《現(xiàn)代控制理論》第三版劉豹_第1頁(yè)
《現(xiàn)代控制理論》第三版劉豹_第2頁(yè)
《現(xiàn)代控制理論》第三版劉豹_第3頁(yè)
《現(xiàn)代控制理論》第三版劉豹_第4頁(yè)
《現(xiàn)代控制理論》第三版劉豹_第5頁(yè)
已閱讀5頁(yè)，還剩48頁(yè)未讀， 繼續(xù)免費(fèi)閱讀

下載本文檔

版權(quán)說明：本文檔由用戶提供并上傳，收益歸屬內(nèi)容提供方，若內(nèi)容存在侵權(quán)，請(qǐng)進(jìn)行舉報(bào)或認(rèn)領(lǐng)

文檔簡(jiǎn)介

K

xxxbx

J

xxKxxx

J

p

K

p

JJJJ

K

xxxpx

xKxxKxxKxKx

n

KK

xux

x

KK

p

p

KKK

xxu

KKK

p

p

p

x

xAx+Bu

y=Cx

K

b

J

K

K

p

JJJJ

p

A=

K

n

K

K

K

K

K

K

p

p

T

K

K

B

C

p

MM

ff

MM

MyfKccByy

MyfKcByKccByy

x=ccyy

T

yyyuff

T

T

xcxcxyxy

xx

xx

KKBB

xxxxf

MMMMM

x

KKKBB

x

B

x

x

x

x

f

M

M

M

M

M

xAx+Bu

y=Cx

KK

MM

BB

MM

A=

KKKBBB

MM

M

M

B

C

M

M

yyyyuu

s

Gs

a

sss

a

a

bbbb

A

aaa

Cbabbab

nn

n

n

B

b

b

b

b

a

aa

aaa

b

b

b

a

aa

a

aaaaaaab

B

T

T

C

yyyyuuu

ss

Gs

sss

bbbb

C

B

WsCSIABd

SIA

S

T

S

S

S

S

T

SSS

S

SSS

S

SS

SSSS

SSS

SS

T

S

SSSSS

IA

APPP

APPP

TPP

T

J

BTB

CCT

IA

APP

P

T

PAP

P

2

P

PAPP

T

T

JTAT

TB

CT

WsWsIWsWs

S

SS

S

S

S

S

S

S

S

S

SS

S

S

S

S

S

S

S

SSS

S

S

S

S

S

S

SSS

S

24.

eAt

01

40

1A

1

1

eAtIAtA2t2A3t3

2!

3!

t

tt

6

t

3

t416

3!

t

2

141664

4

5

,

2!

4!6!

5!

t

3

4t1664

3!

t

t

tt

46

5

2

,141664

2!4!6!

5!

IA0

2j

1,2

APPP12j

1

T

1

1

APPP12j

22

T

2

2

11

T

2j2j

12cos2tsin2t

24sin2t2cos2t

eAtTetT1

S1

S1

1

1

(SIA)1

4SS244S

1

cos2tsin2t

2

eAtL1(SIA)1

2sin2tcos2t

2j

1,2

(t)1

(t)1

1

e

2(e2jte2jt)

j(e2jte2jt)

1

4

t

1

0

1

1

2

e2t

eAt(t)I(t)A

1

0

12cos2tsin2t

24sin2t2cos2t

11

41

2A

11

eAtIAtA2t2A3t3

2!

3!

t213

2!6

7

3

1t5

t,ttt

32

6

4t4t2

28t3,1t5t213t3

2!6

6

11

121

421

T

;T1

22

12(e3tet),e3tet

eAt

4

4(e3tet)2(e3tet)

1111

),()

S3S1S3S1

2(

1

SIA

1

41111

),2()

S3S1S3S1

4(

eAt

(t)

1e3t3et

0

1

(t)4

3t

eet

eAtIT

1

0

2.

010

xu

001

x

y(1,0)x

1

x(0),u(t)

1

.

1t

A

IAtt2...

2

2!

01

xeAtxteA(t)Bu()d

0

0

1t11t0

=

1()d

t

011011

0

=1tt

122tt2

t

d

1

1

2

22t

0

27.23

1.

u(t)K(t)x(0)x

0

x(t)eAtxeAtBK,

0

Ku(t)

xeAtxteA(t)BK()d

0

0

eAtxteA(t)()dBK

0

0

eAtxeA(t)|BK

0

0

eAtxeAtBK

0

2u(t)K1(t)x(0)x

0

xeAtxteA(t)BK1(t)d

0

0

eAtxteA(t)dBK

0

0

eAtxA1eA(t)|tBK

0

0

eAtxA1(eAtI)BK

0

3u(t)k1(t)x(0)x

0

x(t)eAtxteA(t)BK1(t)d

0

0

eAtxteAdeAtBK

0

0

t

(eA)'eAAeA

0

eA|tteAdAteAd

0

0

0

teAdA1eA|tteAd

0

0

0

A1teAA1eA|t

0

A1teAA1(eAtI)

x(t)eAtxA2(eAtI)A1teAteAtBK

0

eAtxA2(eAtI)A1tBK

0

2-9.

2-2

T0.1s1s,uu

1

2

1

xxKu

1

1

xxu

1

2

1

2

y2xx

1

2

x10xK0u

1

1

x10x01u

1

2

2

2

x

1

x

y21

2

10

2

IA

10

1

0;1.

2

1

10

10

IAP

1

P0

1

1

P01

1

T

00

11

IAP

P0

2

2

2

1

01

11

10

P

T

T1

1

11

2

10

eT0

G(T)eATT

T1

0eT

1eT1

K0

et0

H(T)TeAtdtB

T

dt

01

1et1

0

0

K0

1eT0

01

T1eTT

K(1eT)0

K(T1eT)T

x(k1)

1

x(k1)

x(k)

1

x(k)

u(k)

1

u(k)

G(T)

H(T)

2

2

2

x(k)

1

x(k)

eT0

1eT1

2

u(k)

1

u(k)

K(1eT)0

K(T1eT)T

2

x(k)

1

x(k)

y(k)21

2

T0.1s

x(k1)

1

x(k1)

x(k)

1

x(k)

e0.1

0

1e0.11

2

2

u(k)

1

u(k)

K(1e0.1)0

K(e0.10.9)0.1

2

x(k)

1

x(k)

y(k)21

T1s

2

x(k1)

1

x(k1)

x(k)

1

x(k)

e10

1e11

2

2

u(k)

K(1e1)0

1

u(k)

1

Ke1

2

x(k)

1

x(k)

y(k)21

2

x(k1)

1

x(k1)

x(k)

1

x(k)

u(k)

1

u(k)

(TAI)

TB

2

2

2

1T0x(k)KT0u(k)

1

1

T1x(k)0Tu(k)

2

2

T0.1s

x(k1)0.90x(k)

1

1

x(k1)0.11x(k)

2

2

0.1K0u(k)

1

00.1u(k)

2

T1s

x(k1)00x(k)K0u(k)

1

1

1

x(k1)11x(k)01u(k)

2

2

2

1eAt

t

2

ABBA

3-1

1

x

2

u

x

4

y

a,b,c,d

a000

0b00

1

x

x

1

1

0

0u

x

2

x

3

x

4

x

2

x

3

x

4

11

0

c

001d

0

x

1

x

y0010

2

x

3

x

4

MbAbA2bA3b

1aa2

a3

000

0

01aca2acc2

001acd

,

rankM4

a,b,c,d

C

0

0

10

CA

CA2

1

1

c0

N

0

ac

bcc2

CA3a2acc2b2bcc2c30

,

rankN4

a,b,c,d

2

1

1u

x

ab

x

1

1

x

2

cd

x

2

x

y10

1

x

2

1ba

1cd

MbAb

1ba

M

abcd0

1cd

abcd

C10

CAab

b0

N

3

ab0

ab0

cd0

cd0

33

1

1

1

M[bAb]

1

2

M

0

2

1

2

1

11

C

N

CA

N

1

2

0

2

1

2

1

2

1

1

1

2A

b

c10

1

3

2

4

1

1

MbAb

rankM2

1

3

2

4

detM

0

3

4

1

2

1

2

3

4

C10

NCCA

T

CA

1

2

detN

0

2

0

2

1

2

3

4

3MbAbA2b

12

28

2

314

3

3

13

2

3

3

2

41413

3

2

3

2

3

detM14

10

14

25

10

48

66

24

0

2

3

2

2

2

3

2

3

2

3

3

2

2

3

2

3

2

3

C001

NCA014detN0

CA21413

s26s8

2s5

3W(s)

1

s4s3s4s3

2

2

I

01

0

5

cd1

T

A

b

341

II

2

01

50T

bc

3421

T

A

d1

3112

012

MbAbA2b000

101

detM0rankM3

010

R001

c

100

3

001

detR0R100

1

c

c

010

014

A?R1AR122

c

c

002

b?R1b100

T

c

ccR111

c

3

W(s)

1ss

2

W(s)

s3s

1

100100

1

s

s

s3001001

2

0

0

1

2

00

01

01

10

00

,

,

10

0

1

2

0I0

r

r

A00I

r

c

r

r

III

r

0r1r

001000

2r

000100

000010

000001

000000

000000

00

00

00

00

10

01

0

r

r

r

B0

,

c

I

000110

011000

C

c

0

1

2

:

000110

011000

000001

000110

000000

000001

C

c

NCA

cc

CA2

cc

Rank(N)36

R1

o

000110

011000

000001

000010

100000

010000

000010

000001

010001

100100

000100

001000

R1

o

R

o

??

AA

A?R1AR

11

??

12

o

o

AA

21

22

001000

100000

000000

000000

010001

100100

10

00

B?01

1

B?R1B

o

?

10

00

00

B

2

100000

010000

?

??

CCRCC

o1

2

???

ABC

1

11

1

W

001

10

?

?

AA100,BB00,

11

m

1

000

01

100

010

CC?

m

1

W(s)

1ss

2

W(s)

s1

s

3

100100

2

1

s

s

s3001001

0

0

1

2

00

01

01

10

10

00

,

,

0

1

2

000000

000000

100000

010000

001000

000100

00I

m

AI0I

m

0m

o

m

m

0II

1m

m

m

2m

00

01

01

10

10

00

0

1

2

B

,

0

000010

000001

C00I

m

o

m

m

:

MBABA2B

oo

o

oo

000000

010000

010000

100100

100100

001001

rankM36

R1

c

000001

010000

010010

R

100000

c

100100

001000

000100

010000

000001

000110

011000

100000

R1

c

010000

000000

100000

000010

000001

000000

A?R1AR

oc

c

10

01

00

00

00

00

B?R1B

c

o

100100

001000

C?CR

oc

???

ABC

1

11

1

W

010

10

?

?

AA000,BB01,

11

m

1

100

00

100

001

CC?

m

1

1MbAbAn1bn

A

2

33-33

2

3

4-11Q(x)xTPx

PPP

13

P=PPP

11

12

10

21

22

23

1

PPP

31

20

32

33

17.750

20i

17.750

2

1

3

i

2

10

Q(x)

3

2Q(x)xTPx

111

P143

131

10

30

160

1

2

3

Q(x)

4-2

IA

aaaaaa

1122

2

11

22

1221

aa0

1

2

11

aaaa0

22

12

1122

1221

PP

1112

PP

P=

QI

12

22

ATPPAQ

aaPPPPaa

10

11

21

11

12

11

12

aaPPPPaa

11

12

01

12

22

12

22

12

2(aPaP)1

22

21

22

1111

2112

(aa)PaPaP0

11

2112

2111

2(aPaP)1

2122

1212

2122

Aaa

2

2

P

21

2tAA

22

11

r

Aaa

2

2

P

11

2tAA

12

22

r

aaaa

1122

P

2111

2tAA

r

12

tAA

r

tAaa

11

r

22

Aaaaa

1122

1221

P

P

P0

11

PPPP0

11221221

(Aaa2)(Aaa2)

2

2

(aaaa)2

2111

PPPP

11221221

21

22

11

12

1222

4t2AA

r

2

4t2AA

2

r

(aa)2(aa)2

12

0

11

12

21

4t2AA

r

Aaaaa0

11221221

P0

11

A0

tA0

r

aa0

11

22

aaaa0

1122

1221

aa0

11

22

4-31v(x)xx

x0

e

2

2

1

2

v(x)0

v(x)2x6x6xxxTPx

2

2

1

2

12

23

36

30

P

P0

11

1

2

0i

0

i

i

i

P

v(x)0

x

v(x)

ATPPAQ

QI

75

48

53

88

P

00P

2

1

2v(x)xx

2

2

1

2

v(x)2(xx2)0

2

1

2

xv(x)

1

2

0

ATPPAQP

1

2

0

4-2

v(x)xx

22

12

xx

2

1

1

2

5-2

11

0010

Mb,Ab,A2b010110

10100990

rankM3

2Kkkk

1

0

2

f()detI(Abk)

(1110k)

2

(1110k10k)10k

2

3

2

1

0

f*()(10)(1j3)(1j3)

122440

2

3

K41.20.1

21

0010

Mb,Ab,A2b010110

10100990

rankM3

2

I

IA1111

32

0

11

11

0

1

2

100

1000100

1101001110

TA2bAbb

cI

10

2

19901001011111

1

2

1000

0.100

0100T00.10

1

cI

01010

00.10.1

010

AT1AT001

cI

cI

cI

01111

bT1b001

cI

T

2Kkkk

1

0

2

010

AbK001

kk11k11

0

1

2

f()(k11)(k11)k

2

3

2

1

0

f*()122440

2

3

K40131

3KKT41.20.1

1

c

5-51MbAbA2b

240

010

115

detM0RankM3

5-8(1)cA0B10;cA0B00

2

1

cAB10

2

10

10

cA0B

1

cAB

detEdet

det

0

;

2

5-10

(1)

c

N

cA

(2)

II:

detIA2;

aa0;L

01

10

;

0

1

T1LNL;TL;

xT1ATxT1bu

ycTx01x

001

100

u;

g

1

(3)

G,

g

2

:

g

1

1g

f()detI(AGc)

2

gg

2

2

1

(4)

:

f*()(r)(2r)3r2r2

2

(5)f()f*()

:

2r2

3r

g2r2;g3rG

2

1

x:

3r

GTG

2r2

(7)

:

?()?

xAGcxbuGy

2r113r

?

xu

2r2002r2

y

;

()

???

xAxbuGyy

0113r

();

?

yy

?

xu

0002r2

(1)

(2)

A-Gc=01

1

0

g

g

1

g

2

10

1

00

g

2

f()det[I(A-Gc)]

g1

=det

gg

12

1

2

g

2

:

溫馨提示

  • 1. 本站所有資源如無特殊說明，都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請(qǐng)下載最新的WinRAR軟件解壓。
  • 2. 本站的文檔不包含任何第三方提供的附件圖紙等，如果需要附件，請(qǐng)聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
  • 3. 本站RAR壓縮包中若帶圖紙，網(wǎng)頁(yè)內(nèi)容里面會(huì)有圖紙預(yù)覽，若沒有圖紙預(yù)覽就沒有圖紙。
  • 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
  • 5. 人人文庫(kù)網(wǎng)僅提供信息存儲(chǔ)空間，僅對(duì)用戶上傳內(nèi)容的表現(xiàn)方式做保護(hù)處理，對(duì)用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯，并不能對(duì)任何下載內(nèi)容負(fù)責(zé)。
  • 6. 下載文件中如有侵權(quán)或不適當(dāng)內(nèi)容，請(qǐng)與我們聯(lián)系，我們立即糾正。
  • 7. 本站不保證下載資源的準(zhǔn)確性、安全性和完整性, 同時(shí)也不承擔(dān)用戶因使用這些下載資源對(duì)自己和他人造成任何形式的傷害或損失。

評(píng)論

0/150

提交評(píng)論