Колесников А.А. Синергетические методы управления сложными системами: энергетические системы
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U
1i
U
2i
ω
i
= ω
0
•
•
•
•
n
dδ
i
dt
= s
i
;
Tj
i
ds
i
dt
= P
Ti
− E
2
i
y
ii
sin α
ii
− E
i
U
c
y
iN
sin(δ
i
− α
iN
) −
n
j=1,i=j
E
i
E
j
y
ij
sin(δ
ij
− α
ij
)+F (t);
T
di
dE
i
dt
= −E
i
+ U
c
T
d0i
(x
di
− x
di
)
n
j=1,i=j
y
ij
(s
i
− s
j
)sin(δ
ij
− α
ij
)+U
1i
;
T
ci
dP
Ti
dt
= −P
Ti
−
s
i
σ
i
+ U
2i
,
i =1, 2,...,n δ
ij
= δ
i
− δ
j
y
ii
y
ij
y
iN
i α
ii
α
ij
α
iN
T
di
= T
d0i
(1+y
ii
(x
di
−x
di
)cosα
ii
)
i =1, 2 x
1
= δ
1
x
2
= δ
2
x
3
= s
1
x
4
= s
2
x
5
= E
1
x
6
= E
2
x
7
= P
T 1
x
8
= P
t2
a
1
= y
11
sin α
11
a
2
= y
12
a
3
= U
c
y
1N
a
4
= y
22
sin α
22
a
5
= U
c
y
2N
a
6
= U
c
y
12
T
d01
(x
d1
− x
d1
) a
7
= U
c
y
12
T
d02
(x
d2
− x
d2
)
a
8
=1/σ
1
a
9
=1/σ
2
b
1
=1/T j
1
b
2
=1/T
d1
b
3
=1/T
c1
b
4
=1/T j
2
b
5
=1/T
d2
b
6
=1/T
c2
dx
1
dt
= x
3
;
dx
2
dt
= x
4
;
dx
3
dt
= b
1
(x
7
− a
1
x
2
5
− a
3
x
5
sin(x
1
− α
1N
) − a
2
x
5
x
6
sin(x
1
− x
2
− α
12
)+c
1
x
9
);
dx
4
dt
= b
4
(x
8
− a
4
x
2
6
− a
5
x
6
sin(x
2
− α
2N
)+a
2
x
5
x
6
sin(x
1
− x
2
+ α
12
)+c
2
x
9
);
dx
5
dt
= b
2
(−x
5
+ a
6
(x
3
− x
4
)sin(x
1
− x
2
− α
12
)+U
11
);
dx
6
dt
= b
5
(−x
6
+ a
7
(x
3
− x
4
)sin(x
1
− x
2
+ α
12
)+U
12
);
dx
7
dt
= b
3
(−x
7
− a
8
x
3
+ U
21
);
dx
8
dt
= b
6
(−x
8
− a
9
x
4
+ U
22
);
dx
9
dt
= x
10
;
dx
10
dt
= −Ω
2
x
9
.
m =4
ψ
s
=0,s= 1,m,
ψ
1
= b
11
(x
5
− x
0
5
)+b
12
(x
7
+ ϕ
1
+ c
1
x
9
),ψ
2
= b
21
(x
5
− x
0
5
)+b
22
(x
7
+ ϕ
1
+ c
1
x
9
),
ψ
3
= b
31
(x
6
− x
0
6
)+b
32
(x
8
+ ϕ
2
+ c
2
x
9
),ψ
4
= b
41
(x
6
− x
0
6
)+b
42
(x
8
+ ϕ
2
+ c
2
x
9
).
x
5
− x
0
5
=0,x
6
− x
0
6
=0,
x
0
5
x
0
6
T
s
˙
ψ
s
(t)+ψ
s
=0,s= 1,m.
ψ
s
=0,s= 1,m
dx
1ψ
dt
= x
3ψ
;
dx
2ψ
dt
= x
4ψ
;
dx
3ψ
dt
= b
1
(−ϕ
1
− a
1
(x
0
5
)
2
− a
3
x
0
5
sin(x
1ψ
− α
1N
) − a
2
x
0
5
x
0
6
sin(x
1ψ
− x
2ψ
− α
12
));
dx
4ψ
dt
= b
4
(−ϕ
2
− a
4
(x
0
6
)
2
− a
5
x
0
6
sin(x
2ψ
− α
2N
)+a
2
x
0
5
x
0
6
sin(x
1ψ
− x
2ψ
+ α
12
)).
ϕ
1
ϕ
2
ψ
5
=0 ψ
6
=0
ψ
s
=0
ψ
5
= x
3
+ γ
1
(x
1
− x
0
1
),
ψ
6
= x
4
+ γ
2
(x
2
− x
0
2
),
x
1
− x
0
1
=0
x
2
− x
0
2
=0
T
5
˙
ψ
5
(t)+ψ
5
=0,
T
6
˙
ψ
6
(t)+ψ
6
=0,
ψ
5
=0 ψ
6
=0
ϕ
1
ϕ
2
ϕ
1
=
'
x
3
(T
5
γ
1
+1)γ
1
(x
1
− x
0
1
)
(
/(T
5
b
1
) − a
1
(x
0
5
)
2
− a
3
x
0
5
sin(x
1
− α
1N
) − a
2
x
0
5
x
0
6
sin(x
1
− x
2
− α
12
),
ϕ
2
=
'
x
4
(T
6
γ
2
+1)γ
2
(x
2
− x
0
2
)
(
/(T
6
b
4
) − a
4
(x
0
6
)
2
− a
5
x
0
6
sin(x
2
− α
2N
)+a
2
x
0
5
x
0
6
sin(x
1
− x
2
+ α
12
).
ψ
5
= ψ
6
=0
dx
1ψ
56
dt
= −γ
1
(x
1ψ
56
− x
0
1
),
dx
2ψ
56
dt
= −γ
2
(x
2ψ
56
− x
0
2
).
U
11
U
12
U
21
U
22
U
11
=(a
6
(x
3
− x
4
)+p
9
)sin(x
1
− x
2
− α
12
) − p
3
x
3
− p
4
x
7
+
+ x
5
− p
6
(x
1
− x
0
1
) − p
5
(x
5
− x
0
5
)+p
8
sin(x
1
− α
1N
) − p
7
x
9
+ p
10
;
U
12
=(a
7
(x
3
− x
4
)+k
7
)sin(x
1
− x
2
+ α
12
)+k
2
x
4
+ k
4
x
8
+
+ x
6
+ k
5
(x
2
− x
0
2
)+k
3
(x
6
− x
0
6
) − k
8
sin(x
2
− α
2N
)+k
6
x
9
+ k
9
;
U
21
= p
11
(x
3
− x
4
)cos(x
1
− x
2
− α
12
)+x
3
(p
12
cos(x
1
− α
1N
) − p
14
)+p
15
x
7
+
+ p
17
x
2
5
+ p
20
(x
5
− x
0
5
)+(p
22
x
5
x
6
+ p
27
)sin(x
1
− x
2
− α
12
)−
− p
23
(x
1
− x
0
1
)+(p
19
x
5
+ p
26
)sin(x
1
− α
1N
)+p
25
− p
28
x
9
− p
29
x
10
;
U
22
= −k
10
(x
3
− x
4
)cos(x
1
− x
2
+ α
12
)+x
4
(k
11
cos(x
2
− α
2N
) − k
13
) − k
20
x
8
+
+ k
16
x
2
6
− k
18
(x
6
− x
0
6
)+(k
15
x
5
x
6
− k
23
)sin(x
1
− x
2
+ α
12
)−
− k
19
(x
2
− x
0
2
)+(k
17
x
6
+ k
24
)sin(x
2
− α
2N
)+k
22
− k
25
x
9
− k
26
x
10
,
p
0
= b
11
b
22
− b
12
b
21
p
1
= b
12
b
22
(1/T
1
− 1/T
2
)/(b
1
b
2
p
0
) p
2
=(T
5
γ
1
+1)/T
5
p
3
= p
1
p
2
p
4
= p
1
b
1
p
5
=(b
11
b
22
/T
1
− b
12
b
21
/T
2
)/(b
2
p
0
) p
6
= γ
1
p
1
/T
5
p
7
= c
1
p
4
p
8
= a
3
p
4
x
0
5
p
9
=
a
2
p
4
x
0
5
x
0
6
p
10
= a
1
p
4
(x
0
5
)
2
p
11
= a
2
x
0
5
x
0
6
/b
3
p
12
= a
3
x
0
5
/b
3
p
13
=(b
11
b
22
/T
2
− b
12
b
21
/T
1
)/(b
1
b
3
p
0
)
p
14
= p
2
p
13
− a
8
+ γ
1
/(b
1
b
3
T
5
) p
15
= b
1
p
13
+1+1/(T
5
b
3
)+γ
1
/(b
3
) p
16
= a
1
/b
3
p
17
= p
2
p
16
p
18
= a
3
/b
3
p
19
= p
2
p
18
p
20
= b
11
b
21
(1/T
1
−1/T
2
)/(b
3
p
0
) p
21
= a
2
/b
3
p
22
= p
2
p
21
p
23
= γ
1
p
13
/T
5
p
24
= b
1
p
13
p
25
= a
1
p
24
(x
0
5
)
2
p
26
= a
3
p
24
(x
0
5
) p
27
= a
2
p
24
x
0
5
x
0
6
p
28
= c
1
(p
2
/b
3
+ p
24
) p
29
= c
1
/b
3
k
0
= b
31
b
42
−b
32
b
41
k
1
=(T
6
γ
2
+1)/T
6
,k
2
= k
1
b
42
b
32
/(k
0
b
4
b
5
) k
3
=(b
11
b
22
/T
1
−b
12
b
21
/T
2
)/(b
2
p
0
)
k
4
= b
42
b
32
(1/T
4
− 1/T
3
)/(b
5
k
0
) k
5
= γ
2
k
4
/(T
6
b
4
) k
6
= c
2
k
4
k
7
= a
2
k
4
x
0
5
x
0
6
k
8
= a
5
k
4
x
0
6
k
9
= a
4
k
4
(x
0
6
)
2
k
10
= a
2
x
0
5
x
0
6
/b
6
k
11
= a
5
x
0
6
/b
6
k
12
=(b
31
b
42
/T
4
− b
32
b
41
/T
3
)/(b
4
b
6
k
0
) k
13
=
k
1
k
12
−a
9
+γ
2
/(b
4
b
6
T
6
) k
14
=1/b
6
k
15
= a
2
k
1
k
14
k
16
= a
4
k
1
k
14
k
17
= a
5
k
1
k
14
k
18
= b
31
b
41
(1/T
4
−
1/T
3
)/(b
6
k
0
) k
19
= γ
2
k
12
/T
6
k
20
= k
1
k
14
−1+k
12
b
4
k
21
= k
12
b
4
k
22
= a
4
k
21
(x
0
6
)
2
k
23
= a
2
k
21
x
0
5
x
0
6
k
24
= a
5
k
21
x
0
6
k
25
= c
2
(k
1
k
24
+ k
12
b
4
) k
26
= c
2
k
14
T
i
> 0,i= 1, 6; γ
1
> 0,γ
2
> 0,
b
11
b
22
= b
12
b
21
,b
31
b
42
= b
32
b
41
.
x
d1
= x
d2
=2,5
y
1N
= y
2N
=0,37 y
11
=0,37 y
22
=0,41 y
12
=0,22 α
1N
= α
2N
=0,1 α
11
=0,219 α
22
=0,319
α
12
=0,009 x
d1
= x
d2
=0,28 T
d01
= T
d02
=6 Tj
1
= Tj
2
=9,17 T
c1
= T
c2
=4 U
c
=1
ω
0
=1 σ
1
= σ
2
=0,05 c
1
=1 Ω=1,9972 x
0
1
=1,04
x
0
2
=0,698 P
0
=0,85 U
0
=1 x
0
5
=2,387 x
0
6
=3,028 b
11
= b
31
=2 b
22
= b
12
= b
21
= b
42
= b
32
=
b
41
=1 T
1
= T
2
= T
3
= T
4
= T
5
= T
6
=1 γ
1
= γ
2
=0,2507
i =1 i =2
i =1 i =2
U
11
= U
12
= E
qe
U
21
= U
22
= µ
PT
k
0U
=50k
1U
=7
k
0f
=30 k
1f
=8 k
u
=0,6 T
u
=2, 5
•
•
•
•
˙
y(t)=g(y, v, u),
˙
v(t)=h(y, v, u),
y v u
˙
z(t)=R(y, z),
u = u(y, z),
z
•
•
•
•
U
1
,U
2
U
1
,U
2
ψ(t)=ϕ(y, v) − ˆϕ(t),
ϕ(y, v) ˆϕ(t)
ϕ(y, v) ˆϕ(t)=0
ϕ(y, v)− ˆϕ(t)=0 v
y
ψ
˙
ψ(t)=Lψ,
L ψ =0
dim ψ(t)=1
T
i
˙
ψ
i
(t)+ψ
i
=0.
y =[x
1
] v =
[x
2
x
3
x
4
x
5
x
6
]
x
3
ψ
4
= x
3
− ϕ
1
(x
1
) − z
1
.
T
4
˙
ψ
4
(t)+ψ
4
=0,
dx
3
dt
−
∂ϕ
1
(x
1
)
∂x
1
dx
1
dt
−
dz
1
dt
+
1
T
4
(x
3
− ϕ
1
(x
1
) − z
1
)=0
b
2
(−x
3
+ a
3
x
2
sin(x
1
− α
12
)+U
1
) −
∂ϕ
1
(x
1
)
∂x
1
x
2
−
dz
1
dt
+
1
T
4
(x
3
− ϕ
1
(x
1
) − z
1
)=0.
x
2
x
3
ϕ
1
(x
1
)=−a
3
b
2
cos(x
1
−α
12
) T
4
=1/b
2
dz
1
dt
= b
2
(−z
1
+ a
3
b
2
cos(x
1
− α
12
)+U
1
).
ψ
4
=0 x
3
ˆx
3
= z
1
− a
3
b
2
cos(x
1
− α
12
).
x
4
ψ
5
= x
4
− ϕ
2
(x
1
) − z
2
.
T
5
˙
ψ
5
(t)+ψ
5
=0,
dx
4
dt
−
∂ϕ
2
(x
1
)
∂x
1
dx
1
dt
−
dz
2
dt
+
1
T
5
(x
4
− ϕ
2
(x
1
) − z
2
)=0
b
3
(−x
4
− a
4
x
2
+ U
2
) −
∂ϕ
2
(x
1
)
∂x
1
x
2
−
dz
2
dt
+
1
T
5
(x
4
− ϕ
2
(x
1
) − z
2
)=0.
T
5
=1/b
3
,ϕ
2
(x
1
)=−a
4
b
3
x
1
x
2
,x
4
x
4
dz
2
dt
= b
3
(−z
2
+ a
4
b
3
x
1
+ U
2
),
ˆx
4
= z
2
− a
4
b
3
x
1
.
x
2
x
5
x
6
ψ =
⎡
⎢
⎢
⎣
x
2
− ˆx
2
x
5
− ˆx
5
x
6
− ˆx
6
⎤
⎥
⎥
⎦
x
3
=ˆx
3
x
4
=ˆx
4
b
1
(ˆx
4
− a
1
ˆx
2
3
− a
2
ˆx
3
sin(x
1
− α
12
)+c
1
x
5
) −
˙
ˆx
2
(t)=L
11
(x
2
− ˆx
2
)+
+ L
12
(x
5
− ˆx
5
)+L
13
(x
6
− ˆx
6
),
x
6
−
˙
ˆx
5
(t)=L
21
(x
2
− ˆx
2
)+L
22
(x
5
− ˆx
5
)+L
23
(x
6
− ˆx
6
),
− Ω
2
x
5
−
˙
ˆx
6
(t)=L
31
(x
2
− ˆx
2
)+L
32
(x
5
− ˆx
5
)+L
33
(x
6
− ˆx
6
),
L
ij
L
L
12
= c
1
b
1
L
13
=0 L
22
=0 L
23
=1 L
32
= −Ω
2
L
33
=0
L
11
L
21
L
31
L
L
λ = −1
det(pI − L
∗
)=(p +1)
3
= p
3
+3p
2
+3p +1.
det(pI − L)=p
3
− L
11
p
2
+(Ω
2
− L
21
c
1
b
1
)p −L
31
c
1
b
1
− Ω
2
L
11
.
p
L
11
= −3 L
21
=(Ω
2
− 3)/(c
1
b
1
) L
31
=(3Ω
2
− 1)/(c
1
b
1
)
L =
⎡
⎢
⎢
⎣
−3 c
1
b
1
0
(Ω
2
− 3)/(c
1
b
1
)01
(3Ω
2
− 1)/(c
1
b
1
) −Ω
2
0
⎤
⎥
⎥
⎦
.
L
x
1
ˆx
2
ˆx
3
ˆx
4
ˆx
5
ˆx
6
x
2
˙
ˆx
2
(t)
˙
ˆx
5
(t)
˙
ˆx
6
(t)
3x
2
−
˙
ˆx
2
(t)=3ˆx
2
− b
1
(ˆx
4
− a
1
ˆx
2
3
− a
2
ˆx
3
sin(x
1
− α
12
)) − c
1
b
1
ˆx
5
,
−L
21
x
2
−
˙
ˆx
5
(t)=−L
21
ˆx
2
− ˆx
6
,
−L
31
x
2
−
˙
ˆx
6
(t)=−L
31
ˆx
2
+Ω
2
ˆx
5
.