Колесников А.А. Синергетические методы управления сложными системами: энергетические системы
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•
•
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dδ
dt
= ω − ω
0
;
Tj
dω
dt
= M
T
− i
q
E/ω
0
+ F (t);
r
a
i
d
+
1
ω
0
x
d
di
d
dt
+
dE
dt
+
ω
ω
0
x
q
i
q
= −u
d
;
r
a
i
q
+
x
q
ω
0
di
q
dt
+
ω
ω
0
(x
d
i
d
+ E)=−u
q
;
T
d0
dE
dt
= −E − T
d0
(x
d
− x
d
)
di
d
dt
+ E
e
;
T
c
dM
T
dt
= −M
T
−
ω − ω
0
σ
+ U
c
,
δ ω ω
0
E M
T
i
d
i
q
u
d
u
q
r
a
x
d
x
q
x
d
E
e
U
c
T
d0
T
c
Tj
σ
F (t)
R
n
i
d
+
x
n
ω
0
di
d
dt
+
ω
ω
0
x
n
i
q
= u
d
;
R
n
i
q
+
x
n
ω
0
di
q
dt
−
ω
ω
0
x
n
i
d
= u
q
,
x
n
R
n
dδ
dt
= ω − ω
0
;
Tj
dω
dt
= M
T
− i
q
E/ω
0
+ c
1
w
1
;
di
d
dt
= a
1
i
d
+ a
2
i
q
ω + a
3
(E − E
e
);
di
q
dt
= a
4
i
q
+ a
5
i
d
ω + Eω;
dE
dt
= −a
6
(a
1
i
d
+ a
2
i
q
ω) − a
3
(E − E
e
);
T
c
dM
T
dt
= −M
T
−
ω − ω
0
σ
+ U
c
;
dw
1
dt
= w
2
;
dw
2
dt
= −Ω
2
w
1
,
a
1
=
−ω
0
(R
n
+ r
a
)
(x
d
+ x
n
)(1 − a
6
)
a
2
=
−1
1 − a
6
a
3
=
1
T
d0
(1 −a
6
)
a
4
=
−ω
0
(R
n
+ r
a
)
x
q
+ x
n
a
5
=1
a
6
= x
d
− x
d
w
1
,w
2
c
1
Ω
E
e
δ − δ
0
=0.
ω − ω
0
=0
E − E
0
=0.
U − U
0
=0.
δ = δ
0
E = E
0
m =2
ψ
1
= b
11
(E − E
0
)+b
12
(M
T
+ ϕ
3
+ c
1
w
1
)=0;
ψ
2
= b
21
(E − E
0
)+b
22
(M
T
+ ϕ
3
+ c
1
w
1
)=0.
T
1
˙
ψ
1
(t)+ψ
1
=0;
T
2
˙
ψ
2
(t)+ψ
2
=0.
ϕ
3
E = E
0
;
M
T
= −ϕ
3
− c
1
w
1
.
ψ
1
=0 ψ
2
=0
dδ
dt
= ω − ω
0
;
Tj
dω
dt
= −ϕ
3
− i
q
E/ω
0
;
di
d
dt
= a
1
i
d
+ a
2
i
q
ω + a
3
(E
0
− E
e
);
di
q
dt
= a
4
i
q
+ a
5
i
d
ω + E
0
ω,
E
e
= E
e
(δ, ω, i
d
,i
q
)
ψ
1
=0 ψ
2
=0
ϕ
3
ψ
3
= ω − ω
0
+ γ(δ − δ
0
),
γ
T
3
˙
ψ
3
(t)+ψ
3
=0.
ϕ
3
= −E
0
i
q
/ω
0
+
Tj
T
3
(1 + T
3
γ)(ω −ω
0
)+Tj
γ
T
3
(δ − δ
0
).
dδ
dt
= ω − ω
0
;
Tj
dω
dt
= −
Tj
T
3
(1 + T
3
γ)(ω −ω
0
) − Tj
γ
T
3
(δ − δ
0
);
di
d
dt
= a
1
i
d
+ a
2
i
q
ω + a
3
(E
0
− E
e
);
di
q
dt
= a
4
i
q
+ a
5
i
d
ω + E
0
ω,
ω = ω
0
,δ = δ
0
ψ
3
=0
dδ
dt
= −γ(δ − δ
0
);
di
d
dt
= a
1
i
d
+ a
2
i
q
(ω
0
+ γ(δ − δ
0
)) + a
3
(E
0
− E
e
);
di
q
dt
= a
4
i
q
+(a
5
i
d
+ E
0
)ω
0
(ω
0
+ γ(δ − δ
0
)),
E
e
= E
e
(δ, i
d
,i
q
)
ψ
3
=0
E
e
= E + p
1
i
d
+ p
4
i
q
ω + k
1
i
q
− k
0
M
T
− k
2
(δ − δ
0
) − k
3
(ω − ω
0
)+k
4
(E − E
0
) − k
5
w
1
;
U
c
= −p
8
i
d
ω − k
6
i
q
E − p
10
i
q
− p
11
M
T
− k
7
Eω − k
8
(ω − ω
0
) − k
9
(δ − δ
0
)−
− k
10
(E − E
0
) − k
11
w
1
− k
18
w
2
,
p
0
= b
11
b
22
− b
12
b
21
p
1
= a
5
a
1
/a
4
p
2
=1/(a
4
p
0
) p
3
= b
12
b
22
(1/T
1
− 1/T
2
) p
4
= a
5
a
3
/a
4
p
5
=1/T
3
+ γ p
6
= b
12
b
21
/T
2
− b
11
b
22
/T
1
p
7
= T
c
/ω
0
p
8
= p
7
a
8
E
0
p
9
= b
11
b
22
/T
2
− b
12
b
21
/T
1
p
10
= p
7
E
0
(a
6
+ p
9
/p
0
) p
11
= T
c
(p
5
+ p
9
/p
0
) − 1 p
12
= T
c
Tj p
13
= γ/T
3
p
14
= p
12
E
0
(p
13
+
p
5
p
9
/p
0
) p
15
=1/σ p
16
= b
11
b
21
(1/T
2
− 1/T
1
) p
17
= T
c
/p
0
p
18
= T
c
c
1
k
0
= p
2
p
3
k
1
= k
0
E
0
/ω
0
k
2
= k
0
Tjγ/T
3
k
3
= k
0
Tjp
5
k
4
= p
2
p
6
k
5
= k
0
c
1
k
6
= p
7
p
5
k
7
= p
8
/a
8
k
8
= p
14
− p
15
k
9
= p
9
p
12
p
13
/p
0
k
10
= p
16
p
17
k
11
= p
18
(p
5
+ p
9
/p
0
)
T
1
,T
2
,T
3
,γ >0,
b
11
b
22
= b
12
b
21
p
0
=0
U =
u
2
q
+ u
2
d
=
'
−r
a
i
q
+(E + x
d
i
d
)
(
2
+
'
−r
a
i
d
− x
q
i
q
(
2
.
Tj =8,4 T
c
=4 T
d0
=5,12 ω
0
=1 x
d
=0,75 x
d
=0,44 x
n
=0,6 r
a
=0,003 R
n
=0,81
σ =0,05 c
1
=1 Ω=0,02 T
1
= T
2
=5 T
3
=2 E
0
=1,5
δ
0
= π/3 γ =1 b
11
=2 b
12
= b
21
= b
22
=1
x
n
=1,2 R
n
=1,62 50 <t<60
δ = δ
0
U = U
0
m =2
ψ
1
= b
11
'
U
2
− U
2
0
(
+ b
12
(M
T
+ ϕ
4
)=0;
ψ
2
= b
21
'
U
2
− U
2
0
(
+ b
22
(M
T
+ ϕ
4
)=0,
U
2
=(−r
a
i
q
+(E + x
d
i
d
))
2
+(−r
a
i
d
− x
q
i
q
)
2
T
1
˙
ψ
1
(t)+ψ
1
=0,
T
2
˙
ψ
2
(t)+ψ
2
=0.
M
T
ψ
1
=0 ψ
2
=0
dδ
dt
= ω − ω
0
;
Tj
dω
dt
= −ϕ
4
− i
q
ξ/ω
0
;
di
d
dt
= a
1
i
d
+ a
2
i
q
ω + a
3
(ξ − E
e
);
di
q
dt
= a
4
i
q
+ a
5
i
d
ω + ξω,
ξ = −x
d
i
d
+ r
a
i
q
−
−2r
a
x
q
i
d
i
q
− x
2
q
i
2
q
+ U
2
0
− r
2
a
i
2
d
E
e
= E
e
(δ, ω, i
d
,i
q
)
ψ
1
=0 ψ
2
=0
ϕ
4
ψ
3
= ω − ω
0
+ γ
4
(δ − δ
0
),
γ
4
T
3
˙
ψ
3
(t)+ψ
3
=0.
ϕ
4
=
i
q
ω
0
−2r
a
x
q
i
d
i
q
− x
2
q
i
2
q
+ U
2
0
− r
2
a
i
2
d
+
'
i
d
i
q
x
d
− r
a
i
2
q
(
/ω
0
+
+
Tj
T
3
(1 + T
3
γ
4
)(ω − ω
0
)+Tj
γ
4
T
3
(δ − δ
0
).
ψ
3
=0
dδ
dt
= −γ
4
(δ − δ
0
),
di
d
dt
= a
1
i
d
+ a
2
i
q
(ω
0
+ γ
4
(δ − δ
0
)) + a
3
(ξ − E
e
(δ, i
d
,i
q
)),
di
q
dt
= a
4
i
q
+(a
5
i
d
+ ξ)(ω
0
+ γ
4
(δ − δ
0
)),
E
e
= E
e
(δ, i
d
,i
q
)
ψ
3
=0
E
e
U
c
c
1
=0 T
1
= T
2
=5 T
3
=2 U
0
=1 γ
4
=1
δ
0
= π/3 b
11
=2 b
12
= b
21
= b
22
=1
x
n
=0,3 R
n
=0,405 50 <t<60
•
•
•
•
dδ
dt
= ω − ω
0
;
Tj
dω
dt
= M
T
− i
q
ψ
d
+ i
d
ψ
q
;
T
c
dM
T
dt
= −M
T
−
ω − ω
0
σ
+ U
T
;
−u
d
= r
a
i
d
+
dψ
d
dt
+ ωψ
q
;
−u
q
= r
a
i
q
+
dψ
q
dt
− ωψ
d
;
U
fd
= r
fd
i
fd
+
dψ
fd
dt
;
U
fq
= r
fq
i
fq
+
dψ
fq
dt
;
0=r
1d
i
1d
+
dψ
1d
dt
;
0=r
1q
i
1q
+
dψ
1q
dt
,
δ ω ω
0
M
T
U
T
T
d0
T
c
Tj
σ
r
a
r
fd
r
fq
r
1d
r
1q