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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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    Have a look at this this answer to an almost identical question. – Matt L. Nov 01 '14 at 09:35
  • @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? – rapturous Nov 01 '14 at 20:12
  • @MattL. sorry I can't comment on your answer there since I don't have enough rep points. – rapturous Nov 01 '14 at 20:13
  • 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. – Matt L. Nov 01 '14 at 20:29
  • @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) – rapturous Nov 01 '14 at 21:01
  • You simply shift your time reference. You don't always need to sample a sinusoid at $t=0$, $t=T$, $t=2T$, etc., but you can also sample at $t=t_0$, $t=t_0+T$, $t=t_0+2T$, etc. Note that your sample instants of course remain integer multiples of the sampling period $T$, you just shift your sampling grid, and this can be done by arbitrary continuous shifts. Imagine fixed sampling instants but a shifted continuous time signal that's being sampled. – Matt L. Nov 01 '14 at 21:22

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