12
$\begingroup$

What 'exactly' is power spectral density for discrete signal?

I was always under the assumption that taking the Fourier transform of the signal, and then the ratio of desired frequency range magnitude over the entire frequency range gives the power ratio for that frequency range which is the same as power spectral density. Is that wrong?

Reading a student paper got me confused as it says to compute PSD and then 'absolute and relative spectral powers in desired bands' as well. Are they different? If yes, how does one compute it?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
12
$\begingroup$

I have no idea what your calculation of power spectral density gives since I cannot understand it.

If a signal $x(t)$ has Fourier transform $X(f)$, its power spectral density is $|X(f)|^2 = S_X(f)$. The absolute spectral power in the band of frequencies from $f_0$ Hz to $f_1$ Hz is the total power in that band of frequencies, that is, the total power delivered at the output of an ideal (unit gain) bandpass filter that passes all frequencies from $f_0$ Hz to $f_1$ Hz and stops everything else. Thus, $$\text{Absolute Spectral Power in Band} = \int_{-f_1}^{-f_0} S_X(f)\,\mathrm df + \int_{f_0}^{f_1} S_X(f)\,\mathrm df.$$ The relative spectral power measures the ratio of the total power in the band (i.e., absolute spectral power) to the total power in the signal. Thus, $$\text{Relative Spectral Power in Band} = \frac{\displaystyle\int_{-f_1}^{-f_0} S_X(f)\,\mathrm df + \int_{f_0}^{f_1} S_X(f)\,\mathrm df}{\displaystyle\int_{-\infty}^{\infty} S_X(f)\,\mathrm df}.$$

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ As an additional note, you can also define the power spectral density of a wide-sense stationary random process as the Fourier transform of the process's autocorrelation function. This is known as the Wiener-Khinichin theorem. $\endgroup$
    – Jason R
    May 13, 2012 at 14:06
  • 1
    $\begingroup$ @JasonR I didn't get into that aspect of the matter, but of course $S_x(f) = |X(f)|^2 = X(f)X^*(f)$ is the Fourier transform of the autocorrelation function $R_x(\tau)$ of the deterministic signal $x(t)$ where $$R_x(\tau) = \int_{-\infty}^{\infty} x(t)x(t+\tau)\,\mathrm dt = \int_{-\infty}^{\infty} x(t)x(t-\tau)\,\mathrm dt$$ $\endgroup$
    – Dilip Sarwate
    May 13, 2012 at 14:12
  • $\begingroup$ One nitpick is that I think you need conjugates on the $x(t \pm \tau)$ terms, but can you make that case for a general deterministic signal $x(t)$? There's no guarantee that $R_x(\tau)$ is not a function of $t$, which for the stochastic case is provided by the WSS proviso. $\endgroup$
    – Jason R
    May 13, 2012 at 20:16
  • 3
    $\begingroup$ @JasonR Conjugates are needed if $x(t)$ is allowed to be complex, not otherwise, so OK, put in the conjugates and let us agree that that particular nit has been picked. But, $R_x(\tau)$ as defined above cannot be a function of $t$. Note that $t$ is a variable of integration that disappears when the limits are used. For stochastic signals, $R_X(t_1,t_2)$ is defined as $E[X(t_1)X(t_2)]$, not as it is for deterministic signals. $R_X(t_1,t_2)$ is a function of the difference $t_1-t_2$ for WSS processes instead of being a function of the individual values of $t_1$ and $t_2$. $\endgroup$
    – Dilip Sarwate
    May 13, 2012 at 21:18
  • $\begingroup$ Makes sense. I'm on the same page now. $\endgroup$
    – Jason R
    May 13, 2012 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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