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 $x_{N-1}$
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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.
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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.
Best Answer
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:
$f \leftrightarrow \frac{1}{\Delta t} \frac{k}{N}$