I am reading through the PAM transmission scheme and about the power spectral density of the signals. Given that the Average Power Spectral Density of PAM Signals is: $$ \Phi_{ss}(f)=\Phi_{aa}\left(e^{j2\pi ft}\right)\frac{\lvert G(f)\rvert^2}{T} $$
(We consider that the amplitude coefficients of PAM $a[k]$ is a cyclo stationary discrete stochastic process and hence compute average of a PSD).
There are three remarks made after this:
- It becomes obvious that the minimum bandwidth of the pulse $g(t)\overset{\mathcal F}\rightarrow G(f)$ has to be at least $1/T$.
- The PSD $\Phi_{aa}\left(e^{j2\pi fT}\right)$ of the discrete time sequence $a[k]$ is periodic with period $1/T$.
- If $G(f)$ has smaller bandwidth than $1/T$ a full period of the data spectrum is not contained in the PSD of the transmit signal.
I am not sure if I have fully understood the remarks. Here is my understanding on the above:
- Since we have taken an Average PSD, the PSD repeats over the time interval $T$ and the bandwidth of the pulse is $1/T$. (I am not sure how that became the minimum bandwidth).
The PSD of the discrete sequence $a[k]$ is periodic since it's a cyclo stationary process and vary over a time period $T$.
If $G(f)$ would have had less bandwidth than $1/T$, that is the pulse has a higher frequency than $1/T$, the data spectrum would have smeared into each other leading to ISI.
Please let me know if the above understanding is correct/valid. Please add/correct if anything is missing.