2
$\begingroup$

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

enter image description here

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?

$\endgroup$
2
$\begingroup$

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 $\sigma^2$.

| improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ but under the assumption that all samples are drawn iid, this calculation holds? $\endgroup$ – Phinie Jul 13 at 12:24
  • $\begingroup$ this is an example of an old exam I tried to solve as preparation and this is all the information given in order to compute the PSD out of the PDF... $\endgroup$ – Phinie Jul 13 at 12:28
  • 2
    $\begingroup$ Discrete-time white noise is a sequence of zero-mean i.i.d random variables. Nitpickers saying that white noise is a sequence of zero-mean uncorrelated random variables (neither independence nor identical distribution is necessary will be ignored. It is quite correct that there is little connection between the PDF and the PSD except through the mean and variance, but the "AC part" of a white-noise process is the entire process and gives a flat PSD, and it can't be pink or brown or whatever color. The process here is white noise plus DC and the spectrum is what the OP found it to be. $\endgroup$ – Dilip Sarwate Jul 14 at 2:09
  • $\begingroup$ How do you know that it's white noise ? Am I missing something in the question ? $\endgroup$ – Hilmar Jul 14 at 16:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.