# How do we know that CTFT of the autocorrelation function is the PSD?

I know that the Fourier Transform of the autocorrelation function is the Power Spectral Density. But how can we arrive at such a result intuitively? Is it just a theorem?

• It would help to know what you take as the definitions of the autocorrelation function (ACF) and the power spectral density (PSD). Some people just define the PSD as the Fourier transform of the ACF and so there is nothing to prove; others have two different definitions for these concepts and then prove a theorem saying that $\mathbb F(ACF) = PSD$. But to do this, it helps to know the starting point.... – Dilip Sarwate Dec 4 '13 at 17:37

\begin{align} R_x(\tau) &=\int_{-\infty}^\infty x(t)x^*(t-\tau)\,\mathrm dt\\ &= \int_{-\infty}^\infty \left[\int_{-\infty}^\infty X(f)e^{j2\pi ft}\,\mathrm df\right]x^*(t-\tau)\,\mathrm dt\\ &= \int_{-\infty}^\infty X(f) \left[\int_{-\infty}^\infty x^*(t-\tau) e^{j2\pi ft}\,\mathrm dt\right]\,\mathrm df\\ &= \int_{-\infty}^\infty X(f) \left[\int_{-\infty}^\infty x^*(t-\tau) e^{j2\pi f(t-\tau)}\,\mathrm dt\right]e^{j2\pi f\tau}\,\mathrm df\\ &= \int_{-\infty}^\infty X(f) \left[\int_{-\infty}^\infty x^*(\lambda) e^{j2\pi f\lambda}\,\mathrm d\lambda\right]e^{j2\pi f\tau}\,\mathrm df\\ &= \int_{-\infty}^\infty X(f) \left[\int_{-\infty}^\infty x(\lambda) e^{-j2\pi f\lambda}\,\mathrm d\lambda\right]^*e^{j2\pi f\tau}\,\mathrm df\\ &= \int_{-\infty}^\infty X(f)X^*(f) e^{j2\pi f\tau}\,\mathrm df\\ &= \int_{-\infty}^\infty |X(f)|^2 e^{j2\pi f\tau}\,\mathrm df \end{align}
• @TheGrapeBeyond I multiplied by $e^{-j2\pi f\tau}$ and also by $e^{j2\pi f\tau}$ for a net multiplication by $1$ – Dilip Sarwate Dec 5 '13 at 15:37
Here's the story up to constant factors. Correlation is the same as convolution with a time reversed kernel. Autocorrelation is therefore convolution with the time reversed function itself. Convolution can be represented as a product in frequency domain, time reversal as complex conjugation. That means the autocorrelation function of a signal $s(t)$ can be written as $$ACF[s(t)] = \mathcal{F}^{-1}\{\mathcal{F}\{s(t)\}\cdot \mathcal{F}\{s(t)\}^*\}$$
Now power spectral density. It is (roughly) defined as the squared magnitude of the frequeny domain representation of the signal. So $PSD[s(t)]=|\mathcal{F}\{s(t)\}|^2$ which we can write using complex conjugation as in $|z|^2=zz^*$
$$PSD[s(t)] = \mathcal{F}\{s(t)\}\cdot\mathcal{F}\{s(t)\}^*$$
So you can see that $ACF[s(t)]=\mathcal{F}^{-1}\{PSD[s(t)]\}$, or $$\mathcal{F}\{ACF[s(t)]\}=PSD[s(t)]$$ which is what you asked for.