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Superposition law for DFA
For two uncorrelated signals
and
, their root mean
square fluctuation functions are
and
respectively. We want to prove that for the signal
, its rms fluctuation
![\begin{displaymath}
F_{f+g}(n) = \sqrt{F_f(n)^2+F_g(n)^2}
\end{displaymath}](img397.png) |
(27) |
Consider three signals in the same box first. The integrated signals for
,
and
are
,
and
and their corresponding trends are
,
,
(
,
is the box size).
Since
and combine the definition of detrended fluctuation function Eq.3, we have that for all boxes
![\begin{displaymath}
Y_{f+g}(i)=Y_f(i)+Y_g(i),
\end{displaymath}](img409.png) |
(28) |
where
is the detrended fluctuation function for the
signal
,
is for the signal
and
for
. Furthermore, according to the definition of rms
fluctuation, we can obtain
![\begin{displaymath}
F_{f+g}(n) = \sqrt{\frac{1}{N_{max}} \sum\limits_{i=1}^{N_{...
...m\limits_{i=1}^{N_{max}}\left[Y_{f}(i)+Y_{g}(i)\right ]^2},
\end{displaymath}](img413.png) |
(29) |
where
is the number of boxes and
means the
th box.
If
and
are not correlated, neither are
and
and, thus,
![\begin{displaymath}
\sum\limits_{i=1}^{N_{max}}Y_{f}(i)Y_{g}(i)=0.
\end{displaymath}](img416.png) |
(30) |
From Eq.30 and Eq.29, we have
![\begin{displaymath}
F_{f+g}(n)=\sqrt{\frac{1}{N_{max}} \sum\limits_{i=1}^{N_{ma...
...}
=\sqrt{\left[F_{f}(n)\right ]^2+\left[F_{g}(n)\right ]^2}.
\end{displaymath}](img417.png) |
(31) |
Next: DFA-1 on linear trend
Up: Appendix
Previous: Noise
Zhi Chen
2002-08-28