# Non-uniqueness of continuous time sinusoids?

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.

• 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. – Jazzmaniac Jun 22 '14 at 9:30
• Why not to make it an answer? – jojek Jun 22 '14 at 12:53
• Too lazy to fill in the details to make it a good answer. I'll leave that honour to you if you like, jojek. – Jazzmaniac Jun 22 '14 at 13:13
• exp(2πm) doesn't equal 1 :) – geometrikal Jun 22 '14 at 14:05
• @geometrikal, yep, I forget the $i$! So it should have read $\exp(2\pi i m)=1$ for integer $m$. – Jazzmaniac Jun 22 '14 at 19:41

## 1 Answer

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.