Let's consider a continuous signal, $f(t)$, which has been sampled $N$ times, with spacing $T$ between samples. We denote the $N$ samples $f[0], f[1], ..., f[N-1]$.

The Fourier transform of the continuous signal by definition is:

$$F(\omega) = \int_{-\infty}^{\infty}f(t)e^{-i\omega t}\cdot dt.$$

If we regard each sample $f[n]$ as an impulse having area $f[n]$, then in looking at the sampled function - which I'll denote $\tilde{f}(t)$ - we see that:

$$\tilde{F}(\omega) = \int_{-\infty}^{\infty}\tilde{f}(t)e^{-i\omega t}\cdot dt$$ $$ = f[0]+f[1]e^{-i\omega T}+...+f[n]e^{-i\omega nT}+...+f[N-1]e^{-i\omega (N-1)T}.$$

Thus we see that the Fourier transform of the sampled function is such that:

$$\tilde{F}(\omega) = \sum_{n=0}^{N-1} f[n]e^{-i\omega nT}.$$

So far, we have placed no constraints on the values that $\omega$ can take on. Indeed, it appears as if $\tilde{F}(\omega)$ is a continuous function of $\omega$.

Despite the lattermost observation, it is the case that the discrete Fourier transform is itself discrete. My question is simply this: why?


Your Answer

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

Browse other questions tagged or ask your own question.