-
Control System(1)
控制系統
‧清雲科技大學電機工程系
• Ya-Fu Peng• February, 2009
1
Control Systems
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Control System(1) 2
內容
簡介
頻域模型
時域模型
時間響應
互聯子系統之簡化
穩定度
穩態誤差穩態誤差
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Control System(1)
穩態誤差(steady-state error):用一規定的測試輸入,當 時輸入與輸出之間的差。
∞→t
Chapter 7 Steady-state error
穩態誤差的來源
非線性源產生-齒隙,馬達的驅動電源…
本身結構與應用的輸入型態而來
3
-
Control System(1)
測試輸入信號:
4
Chapter 7
-
Control System(1)
應用於穩定系統-“穩態"誤差的計算
方塊圖表示
5
Chapter 7
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Control System(1)
)()(1
1)()](1[
)()()( )()()(
)()(1
)()()()(
sRsG
sRsT
sRsTsRsCsRsE
sRsG
sGsRsTsC
+=−=
−=−=+
==
)()(1
1lim)()](1[lim
)(lim)(lim)(
00
0
sRsG
ssRsTs
sEsteee
ss
stss
+=−=
⋅==∞=
→→
→∞→
單位負回授系統的穩態誤差
利用終值定理
6
Chapter 7
-
Control System(1)
若輸入為單位步階之系統試求穩態誤差。 107
5)( 2 ++=
sssT
21
110757lim
)(lim)(lim
2
2
0
0
=
⋅++++
⋅=
⋅==
→
→∞→
ssssss
sEstee
s
stss
[ ]107571)()(1)( 2
2
++++
⋅=⋅=ssss
ssRs - TsE
7
Chapter 7
-
Control System(1)
以G(s)表穩態誤差
)()(1
1
)()(1
)()(
)()()(
)()(1
)()(
sRsG
sRsG
sGsR
sYsRsE
sRsG
sGsY
+=
+−=
−=+
=
)(1)(lim)(lim
00 sGsRssEse
ssss +⋅
=⋅=→→
+)(sR)(sG
)(sY-
8
)(sE
Chapter 7
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Control System(1)
sAsR =)(
LL
LL
))(())(()(:21
21
pspsszszssGif n ++
++=
)(lim1)(1lim
00 sG
AsG
Aes
sss
→→ +
=+
=
( ) ∞=⇒≥→
)(lim 1 b0
sGns
0)( =∞
=∞⇒Ae( ) k
ppzzsGn
s==⇒=
→ L
L
21
21
0)(lim 0 a
kAe+
=∞⇒1
)(
步階輸入:
9
Chapter 7
-
Control System(1)
)(lim)(lim
)(1lim
0020 ssG
AssGs
AsA
sGse
sssss
→→→
=+
=+
=
∞=⇒≥→
)(lim 20
ssGns
kAe =∞⇒ )(k
ppzzssGn
s==⇒=
→ L
L
21
21
0)(lim 1
0)( =∞
=∞⇒Ae
0)(lim 00
=⇒=→
ssGns
∞==∞⇒0
)( Ae
2)( sAsR =斜坡輸入:
10
Chapter 7
-
Control System(1)
∞=++
⋅=⇒≥→→ L
L
)()(lim)(lim 31
12
0
2
0 psszsssGsn nss
kAe 2)( =∞⇒k
ppzzsGsn
s==⇒=
→ L
L
21
212
0)(lim 2
02)( =∞
=∞⇒Ae0)(lim 2 2
0=⇒<
→sGsn
s
∞==∞⇒0
2)( Ae
3
2)(sAsR =拋物線輸入:
)(lim2
)(2lim2
)(1lim 2
0
22030 sGsA
sGssA
sA
sGse
sssss
→→→
=+
=+
=
11
Chapter 7
-
Control System(1)
+( )sR ( )( )( )43
2120+++ss
s( )sE ( )sC ( ) ( )( )( )tutttututr
2555分別為
-
( ) ( ) ( ) sCsRsE −=
( ) ( ) ( )sGs - EsR ⋅=
( ) ( )( )sG sRsE
+=
1
Ex:系統無任何積分,求各自之穩態誤差
12
Chapter 7
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Control System(1)
ssR tr 5)(5)( (a) =⇒=
215
2015
)(lim15
)(15lim
)(1)(lim)(lim
0
000
=+
=+
=
+=
+⋅=⋅=
→
→→→
sG
sGsGsRssEse
s
sssss
2
5)(5)( (b)s
sRt tr =⇒=
∞==
+++
⋅==
+=
+⋅=
→→
→→
05
)4)(3()2(120lim
5)(lim
5)(
5lim)(1
5
lim
00
0
2
0
ssssssG
ssGssGsse
ss
ssss
13
Chapter 7
-
Control System(1)
32 10)()(5)( (c)
ssR tuttr =⇒=
)(1
10
lim3
0 sGse
sss +=
→∞==
⋅=
→0
10)(lim
102
0sGs
s
14
Chapter 7
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Control System(1)
步階輸入 位置常數(position constant)
斜坡輸入 速度常數(velocity constant)
拋物線輸入 加速度常數(Acceleration constant)
⇒)(A
⇒)(At
⇒)( 2At
靜態誤差常數(單位回授系統)
ass k
Ae 2 =⇒
pss k
Ae+
=⇒1
)(lim 20
sGsksa →
=
)(lim0
sGksp →
=
)(lim0
ssGksv →
=v
ss kAe =⇒
pk
ak
vk
15
Chapter 7
-
Control System(1)
+)(sR)(sG
)(sE)(sY
-
例題:對下圖求其穩態誤差常數.並在標準的步階、斜坡、拋物線輸入下求穩態誤差。
16
Chapter 7
-
Control System(1)
161
1115
1210852540)(lim
0=
+=⇒=
××××
==→
psssp k
esGk
∞==⇒==→
vsssv k
essGk 10)(lim0
∞==⇒==→
asssa k
esGsk 10)(lim 20
)12)(10)(8()5)(2(540)()a(
+++++
=sss
sssG
17
Chapter 7
-
Control System(1)
01
11
1)(lim0
=∞+
=+
=⇒∞==→
psssp k
esGk
∞==⇒==→
asssa k
esGsk 10)(lim 20
)12)(10)(8()5)(2(540)()b(+++
++=
sssssssG
151115
1210852540)(lim
0==⇒=
××××
==→
vsssv k
essGk
18
Chapter 7
-
Control System(1)
01
11
1)(lim0
=∞+
=+
=⇒∞==→
psssp k
esGk
)12)(10)(8()5)(2(540)()c( 2 +++
++=
sssssssG
011)(lim0
=∞
==⇒∞==→
vsssv k
essGk
152215
1210852540)(lim 2
0==⇒=
××××
==→
asssa k
esGsk
19
Chapter 7
-
Control System(1)
+( )sR ( )( )( )( ) LL
LL
21
21
pspsszszsk
n ++++( )sE ( )sC
-
單位負回授系統之系統組態(System Type):順向路徑的純積分次數,即為n值。
n=0則type 0; n=1則type 1;依此類推…
20
Chapter 7
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Control System(1) 21
Chapter 7
-
Control System(1)
EX:試求單位負回授系統 之型態、靜態誤差常數及穩態誤差。
)9)(7()8(1000)(
+++
=ss
ssG
∞=⇒=∞=⇒=
ssa
ssv
ekek
00
0079.0128
11
1
98.12663
800097
81000
≅=+
=⇒
==××
=
pss
p
ke
k
Sol:
type 0
22
Chapter 7
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Control System(1)
Sol:
type 1 斜坡輸入
EX:試求k值使系統在穩定狀態有10%穩態誤差。
+( )sR ( )( )( )( )876
5+++
+ssss
sk( )sE ( )sC-
)(lim11
0ssGk
esv
ss
→
==
6725
87610876
5)8)(7)(6(
)5(lim100
=⋅⋅⋅
=⇒
⋅⋅⋅
=+++
+⋅==
→
k
kssss
sksksv
23
-
Control System(1)
非單位回授的穩態誤差
24
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Control System(1)
)(sR )10(100+ss
)(sE)(sY_
)5(1+s
EX:試求型態及適當之誤差常數和對一單位步階輸入之穩態誤差。
+
25
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Control System(1)
0 type⇒−−+
+=
−−++++
=
+−
+⋅
++
+=−+
=
4005015)5(100
5001001005015)5(100
)10(100
)5(1
)10(1001
)10(100
)()()(1)(
23
23
sss s
ssss s
sssss
sssGsHsG
sGGe
45
400500)(lim
0−=
−==⇒
→sGk esp
4)
45(1
11
1−=
−+=
+=⇒
pss k
e
Sol:
26
-
Control System(1)
PF
FP
PF
FPS
P
PPPPF
ΔΔ
=
ΔΔ⋅=
ΔΔ
==
→Δ
→Δ→Δ→Δ
0
000:
lim
limPPFFlimlim
參數的比例變化
函數的比例變化
PF
FPS PF ∂
∂⋅=:
靈敏度:函數內的比例變化與參數比例變化之比值。
27
-
Control System(1)
+)(sR)( ass
k+
)(sE)(sY
-
kassksT++
= 2)(
EX:算出改變參數a對系統之靈敏度,又如何降低?
kassas
kassks
kasska
aT
TaS aT ++
−=
++⋅−
++
=∂∂⋅= 222
2
: )()(
28
Sol:
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Control System(1)
EX:求上圖參數 k 和 a 對穩態誤差之靈敏度。
ka
assksk
e
sv
ss =
+⋅
==
→ )(lim
11
0
11: =⋅=∂∂⋅=
kkaa
ae
eaS ae
12: −=−⋅=
∂∂⋅=
ka
kak
ke
ekS ke
29
Sol:
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Control System(1)
+
)(sR))(( bsas
k++
)(sE)(sY
-
EX:試求參數 k 和 a 對穩態誤差之靈敏度。
kabab
abk
bsaskk
e
sp
ss +=
+=
+++
=+
=⇒
→1
1
))((lim1
11
1
0
Sol:
30
-
Control System(1)
kabk
kababkabb
kababa
ae
eaS ae +
=+−+
⋅
+
=∂∂⋅= 2
2
: )()(
kabk
kabab
kababk
ke
ekS ke +
−=
+−
⋅
+
=∂∂⋅= 2: )(
1 , :: 小於keae SSQ 敏度比例回授系統會降低靈⇒
31
-
Control System(1)
)()()( )()()( sRsTsYsYsRsE =−= 且
DuCxy BuAxx
+=+=&
( ) )(]1[)()](1[)( 1 sRssRsTsE BAIC −−−=−=⇒
狀態空間
( ) )(]1[lim)(lim 100
sRsssEsessss
BAIC −→→
−−⋅=⋅=⇒
32
-
Control System(1)
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
100
B⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
110-2012-0015-
A [ ]011−=C
)()2013616126(lim
)()20136
41(lim
23
23
0
230
sRsssssss
sRsss
sse
s
sss
++++++
⋅=
++++
−⋅=
→
→
∞==⇒=
==⇒=
→
→
016lim1)(
54
2016lim1)(
02
0
sss
sss
es
sR
es
sR
EX:
Sol:
33