# Analytical expression for PSD

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.