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Hi All: This is very basic but I've always wondered about it and now I see it in print in a textbook so I may as well ask. In Signals and Systems on page 26, it says

$$e^{j(\omega_0 + 2\pi)n} = e^{j2\pi n} e^{j\omega_0 n} = e^{j\omega_0 n} \tag{1.51} $$

Then it says:

From eq. (1.51), we see that that the exponential at frequency $\omega_{0} + 2\pi$ is the same as that at frequency $\omega_{0}$. Thus, we have a very different situation from the continuous time case in which the signals $e^{j\omega_{0} t}$ are all distinct for distinct values of $\omega_{0}$. In discrete time, these signals are not distinct, as the signal with frequency $\omega_{0}$ is identical to the frequencies with $\omega_{0} \pm 2\pi$, $\omega_{0} \pm 4\pi$ and so on.

The reason why the concept above has always confused me is because if we replace $n$ with $t$, then it is true for integer values of $t$ also. Why does this not matter ? I know that concept also has to do with why aliasing cannot exist in the continuous case so, if I can understand this, a lot of things would get clear for me. Thanks for your help and the great answers in general from many.

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  • $\begingroup$ Does this help? $\endgroup$
    – Matt L.
    May 1 at 7:17
  • $\begingroup$ Thanks Matt. I'm just seeing above now. I will print out and read carefully a few times because it will take me some time to digest. But I dio get what you're saying about the integer values of $t$ not causing periodicity. That was a real eye opener to a long time standing confusion. $\endgroup$
    – mark leeds
    May 2 at 14:40
  • $\begingroup$ Okay, great that it helped! $\endgroup$
    – Matt L.
    May 2 at 14:42
  • $\begingroup$ Matt: I read does this help which then has a link to another one which is was even more useful. Really great learning from these. It's finally hitting me that the whole periodicity phenomena in the discrete case is really due to aliasing rather than integers versus reals. The periodicity arises because of the limit in the frequencies that can be backed out given the sampling rate. Note to anyone who doesn't understand discrete periodicity: Read "does this help" and then the various links there.. Thanks. $\endgroup$
    – mark leeds
    May 2 at 19:06
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The simple answer is that

$$e^{j\omega_0t}\neq e^{j(\omega_0+2\pi)t}\tag{1},\qquad t\notin\mathbb{Z}$$

Since $t$ is a real variable, the inequality is true for uncountably many values of $t$. Equality is only achieved for countably many integer values of $t$. From $(1)$ it follows that for real $t$, the function $e^{j\omega_0t}$ is not $2\pi$-periodic in $\omega_0$, unlike in the discrete-time case.

Two functions of a real variable can be equal on a countably infinite set of values of the independent variable, even though they're not the same function.

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  • $\begingroup$ Matt: I see it now. The numbers in between the integers will still maintain the lack of periodicity. you explained that nicely. It's not intutitve to my brain but I get it now. thanks. $\endgroup$
    – mark leeds
    May 1 at 9:18
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But $t$ can also have fractional values. If $t=n=$ integer then $\omega_0$ and $\omega_0+2\pi$ are aliases for the same frequency.

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  • $\begingroup$ Hi : I understand that $t$ can have fractional values but, when $t$ is integer, doesn't that present the same problem as the discrete case because $t=1,2,3, \ldots$ will all result in the same frequencies. $\endgroup$
    – mark leeds
    May 1 at 4:29
  • $\begingroup$ yes, it's the same as the discrete case. we can choose our unit time to be the sampling period. then a frequency of $\pm2\pi$ or $\pm4\pi$ are the same as DC. $\endgroup$ May 1 at 4:31
  • $\begingroup$ Robert: I see what you're saying but I think matt hit my confusion on the button. I was thinking there would be periodicity even in the continuous time case but that's not correct and now I see why. thanks for trying to un-confuse my feeble mind. $\endgroup$
    – mark leeds
    May 1 at 9:15
  • $\begingroup$ Yes, uniform sampling in one domain causes periodic copying and overlap-adding in the reciprocal dimension. $\endgroup$ May 1 at 12:58

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