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I am trying to make my own demonstration in order to find what is equal to the FT of the Autocorrelation function. The autocorrelation function in some book about turbulence is defined as:

r($\tau$)=$\lim_{T \to \infty}\frac{1}{T} \int_{t0}^{t0+T} u(t)u(t+\tau)dt$ (1)

where T is the time interval, t0 is the starting time of measurement of the parameter. The power spectral density in this case is defined:

S(f)=$\frac{1}{2\pi} \int_{-\infty}^{\infty} exp(-i\tau 2 \pi f)r(\tau) d \tau $ (2)

If we substitute (1) in (2) we obtain:

S(f)=$\frac{1}{2\pi} \int_{-\infty}^{\infty} exp(-i\tau 2 \pi f) [\lim_{T \to \infty}\frac{1}{T} \int_{t0}^{t0+T} u(t)u(t+\tau)dt] d \tau $ (3)

Now I would like to do a similar demonstration like this the demonstration that states that the FT of the ACF function is the square of the DTFT of the signal , but there are some differences and unconsistency, that I would like to help me to fix it (because I am getting mad trying to demonstrate it)- First of all, my autocorrelation definition is different and it hasn’t an integration from -$\infty$ to +-$\infty$ , and furthermore, in my definition of autocorrelation is divided by the total time interval. So do you think that I should, redefine the integral domain in the power spectral density? If yes how?

Anyway I tried to continue this “demonstration” in order to explain all my doubts (maybe I made a mistake, but I hope that you can help me to fix my error). I rearranged (3) in this way

S(f)= $lim_{T \to \infty} \frac{1}{T} \int_{-\infty}^{\infty} u(t) [\frac{1}{2\pi} \int_{t0}^{t0+T}u(t+\tau)exp(-i\tau 2 \pi f) d \tau] dt $ (4)

I am not sure about that, but I thought that the part in the square brackets can be considered, as happend in the case in the link the demonstration that states that the FT of the ACF function is the square of the DTFT of the signal, to be:

$\frac{1}{2\pi} \int_{\color{red}t\color{red}0}^{\color{red}t\color{red}0\color{red}+\color{red}T}u(t+\tau)exp(-i\tau 2 \pi f)d \tau =U (f) exp(i 2 \pi ft)$ (5)

where U(f) is the fourier transform of u(t). An other doubt about the equation (5) is the fact that the coefficient $\frac{1}{2\pi}$ does not appear in the demonstration of the book that I cited in the link.

If we substitute (5) in (4) we obtain:

S(f)= $lim_{T \to \infty}\frac{1}{T} \int_{-\infty}^{\infty} u(t) [U (f) exp(i 2 \pi ft)] dt $ (6)

If you rearranged this:

S(f)= $lim_{T \to \infty}\frac{1}{T} U (f) [\int_{-\infty}^{\infty} u(t) exp(i 2 \pi ft)] dt $ (7)

if we consider that the part in the square bracket is equal to U(f)* which is the conjugate of U (f):

S(f)= $lim_{T \to \infty}\frac{1}{T} U (f) U (f)*$ (8)

which is equal

S(f)= $lim_{T \to \infty}\frac{1}{T} |U (f)|^2$ (9)

Now in this demonstration there are a lot of uncertainities and "bug". I hope you can kindly explain in what I am wrong and which could be the best strategy to reach the right result.

If you find references of book where similar demonstration are done with this definition of autocorrelation, it will be amazing I need a formulation without citing the expected value and the $\delta (t-\tau -t')$

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    $\begingroup$ Careful with limits and integrals, in general these cannot be exchanged so easily. In your equation (4): the integrand depends on $T$ (which the outer limit goes over). After you substitute (5) into (4), your integrand does not depend on $T$ anymore. So you have $\lim_T \frac 1T$ times a constant integral. If the integral is finite, this should be zero. If it's not, there should be some dependence on $T$ for things to make any sense. $\endgroup$ – Florian Jul 12 at 11:10
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    $\begingroup$ Does this related answer answer your question? $\endgroup$ – Matt L. Jul 12 at 11:14
  • $\begingroup$ @MattL. No. I need a formulation without citing the expected value and the $\delta (t-\tau -t')$ . I updated the question in orther to clarify this $\endgroup$ – Ashish Bhigah Jul 12 at 11:20
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    $\begingroup$ I think you have a non-standard definition of the auto correlation. Starting the integration at $t_0$ instead $t_0 - T_0$ feels strange to me. I wouldn't expect to get the same result with a different definition $\endgroup$ – Hilmar Jul 12 at 12:23
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    $\begingroup$ @AshishBhigah: In your equations (6)ff, the integrand still does not depend on $T$. How can this dependence vanish when you insert (5) into (4)? And no (5) does not make sense to me. You're saying the LHS is equal to the RHS for every $T$? What if I set $T=0$ in (5)? $\endgroup$ – Florian Jul 12 at 13:09

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