My teacher told me that DFT is DTFT sampled, i.e.:

$$X[k] = X(e^{j \omega})\Bigg|_{\omega = \frac{2\pi k}{N}}$$

But, if I have the sine

$$ x[n] = \sin(\omega_0 n) $$

the DTFT is:

$$X(e^{j \omega}) = \frac{\pi}{j} \big(\delta(\omega - \omega_0) - \delta(\omega + \omega_0)\big)$$

(periodic in $\omega$ with period $2\pi$). How is it possible to reconcile this with the DFT for $N$ values for this sine or any signal?

Thanks people. Sorry for the English.

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    $\begingroup$ this is a good question. and Kaorthraya, your English is fine. $\endgroup$ – robert bristow-johnson Nov 16 '18 at 4:11
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    $\begingroup$ them Dirac delta "functions". them's a bitch! $\endgroup$ – robert bristow-johnson Nov 16 '18 at 4:13
  • $\begingroup$ Robert, I'd like to thank you for the help editing the question, now it is much better. And yes, Dirac delta are some sort of crazy things! $\endgroup$ – Kaorthraya Nov 16 '18 at 6:05
  • $\begingroup$ Could you please clarify which reconcilement you are looking for? $\endgroup$ – AlexTP Nov 16 '18 at 9:14
  • $\begingroup$ Hi AlexTP. I'll try to be more clear. If I have the DTFT of a signal $x[n]$, is it possible that I find the coefficients $X[k]$ for $k = [0, N-1]$ of DFT of this function, only by his DTFT? In other form, can I obtain the values of DFT starting from a DTFT of a signal? $\endgroup$ – Kaorthraya Nov 16 '18 at 13:39

The DFT is a sampled version of the DTFT only for finite length signals. Otherwise, there is no point in comparing the DTFT with the DFT because you can only compute the DFT for finite length (or periodic) signals. Your example $x[n]=\sin(\omega_0n)$ is an infinitely long sequence. For specific choices of $\omega_0$ it is periodic, but nevertheless, you can't obtain its DFT by sampling its DTFT. The operation of sampling a Dirac impulse is undefined because $\delta(\omega)$ cannot be evaluated at $\omega=0$ since the Dirac impulse is not an ordinary function but a distribution.

  • $\begingroup$ Thanks, Matt, for your answer, So, I only can make the process of sampling the DTFT for signal who have a finite lenght, ok? i.e: $x [n] = \alpha^n[u[n] - u[n-n_0], n_0 > 0 $ have, $X(e^{j\omega}) = \frac{1 - (\alpha e^{-j\omega})^{n_{0}}}{1 - \alpha^{-j\omega}}, |\alpha e^{-j \omega}|<1$ So, in this case I can substitute the $\omega = \frac{2\pi k}{N}$? $\endgroup$ – Kaorthraya Nov 16 '18 at 14:04
  • $\begingroup$ @Kaorthraya: yes, for finite length signals that should work. $\endgroup$ – Matt L. Nov 16 '18 at 14:06
  • $\begingroup$ @MattL., isn't $x[n]=\sin(\omega_0 n)$ periodic? at least if $\frac{2\pi}{\omega_0} = \frac{N}{k} \qquad N,k \in \mathbb{Z}$, isn't it even periodic for discrete-time $x[n]$? how is it that $$X[k] = X(e^{j \omega})\Bigg|_{\omega = \frac{2\pi k}{N}}$$ even with the Diracs in there? That question remains unanswered. $\endgroup$ – robert bristow-johnson Nov 16 '18 at 20:19
  • $\begingroup$ @robertbristow-johnson: $\sin(\omega_0n)$ is periodic if $\omega_0$ is a rational multiple of $2\pi$, as you suggested. But that has no bearing on the relation between the DTFT and the DFT. The signal $x[n]=\sin(\omega_0n)$ is infinitely long, and that relation between the DTFT and the DFT only holds for signals of finite length. $\endgroup$ – Matt L. Nov 16 '18 at 20:46
  • $\begingroup$ @MattL. I had tested this concepts in MatLab, i tried to sample in $\omega$, the location of samples makes sense considering my tests, but the amplitude of samples is totally wrong, as expected. In truly, it's can be predicted, because the delta of Dirac in $\omega$ domain is multiplied by a $\pi$, but when I do the DFT the amplitude is $\frac{N}{2}$ $\endgroup$ – Kaorthraya Nov 18 '18 at 1:12

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