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


1 Answer 1


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.

  • $\begingroup$ 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 $\endgroup$ Jul 12, 2019 at 7:43
  • $\begingroup$ I edited my reply regarding your second comment about $t$ vs. $\tau$. $\endgroup$
    – Florian
    Jul 12, 2019 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 $\endgroup$ Jul 12, 2019 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, 2019 at 9:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.