- #1
Jano L.
Gold Member
- 1,333
- 75
Consider continuous function [itex]x(t)[/itex], which has zero time average:
[tex]
\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T/2}^{T/2} x(t)\,dt = 0
[/tex]
and exponentially decaying autocorrelation function:
[tex]
\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T/2}^{T/2} x(t)x(t-\tau)\,dt = C_0e^{-\gamma |\tau|}.
[/tex]
Is it possible to write such function as the Fourier integral
[tex]
\int_{-\infty}^{\infty} x(\omega) e^{i\omega t} d\omega/2\pi
[/tex]
for some function/distribution [itex]x(\omega)[/itex]? It seems the Fourier analysis is used in the theory of random functions a lot, but on the other hand, when I insert the integral representation into the time averaging integral, I'm getting autocorrelation function equal to 0.
[tex]
\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T/2}^{T/2} x(t)\,dt = 0
[/tex]
and exponentially decaying autocorrelation function:
[tex]
\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T/2}^{T/2} x(t)x(t-\tau)\,dt = C_0e^{-\gamma |\tau|}.
[/tex]
Is it possible to write such function as the Fourier integral
[tex]
\int_{-\infty}^{\infty} x(\omega) e^{i\omega t} d\omega/2\pi
[/tex]
for some function/distribution [itex]x(\omega)[/itex]? It seems the Fourier analysis is used in the theory of random functions a lot, but on the other hand, when I insert the integral representation into the time averaging integral, I'm getting autocorrelation function equal to 0.