# Extremes of integral of the autocorrelation theorem

I am trying to apply the correlation theorem (here attached) to this definition of correlation of a finite continuous signal $$g(t)$$:

$$\rho(\tau)=\frac{1}{T} \int_{t_{0}}^{ t_{0}+T}g(t)^{.}g(t+\tau)dt \tag{1}$$

where $$t$$ is the time, T is the time duration of the signal, $$t_{0}$$ is the starting time, $$\tau$$ is the time shift of the correlation. As you can see, in the attached demonstration the correlation is defined from $$[-\infty,+\infty]$$. Obviously, the signals we work with are always finite. So I defined the interval as $$[t_{0},t_{0}+T]$$, as I also see in a book of Turbulence. If I do the fast Fourier transform of both the member of the Eq (1) I obtain

$$E(f)= \frac{1}{T} \int_{\color{red}{t_{0}}}^{\color{red}{t_{0}+T}}\int_{\color{red}{t_{0}}}^{ \color{red}{t_{0}+T}}g(t)^{.}g(t+\tau) e^{-i2\pi f \tau}d\tau dt \tag{2}$$

where $$E(f)$$ is the Fourier Transform of $$\rho(\tau)$$. Usually the Fourier transforms, that I have seen in books, are defined in the interval $$[-\infty,+\infty]$$, but, as I said before, the signal we work with are defined in a time window such as $$[t_{0},t_{0}+T]$$, so I tried to apply them to finite signal

Introducing the variable

$$y=t+\tau$$

The Eq. (2) can be rewritten:

$$E(f)= \frac{1}{T} \int_{\color{red}{t_{0}}}^{\color{red}{ t_{0}+T}}\int_{\color{red}{t_{0}}}^{ \color{red}{t_{0}+T}}g(t)^{.}g(y) e^{-i2\pi f (y-t)}dy dt \tag{3}$$

Furthermore the Eq. (3) can be rewritten as:

$$E(f)= \frac{1}{T} \int_{\color{red}{t_{0}}}^{\color{red}{ t_{0}+T}}g(t) e^{+i2\pi f t}dt\int_{\color{red}{t_{0}}}^{ \color{red}{t_{0}+T}}g(y) e^{-i2\pi f y}dy \tag{4}$$

So E(f) can be rewritten as:

$$E(f)= \frac{1}{T} G^{*}(f)^{.}G(f)\tag{5}$$

I am not sure about the (red) extremes of the integral, I think they are wrong can you help me? Is it possible to apply the correlation and the Fourier transform to finite signals with interval of existence $$[t_{0},t_{0}+T]$$?

• We were talking about this earlier ina different thread but if you are actually doing an integration, then the integral limits are the same as wherever the signal starts and ends. But, usually ( your case could be an exception ) autocorrelation is a summation because you end up calculating an empirical estimate of the actual autocorrelation. So, if you are doing something empirically in the time domain, my guess is that you should be using two summations in order to calculate an estimate of the autocorrelation at each time lag $\tau$. Then, after that, you can do fourier transform etc. Commented Apr 19, 2021 at 19:14
• I don't think so. The fact that is random does not mean that is discrete. The Eq. (1) is taken from a book of turbulence. My problem is about the extremes. I remember thread where we talked about this, but I preferred to ask contestualising it Commented Apr 19, 2021 at 19:32
• @Luca Marianini: How are you obtaining the turbulence data ? Is it discrete or continuous. Commented Apr 20, 2021 at 0:11
• @Luca Marianini: I'm not saying that there's anything wrong with equation (1). The only problem with it is that it assumes that A) there is a known deterministic function, $g(t)$ that produces the turbulence values for any value of $t$ and B) that the signal only exists over the interval $(t_{0}, t_{0} + T)$. A) is usually not the case because, in an actual application, one is given the actual data at discrete times rather than a function $g(t)$. Commented Apr 20, 2021 at 12:59
• sorry @markleeds I cannot understand you. g(t) is a velocity component and I suppose that is continuous. It is a continuous and finite signal. If you use an analogic system of acquirement, the signal is continuous and finite. Commented Apr 20, 2021 at 16:08

Maybe your signal is finite, but the function you are transforming is infinite. So you can do your demonstration using the interval $$[0,T]$$ or $$[-\infty,+\infty]$$, the result should be the same, because if your signal is finite the FT threats it as an infinite signal with period T.