Колесников А.А. Синергетические методы управления сложными системами: теория системного синтеза
Подождите немного. Документ загружается.
ψ
m
= x
2
+ u
m+1
(x
1
)
x
1
,x
2
,...,x
n
u
1
(x
1
,...,x
n+2
)
T
k
˙
ψ
k
(t)+ψ
k
=0; k =1, 2,...,m<n− 1.
˙x
1
(t)=x
2
;
˙x
2
(t)=u = x
3
;
˙x
3
(t)=−A
1
tg b
1
x
3
+ x
4
;
˙x
4
(t)= ˙y(t)=u
1
.
ψ
1
= x
4
+ u
2
(x
1
,x
2
,x
3
)
k =1
T
1
˙
ψ
1
(t)+ψ
1
=0,
u
1
= −
∂u
2
∂x
1
x
2
−
∂u
2
∂x
2
x
3
−
∂u
2
∂x
3
x
4
− A
1
tg b
1
x
3
−
1
T
1
x
4
−
1
T
1
u
2
(x
1
,x
2
,x
3
).
u
1
ψ
1
=0
˙x
1ψ
1
(t)=x
2ψ
1
;˙x
2ψ
1
(t)=x
3ψ
1
;˙x
3ψ
1
(t)=−A
1
tg b
1
x
3
− u
2
.
ψ
2
= x
3
+ β
1
x
1
+ β
2
x
2
T
2
˙
ψ
2
(t)+ψ
2
=0,
u
2
=
β
1
T
2
x
1
+
β
1
+
β
2
T
2
x
2
+
β
2
+
1
T
2
x
3
− A
1
tg b
1
x
3
.
ψ
2
=0
˙x
1ψ
2
(t)=x
2ψ
2
;˙x
2ψ
2
(t)=−β
1
x
1ψ
2
− β
2
x
2ψ
2
.
β
1
=
1
T
2
β
2
=
2
T
p
1,2
= −
1
T
u
2
T
1
= T
2
= T
u
1
= −
x
1
T
4
−
4x
2
T
3
−
6x
3
T
2
−
4
T
−
b
1
A
1
cos
2
A
1
x
3
x
4
−
4
T
−
b
1
A
1
cos
2
b
1
x
3
A
1
tg b
1
x
3
.
u
1
|u| u
max
=
π
2b
1
x
1
(t) x
2
(t) T
T =0, 5 b
1
=
π
2
A
1
=0, 05 x
10
=1 x
20
= x
30
= x
40
=0
t
p
≈ 4, 5 x
3max
= u
|0, 5| <u
max
=1 T =1
x
3max
= |0, 13| < 1
u
1
u
1
|u| u
max
T =0, 5 T =1
T =1 T =1, 5
u
1
T =1 |u| 1 x
10
=5 x
20
= x
30
= x
40
=0
T =1, 5
u(t)
|u| u
max
u
1
(t)
|x
n+1
| x
n+1,max
A
1
tg b
1
x
n+1
ψ
s
(x
1
,...,x
n
)=0,s=1, 2,...,m n,
x
1
,...,x
n
R(x
1
,...,x
n
)
ϕ
j
(x
1
,...,x
q
)=0; j =1, 2,...,p n; q<n.
p s
m = s
ψ
s
=
p
k=1
β
sk
ϕ
k
(x
1
,...,x
q
)+
n
p=p+1
α
sp
x
p
.
u
s
ψ
s
=0
x
j
<a
j
.
y
j
(t)=x
j
(t) − a
j
,j=1, 2,...,p,
ψ
s
=
p
k=1
β
sk
y
k
(x
1
,...,x
q
)+
n
p=p+1
α
sp
x
p
.
ψ
s
=0
y
j
(t) → 0
ϕ
j
(x
1
,...,x
q
) 0; j =1, 2,...,p n; q<n.
F
j
(x
1
,...,x
q
)=
ϕ
j
(x
1
,...,x
q
)
− ϕ
j
(x
1
,...,x
q
)=0.
ψ
s
(x
1
,...,x
n
)=
p
k=1
β
sk
F
k
(x
1
,...,x
q
)+
n
p=p+1
α
sp
x
p
,
˙x
1
(t)=x
2
;
˙x
2
(t)=
a
x
2
1
+ x
1
x
2
4
+
1
m
u
1
;
˙x
3
(t)=x
4
;
˙x
4
(t)=−
2x
2
x
4
x
1
+
1
bx
1
u
2
.
x
1
(t) x
3
(t)
x
2
(t) x
4
(t)
a m
ψ = A th(β
1
x
1
+β
2
˙x
1
)+x
3
u
1
(x
1
,...,x
n
) u
2
(x
1
,...,x
n
)
r ω
x
2
+ δx
1
= δr; δ>0,
x
4
= ω.
x
1
→ r
˙x
1
(t)+δx
1
= δr,
x
1
(t)=(x
10
− r)e
−δt
+ r → r t →∞.
ψ
1
= β
1
y
1
+ β
2
y
2
; ψ
2
= α
1
y
1
+ α
2
y
2
,
y
1
= x
2
+ δx
1
− δr y
2
= x
4
− ω
T
s
˙
ψ
s
(t)+ψ
s
=0; s =1, 2,
u
1
m
= −
1
B
β
1
α
2
T
1
−
α
1
β
2
T
2
(x
2
+ δx
1
−δr) −δx
2
+
a
x
2
1
−
α
2
β
2
B
1
T
1
−
1
T
2
(x
4
−ω) −x
1
x
2
4
;
u
2
m
= −
α
1
β
2
B
1
T
2
−
1
T
1
(x
2
+ δx
1
− δr)x
1
+2x
2
x
4
−
1
B
β
1
α
2
T
1
−
α
1
β
2
T
2
(x
4
− ω).
B =0
β
1
β
2
=
α
1
α
2
ψ
1
=0
ψ
2
=0
ψ
12
= By
1
= B(x
2
+ δx
1
− δr)=0
ψ
12
= By
2
= B(x
4
− ω)=0,
ψ
12
=0
˙x
1ψ
(t)=−δ(x
1ψ
− r) ˙x
3ψ
(t)=ω,
t →∞
x
1s
= r x
3
(t)=x
30
+ ωt
T
1
T
2
T
1
= T
2
= T
u
1
= −m
1
T
1
(x
2
+ δx
2
− δr)+δx
2
−
a
x
2
1
+ x
1
x
2
4
;
u
2
= −m
x
1
T
(x
4
− ω) − 2x
2
x
4
.
β
1
β
2
α
1
α
2
u
1
u
2
λ
1
= λ
2
= −
1
T
λ
1
λ
2
β
1
β
2
α
1
α
2
ψ(x
1
,...,x
n
) th ψ
(n − 1)
T
1
˙x
1
(t)=k
1
x
2
;
T
i
˙x
i
(t)+x
i
= k
i
x
i+1
,i=2,...,n− 1;
T
n
˙x
n
(t)+x
n
= k
n
u.
ψ =
n
i=1
β
i
x
i
,
˙
ψ(t)
˙x
i
(t)
˙
ψ(t)=
n−1
i=1
β
i
k
i
T
i
−
β
i+1
T
i+1
x
i+1
+
β
n
k
n
T
n
u.
β
i
k
i
T
i
=
β
i+1
T
i+1
,
˙
ψ(t)=
β
n
k
n
T
n
u.
β
i
=
T
i
k
i
...k
n−1
T
n
; i =1, 2,...,n− 1; β
n
=1,
β
i
ψ
ψ
˙
ψ(t)
J
Σ
=
∞
0
th
2
ψ + T
2
˙
ψ
2
(t)
dt,
J
1
=
∞
0
th
2
n
i=1
β
i
x
i
+
T
2
β
2
n
k
2
n
T
2
n
u
2
dt,
J
1sup
∼
t
k sup
t
0sup
1+
T
2
β
2
n
k
2
n
T
2
n
u
2
dt.
ψ
T
˙
ψ(t)+thψ =0,
u = −
T
n
k
n
β
n
T
th
n
i=1
β
i
x
i
,
u
max
=
T
n
k
n
β
n
T
,
u
inf
= −
T
n
k
1
k
2
...k
n
T
x
1
+
1
T
1
n
i=2
k
1
k
2
...k
i−1
T
i
x
i
.
J
2
= M
⎡
⎣
∞
0
x
2
1
dt
⎤
⎦
1
k
M
⎡
⎣
∞
0
u
2
dt
⎤
⎦
+
⎡
⎣
∞
0
n
i=1
α
i
x
i
2
dt
⎤
⎦
= const,
k T
k =
T
1
T
2
2
...T
2
n−2
T
n
k
1
k
2
...k
n−2
T
,
α
i
β
i
ψ
α
i
=
k
1
k
2
...k
n−2
T
1
T
2
2
...T
2
n−2
T
n
β
i
.
T
k
β
i
α
i
u
T
1
=1 T
2
=2 T
3
=1 k
1
= k
2
=1 β
1
=1 β
2
=2
˙x
1
(t)=x
2
;
˙x
2
(t)+0, 5x
2
=0, 5x
3
;
˙x
3
(t)+x
3
= u.
u = −
1
T
th (x
1
+2x
2
+ x
3
).
x
10
=
10
x
20
= x
30
=0 T =0, 5
J
sup
∼
t
k sup
t
0sup
(1 + 0, 25u
2
)dt,
x
10
=10 x
20
= x
30
=0 T =1
J
sup
∼
t
k sup
t
0sup
(1 + u
2
)dt.
T =0, 5 T =1
u
T =0, 5 T =1
˙
x(t)=A(x)+B(x)u; y = h(x),
A(·) B(·) h(·) u
y
z =Γ(x)=
col
γ
1
(x),...,γ
n
(x)