Module 8
System Type:
Steady-State Error and Muriple Control
(3 hours)
Definitionof the system type
The kind of input of system
How to identify steady-state error quickly
How to eliminatesteady-state error
8.1
8.1.1 Why we should consider the
actuating error Ea (P125)
The true error E=R-C is not
observable,since the output
has to be measured by H(s)
before its value is known.
The actuating error Ea =R-C’
is observable,since C’=HC
can be measured.
For the study convenience,we
now make a important
assumption,H=1,that E= Ea
aE CR
'CHC?
G
H
RGHGCE 1 1
R
G?
1
1
8.1.2 Errors of non-unity and unity
feedback
aE CR
'CHC?
G
H
CR
H
1 GH
'1 RRH? 'E
Non-unity feedback sys.
Unity feedback system
HCRE a
)1( CRHH
'HE?
aEHE
1'
8.2 The steady-state error ess ( ))(?e
)()(1 1)()()( sRsGsCsRsE
)(1
1)(lim)(lim)(
00 sG
ssRssEe
ss?
)(?e
is determined by two factors:
)(tr )(sG
and
8.3 The sys,type n – the integral elements
number in the forward path of sys.
The system type indicates what order of input signals can
the given system ―track‖ with zero steady state error,The
―order‖ here refers to the power of s in the Laplace
transform of the input signal,
To see this,we will investigate the steady state error of
various system types due to impulse R(s)=1,step R(s)=1/s,
ramp R(s)=1/s2 and acceleration R(s)=1/s3 inputs.
系统类型表明什么阶次的输入信号能为系统无稳态误差的跟踪 。 这里的阶次引用的是 Laplace变换中输入信号的幂次 。
为了看清这一点,我们来观察由输入信号为脉冲 R(s)=1,
阶跃 R(s)=1/s,斜坡 R(s)=1/s2 和加速度 R(s)=1/s3 时,不同系统的稳态误差 。
在 … 条件下
)1(
)1('
1
1
sTs
sK
j
q
j
n
i
m
i
j
q
j
i
m
j
p
z
KK
1
1
'
Where,–––– the open-loop gain
given by this formula,
)(lim'
0
sGsK n
s?
8.4 Errors and the error constants
(Steady-State Error Coefficients)
误差和误差系数（稳态误差系数）
How to calculate steady-state error simply? –
Whether or not we must calculate steady-
state error by using Laplace Inverse
Transform? ss
e
1 Step Input
ssR
1)(?
)(lim1
1
)(1
1lim
)(1
11lim)(lim
0
000 sGsGsGs
sssEe
s
sssss
Define
)(lim 0 sGK sp
–––the position error constant
For the type 0 sys.,'KK
p?
–––the open-loop gain
For the type 1 sys.,
pK
For the type 1 sys.,
pK
.,2,1,
1
1
0
1
1
.0,
1
1
'1
1
s y st y p e
K
s y st y p e
KK
e
p
p
ss
2 Ramp Input
2
1)(
ssR?
Define –––the velocity error constant
)(lim 0 ssGK sv
)(lim
1
)(
1lim
)(1
11lim)(lim
0
0200 ssGssGssGs
sssEe
s
sssss
For the type 0 sys.,0?
vK
–––the open-loop gainFor the type 1 sys.,'KK
v?
.,2,0
.1,
'
1
.0,
1
s y sty p e
s y sty p e
K
s y sty p e
K
e
v
ss
For the type 2 sys.,vK
3 Acceleration Input
.,3,0
.2,
'
1
.1,0,
1
s y sty p e
s y sty p e
K
s y sty p e
K
e
a
ss
3
1)(
ssR?
)(li m
1
)(
1li m
)(1
11li m)(li m
2
0
220300 sGssGsssGssssEe
s
sssss
Define – the acceleration error constant
)(lim 20 sGsK sa
For the type 0 sys.,0?aK
For the type 1 sys.,0?
aK –––
the open-loop gainFor the type 2 sys.,'KK
a?
Sample Problem,A unity-feedback system,
when its inputs are
)0(,21)(1)( 332210 3 seRtRtRtRtr ts
sse
Solution,From the linear theorem,we know that:
avp
ss KRKRKRe
11
1
1
210
),0:( 3 tw h eneN o t e ts
For Example,For a type 2 sys.,
)1(
)1)(1(')(
2
21
1?
Tss
ssKsG
',,KKKK avp
''
11
1
1 2
210 K
R
KRRRe ss
For Example,For a type 1 sys.,
)1(
)1)(1(')( 21
2?
Tss
ssKsG
0,', avp KKKK
01'11 1 210 RKRRe ss
For Example,For a type 0 sys.,
0,0,' avp KKKK
0101'1 1 210 RRKRe ss )1)(1)(1(
)1)(1(')(
321
21
3
sTsTsT
ssKsG
8.5 The equivalent unity-feedback sys,open-loop
T.F,G(s)’ derived from non-unity-feedback sys,
open-loop T.F,G(s)
8.6 Steady-state Error caused by disturbance input
CR E
D
1G 2G
DGGGRGG GGC
21
2
21
21
11 DGG
GR
GGCRE 21
2
21 11
1
For analysis convenience,assume that:
)(
)('
)1(
)1('
)(
1
11
1
1
1
1
11
1
1
sNs
sMK
sTs
sK
sG
n
i
q
i
n
j
m
j
)(
)('
)1(
)1('
)(
2
22
1
1
2
2
22
2
2
sNs
sMK
sTs
sK
sG
n
i
q
i
n
j
m
j
8.6.1 For Command input R:
212121
21
0
2
22
1
110 ''
lim
)(
)('
)(
)('1
1lim
21
21
21
MMKKNNs
NNsRs
sNs
sMK
sNs
sMKRse nn
nn
s
nn
sss
.),2,1(,2,1,0
.)0(0,
'1
1
21
21
1
s y sty p ennn
s y sty p ennn
K
sR
Where,nnn
21
is sys,type — integral elements
number in forward path.
Conclusion 1,The error caused by command input R is
determined by the integrators in R(s)-C(s) forward path.
8.6.2 For Disturbance input D:
212121
221
0
2
22
1
11
2
22
0 ''
lim
''
1
'
lim
21
1
21
2
MMKKNNs
MKNs
Ds
Ns
MK
Ns
MK
Ns
MK
Dse
nn
n
s
nn
n
s
ss
,2,1,0
0,
'1
1
1
1
1
n
n
K
sD
Where,
1n
is the sys,type for disturbance input D(s)—the
integrators number in D(s)-C(s) feedback path.
Conclusion 2,The error caused by disturbance input D(s)
is determined by the integrators in D(s)-C(s) feedback path.
Conclusion on 8.6
The integrator must be set in front of the action
point of the disturbance input D if possible,in
order to eliminate the total error (the error
caused by the command input R and the error
caused by the disturbance D)
积分环节应该尽可能放置在扰动作用点的前面，以便消除总误差（由指令输入 R引起的误差和扰动引起的误差）
Sample Problems:
P8.1 (Page 141-142)
SP,We must be all attention!
There are two unity-feedback systems:
)1(
4
21 ssG )1(
)12(4
22?
ss
sG
If input 25)( ttr?
21 ssss ee
Solution:
System 1,4',,.2 KKKKs y sT y p e
avp
5.2410125,1
a
ss Keg e tc a nweSo
Is that result true? Not !
From the closed-loop characteristic equation:
)1(
4
21 ssG
04)1(2ss 0423 ss
2
7
2
1,2
3,21 jss
At this case,no steady-state
exists,is meaningless.sse
System 2:
4',,.2 2222 KKKKs y sT y p e avp
5.2410125,2
a
ss Keg e tc a nweSo
We can prove this system is stable,
that is,all closed-loop poles are in
the left half s plane,So,)1( )12(4 22 ss sG
The Basis of the Steady-State
Recall Laplace Transform:
8.7 Muriple Control
Recall
the
example
in
Module
1—
Muriple
Control
Forward Control
8.7.1 Eliminate the the error caused by the
command input R
CR E
1G 2G
RGGGGC
21
21
1
RGGCRE
211
1
E CR
1G 2G
cG
RGG GGGGGRGG GGGC cc
21
2121
21
21
11)1(?
Select:
21
1 GGG c?
We can get
0,1/ ERC
0?sse
E CR
1G 2G
cG
R
GG
GGGGC c
21
221
1?
Select:
2
1 GG c?
We can get
0,1/ ERC
0?sse
8.7.2 Eliminate the the error caused by the
disturbance input D
R CE
D
1G 2G
DGGGC R
21
2
0 1
DGGGCRE
21
2
1
CE
D
1G 2G
cG
D
GG
GGGE c
R
21
21
0 1
)1(
Select:
1
1 GG c
We can get
0?E
00Rsse
SP,Two voltage control systems—which one has error?
How to eliminate this error for the system with error?
图 CP1-1 自动调压系统
LOAD
LOAD
Ru
Ru
Cu
Cu
Eu Eu
SP,Find C(s)/R(s)
R C
s
2s
eR
g s1
1K
2K
s
K3
s
1
R C
2s
s?
eR
g
s
1
1K?
1?
s
K3
2K s
1
11 121 sKsP
11 232 ssKsP
1111 323 sssP
11111 44 ssRgP
e
)1111111(1 321 ssRgssKsRgsKsRgsK
eee
ssR
g
ss
K
sR
g
sKsR
g
s
K
eee
11111111 3
2
1
R C
2s
s?
eR
g
s
1
1K?
1?
s
K3
2K s
1
1111 5225 sKssP
1111 6326 ssKssP
11111 727 sKsRgP
e
11111 838 ssKsRgP
e
1111)()(1 919 ssKsP
1/
9
1?
i
iiP
RC
R C
2s
s?
eR
g
s
1
1K?
1?
s
K3
2K s
1
Instructional objectives:
At the end of this lecture students should
be able to
Identify the system type
Calculate the steady state error
Select the way to eliminate steady-state
error
System Type:
Steady-State Error and Muriple Control
(3 hours)
Definitionof the system type
The kind of input of system
How to identify steady-state error quickly
How to eliminatesteady-state error
8.1
8.1.1 Why we should consider the
actuating error Ea (P125)
The true error E=R-C is not
observable,since the output
has to be measured by H(s)
before its value is known.
The actuating error Ea =R-C’
is observable,since C’=HC
can be measured.
For the study convenience,we
now make a important
assumption,H=1,that E= Ea
aE CR
'CHC?
G
H
RGHGCE 1 1
R
G?
1
1
8.1.2 Errors of non-unity and unity
feedback
aE CR
'CHC?
G
H
CR
H
1 GH
'1 RRH? 'E
Non-unity feedback sys.
Unity feedback system
HCRE a
)1( CRHH
'HE?
aEHE
1'
8.2 The steady-state error ess ( ))(?e
)()(1 1)()()( sRsGsCsRsE
)(1
1)(lim)(lim)(
00 sG
ssRssEe
ss?
)(?e
is determined by two factors:
)(tr )(sG
and
8.3 The sys,type n – the integral elements
number in the forward path of sys.
The system type indicates what order of input signals can
the given system ―track‖ with zero steady state error,The
―order‖ here refers to the power of s in the Laplace
transform of the input signal,
To see this,we will investigate the steady state error of
various system types due to impulse R(s)=1,step R(s)=1/s,
ramp R(s)=1/s2 and acceleration R(s)=1/s3 inputs.
系统类型表明什么阶次的输入信号能为系统无稳态误差的跟踪 。 这里的阶次引用的是 Laplace变换中输入信号的幂次 。
为了看清这一点,我们来观察由输入信号为脉冲 R(s)=1,
阶跃 R(s)=1/s,斜坡 R(s)=1/s2 和加速度 R(s)=1/s3 时,不同系统的稳态误差 。
在 … 条件下
)1(
)1('
1
1
sTs
sK
j
q
j
n
i
m
i
j
q
j
i
m
j
p
z
KK
1
1
'
Where,–––– the open-loop gain
given by this formula,
)(lim'
0
sGsK n
s?
8.4 Errors and the error constants
(Steady-State Error Coefficients)
误差和误差系数（稳态误差系数）
How to calculate steady-state error simply? –
Whether or not we must calculate steady-
state error by using Laplace Inverse
Transform? ss
e
1 Step Input
ssR
1)(?
)(lim1
1
)(1
1lim
)(1
11lim)(lim
0
000 sGsGsGs
sssEe
s
sssss
Define
)(lim 0 sGK sp
–––the position error constant
For the type 0 sys.,'KK
p?
–––the open-loop gain
For the type 1 sys.,
pK
For the type 1 sys.,
pK
.,2,1,
1
1
0
1
1
.0,
1
1
'1
1
s y st y p e
K
s y st y p e
KK
e
p
p
ss
2 Ramp Input
2
1)(
ssR?
Define –––the velocity error constant
)(lim 0 ssGK sv
)(lim
1
)(
1lim
)(1
11lim)(lim
0
0200 ssGssGssGs
sssEe
s
sssss
For the type 0 sys.,0?
vK
–––the open-loop gainFor the type 1 sys.,'KK
v?
.,2,0
.1,
'
1
.0,
1
s y sty p e
s y sty p e
K
s y sty p e
K
e
v
ss
For the type 2 sys.,vK
3 Acceleration Input
.,3,0
.2,
'
1
.1,0,
1
s y sty p e
s y sty p e
K
s y sty p e
K
e
a
ss
3
1)(
ssR?
)(li m
1
)(
1li m
)(1
11li m)(li m
2
0
220300 sGssGsssGssssEe
s
sssss
Define – the acceleration error constant
)(lim 20 sGsK sa
For the type 0 sys.,0?aK
For the type 1 sys.,0?
aK –––
the open-loop gainFor the type 2 sys.,'KK
a?
Sample Problem,A unity-feedback system,
when its inputs are
)0(,21)(1)( 332210 3 seRtRtRtRtr ts
sse
Solution,From the linear theorem,we know that:
avp
ss KRKRKRe
11
1
1
210
),0:( 3 tw h eneN o t e ts
For Example,For a type 2 sys.,
)1(
)1)(1(')(
2
21
1?
Tss
ssKsG
',,KKKK avp
''
11
1
1 2
210 K
R
KRRRe ss
For Example,For a type 1 sys.,
)1(
)1)(1(')( 21
2?
Tss
ssKsG
0,', avp KKKK
01'11 1 210 RKRRe ss
For Example,For a type 0 sys.,
0,0,' avp KKKK
0101'1 1 210 RRKRe ss )1)(1)(1(
)1)(1(')(
321
21
3
sTsTsT
ssKsG
8.5 The equivalent unity-feedback sys,open-loop
T.F,G(s)’ derived from non-unity-feedback sys,
open-loop T.F,G(s)
8.6 Steady-state Error caused by disturbance input
CR E
D
1G 2G
DGGGRGG GGC
21
2
21
21
11 DGG
GR
GGCRE 21
2
21 11
1
For analysis convenience,assume that:
)(
)('
)1(
)1('
)(
1
11
1
1
1
1
11
1
1
sNs
sMK
sTs
sK
sG
n
i
q
i
n
j
m
j
)(
)('
)1(
)1('
)(
2
22
1
1
2
2
22
2
2
sNs
sMK
sTs
sK
sG
n
i
q
i
n
j
m
j
8.6.1 For Command input R:
212121
21
0
2
22
1
110 ''
lim
)(
)('
)(
)('1
1lim
21
21
21
MMKKNNs
NNsRs
sNs
sMK
sNs
sMKRse nn
nn
s
nn
sss
.),2,1(,2,1,0
.)0(0,
'1
1
21
21
1
s y sty p ennn
s y sty p ennn
K
sR
Where,nnn
21
is sys,type — integral elements
number in forward path.
Conclusion 1,The error caused by command input R is
determined by the integrators in R(s)-C(s) forward path.
8.6.2 For Disturbance input D:
212121
221
0
2
22
1
11
2
22
0 ''
lim
''
1
'
lim
21
1
21
2
MMKKNNs
MKNs
Ds
Ns
MK
Ns
MK
Ns
MK
Dse
nn
n
s
nn
n
s
ss
,2,1,0
0,
'1
1
1
1
1
n
n
K
sD
Where,
1n
is the sys,type for disturbance input D(s)—the
integrators number in D(s)-C(s) feedback path.
Conclusion 2,The error caused by disturbance input D(s)
is determined by the integrators in D(s)-C(s) feedback path.
Conclusion on 8.6
The integrator must be set in front of the action
point of the disturbance input D if possible,in
order to eliminate the total error (the error
caused by the command input R and the error
caused by the disturbance D)
积分环节应该尽可能放置在扰动作用点的前面，以便消除总误差（由指令输入 R引起的误差和扰动引起的误差）
Sample Problems:
P8.1 (Page 141-142)
SP,We must be all attention!
There are two unity-feedback systems:
)1(
4
21 ssG )1(
)12(4
22?
ss
sG
If input 25)( ttr?
21 ssss ee
Solution:
System 1,4',,.2 KKKKs y sT y p e
avp
5.2410125,1
a
ss Keg e tc a nweSo
Is that result true? Not !
From the closed-loop characteristic equation:
)1(
4
21 ssG
04)1(2ss 0423 ss
2
7
2
1,2
3,21 jss
At this case,no steady-state
exists,is meaningless.sse
System 2:
4',,.2 2222 KKKKs y sT y p e avp
5.2410125,2
a
ss Keg e tc a nweSo
We can prove this system is stable,
that is,all closed-loop poles are in
the left half s plane,So,)1( )12(4 22 ss sG
The Basis of the Steady-State
Recall Laplace Transform:
8.7 Muriple Control
Recall
the
example
in
Module
1—
Muriple
Control
Forward Control
8.7.1 Eliminate the the error caused by the
command input R
CR E
1G 2G
RGGGGC
21
21
1
RGGCRE
211
1
E CR
1G 2G
cG
RGG GGGGGRGG GGGC cc
21
2121
21
21
11)1(?
Select:
21
1 GGG c?
We can get
0,1/ ERC
0?sse
E CR
1G 2G
cG
R
GG
GGGGC c
21
221
1?
Select:
2
1 GG c?
We can get
0,1/ ERC
0?sse
8.7.2 Eliminate the the error caused by the
disturbance input D
R CE
D
1G 2G
DGGGC R
21
2
0 1
DGGGCRE
21
2
1
CE
D
1G 2G
cG
D
GG
GGGE c
R
21
21
0 1
)1(
Select:
1
1 GG c
We can get
0?E
00Rsse
SP,Two voltage control systems—which one has error?
How to eliminate this error for the system with error?
图 CP1-1 自动调压系统
LOAD
LOAD
Ru
Ru
Cu
Cu
Eu Eu
SP,Find C(s)/R(s)
R C
s
2s
eR
g s1
1K
2K
s
K3
s
1
R C
2s
s?
eR
g
s
1
1K?
1?
s
K3
2K s
1
11 121 sKsP
11 232 ssKsP
1111 323 sssP
11111 44 ssRgP
e
)1111111(1 321 ssRgssKsRgsKsRgsK
eee
ssR
g
ss
K
sR
g
sKsR
g
s
K
eee
11111111 3
2
1
R C
2s
s?
eR
g
s
1
1K?
1?
s
K3
2K s
1
1111 5225 sKssP
1111 6326 ssKssP
11111 727 sKsRgP
e
11111 838 ssKsRgP
e
1111)()(1 919 ssKsP
1/
9
1?
i
iiP
RC
R C
2s
s?
eR
g
s
1
1K?
1?
s
K3
2K s
1
Instructional objectives:
At the end of this lecture students should
be able to
Identify the system type
Calculate the steady state error
Select the way to eliminate steady-state
error
