Power Spectral Density from Probability Density Function

Question

The samples of a signal $x[n]$ are i.i.d. and follow a triangular pdf with $a = 0,\ b = 2,\ c = 1$:

The DC-power of the signal is

$$\mu_x^2 = \big(\mathbb{E}(X)\big)^2 = \left(\int_{-\infty}^{\infty} x f_x(x)dx\right)^2 = 1$$

the total power of the signal

$$\mathbb{E}(X^2) = \int_{-\infty}^{\infty} x^2 f_x(x)dx = \frac{7}{6}$$

and the variance (AC-power) as

$$\sigma^2 = \mathbb{E}\big((X-\mu_x)^2\big)= \mathbb{E}\left(X^2\right)-\mu_x^2 = \frac{1}{6}$$.

Thus I thought, that the PSD of the signal has the following form:

$$ S_{xx}(e^{j\omega}) = 2\pi \delta(\omega) + \frac{1}{6}$$

as the AC-power results in a constant value over the whole spectrum and the DC power in a Dirac-Delta at $\omega = 0$. Furthermore, the integration over the PSD should return the total power of the signal: $$ \frac{1}{2\pi} \int_{-\pi}^{\pi} S_{xx} dx = \mathbb{E}(X^2) $$

Is my solution correct or did I miss something? Can this procedure be done for every probability density function under the assumption that all samples of $x[n]$ are iid?

Answer 1

If the $x[n]$ are i.i.d. random variables with mean $\mu$ and variance $\sigma^2$, then the power spectral density of this discrete-time random process (which is effectively white noise plus a possibly nonzero mean $\mu$) is what you have calculated. The shape of the common pdf of the random variables is irrelevant in all this except insofar as the shape determines the mean and variance; for example. you would get the same PSD if the $x[n]$ were Gaussian random variables with mean $\mu$ and variance $\sigma^2$.

Answer 2

as the AC-power results in a constant value over the whole spectrum

Why would it ? In general you can't make a connection between the PDF and the PSD. They tend to be independent. I.e. your AC noise signal could be white (your assumption) but it could also be pink or brown or many other spectral shapes.