1
$\begingroup$

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?

enter image description here

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?

$\endgroup$
1
$\begingroup$

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.

|improve this answer|||||
$\endgroup$

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.