# 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

## 2 Answers

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.

Ah okay thank you. I think what irritated me was that I didn’t consider that the sampled function is a periodic or periodically continued function so that the Fourier transform delivers amplitude values for the base frequency and multiples of it only (so the Fourier transform is discrete), which are indexed as n*(base frequency), so that it is obvious why F „depends“ on n. The amplitude value for n can then be converted into an amplitude value of a frequency given by n*(base frequency) (and the base frequency can be obtained by the period since a signal is only sampled between its starting point and endpoint, whose distance is the period).