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

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

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

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

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

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

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

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

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

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

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:

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:

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