We know that for a signal $x(t)$, it is related to $R_x(\tau)$ as, $$R_x(\tau) = \int_{-\infty}^{\infty}x(t)x(t-\tau)dt$$. $\\$We also know that $$R_x(\tau) \rightleftharpoons S_x(f)$$ $\\$How do we mathematically relate $X(f)$ with $S_x(f)$, where $X(f) \leftrightharpoons x(t) $?

  • $\begingroup$ you want a derivation? $\endgroup$ – robert bristow-johnson Jul 19 at 1:34
  • $\begingroup$ A derivation would be great, although just having the result would be fine too, I would try to work on the derivation myself. $\endgroup$ – helloworld1e. Jul 19 at 1:40
  • $\begingroup$ Okay, this is about the autocorrelation and energy spectral density function of a finite energy signal, not a "power signal". And, it appears to me that $x(t)$ is deterministic, not random. Is that correct? $\endgroup$ – robert bristow-johnson Jul 19 at 2:34
  • $\begingroup$ Yes, it is for finite energy signal(deterministic). $\endgroup$ – helloworld1e. Jul 19 at 3:29

$x(t)$ is a real-valued finite energy signal with Fourier transform $X(f)$. Its autocorrelation function is \begin{align} R_x(\tau) &= \int_{-\infty}^{\infty}x(t)x(t-\tau) \,\mathrm dt\\ &= \int_{-\infty}^{\infty}x(t)y(\tau-t) \,\mathrm dt & {\scriptstyle{\text{Define}~y(t)~\text{as the time-reversal}~x(-t)~\text{of}~x(t)}}\\ &= x\star y\big|_\tau \end{align} so that $S_x(f) = \mathcal F\{R_x(\tau)\} =\mathcal F\{x(t)\}\mathcal F\{y(t)\}=X(f)Y(f)$. But, \begin{align} Y(f) &= \int_{-\infty}^{\infty}y(t)\exp(-j2\pi ft) \,\mathrm dt\\ &= \int_{-\infty}^{\infty}x(-t)\exp(-j2\pi ft) \,\mathrm dt\\ &= \int_{-\infty}^{\infty}x(\lambda)\exp(j2\pi f\lambda) \,\mathrm d\lambda &{\scriptstyle{\text{Substitute}~t=-\lambda, \mathrm dt = -\mathrm d\lambda~\text{ etc}}}\\ &= \left[\int_{-\infty}^{\infty}x(\lambda)\exp(-j2\pi f\lambda) \,\mathrm d\lambda\right]^*\\ &= X^*(f) \end{align} which leads to $$S_x(f) = X(f)Y(f) = |X(f)|^2.$$


$$ \mathscr{F} \Big\{ x(t) \Big\} \triangleq X(f) \triangleq \int\limits_{-\infty}^{\infty} x(t) \, e^{-j2 \pi f t} \ \mathrm{d}t $$

$$ \mathscr{F}^{-1} \Big\{ X(f) \Big\} \triangleq x(t) = \int\limits_{-\infty}^{\infty} X(f) \, e^{j2 \pi f t} \ \mathrm{d}f $$

Assuming $x(t)$ is real (which means that $R_x(\tau)$ is also real).

$$\begin{align} S_x(f) &\triangleq \mathscr{F} \Big\{ R_x(\tau) \Big\} \\ \\ &= \int\limits_{-\infty}^{\infty} R_x(\tau) \, e^{-j2 \pi f \tau} \ \mathrm{d}\tau \\ \\ &= \int\limits_{-\infty}^{\infty} \left( \int_\limits{-\infty}^{\infty}x(t)x(t-\tau) \mathrm{d}t \right) \ e^{-j2 \pi f \tau} \ \mathrm{d}\tau \\ \\ &= \int\limits_{-\infty}^{\infty} \int_\limits{-\infty}^{\infty}x(t)x(t-\tau) \ e^{-j2 \pi f \tau} \,\mathrm{d}t \,\mathrm{d}\tau \\ \\ &= \int\limits_{-\infty}^{\infty} \int_\limits{-\infty}^{\infty}x(t)x(t-\tau) \ e^{-j2 \pi f \tau} \,\mathrm{d}\tau \,\mathrm{d}t \\ \\ &= \int\limits_{-\infty}^{\infty} x(t) \int_\limits{-\infty}^{\infty}x(t-\tau) \ e^{-j2 \pi f \tau} \,\mathrm{d}\tau \,\mathrm{d}t \\ \\ &= \int\limits_{-\infty}^{\infty} x(t) e^{-j2 \pi f t} \int_\limits{-\infty}^{\infty}x(t-\tau) \ e^{-j2 \pi f (\tau-t)} \,\mathrm{d}\tau \,\mathrm{d}t \qquad \text{let } u \triangleq \tau-t\\ \\ &= \int\limits_{-\infty}^{\infty} x(t) e^{-j2 \pi f t} \int_\limits{-\infty}^{\infty}x(-u) \ e^{-j2 \pi f u} \,\mathrm{d}u \,\mathrm{d}t \\ \\ &= \int\limits_{-\infty}^{\infty} x(t) e^{-j2 \pi f t} \,\mathrm{d}t \int_\limits{-\infty}^{\infty}x(-u) \ e^{-j2 \pi f u} \,\mathrm{d}u \\ \\ &= \mathscr{F} \Big\{ x(t) \Big\} \cdot \mathscr{F} \Big\{ x(-t) \Big\} \\ \\ &= X(f) \cdot X(-f) \\ \\ &= X(f) \cdot \big( X(f) \big)^* \\ \\ &= \Big|X(f)\Big|^2 \\ \end{align}$$

  • $\begingroup$ i was playing a little fast-and-loose about swapping the order of integration with the double integral. $\endgroup$ – robert bristow-johnson Jul 19 at 3:01

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