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Is it possible to obtain an analytical expression for the PSD?

The PSD is defined as follows,

$S(\omega) = lim_{T \rightarrow \infty} \frac{1}{T} |Y(\omega)|^2 $

Assuming there is $Y(\omega)$ that already assumes an infinite observation time (pure unit pulse, for example), how to account for the $\frac{1}{T}$ term? This would force the power to 0, which makes no sense.

Relating the periodogram to the infinite continuous case by using $N_{samples} << $ and $f_s >>$ still requires the sample time as a correction, but shed no light how to implement this for the analytical case.


footnote

The periodogram estimates the PSD by,

$I_n = \frac{1}{N f_s} |Y[k]| $

However, when approximating the infinite continuous case ($n>>$ and $f_s <<$a), $I_n$ does not equate to the same as $|Y(\omega)|^2$, so I guess this is wrong.

Knowing that $I_n$ approximates the PSD.

The sampling frequency $f_s$ corrects the DFT, since $\frac{1}{f_s} > Y[k] \approx Y(\omega)$. Rewriting $I_n$ by multiplying it with $f_s/f_s$

$I_n = \frac{1}{N f_s} | Y[k]|^2 = \frac{f_s}{N f_s^2} | Y[k]|^2 > \approx \frac{f_s}{N} | Y(\omega)|^2 $

$N/f_s$ = T, thus:

$I_n \approx \frac{f_s}{N} | Y(\omega)|^2 $

The periodogram is a a unbiased estimate if $N\rightarrow \infty$, thus $T \rightarrow \infty$, which equates to the same as the definition of the PSD. However, for a discrete set of observations that is large, and a $f_s$ that is sufficiently small,

Using $I_n \approx \frac{f_s}{N} | Y(\omega)|^2 $, in the continuous case $f_s \rightarrow 0$. If N goes to infinity (infinite observation time) then this would be an infinitessimal small number divided by a very big number. Implicitly stating that there is almost no power throughout. This makes no sense to me.

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