I have read the DSP literature and I get why: $$e^{j(\omega+2 \pi m)n}=e^{j\omega n}$$

But can't we prove the same in continuous domain also in the following way: $$e^{j\omega t}$$ $$e^{j(\omega+2 \pi m)t}$$ $$\left( e^{j\omega} \cdot e^{2j\pi m}\right) e^t$$ $$e^{j\omega t}$$

If my above reasoning is correct, then does not it mean that analog frequencies separated by $2\pi$ are same? But I know this is not the case.

Please help me in this case.

  • 2
    $\begingroup$ The proof in the discrete domain relies on the fact that for integer $m$ we always have $\exp(2\pi m ) = 1$. In the continuous domain you don't have the restriction to integers and the relationship does not hold. $\endgroup$ – Jazzmaniac Jun 22 '14 at 9:30
  • $\begingroup$ Why not to make it an answer? $\endgroup$ – jojek Jun 22 '14 at 12:53
  • $\begingroup$ Too lazy to fill in the details to make it a good answer. I'll leave that honour to you if you like, jojek. $\endgroup$ – Jazzmaniac Jun 22 '14 at 13:13
  • $\begingroup$ exp(2πm) doesn't equal 1 :) $\endgroup$ – geometrikal Jun 22 '14 at 14:05
  • $\begingroup$ @geometrikal, yep, I forget the $i$! So it should have read $\exp(2\pi i m)=1$ for integer $m$. $\endgroup$ – Jazzmaniac Jun 22 '14 at 19:41

In discrete-time we have

$$e^{j\omega_0n}=e^{j(\omega_0+2\pi k)n},\quad \forall n$$

because $2\pi kn$ is an integer multiple of $2\pi$ if $k$ and $n$ are integers, and, consequently, $e^{j2\pi kn}=1$. In continuous time you generally have

$$e^{j\omega_0t}\neq e^{j(\omega_0+2\pi k)t}$$

because $2\pi k t$ is (for arbitrary $t$) not an integer.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.