# DFT: a function of $n$?

I‘m a high school student and I haven’t studied physics or anything.

Why does the DFT depend on an integer, say $$k$$ or $$n$$ (it’s usually expressed like $$F(n)=...$$ or $$F(k)$$ or $$F_k$$, etc.) if it is supposed to deliver a frequency information of a sampled signal?

Can the frequency content of the signal be expressed as a multiple of the integer?

• Your "answer" to my answer looks more like a comment, I'd suggest to edit it – Laurent Duval Jun 1 '19 at 13:55

Let us assume that you have a finite length discrete signal $$x$$, denoted by its samples $$x_n$$, $$0\le n; $$x$$ does not depend on $$n$$, but its is values are indexed by $$n$$. Once you index a signal with integers, it somehow "looses" its dependence to an "actual time" in seconds. In other words, one does not know how much time actually elapsed between $$x_{13}$$ and $$x_{14}$$. And, in a relative way, one does not care, when it comes to understanding which (relative) frequencies compose $$x$$.
When we compute the DFT of $$x$$, we turn its $$N$$ values onto $$K$$ other values $$F_k$$ (most often $$K=N$$), indexed by $$0\le k. The $$F_k$$'s are Fourier amplitudes, relatively indexed by integers, but the Fourier transform, globally, does not depend on an integer.