# Why must the length of a sequence under the discrete Fourier transform be equal to the input sequence length?

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

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:

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

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?