3-5 steady-state error calculation the steady-state performance is an important characteristic of...
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3-5 steady-state error calculation The steady-state performance is an important ch
aracteristic of control system , it represses the ability to follow input signal and resist interference 。
一 .error and steady-state error
1.definition:
(1) according to output
e(t): system error,Cr(t): requested output, c(t) :practical output 。
)()()( tctCte rc
)(sG
)(sH
)(sC
-
)(sR
Steady-state error:
The steady-state error relies on the system's construction ,still the form of the input signal.But stability depends only on system’s construction.
)]()([lim)(lim tctCtee rt
ct
css
0 t
C(t)
essCr(t)
(2) according to input
)()()( tbtrte
)]()([lim)(lim tbtrteett
ss
)(sG
)(sH
)(sC
-
)(sR
)(sB
)(sE
2.calculation
(1).final theorem
when input is , final theor
em can be used.but when input is sine or co
sine signal,cannot used.
(2).from the definition
a.solve error TF
)]()([lim)(lim)(lim00
sBsRssEsteesst
ss
2
2
1,),(1),( tttt
)(
)(
sR
sE
b.solve
c. solve
d. Solve limit that is steady-state error.
If the system exists at the same time
input and interference,method is as
follows ：
)()(
)()( sR
sR
sEsE
)(te)(lim te
t
R(s)
N(s)
E(s)+
+
)(ser
)(sen
TF between error and input,
TF between error and interference 。
so:
)(ser
)]()()()([lim
)(lim
0
0
sNssRss
ssEe
eners
sss
)(sen
2
n(t)=-1(t)r(t)=t E(s)
-C(s)
12.0
1
s )1(
2
ss
exam:given a block diag. As follows.when input
r(t)=t and interference n(t)=-1(t),ess=?
to interference N(s),we hope it produce output 0,that is ,output is independent of the interference N(s) 。
when input and interference acts on a system,output is respectively:
G1(s)
N(s)R(s) E(s)
-
C(s)
H(s)
G2(s)
error produced by R(s) and N(s) :
)()()(1
1)(
)()()(
21
1
sHsGsGs
sRssE
er
er
)()()()()()()(
)()()(1
)()()(
)()()(
21
21
2
2
sNssRssEsEsE
sHsGsG
sHsGs
sNssE
ener
en
en
in this exam.,stability must be assured first,or else,steady-state error cannot exist. 。
Characteristic is:
02)1(
2
12.0
11
0)()()(1 21
sss
sHsGsG
That is
System is stable 。from:
0321
23
8.02.1
042.12.0
aaaa
sss
，即
)1(2
12.01
1
2)1(
2
)(
2)1(
212.0
11
1)(
sss
sss
sss
s
en
er
The steady-state error relies on the system's construction ,still the form of the input signal.
4
7)]()()()([lim
)(lim
1)()(1)(
1)()(
0
0
2
sNssRss
ssEes
sNttn
ssRttr
eners
sss
二 . steady-state error analysis when input acts on a system
)()()(1
1)( sR
sHsGsE
1)()(,0
)()()()(
00
00
sHsGswhen
sHsGs
ksHsG
v---open-loop TF
whenγ=0,1,2 called respectively 0 type system,Ⅰtype system,Ⅱ type system (in generalγ<2)
so
)12()1(
)12()1()()(
2222
21
2122
2100
sTsTsT
ssssHsG
)()()(1
1lim)(lim
)()()(1
1)(
00
00
00
sRsHsG
sk
sssEe
sRsHsG
sk
sE
ssss
Define kp,kv,ka error coefficient 。
Step input ,represented by kp
position error coefficient 。Speed input, represented by kv
velocity error coefficient 。
Acceleration input, represented by ka
Acceleration error coefficient 。
,lim)()(lim0
000 s
ksHsG
s
kk
ssp
,lim)()(lim10
0010 s
ksHsG
s
kk
ss
,lim)()(lim20
0020 s
ksHsG
s
kk
ssa
20
10
01
1
lim
)()(1
1lim
1
)()(1
1lim
/1)(
)()()(1
1lim)(lim
)()()(1
1)(
0
00
0
00
0
00
00
00
k
ks
s
sHsGskssHsG
sk
se
ssRwhen
sRsHsG
sk
sssEe
sRsHsG
sk
sE
s
ssss
ssss
20
1/1
0
lim
)]()(1[
1lim
1
)()(1
1lim
/1)(
1
0
00
0
2
00
0
2
kks
s
sHsGsk
s
ssHsGsk
se
ssRwhen
s
s
sss
30
2/11
0
lim
)]()(1[
1lim
1
)()(1
1lim
/1)(
2
0
0020
3
00
0
3
kks
s
sHsGsk
s
ssHsGsk
se
ssRwhen
s
s
sss
system
Stationary error coefficient
Steady state error
type
0
Ⅰ
Ⅱ
)(1)( 0 tRtr tVtr 0)( 2)( 20tAtr
k
R
10
k
V0
k
A0
0
00
0
00k
k
k
pk vk ak
sse
to increase system type and open-loop gain can increase system's accuracy, but would lower stability, and must consider completely.
exam.: given a system.its block diag. is
when H(s)=1and H(s)=0.5,solve ess 。
12
102 ss
)(sH
)(sC
-
)(15)( ttr
When H(s)=1,open-loop TF
When H(s)=0.5,
0,10,12
10)()(
2
k
sssHsG
11
5
101
5
10
k
Ress
0,5,12
5.010)()(
2
kss
sHsG
6
5
51
5
10
k
Ress
if in exam. above, H(s)=1,ess=0.2,solve k=?
when , solve ess i
n exam. above.
Because 0 type system ess = ∞ under the speed input and acceleration input, from addition rule,ess=∞
2412.0
51,
100
ssss e
Rkso
k
Re
1)(,2
1)(1)( 2 sHttttr
三 .steady state error analysis under interference
We hope ess=0 under interference,but it is impossible 。
if input R(s)=0, under interference N(s), ess :
)()()()(1
)()(lim
)()()()(1
)()()(
21
2
0
21
2
sNsHsGsG
sHsGse
sNsHsGsG
sHsGsE
sss
G1(s)
N(s)R(s)
E(s)
-
C(s)
H(s)
G2(s)
)()()()(1
)()(lim
)()()()(1
)()()(
21
2
0
21
2
sNsHsGsG
sHsGse
sNsHsGsG
sHsGsE
sss
under interference N(s),steady state output changes, it is connected with open-loop TF
G(s)=G1(s)G2(s)H(s) ,interference N(s) and its action position.
r(t)=0
-C(t)
12
Ts
k
)(1)( 0 tMtn
0k s
k1
(a)
r(t)=0
-
C(t)1
2
Ts
k
)(1)( 0 tMtn
0k s
k1
(b)
0
00
210
21
0
0
210
2
0
)1(1
)1(lim:)(
0
)1(1
1lim:)(
k
M
s
M
Tsskkk
Tsskk
seb
s
M
Tsskkk
Tsk
sea
sss
sss
The action position is different, the steady state error is different too.
in front of action position,if increase an integer( proportion) to replace
)1
1(0
0 sTk 0k
r(t)=0
-
C(t)1
2
Ts
k
)(1)( 0 tMtn
)1
1(0
0 sTk
s
k1
(b)
improving integer No.in front of interference action point can decrease or cancel steady state error under interference , but lower the system's stability.
0
)1()
11(1
)1(lim 0
21
00
21
0
s
M
Tsskk
sTk
Tsskk
ses
ss
in a word, to decrease steady state error under input, we can increase open-loop TF integer No. and gain.
in order to decrease interference error,we must improve integer No.and gain in front of interference action point.this can lower the system's stability,and even transient response goes bad.
Following is a good method.
四 . method of decreasing or canceling steady state error
1. compensation according to interference If interference can be measured ,at the same
time,the influence to system is clear and definite, compensation can then improve the steady state accuracy.
G2(s)
Gn(s)
G1(s)
C(s)
R(s) E(s)
N(s)
-
+
Under interference,output:
completely cancel the influence to system output.
adding compensating equips makes the system's steady state output independent of interference, that is to say, steady state error 0.
,0)(,)(1)(
)()()(1
)()()()()(
1
21
212
sCthensGsGif
sNsGsG
sGsGsGsGsC
n
n
Exam.:
System output:
)1( sTs
k
m
m
1
1
1 sT1k
)(sGn
-
R(s)=0
N(s)
C(s)
Compensation equip.
amplifier filter
)(
)1)(1(1
]1
)(1[)1(
)(
1
1
1
1
sN
sTsTs
kksTk
sGsTs
k
sC
m
m
nm
m
n
if select the system's output
independent of interference,physically too difficult to
realize, because the order of denominator of a TF is higher
than numerator.
if select when steady state,
This is completely steady state compensated,and easily realized.
),1(1
)( 11
sTk
sGn
,1
)(1k
sGn
0)(
)1)(1(1
]1
)1
(1[)1(
lim
)(lim)()(lim
1
1
1
1
1
0
0
sN
sTsTskk
sTk
ksTsk
s
scsctc
m
m
m
m
s
ns
nnt
2. Compensation according to given input.
If completely compensated,
R(s) G1(s) G2(s)
Gr(s)
-
C(s)
Compensation equipment
Similarly, complete compensation is too difficult to realize. usually adopt the steady-state compensation to cancel the steady-state error to some inputs.
)(
1)()()(
0)()()(
)()()(1
)()]()([)(
2
21
21
sGsGsCsR
sCsRsE
sRsGsG
sGsGsGsC
r
r
No compensation ， open-loop TF
typeⅠ,input is a ramp,there are
some steady-state error.
tVtrsTs
ksG
sT
ksGif 0
2
22
1
11 )(,
)1()(,
1)(
)1)(1()()()(
21
2121
sTsTs
kksGsGsG
21
0
kk
Vess
Introduce compensation Gr(s) to input
If select so
but is difficult to realize physically 。
)()()(1
)()(1
)()()(1
)()]()([)()(
21
2
21
21
sRsGsG
sGsG
sRsGsG
sGsGsGsRsE
r
r
2
2
2
)1(
)(
1)(
k
sTs
sGsGr
0)( sE
2
2 )1()(
k
sTssGr
If take ,so
Thus steady-state compensated 。
2
)(k
ssGr
)(lim
)()(
)(
)1)(1(1
)1(1
)()()(1
)(1
)(
0
20
0
21
21
2
2
2
21
22
ssEes
VsRtVtr
sR
sTsTskksTsk
ks
sRsGsG
sGks
sE
sss
3-8 using Matlab analyze in time domain
3.8.1 stability analysisPzmap :plot zeros and poles diag.Tf2zp:transform transfer function to zero-pole typeRoots:solve roots of a polynominal
27243
64523)(
.
235
23
4
4
sssss
sssssG
systemfollowingofstabilitytheupcheck
>> num=[3 2 5 4 6]
num =
3 2 5 4 6
>> den=[1 3 4 2 7 2]
den =
1 3 4 2 7 2
>> [z,p,k]=tf2zp(num,den)
z =
0.4019 + 1.1965i
0.4019 - 1.1965i
-0.7352 + 0.8455i
-0.7352 - 0.8455i
p =
-1.7680 + 1.2673i
-1.7680 - 1.2673i
0.4176 + 1.1130i
0.4176 - 1.1130i
-0.2991
k =
3
>> pzmap(num,den)
-2 -1.5 -1 -0.5 0 0.5-1.5
-1
-0.5
0
0.5
1
1.5Pole-Zero Map
Real Axis
Imag
inar
y A
xis
roots(den)
ans =
-1.7680 + 1.2673i
-1.7680 - 1.2673i
0.4176 + 1.1130i
0.4176 - 1.1130i
-0.2991
>> roots(num)
ans =
0.4019 + 1.1965i
0.4019 - 1.1965i
-0.7352 + 0.8455i
-0.7352 - 0.8455i
To illustrate the meaning of the following command.
3.8.2 transient characteristic analysis
Step:solve unit step response
num=[25]
num =
25
>> den=[1 4 25]
den =
1 4 25
>> step(num,den)
>> grid0 0.5 1 1.5 2 2.5 3
0
0.2
0.4
0.6
0.8
1
1.2
1.4Step Response
Time (sec)
Am
plitu
de
[num,den]=zp2tf([-2],[-1,-0.2+10*i,-0.2-10*i],100)
num =
0 0 100 200
den =
1.0000 1.4000 100.4400 100.0400
>> [y,x,t]=step(num,den)
subplot(2,1,1),plot(t,y)
[yi,xi,t]=impulse(num,den);
>> subplot(2,1,2),plot(t,yi)
0 5 10 15 20 250
1
2
3
0 5 10 15 20 25 30-10
-5
0
5
10
15
homework:
3-15
3-18
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