Bài 1 - Chương 3 - Điều khiển liên tục trong miền thời gian - Mô hình toán học

8/10/2019 Bi 1 - Chng 3 - iu khin lin tc trong min thi gian - M hnh ton hc

1/16

III. !I"U KHI#N LIN T$CTRONG MI"N TH%I GIAN

BM!i"u Khi#n T$!%ngTh.S.!&ng V'n M(

1 Bai 1 - Chuong 3 - DKLT mien t - Mo hinh toan hoc.key - November 12, 2014


2/16

.

3.1 M&T S'CNG C$TON H(C

!)nh ngh*a ma tr+n

Ma tr+n hng, ma tr+n c,t, ma tr+n -.n v), ma tr+n -/0ng cho

Cc php ton ma tr+n: c,ng, tr1, nhn, chia

Ma tr+n chuy2n v)v cc tnh ch3t

H4ng c5a ma tr+n

!)nh th6c ma tr+n

Ma tr+n ngh)ch -7o

V8i ma tr+n b c cc ph t9

!:I S'MA TR;N

Amxn

I3x 3

=

1 0 0

0 1 0

0 0 1

!

"

##

$

%

&&

Cmxn =Amxn Bmxn = (aij+ bij) AB =C! cij = aikbkjk=1

p

"

AT (AB)

T= A

TB

T

A(B+C) = AB+ AC AI = IA = A

A!1=

Aadj

det(A)

aij'=(!1)

i+jdet(Aji)Aadj



3/16

.

3.2 XY Dth?ng -i@u khi2n c nhi@u -Au vo - nhi@u -Au ra (MIMO) th ph/.ng php tBng hCp h>th?ng trong khnggian tr4ng thi th/0ng -/Cc s9dDng. Ph/.ng php ny cho php ng/0i ta tnh -/Cc c7cc -i@u ki>n khEi t4o -2tBnghCp h>th?ng khi cAn thiFt.

Qung !"#ngd$ch chuy%n

V!n t"c kh"i v!t

&x(t) = Ax(t)+Bu(t)

y(t) = Cx(t)+ Du(t)

!"#

B



4/16

.3.2.1 M HNH TR:NG THI

Xt h!g"m c:

mtn hi!u vo

rtn hi!u ra

nbi#n tr$ng thi

u(t) = {u1(t),...,u

m(t)}

y(t) = {y1(t),...,y

r(t)}

x(t) = {x1(t),...,xn(t)}

!u "i#m:So v%i ph&'ng trnh hm truy(n, h!ph&'ng trnh tr$ng thi c th)s*d+ng,)m t-h!MIMO. Ngoi ra, MHTT cn gip ta kh-o st ,&.c tr/c ti#p cc tr$ngthi bn trong h!th0ng.

&x(t) = Ax(t)+Bu(t)

y(t) = Cx(t)+ Du(t)

!"#

A: Ma tr1n h!th0ng (nxn)

B: Ma tr1n ,2u vo (nxm)

C: Ma tr1n ,2u ra (rxn)

D: Ma tr1n lin thng (rxm)

&x(t) = A(t)x(t)+B(t)u(t)

y(t) = C(t)x(t)+ D(t)u(t)

!"#

&x(t) = A(v)(x(t)+B(v)u(t)

y(t) = C(v)x(t)+ D(v)u(t)

!"#

H>tham s?phDthu,c th0i gian H>tham s?r7i (phDthu,c khng gian)



5/16

.3.2.1 M HNH TR:NG THI

Xt h!SISO c m3t tn hi!u vo u(t) v m3t tn hi!u ray(t):

M hnh ,&.c vi#t l$i thnh:

45t:

XC !GNH M HNH TR:NG THI THPHIJNGTRNH VI PHN M TKQUAN HLVO RA

a0y + a

1

dy

dt+ ...+ an!1

dn!1y

dtn!1

+d

ny

dtn = b

0u + b

1

du

dt+ ...+ bn!1

dn!1u

dtn!1

+ bnd

nu

dtn

G(s) =Y(s)

U(s)=

b0 + b1s + ...+ bnsn

a0 + a1s + ...+ an!1sn!1

+ sn =B(s)

A(s)

X1 =

U(s)

A(s),X2 =

sU(s)

A(s),...,X

n =

sn!1U(s)

A(s)

sX1 = X

2,K,sX

n!1 = X

n

X1 =

U(s)

A(s)! A(s)X1 =U(s) =a0X1 + a1X2 +K+ an!1Xn + sXn

= a0

X1

+ a1

X2

+K+ an!1

Xn

+ L{dx

n

dt

}! (*)

!dx

1

dt= x

2,dx

2

dt= x

3,K,

dxn"1

dt= x

n

(*)#dx

n

dt= "a

0x1" a

1x2"K" a

n"1x

n+ u

dx

dt=

0 1 0 K

0

0 0 1 K 0

M M M O 0

0 0 0 K 1

!a0 !a

1 !a

2 K !a

n!1

!

"

#####

$

%

&&&&&

x1

x2

M

xn!1

xn

!

"

######

$

%

&&&&&&

+

0

0

M

0

1

!

"

#####

$

%

&&&&&

u

Y(s) =U(s)(b0 + b1s + ...+ bns

n)

A(s)= b0X1 + b1X2 + ...+ bn!1Xn + bnsXn

"y = (b0! a0bn )x1 + (b1! a1bn )x2 + ...+ (bn!1! an!1bn )xn + bnu

"y = (b0! a0bn ),(b1! a1bn ),...,(bn!1! an!1bn )( )

x1

M

xn

#

$

%%%

&

'

(((+ bnu

A(nxn) B(1xn)

C(nx1)D(1x1)



6/16

.3.2.1 M HNH TR:NG THI (TIMP)

XC !GNH M HNH TR:NG THI THHM TRUY"N !:T

M HNH TR:NG THI D:NG CHUNN !I"U KHI#N

S!"#c$u trc d%ngchu&n "i'u khi(n

G(s) =Y(s)

U(s)=

b0 + b1s + ...+ bns

n

a0+ a

1s + ...+ a

n!1sn!1

+ sn =B(s)

A(s)

X1 =

U(s)

A(s),X2 =

sU(s)

A(s),...,X

n =

sn!1U(s)

A(s)

V8i h>c PTHT:

!Ot:

dx

dt=

0 1 0 K 0

0 0 1 K 0

M M M O 0

0 0 0 K 1

!a0 !a

1 !a

2 K !an!1

"

#

$$$$$

%

&

'''''

x1

x2

Mxn!1

xn

"

#

$$

$$$$

%

&

''

''''

+

0

0

M0

1

"

#

$$$$$

%

&

'''''

u

y = (b0! a0bn ),(b1! a1bn ),...,(bn!1! an!1bn )( )

x1

M

xn

"

#

$$$

%

&

'''+bnu

(

)

**

****

+

******

G(s) =Y(s)

U(s)=

b0+ b

1s + b

2s2

a0+ a

1s + a

2s2+ s

3 =B(s)

A(s)V dD: Cho h>c PTHT

!OtX1 =

U(s)

A(s),X2 =

sU(s)

A(s),

X3 =s2U(s)

A(s)



7/16



M HNH TR:NG THI D:NG CHUNN QUAN ST

G(s) =Y(s)

U(s)=

b0+ b

1s + ...+ b

nsn

a0 + a1s +...

+ an!1sn!1

+ sn

x1=y + bnu

&x1 =x

2+ bn!1u ! an!1x1

&

x2 =

x3+

bn!2u!

an!2x1M

&xn!1 =xn + b1u ! a1x1

&xn = b0u ! a0x1

!

"

###

$

###

!a0Y+ a1sY + a2s

2Y+K+ a

n"1sn"1Y+ s

nY = b0U+ b1sU+ b2s

2U+K+ b

nsnU# (*)

X1 =Y" bnU

X2 =a

n"1Y" bn"1U+ s(Y" bnU)

X3= (a

n"2Y + an"1sY)" (bn"2U+ bn"1sU)+ s2(Y" b

nU) = (a

n"2Y" bn"2U)+ s(an"1Y" bn"1U+ s(Y" bnU))

M

Xn = (a

1Y+ a

2sY +K+ a

n"1sn"2Y)" (b

1U+ b

2sU+K+ b

n"1sn"2U)+ s

n"1(Y" b

nU)# (**)

$

%

&&&

'

&

&&

# (*)&(**)! b0U" a

0Y = s (a

1Y+ a

2sY+K+ a

n"1sn"2Y)" (b

1U+ b

2sU+K+ b

n"1sn"2U)+ s

n"1(Y" b

nU)( ) = sXn

H!s0an=1

&x =

!an!1

1 0 K 0

!an!2

0 1 K 0

M M M O 0

!a1

0 0 K 1

!a0 0 0 K 0

!

"

####

##

$

%

&&&&

&&

x1

x2

M

xn!1

xn

!

"

####

##

$

%

&&&&

&&

+

bn!1

bn!2

M

b1

b0

!

"

####

##

$

%

&&&&

&&

u

y = (1,0,K,0( )

x1

M

xn

!

"

###

$

%

&&&+ b

n.u

!

"

#####

#

$

######

Cho h>c PTHT



8/16



S!"#c$u trc d%ngchu&n quan st

M hnhtr!ng thid!ng chu"nquan st

G(s) =Y(s)

U(s)

=b0 + b1s + b2s

2

a0+ a

1s + a

2s2+ s

3 =B(s)

A(s)

V dD: Cho h>c PTHT



9/16



G(s) =Y(s)

U(s)=

b0+ b

1s + ...+ b

nsn

a0 + a1s +...

+ an!1sn!1 + sn

Chu!n

"i#u khi$n

Chu!n

quan st

&x =

!an!1

1 0 K 0

!

an!2 0 1 K 0

M M M O 0

!a1

0 0 K 1

!a0

0 0 K 0

!

"

######

$

%

&&&&&&

x1

x2

M

xn!1

xn

!

"

######

$

%

&&&&&&

+

bn!1

bn!2

M

b1

b0

!

"

######

$

%

&&&&&&

u

y = (1,0,K,0( )

x1

M

xn

!

"

###

$

%

&&&

+ 0.u

!

"

#

#####

$

####

##

dx

dt=

0 1 0 K 0

0 0 1 K

0

M M M O 0

0 0 0 K 1

!a0 !a

1 !a

2 K !an!1

"

#

$$$$$

%

&

'''''

x1

x2

M

xn!1

xn

"

#

$$$$$$

%

&

''''''

+

0

0

M

0

1

"

#

$$$$$

%

&

'''''

u

y = (b0! a0bn ),(b1! a1bn ),...,(bn!1! an!1bn )( )

x1

M

xn

"

#

$

$$

%

&

'

''+ bnu

(

)

******

+

****

**

Lm sao "#"$a m hnh tr%ng thi b&t k v'd%ng chu(n "i'u khi#n ho)c chu(n quan st?



10/16


S


11/16


XC !GNH PHIJNG TRNH HM TRUY"N THM HNH TR:NG THI

*+nh l,: Cho h!SISO tuy#n tnh v%i m hnh tr$ng thi:

Khi , h!c ph&'ng trnh hm truy(n:

b)N#u th

v%i l ma tr1n b c6a ma tr1n (sI-A) v%i

a)G(s) = C(sI! A)!1B+ D

G(s) =B(s)

A(s)

A(s) =a0+ a

1s + ...+ an!1s

n!1+ ans

n=det(sI!A)

B(s) = b0+ b

1s + ...+ bms

m= C

)

AadjB+D.det(sI!A)

)Aadj

&x(t) = Ax(t)+Bu(t)

y(t) = Cx(t)+ Du(t)

!"#

(sI! A)!1=

)

Aadj

det(sI! A)



12/16

.3.2 XY Dth?ng

!Sth)qu*-4o tr4ng thi l -/0ng cong bi2udiTn khi cho t ch4y t1 trongkhng gian tr4ng thi

x(t) 0!"n

R

QUi -4o tr4ng thi l nghi>m c5a h>ph/.ng trnh vi phn: dx(t)

dt

= Ax(t)+ Bu(t)

6ng v8i m,t kch thch v tr4ng thi ban -Au cho tr/8c( )u t 0(0)x x=



13/16

.3.2.1 QUR!:O TR:NG THI (TIMP)

T!"ng t#v$i ph!"ng trnh vi phn th%2

V&y 'nh Laplace c(a h)ph!"ng trnh vi phn l



14/16


L!y "nh Laplace ng#$c %&tm nghi'm x(t)

Ma tr(n chuy

&n tr

)ng thi



15/16


Ma tr+n chuy2n tr4ng thi:



16/16


Gi!s"v#i tn hi$u vo b%#c nh!y &'n v(u(t)=1, ta c

Suy ra nghi!m c"a h!ph#$ng trnh vi phn:


Documents

Bài 1 - Chương 3 - Điều khiển liên tục trong miền thời gian - Mô hình toán học