Голяндина Н.Э. Метод Гусеница-SSA: анализ временных рядов
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F
(1)
N
F
(2)
N
L rank
L
(F
(i)
N
) < min(L, K)
i = 1, 2
rank
L
(F
N
) = rank X = rank XX
T
= rank X
T
X.
{U
i
}
d
i=1
λ
1
≥ . . . ≥ λ
d
XX
T
L
(L)
rank
L
(F
N
)
X
F =
(. . . , f
−1
, f
0
, f
1
, . . .)
F
d
a
1
, . . . , a
d
i
f
i+d
=
d
X
k=1
a
k
f
i+d−k
, a
d
6= 0.
d
d
F
N
f
n
= Ae
αn
cos(2πωn + φ),
ω ∈ [0, 1/2] φ ∈ [0, 2π) T = 1/ω
ω ∈ (0, 1/2)
ω ∈ (0, 1/2) L ≥ 2 N ≥ L +1
ω α
α ≤ 0 L, K → ∞
ω = 1/2
ω = 1/2 sin φ 6= 0 f
n
(−e
α
)
n
L
(ω = 0)
ω = 0 cos φ 6= 0 f
n
e
αn
N L
α
(α = 0)
ω < 1/2
ω = 1/2
ω ω < 1/2 K L T = 1/ω
Lω Kω
π/2
C cos(2πωn +
e
φ) C sin(2πωn +
e
φ)
F
N
f
n
=
m
X
k=0
a
k
n
k
, a
m
6= 0.
F
N
m
L ≥ m + 1
N ≥ L + m m + 1
m
L
L
L
L ¿ K
[a, b]
[a, b]
F
F
F
(1)
F
(2)
= F −F
(1)
F F
(1)
F
(2)
ω
F
(1)
F
(2)
= F −F
(1)
L
ω = 1/2
F F
(1)
ω < 1/2
α = 0
ω
Lω Kω
K = N − L + 1
T
T = 1/ω T
T
ω = q/p < 1/2 p q
p
1/12 10/53 2/5
5/12
L, K → ∞
L K
1/ω
F
ω
0
x(n) = r(n) cos(2πω(n)n + φ(n)) ,
y(n) = r(n) sin(2πω(n)n + φ(n)) ,
r ω φ ω(n) ≈ ω
0
(r(n), δ(n)) δ(n)
δ(n) = 2πω ( n)n + φ(n)
∆
n
def
= δ(n + 1) −δ(n) ≈ 2πω
0
ω
0
∆
n
n = 0, . . . , L − 1
ω = 1/2
F
(1)
F T
f
(1)
n
=
[T/2]
X
k=1
q
a
2
k
+ b
2
k
cos(2πkn/T + φ
k
).
T −1
k 6= T/2 k = T/2 L
a
2
k
+ b
2
k
F
(1)
F
(2)
F = F
(1)
+ F
(2)
L
a
2
k
+ b
2
k
L
F
(1)
F
k/T
F
1/12 1/6 1/4 1/3
1/2.4 1/2
a
2
1
+b
2
1
= a
2
2
+b
2
2
L
F
ω
1
= 1/T ω
2
= 2/T
ω
1
ω
2
F
(1)
α 6= 0
ω 1/2 L
(−a)
n
a = e
α
F
(1)
L K = N − L + 1
F
(1)
F
(2)
F = F
(1)
+ F
(2)
F
(2)
F
(1)
F
(1)
k/T
F
(1)
F
(2)