Power Spectral Density From Variance

Question

For an exercise, we made a noise signal by generating a normally-distributed random vector in MATLAB. The mean is defined to be equal to zero and the variance is set to 0.5.

I know that the power spectral density is just the fourier transform of the variance in frequency domain. In this case, does the PSD just equals to 0.5 since the variance is constant with time?

Answer 1

The power spectral density is NOT the Fourier transform of the variance. For wide-sense-stationary (WSS) signals, the power spectral density is the Fourier transform of the Autocorrelation function (also called the auto covariance function perhaps leading to the OP’s confusion), and the variance is the autocorrelation at time offset $\tau = 0$ (when the WSS process is zero-mean).

Consider for example a white noise process (which is done if the OP's randomly generated samples were completely uncorrelated from sample to sample), to a tightly filtered process, both can have the same variance (which is the power if WSS and zero-mean), but VERY different spectrums. A white noise spectrum is constant over all frequencies while the filtered process would be limited to a narrow bandwidth. The relationship between the two is given by Parseval's theorem; the sum of the squares of each frequency component will be the same in both cases.

See this answer from Matt L further detailing variance with relation to the signal power: variance in the time domain versus variance in frequency domain

Update: Dilip stated this all much more elegantly and concisely in the comments, paraphrased here: The area under the power spectral density function of a WSS process equals the value of the autocorrelation function at offset $\tau=0$. For a zero-mean WSS process, this value is the variance. Therefore deducing the power spectral density from knowledge of the variance is equivalent to asking for a curve given that we only know the area under the curve, which we know is not possible.