In Alan Oppenheim's book Signals and Systems a comparison is made between the properties of discrete-time and continuous-time complex exponential signals in section 1.3 pg. 26. Specifically it says:

The continuous-time complex exponential $e^{j\omega_0t}$ has two properties: 1. the larger the magnitude of $\omega_0$, the higher is the rate of oscillation in the signal 2. $e^{j\omega_0t}$ is periodic for any value of $\omega_0$

Consider the discrete-time signal $e^{j(\omega_o+2\pi)n}= e^{j2\pi n} e^{j\omega_0 n}=e^{j\omega_0 n}$. From this we see that the exponential at the 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 signals $e^{j\omega_ot}$ are all distinct for distinct values of $\omega_0$

The part I don't understand is the one in bold . It seems to me I can do exactly the same thing for continuous-time complex exponential by replacing $n$ with $t$:

$e^{j(\omega_o+2\pi)t}= e^{j2\pi t}e^{j\omega_0 t}=e^{j\omega_0 t}$

So there we have it. I have produced the same signal in the continuous-time complex exponential. So what are they talking about? I see no difference.

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    $\begingroup$ But $\textrm{exp}(j2\pi t) \neq 1$, except when $t$ is an integer (and then you have the discrete-time formula). $\endgroup$ – MBaz Sep 3 '17 at 16:38
  • $\begingroup$ @MBaz - Thanks. If you answer the question below I can mark it correct. $\endgroup$ – user1068636 Sep 3 '17 at 17:28
  • $\begingroup$ Done :) Glad to be of help. $\endgroup$ – MBaz Sep 3 '17 at 17:32

The problem is that $$e^{j 2 \pi t} \neq 1,$$ except in the case when $t$ is an integer. However, in that case, you have the discrete-time formula: $n$ is assumed to be an integer. In most DSP literature, wherever you see $n$ used as in this case, you can safely assume it's an integer.

  • $\begingroup$ i might add that what is meant by the bold statement is to point out the existence of the possibility of aliasing in discrete signals. if one increases the frequency to $\omega_0 + 2\pi$, it appears exactly as if the signal was at frequency $\omega_0$. $\endgroup$ – robert bristow-johnson Sep 3 '17 at 18:05

I did the same thing you did at first and I could not understand it. But then I realized that the continuous time case was adding $2 \pi$ to $t$, not $w_0$.


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