# Why is DTFT of $e^{jn\omega_0}$ an impulse train?

update : After asking the question, I figured out that DTFT result is an impulse train. Now my question evolved to, how it is derived in this way? Using the DTFT formula seems not to be working, because of "sum does not converge".

$$e^{j\Omega_{0}n}=2\pi \sum_{k=-\infty}^{\infty} \delta\left(\Omega-\Omega_0-2\pi k\right)$$

However, that DTFT is defines as above in most places.

Why cannot I find the transform using the forward DTFT formula?

• need to learn how to do $\LaTeX$ here. this is about the sampling theorem. i have an answer somewhere here. – robert bristow-johnson Jan 10 '20 at 2:44
• – robert bristow-johnson Jan 10 '20 at 2:46

First of all, note that the discrete-time Fourier transform (DTFT) of a sequence is always periodic with period $$2\pi$$:

$$X(\omega)=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\tag{1}$$

This is obvious because $$e^{-jn\omega}$$ is $$2\pi$$-periodic in $$\omega$$.

Next, consider the inverse DTFT:

$$x[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(\omega)e^{jn\omega}d\omega\tag{2}$$

For $$X(\omega)=\delta(\omega-\omega_0)$$, we get from $$(2)$$

$$x[n]=\frac{1}{2\pi}\int_{-\pi}^{\pi}\delta(\omega-\omega_0)e^{jn\omega}d\omega=e^{jn\omega_0}\tag{3}$$

[Note that without loss of generality we assume that $$-\pi\le\omega_0\le\pi$$ because $$X(\omega)$$ is $$2\pi$$-periodic. $$X(\omega)=\delta(\omega-\omega_0)$$ is a shorthand for $$X(\omega)=\sum_k\delta(\omega-\omega_0-2k\pi)$$. That shorthand can be used here because we know that $$X(\omega)$$ is the DTFT of some sequence, and hence $$2\pi$$-periodicity is understood].

Since $$\textrm{IDFT}\{\delta(\omega-\omega_0)\}=e^{jn\omega_0}$$ we're inclined to accept

$$\textrm{DTFT}\{e^{jn\omega_0}\}=\sum_{n=-\infty}^{\infty}e^{jn\omega_0}e^{-jn\omega}=\delta(\omega-\omega_0)\tag{4}$$

(where $$2\pi$$-periodic continuation is understood), even though the sum

$$\sum_{n=-\infty}^{\infty}e^{jn\omega_0}e^{-jn\omega}\tag{5}$$

doesn't converge in the conventional sense. However, this can be made mathematically sound by interpreting the sum $$(5)$$ as a distribution.