response
r(t) = u(t) , R(s)=1/s
1 1( ) ( ) ( )( 1)
11
C s G s R sTs s
Ts Ts
= = ⋅+
= −+
Tt
etc−
−= 1)( t ≥ 0
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 15
29
0
1
0.632
( )c t
tT 2T 3T 4T 5T
63.2%
86.5% 95% 98.2% 99.3%
slope
30
Impulse response:
TssR
TssC
+=•
+=
11)(
11)(
Tt
eT
tctg−
==1)()(
( )R s +−
( )C s
Ts1
0 T 2T 3T 4T t
( )c t
1T
/1( ) t Tc t eT
−=
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 16
31
Ramp response :
TsT
sT
ssTssRsGsC
/111
11)()()( 22 +
+−=⋅+
==
( )R s +−
( )C s
Ts1
TtTeTttc /)( −+−=
32
)(tr
)(tc
t
)(tr)(tc
Tess =TtTeTttc /)( −+−=
TtTeTttrtc Tt
tt−=−+−=− −
∞→∞→][lim)]()([lim /
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 17
33
4.2 the transient response of the second-order system
4.2.1 the mathematical model
( 1)K
s Ts +( )R s ( )C s+
−
34
( 1)K
s Ts +( )R s ( )C s+
−TKs
T1s
TK
)s(R)s(C)s(
2 ++==Φ
letTK
n =2ωTn12 =ςω
2nn
2
2n
s2s)s(
ωζωω
Φ++
=
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 18
35
The characteristic equation :
0s2s 2nn
2 =++ ωζω
The roots of the characteristic equation are
1s 2nn2,1 −±−= ζωζω
36
djs ωσ ±−=2,11s 2nn2,1 −±−= ζωζω
1. underdamped )10( << ζ
If input is a unit step u(t):
222222
2
)()(11
)2()(
dn
n
dn
n
nn
n
sss
sssssC
ωζωζω
ωζωζω
ωζωω
++−
+++
−=⋅++
=
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 19
37
)tsin1
t(cose1)t(c d2dtn ω
ζ−
ζ+ω−=
ζω−
)sincos1(1
11 2
2tte dd
tn ωζωζζ
ζω +−−
−= −
)tsin(1e1 d2
tnθω
ζ
ζω
+−
−=−
ζ=θ arccos
38
0
1
2
111 ζ
+−
2
111 ζ
−−
21
1
nte ζω
ζ
−+
−1
nT
ζω=
T 2T 3T 4T t
( )c t
21
1
nte ζω
ζ
−−
−
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 20
39
1s 2nn2,1 −±−= ζωζω
ns ω−=2,1
nn
n
nn
n
sssssssc
ωωω
ωωω
+−
+−=⋅
++=
1)(
11)2(
)( 222
2
)1(1)( tetc ntn ω
ζω+−=
−( )c t
( )r t
0
1
t
( )c t
2. Critically damped response 1=ζ
u(t)
40
1>ζ
122,1 −±−= ζωζω nns
1
]11(2[
1
)]11(2[1)(2
122
2
122
−++
−−++
−−+
−−−+=
−−
ζωζω
ζζζ
ζωζω
ζζζ
nnnn ssssc
)T
eT
e(12
11)t(c2
t2T
1
t1T
2 −−
−−+=
ζ
n
n
T
T
ωζζ
ωζζ
)1(
)1(2
2
21
−−−=
−+−=
3. Over damped case
Let
1
122
2
122 ]11(2[)]11(2[1)(TsTss
sc−
−−++
−−−−
+=−− ζζζζζζ
( )r t
t
( )c t
0
c(t)
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 21
41
njs ω±=2,11s 2nn2,1 −±−= ζωζω
tcos1)t(c nω−= )0t( ≥
( )c t
0
1
2
t
)0( =ζ
4. undamped case
2222
2 11)(
)(nn
n
ss
ssssc
ωωω
+−=⋅
+=
)0( =ζ
42
1. Rise time tr
2. The peak time tp
4.3 Time domain performance specifications
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 22
43
3. Percentage overshoot σp
%100)(c
)(c)t(c PP ×
∞∞−
=σ
4. Steady-state error ess
( ) ( ) 100%( )ss
c rer
∞ − ∞= ×
∞
44
5. Settling time ts
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 23
45
4.4 The relationship between the performance specifications and systemparameters
1. Rise time-system parameter
1)( =rtcAccording to definition:
1)sin(1
1)(2
=+−
−=−
θωζ
ωζ
rd
t
r tetcrn
πθω =+rdt
21 ζω
θπω
θπ
−
−=
−=
ndrt
)10( << ζ
)sin(1
1)(2
θωζ
ζω
+−
−=−
tetc d
tn
46
2. Peak time -system parameter
21 ζω
πωπ
−==
ndPt
( )c t
rt
pt
st
1
00.1
0.9
pσ
t
2
2 221
Pd n
t π πω ω ζ
= =−
(±0.05 or±0.02 ) × r (∞)
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 24
47
3. Percentage overshoot - system parameter
)sincos1(1
11)( 22
ttetc ddtn ωζωζ
ζ
ζω +−−
−= −
dPt ω
π=
21
2( ) 1 (cos sin )
1Pc t e
ζ π
ζ ζπ π
ζ
−−= − +
−
2 21 1n n nPd n
t π π πω ω ω
ω ω ξ ξ= = =
− −
2
2
1( ) 1 ( 1 cos sin )1
n ptp d p d pc t e t tζ ω ζ ω ζ ω
ζ
−= − − +−
48
2 21 1
2( ) 1 (cos sin ) 1
1Pc t e e
ζπ ζπ
ζ ζζπ π
ζ
− −− −= − + = +
−
( ) ( ) ( ) 1100% 100%( ) 1
P PP
c t c c tc
σ− ∞ −
= × ×∞
=
21P e
ζπ
ζσ−
−=
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 25
49
21 ζ
ζπ
σ −−
= ep
σp is only related to ζ.
0.5 1.0 1.50
2
22( )2
n
n ns
s sω
ζω ωΦ =
+ +
pσ
20
40
60
80
100%
ζ
pσPercentage overshoot
50
4. The settling time ts- - system parameter
)()()( ∞⋅∆=∞− cctc s
0.02 0.05or∆ = ±
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 26
51
0
1
2
111 ζ
+−
2
111 ζ
−−
21
1
nte ζω
ζ
−+
−1
nT
ζω=
T 2T 3T 4T t
( )c t
21
1
nte ζω
ζ
−−
−
∆ζ
ζω
=−
−
2
stn
1e
2n
s11ln1t
ζ∆ζω −=
21
n ste ζω
ζ
−
= ∆−
52
9.00 << ζ
ns
4tζω
=
ns
3tζω
=
02.0=∆
05.0=∆
approximately:
for
for
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 27
53
Example: ( )R s ( )C s+
−( 1)
Ks s +
01 K s+
The desired specifications are: 2 0 % , 1P Pt sσ = =
What should the value of K and K0 be? Determine the value of tsand tr .
Solution: The transfer function of the closed loop is
KsKKsK
sKKKssK
sRsC
++=
+++=
)1()1()()(
02
0 +
012, KKK nn +== ζωω
54
(1)2.0
21 =−
−
ζ
ζπ
e 456.0=ζ
(2)
11 2
=ζω
πωπ
−==
ndPt
ωn
sradn /53.3=ω
5.12K 2n == ω
nKK ζω21 0 =+ 178.00 =K
KsKKsK
sKKKssK
sRsC
++=
+++=
)1()1()()(
02
0 +
ζ
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 28
55
(3) tr
s65.01
t2
n
r =−
−=
ζω
θπ
(4) ts
stn
s 86.13==
ζω
rad1.11
tg2
1 =−
= −
ζζ
θ
for ∆=5%
56
sjsjsps
zsksC q
j
r
kkkkkj
m
ii 1
)]()][(([)(
)()(
1 1
1 ⋅−++++
+=
∏ ∏
∏
= =
=
ωσωσ
( ) ( )0
22 21 1
( )1
q rk kj
j kjk k k k
saaC ss s p s
α β
ζ ω ω ζ= =
+= + +
+ + + −∑ ∑
1. Step response of a higher order system
4.5 Transient response of higher-order systems
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 29
57
( )c t)(tr
1
0 t t
( )c t
0
1)(tr
( )c t
t0
1)(tr
2 20
1 1
20
1 1
( ) ( cos 1 sin 1 )
sin( 1 )
j k k
j k k
q rp t t
j k k k k k kj k
q rp t t
j k k k kj i
C t a a e e B t C t
a a e D e t
ζ ω
ζ ω
ω ζ ω ζ
ω ζ φ
− −
= =
− −
= =
= + + − + −
= + + − +
∑ ∑
∑ ∑
58
3. The dominant poles of higher-order systems
2 20
1 1
20
1 1
( ) ( cos 1 sin 1 )
sin( 1 )
j k k
j k k
q rp t t
j k k k k k kj k
q rp t t
j k k k kj i
C t a a e e B t C t
a a e D e t
ζ ω
ζ ω
ω ζ ω ζ
ω ζ φ
− −
= =
− −
= =
= + + − + −
= + + − +
∑ ∑
∑ ∑
Dr.-Ing. Wei Wang, Dept. of Automation, Shanghai Jiao Tong University
Automatic Control System 30
59
σ
j[ ]s
主导极点
Re[ ] 1Re[ ] 5
poles of closed loopdominant poles
<
there is no zeros of the closed loop near the dominant poles
dominant closed loop poles