bài 1 - chương 3 - Điều khiển liên tục trong miền thời gian - mô hình toán học
TRANSCRIPT
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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(
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.
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
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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$ch chuy%n
V!n t"c kh"i v!t
&x(t) = Ax(t)+Bu(t)
y(t) = Cx(t)+ Du(t)
!"#
B
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.3.2.1 M HNH TR:NG THI
Xt h!g"m c:
mtn hi!u vo
rtn hi!u ra
nbi#n tr$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)
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.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)
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.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)
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.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 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
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.3.2.1 M HNH TR:NG THI (TIMP)
XC !GNH M HNH TR:NG THI THHM TRUY"N !:T
S!"#c$u trc d%ngchu&n quan st
M hnhtr!ng thid!ng chu"nquan st
G(s) =Y(s)
U(s)
=b0 + b1s + b2s
2
a0+ a
1s + a
2s2+ s
3 =B(s)
A(s)
V dD: Cho h>c PTHT
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.3.2.1 M HNH TR:NG THI (TIMP)
XC !GNH M HNH TR:NG THI THHM TRUY"N !:T
G(s) =Y(s)
U(s)=
b0+ b
1s + ...+ b
nsn
a0 + a1s +...
+ an!1sn!1 + sn
Chu!n
"i#u khi$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 "#"$a m hnh tr%ng thi b&t k v'd%ng chu(n "i'u khi#n ho)c chu(n quan st?
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.3.2.1 M HNH TR:NG THI (TIMP)
S
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.3.2.1 M HNH TR:NG THI (TIMP)
XC !GNH PHIJNG TRNH HM TRUY"N THM HNH TR:NG THI
*+nh l,: Cho h!SISO tuy#n tnh v%i m hnh tr$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)
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.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=
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.3.2.1 QUR!:O TR:NG THI (TIMP)
T!"ng t#v$i ph!"ng trnh vi phn th%2
V&y 'nh Laplace c(a h)ph!"ng trnh vi phn l
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.3.2.1 QUR!:O TR:NG THI (TIMP)
L!y "nh Laplace ng#$c %&tm nghi'm x(t)
Ma tr(n chuy
&n tr
)ng thi
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.3.2.1 QUR!:O TR:NG THI (TIMP)
Ma tr+n chuy2n tr4ng thi:
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.3.2.1 QUR!:O TR:NG THI (TIMP)
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:
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