0

I am working with the Nyquist criterion for zero ISI when I find something that makes me think that I misunderstood this criterion.

What Nyquist says is that to avoid ISI in the sampling we must ensure that the sample frecuency is less than 2 times the bandwith of the signal, so if the channel filter is something like a rise cosine the ISI in the sampling will be $0$.

At the same time, we have (another) Nyquist criterion which says that to be able to perfect reconstruct a signal, the sample frequency must be at least $2$ times the bandwith of the signal.

So... taken into account that the case of $F_s=2B$ is not implementable, how can we assure no ISI at the same time that we have perfect reconstruction? Aren't this two criteria contradictory? Am I missing something?

Thanks!!!

Alma Cantu
  • 64
  • 4
Rodrigo Serna Pérez
  • 21
  • 1
  • 3
  • The Nyquist ISI criterion is not directly related to Nyquist sampling rate. Also, the "twice the bandwidth" condition is only true for real-valued sampling. If you simultaneously sample I and Q of an equivalent baseband signal, one time the bandwidth is sufficient as sample rate. – Marcus Müller Dec 16 '16 at 21:38
  • What Nyquist says is that to avoid ISI in the sampling we must ensure that the sample frecuency is less than 2 times the bandwith of the signal, <-- where do you take that from? – Marcus Müller Dec 16 '16 at 21:39

1 Answers1

5

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:

enter image description here

The equivalent baseband signal has bandwidth limited in $[-1/2T,1/2T]$ and a sampling rate $F_s > 1/T$ 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. enter image description here

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 $F_s > 1/T$ that satisfies Shannon-Nyquist sampling theorem to avoid aliasing ? We still have ISI.

OK another one can say that the criterion $F_s > 1/T$ 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 ?

Community
  • 1
AlexTP
  • 6,585
  • 2
  • 21
  • 38