Skip to main content
Signal Processing

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

Required fields*

10
  • $\begingroup$ I understand all of this. For a discrete signal, the phase is an integer. How can we be sure that $\phi = \Omega_o n_o$ is an integer ? $\endgroup$ Commented Apr 23, 2014 at 8:52
  • $\begingroup$ @curryage That's exactly the point. You can't be sure, that's why an arbitrary phase shift does not in general correspond to a time shift of the signal, unlike with time-continuous signals. $\endgroup$ Commented Apr 23, 2014 at 8:55
  • $\begingroup$ I am confused. My previous comment was referring to the first part of your answer where you supposedly show how a time shift corresponds to a phase shift. But the computed phase is not necessarily an integer,as it must be for discrete signals. I understand the second part of your answer which shows an arbitrary phase shift does not, in general, correspond to a time shift. $\endgroup$ Commented Apr 23, 2014 at 9:14
  • 1
    $\begingroup$ @curryage I think the problem is that you think $\phi$ is an integer. In general it isn't. Note that $\Omega_0$ and $\phi$ are just real number, whereas $n$ and $n_0$ are integers. $\endgroup$ Commented Apr 23, 2014 at 9:17
  • 2
    $\begingroup$ @David Your remark on the linear phase is not true. Consider a symmetric (linear phase) FIR filter with an even number of coefficients. Despite the exact linear phase the filter's delay is non-integer (it is $(N-1)/2$). $\endgroup$ Commented Apr 23, 2014 at 13:32