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I'm going through a Signal Processing lecture where the professor mentions this fact and the argument given is: Suppose you have a sinusoidal signal: $Acos(\omega t)$

Now if you change the phase of the signal: $Acos(\omega t + p)$ then if it would correspond to a time shift, then $Acos(\omega t + \omega t_0) = Acos(\omega t + p)$

So, $t_0 = p/\omega $

And this can't always have integral values, which is contradictory since we're considering the discrete case.

My argument is that, since we're considering the discrete case, the value of 'p' will be restricted to a certain set of values, right? And these values will only be those for which p/w is an integer.

According to the definition here the phase ($p$ here) gives you how far the signal is in it's cycle. So by default the value of $p$ would only be those values that allow p/w to be integral, since our signal is discrete right? What am I missing here?

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    $\begingroup$ Have a look at this this answer to an almost identical question. $\endgroup$ – Matt L. Nov 1 '14 at 9:35
  • $\begingroup$ @MattL., referring to the part after "EDIT" in your answer, when you say you have a discrete time signal x(t) = cos(ω0t+ϕ) where ϕ can be real, what is the interpretation of the RHS in that equation then? For continuous case, that would have meant that the signal cos(ω0t) has shifted some amount in time. What does this mean for the discrete case? $\endgroup$ – rapturous Nov 1 '14 at 20:12
  • $\begingroup$ @MattL. sorry I can't comment on your answer there since I don't have enough rep points. $\endgroup$ – rapturous Nov 1 '14 at 20:13
  • $\begingroup$ Note that $x(t)$ and $y(t)$ are continuous-time signals. If you sample two time-shifted versions of the same continuous-time signal, in general the resulting discrete-time sequences are not shifted versions of each other. This is what the example in that answer is about, and there's also the condition under which the two discrete-time signals are shifted version of one another. $\endgroup$ – Matt L. Nov 1 '14 at 20:29
  • $\begingroup$ @MattL. ok, I get that. My question is, what does it mean when you add a phase ϕ that's real and not an integer to a discrete signal cos(ω0t) $\endgroup$ – rapturous Nov 1 '14 at 21:01

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