Nyquist criterion for zero isi
My answer is related to this question
Why root raised cosine filter can eliminate intersymbol interference (ISI) ?

A classical system with Nyquist pulse $p(t)$ is:



The equivalent baseband signal has bandwidth limited in $[-1/2T,1/2T]$ and a sampling rate $F_s &gt; 1/T$ is enough to avoid aliasing.

We normally sample at rate $t=kT$. I wonder what would happen if suddenly I sample faster than the normal sampling rate, for example at $t = kT/2$ instead of $t = kT$ ?


In the frequency domain, the composite filter $g(t) = p(t) \star p^*(-t)$ has Fourier transform $G(f)$ satisfies $\sum_k G(f-k/T) = T, \forall f$. Double the sampling rate $\sum_k G(f-2k/T)$ is not constant any more. This is true at least for typical pulses such as sinc or raised cosine.
In the time domain, thank Dan Boschen for the visualization below, the sample at impair $k$ of $kT/2$ is contributions of two adjacent symbols.



Thus, is it true that for free-ISI-Nyquist-pulse-shaped signal, the largest sampling rate is the value of nominal (Nyquist) bandwidth $1/T$ ?

One can say that we still have correct data by taking only one per two samples (downsampling). What if I sample at rate $F_s &gt; 1/T$ that satisfies Shannon-Nyquist sampling theorem to avoid aliasing ? We still have ISI.

OK another one can say that the criterion $F_s &gt; 1/T$ is in the sense that the entire information after being sampled is intact, just take a baseband filter to get back the continous time spectrum and then sample it at the correct sampling rate at $t=kT$. Have I answered my questions by myself ?