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I have a very silly doubt:

If we define the power spectral density:

S(f)=$\frac{1}{2\pi}\int exp(-i\tau2\pi f)r(\tau)d\tau$ (1)

where $r(\tau)$ is the correlation coefficient.

If we do the Fourier anti-transform, we obtain $r(\tau)=\int exp(i\tau2\pi f)S(f)df$ (2)

Now my doubt is: if I substitute in the second equation the first equation, it seems to me that I don't find the identity $r(\tau)=r(\tau)$

I hope you can help me, maybe I am missing something

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Different ways of showing it, depends where you start. Are you willing to accept that the Fourier transform of $\delta(t)$ is $1$ and vice versa (i.e., $\int_{-\infty}^\infty {\color{red}1} \cdot {\rm e}^{\jmath 2\pi f t} = \delta(t)$)? If so, it's easy:

$$\begin{align} r(\tau) & = \int_{-\infty}^\infty {\rm e}^{\jmath 2\pi f \tau} S(f) {\rm d}f \\ & = \int_{-\infty}^\infty {\rm e}^{\jmath 2\pi f \tau} \int_{-\infty}^\infty {\rm e}^{-\jmath 2\pi f t} r(t) {\rm d}t {\rm d}f \\ & = \int_{-\infty}^\infty \int_{-\infty}^\infty {\rm e}^{\jmath 2\pi f (\tau-t)} r(t) {\rm d}t {\rm d}f \\ & = \int_{-\infty}^\infty r(t) \int_{-\infty}^\infty {\color{red}1} \cdot {\rm e}^{\jmath 2\pi f (\tau-t)} {\rm d}f {\rm d} t \\ & = \int_{-\infty}^\infty r(t) \delta(\tau-t) {\rm d} t \\ & = \int_{-\infty}^\infty r(\tau) \delta(\tau-t) {\rm d} t \\ & = r(\tau) \int_{-\infty}^\infty \delta(\tau-t) {\rm d} t \\ & = r(\tau) \end{align}$$

  • Step 1: Insert (2) into (1). Note that the inner integration variable is a new one, different from $\tau$. I call it $t$.
  • Step 2: Pulling the first exp inside the integral, combining the exps.
  • Step 3: Changing integration order (PSD and ACF are absolutely integrable), pulling out what does not depend on the inner integration variable $f$.
  • Step 4: Using the fact that the inverse Fourier of a constant is a delta (think $\tau-t$ as one variable here, then it's the inverse Fourier of 1).
  • Step 5: Using the sifting property of the delta.
  • Step 6: Moving out the constant term
  • Step 7: Area under the delta is one.

Of course, step 4 is the critical one. If you don't buy it, you need a different, more fundamental/mathematical approach. This is more the engineering point of view I'm presenting here.

Regarding your reply with $\tau$ vs. $t$: What you say is not true. See, we're computing $r(\tau)$ via $\int_{-\infty}^\infty {\rm e}^{\jmath 2\pi f\tau} S(f) {\rm d}f$, which means it's an integral over frequency and the integration kernel depends on $\tau$. The function $S(f)$ does not depend on $\tau$ of course. Now, you are replacing $S(f)$ by the inverse Fourier transform of the autocorrelation, which is an integral over time. But it's a different time variable, which I called $t$. It must be, since it if were $\tau$ it would mean that $S(f)$ somehow depends on $\tau$.

Another way to think about it: The variable $\tau$ is the independent one. Don't forget that all our integrals are definite ones (from $-\infty$ to $\infty$). We sometimes drop that for laziness, but I added them now to be more clear. Now, the integration variables on the right-hand side are the ones we integrate over, hence the result cannot depend on it. Imagine something like $x(\tau) = \int_{-\infty}^\infty \int_{-\infty}^\infty X(f) g(\tau) {\rm d}f {\rm d}\tau$. This does not make sense as the right-hand side integrates over $f$ and $\tau$ (the result is a number) whereas the right-hand side depends on $\tau$. This is why we need a new time variable.

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  • $\begingroup$ Thank for your answer, I will look into it, but I was searching in something without $\delta(\tau)$. In the first step you placed a formula of the autocorrelation that is r(t), but it is $r(\tau)$ instead. So you will not have dt but d$\tau$ and exp(2$\pi f \tau$) . What I do not understand is why placing directly the $r(\tau)$ you don't find the identity, it seems very strange to me $\endgroup$ – Ashish Bhigah Jul 12 at 7:43
  • $\begingroup$ I edited my reply regarding your second comment about $t$ vs. $\tau$. $\endgroup$ – Florian Jul 12 at 8:11
  • $\begingroup$ Ok, I perfectly agree when you say "this does not make sense as the right-hand side integrates over f and $\tau$ (the result is a number) whereas the right-hand side depends on $\tau$." That's why I asked it. I would like to know which properties should I invoke in order to say "you cannot substitute r($\tau$) in his anti-transform. Becouse normally if you have : x=b and b=x, if you substitute the first with the second you reach the identity, it is that that confuse me $\endgroup$ – Ashish Bhigah Jul 12 at 8:35
  • $\begingroup$ And that's exactly what I did. It's just that in order to properly substitute $b$ in $x$ we need to watch out to get it right. $r(\tau)$ means it's a function of an independent time variable. In the equation, $\tau$ is already being used on the left-hand side hence it's not an independent time variable anymore. To properly substitute in $r(\tau)$ we hence need to give the independent time variable a new name. $\endgroup$ – Florian Jul 12 at 9:22

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