# Discrete Fourier transform - period and other questions

I am trying to get understand few things about DFT, from wikipedia:

$X_k=\sum_{n=0}^{N-1}x_ne^{-i2\pi\frac{k}{N}n}$, we have N numbers $x_0$ to $x_{N-1}$

1. Why there is $\frac{k}{N}*n$ in the exponent, in normal Fourier transform there was $e^{-i2\pi ft}$ where $t$ - time and $f$ - frequency, what is the meaning of $k$, does frequency is $\frac{k}{N}$ - if yes why also where is the time in which $x_n$ measured. I just can't see analogy between the equation for DFT and normal Fourier transform.

2. How do we specify the peroid of our data, we have discrete points, do we assume that if we would have 100 points then $T$ = 100?

I would be very grateful for answers.

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You see that the correspondence between continuous and discrete is $f \; t \leftrightarrow \frac{k}{N} n$ . Now, imagine we take $n$ to be the "discrete time", asuming a unit time-step. We have then a "frequency" of $\frac{k}{N}$;its lowest value (apart from zero) is for $k=1$, which correspond to a maximum period $T_{max}=N$, which makes sense, because that's the length of our data (we can't consider a period larger than that). And the maximum frequency is at $k=N/2$ (not at $k=N-1$: this would rather correspond to a "negative" frequency), and that corresponds to a minimum period $T_{max}=2$. Again, this makes sense: because our assumed time step is 1, and hence the smallest period (maximum frequency) we can consider is $T=2$.
The equivalence $f \leftrightarrow \frac{k}{N}$ assumes a adimensional time, i.e, a unit time step: $\Delta t = 1$. You'd multiply this for the real sampling frequency if you want the "real" frequency:
$f \leftrightarrow \frac{1}{\Delta t} \frac{k}{N}$