Fourier transform and anti-trasform--identity missing

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

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.

• 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 – Ashish Bhigah Jul 12 '19 at 7:43
• I edited my reply regarding your second comment about $t$ vs. $\tau$. – Florian Jul 12 '19 at 8:11
• 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 – Ashish Bhigah Jul 12 '19 at 8:35
• 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. – Florian Jul 12 '19 at 9:22