Why Does the Fourier Transform of Autocorrelation Equal the ESD?

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Does anyone here have a good explanation of why the Fourier transform of the autocorrelation function equals the ESD of the the original signal. It kind of make sense intutively because functions that have a autocorrelation that drops of quickly are high frenquency and the Fourier transform of that resulting function will obviuosly have a wide bandwidth but it seems like there should but a analytic derivation of this.
 
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dionysian said:
Does anyone here have a good explanation of why the Fourier transform of the autocorrelation function equals the ESD of the the original signal. It kind of make sense intutively because functions that have a autocorrelation that drops of quickly are high frenquency and the Fourier transform of that resulting function will obviuosly have a wide bandwidth but it seems like there should but a analytic derivation of this.
Please be careful: ESD and PSD (energy and power spectral density) are not interchangeable.
You are inquiring about the Wiener-Khinchin theorem, which states that the PSD is the Fourier Transform (FT) of the autocorrelation function (and vice versa). Here is an online mathematical derivation:
http://mathworld.wolfram.com/Wiener-KhinchinTheorem.html"
The W-K is intuitively reasonable. It predicts, for instance, that a sharply peaked autocorrelation function R transforms to a broad power density spectrum. Think about a random noise signal for which R is a delta function; the FT of a delta is a uniform spectrum, and indeed random noise has a white spectrum. Other examples are also easily imagined.
I recommend that you check out the discussion in a textbook for more details. The W-K theorem is discussed in every text on Fourier transforms, signal processing, and if you are a physicist, statistical mechanics (because of the interest in characterizing random fluctuations).
 
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Ahhh haa... I just found that in one of my books. Thanks. I just needed someone to point me in the right direction.