# Is autospectrum the same as power spectrum density?

In my slides about signal processing, there is one that mentions the same thing as the beginning of this and this answer, namely that the Fourier transform of a signal, squared is the power spectrum density of the signal.

In this talk, it is mentioned that coherence is calculated by dividing the squared cross-spectrum by the product of the two auto-spectra.

However, the formula in my slides divides the squared cross-spectrum by the product of the formula of our previously seen PSD and another PSD.

So, is autospectrum the same as PSD? I can find a lot of information about PSD, but not on autospectrum.

• I don't know yet how to put in formulas nicely. I think it'll become clearer what I mean. I'll try to edit those in tomorrow.
– Mien
Jun 24, 2012 at 22:04
• Neither answer that you link to says "the Fourier transform of a signal, squared is the power spectrum density of the signal." as you assert. The power spectral density is the Fourier transform of $R_x(\tau)$, the autocorrelation function of a signal $x(t)$ which is not the same thing as Fourier transform of the signal squared, i.e. $\mathcal F[x^2(t)]$$, **nor** the Fourier transform squared, i.e.$[X(f)]^2$. The power spectral density is equal to$|X(f)|^2$, but not equal to$[X(f)]^2\$ in general. Jun 25, 2012 at 0:40