Методы анализа сложных сигналов. Павлов А.Н. - 98 стр.

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µ
0
= 1
µ
1
= p
1
µ
0
µ
2
= (1 p
1
)µ
0
p
1
p
2
1
n=1
n=0
n=2
... ... ... ... ... ... ... ... ...
p p
p p p p p p p p
1 2
1 1 1 2 2 1 2 2
ε = 3
n
x
0
= 0 µ
0
=
p
n
1
µ
0
= p
n
1
α(0) = ln p
1
/ ln(1/3)
x
0
= 1
α(1) = ln p
2
/ ln(1/3) p
1
6= p
2
α(0) 6= α(1) f(α)
0.2 0.4 0.6 0.8 1.0
α
0.0
0.2
0.4
0.6
0.8
f( )
α
f(α)