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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
Control System(1) 2
內容
簡介
頻域模型頻域模型
時域模型
時間響應
互聯子系統之簡化
穩定度
穩態誤差
Control System(1) 3
從物理系統的示意圖推導數學模型
Chapter 2 Modeling in the Frequency Domain
頻域:轉移函數 Ch2
時域:狀態方程式 Ch3
Control System(1) 4
Chapter 2
為何採用頻域分析-便利性
Control System(1) 5
∫∞ −−
==0
)()()]([ dtetfsFtf stL
ωσ js +=
Chapter 2
拉氏轉換(Laplace Transform)
定義
其中
Control System(1) 6
Chapter 2
拉氏轉換表
Control System(1) 7
Chapter 2
拉氏轉換表
Control System(1) 8
Chapter 2
拉氏轉換表
Control System(1)
試求 的拉氏轉換
9
)()( tuAetf at−=
∫∞ −=
0)()( dtetfsF st
asA+
=
( )∫∞ +−=
0dteA tas ( )
∞
=
+−
+−=
0t
taseas
A
)(0as
A+
−−=
∫∞ −−=
0dteAe stat
Chapter 2
Ex:
Control System(1)
試求 的反拉氏轉換
10
( )( )23
1+
=s
sF
→ 12s
( )ttue t3−
( )ttu
( )=⎥
⎦
⎤⎢⎣
⎡
+ 21
31
s-L
( ) →+ asF ( )tfe at−
Chapter 2
Ex:
Control System(1) 11
)()()()(
)()())(()(
)()()(
2
2
1
1
21
n
n
m
m
nm
psk
psk
psk
psk
pspspspssN
sDsNsF
++
+++
++
+=
++++=
=
LL
LL
Chapter 2
部分分式展開
狀況1: F(s)分母含有不同的實數根
Control System(1)
Chapter 2
部分分式展開
狀況2:F(s)分母的根為實根且重根
)()()()(
)()()()(
)()()(
1
11
1
2
1
1
21
n
nrrrr
nr
psk
psk
psk
psk
pspspssN
sDsNsF
++
+++
++
+=
+++=
=
−+− LL
L
12
Control System(1)
Chapter 2
部分分式展開
狀況3:F(s)分母的根為複數根或虛根
L
L
+++
++
+=
+++==
)()(
))(()(
)()()(
232
1
1
21
rqssksk
psk
rqsspssN
sDsNsF
13
Control System(1)
,其中
14
)(32)(32)(12)( tutytyty =+′+′′0)0()0( ==′ −− yy
ssYssYsYs 32)(32)(12)(2 =++
)0()0()()( 2 −− ′−−→′′ ysysYsty
)0()()( −−→′ yssYty
stu 1)( →
Chapter 2
Ex:
)3212(32)( 2 ++
=sss
sY
Control System(1)
( ) ( )81
421
)8()4(
)8)(4(32
)3212(32)(
321
2
++
+−=
++
++=
++=
++=
ssssk
sk
sk
sssssssY
Chapter 2
)8(1
)4(21)(
++
+−=
ssssY
)()(2)()( 84 tuetuetuty tt −− +−=
1 -L
)()21()( 84 tueety tt −− +−=
15
Control System(1)
,其中
16
6)(7)(9)(2 =+′+′′ tytyty0)0()0( ==′ −− yy
Chapter 2
練習題
Control System(1) 17
)2()2()1()2)(1(2)( 32
212 +
++
++
=++
=sk
sk
sk
sssF
( ),2
22
121 =+
=−→ss
k ( ) 212
22 −=+=
−→ssk
Chapter 2
EX:
2)1(
2)(!)12(
1
22
2
13 −=+
−=
−=
−→−→ ss sdssdFk
)2(2
)2(2
12)( 2 +
−+
−+
=sss
sY1 -L
)()222()( 22 tueteety ttt −−− −−=
Control System(1) 18
)52(
)52(3)(
2321
2
+++
+=
++=
ssksk
sk
ssssF
53
)52(3
021 =++
=→sss
k
skskss
skskssk
32
22
32
22
1
)52(533
)52(3
++++=
++++=
56 ,
53
32 −=−=⇒ kk
Chapter 2
Ex:
接著兩邊同乘分母可得:
Control System(1) 19
( )( )
( )( ) ( ) 2222
22
2
21
221
53
21
153
53
21531
53
53
5256
53
53
)(
++
⋅⋅−+
++
+−+=
++
−+−+=
++
−−+=
ss
s
s
s
s
s
ss
s
ssF
tetety tt 2sin1032cos
53
53)( −− −−=
Chapter 2
1 -L
( ) )(
22232
ωω
++++
⇒++
+as
BasArqss
ksk
tAeas
asA at ωω
cos)(
)(22
−⇒++
+
tBeasB at ω
ωω sin)( 22
−⇒++
Control System(1) 20
2
24321
2
)3(3
10
)3(9
40
)2(59
5)3()3()2(
)3)(2(10)(
++
++
+−
+=
++
++
++=
++=
ssss
sk
sk
sk
sk
ssssF
ttt teeety 3323
109
40595)( −−− ++−=
Chapter 2
EX:
1 -L
Control System(1) 21
系統輸入 輸出
)(sR )(sY 0IC)()()(
=
≡sRsYsT
Chapter 2
轉移函數(Transfer Function):T(s)
n 階線性非時變微分方程
)()()(
)()()(
01
1
1
01
1
1
trbdt
trdbdt
trdb
tyadt
tydadt
tyda
m
m
mm
m
m
n
n
nn
n
n
+++=
+++
−
−
−
−
−
−
L
L
)()()(
)()()(
01
1
01
1
sRbsRsbsRsb
sYasYsasYsam
mm
m
nn
nn
+++=
+++−
−
−−
L
L
L
Control System(1) 22
)()()()( 01
101
1 sRbsbsbsYasasam
mm
mn
nn
n +++=+++−
−−
− LL
Chapter 2
011
1
011
1)()()(
aasasabbsbsbsT
sRsY
nn
nn
mm
mm
++++++++
== −−
−−
L
L
)(sY
Control System(1) 23
)()(2)( trtydt
tdy=+
)()(2)( sRsYssY =+2
1)()()(
+==⇒
ssRsYsT
Chapter 2
Ex:試求轉移函數T(s),若r(t)=1試求y(t)Sol:
25.05.01
21)()()(
+−
+=⋅+
=⋅=ssss
sRsTsY
tety 221
21)( −−=
1 -L
L
Control System(1) 24
57334
)()()(
)()34()()573()(3)(4)()(5)(7)(3)(
23
2
223
223
+++++
==⇒
++=+++⇒
++=+++
sssss
sRsYsT
sRsssYssssRssRsRssYssYsYssYs
)(3)(4)()(5)(7)(3)( 22
2
2
3
3
trdt
tdrdt
trdtydt
tdydt
tyddt
tyd++=+++
Chapter 2
Ex:試求轉移函數T(s)
Sol:
Control System(1) 25
2612
)()()( 2 ++
+==
sss
sRsYsT
)()(2)(2)(6)()()(2)(2)(6)(
)()12()()26(
2
2
2
2
trdt
tdrtydt
tdydt
tydsRssRsYssYsYs
sRssYss
+=++⇒
+=++⇒
+=++
Chapter 2
Ex:試由轉移函數T(s)求出對應的微分方程式
Sol:
Control System(1)
Chapter 2
電路之轉移函數
)()( sRIsVR =⇒電阻
)(1)( sICs
sVC =⇒電容
)()( ssLIsVL =⇒電感
)()()(Z
sIsVs =⇒阻抗
26
Control System(1)
Chapter 2
Ex:試求輸入電壓與電容電壓之轉移函數
)()(1)()( sVsIsC
sRIssLI =++
)()(1)()(0
tvdiC
tRidt
tdiLt
=++ ∫ ττL
27
Control System(1) 28
Chapter 2
11
)()(
)(1
1)(1)(
)(1
)(11)(
)()()1(
2
2
2
++=⇒
++==⇒
++=
++=⇒
=++
RCsLCssVsV
sVRCsLCs
sIsC
sV
sVRCsLCs
sCsV
sCRsL
sI
sVsIsC
RsL
C
C
11
2 ++ RCsLCs)(sV
)(sVC
Control System(1)
)()()( 2
sVsIsT =
Chapter 2
Ex:試求輸入電壓與電容上電流之轉移函數
29
Control System(1)
⎪⎩
⎪⎨⎧
=++−
=−+⇒ 0)(1)()]()([
)()]()([)(
22212
2111
sIsC
sIRsIsIsL
sVsIsIsLsIR
Chapter 2
30
Control System(1)
)()(1
0)(
)(1
2121
2
1
1
2 sV
CL
sCRsRRLRR
sL
sCsLRsL
sLsLRsL
sVsLR
sI++++
=
++−
−+−+
=
Chapter 2
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
++−
−+⇒
0)(
)()(
12
1
2
1 sVsIsI
sCsLRsL
sLsLR
)()()()(
)()(121
221
22 sV
RsLCRRsRRLCLCs
sVsIsT
++++==⇒
利用Cramer 法則
31