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There is a nice paper on explaining DFT from the 1960s in IEEE A guided tour of the fast Fourier transform. The author uses the following definitions of DFT

DFT $$ X(j)=\sum_{k=0}^{N-1} x(k) \exp \left(-i 2 \pi\left(\frac{j}{N}\right) k\right) $$

Inverse $$ x(k)=\frac{1}{N} \sum_{j=0}^{N-1} X(j) \exp \left(i 2 \pi\left(\frac{j}{N}\right) k\right) $$

where the indexes j = 0, 1, 2, ..., N-1 and similarly k=0, 1, 2,..., N-1.

Now the authors show a figure, where the j and k indices run from 0 to N not N-1. Let us say we had 10 data points, so N=10; and j and k should run from 0 to 9 not 10. Is this a typographical error in the figure?

It seems that his N also starts from zero, then the figure is consistent but the summation formula has N-1.

enter image description here

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The figures are correct. You can see that there are samples at indices $0,1,\ldots,N-1$. There are no samples shown at index $N$, neither in the time domain nor in the frequency domain. The value $N$ is only shown on the abscissa because in the frequency domain it corresponds to the sampling frequency, and in the time domain it corresponds to the end of the continuous-time signal represented by $N$ samples (if we assume that each sample represents a portion of length $\Delta T$).

For a very detailed discussion on the definition of the actual duration of a discrete-time sequence have a look at the answers to this question.

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  • $\begingroup$ Thanks for the link. With specific ref. to the figure, you mean the last vertical line in the figure refers to k= N-1 in the time domain and j=N-1 in the frequency domain? My field is chemistry and I wanted to use DFT for deconvolution, hence looking at some fundamental early papers for self-teaching. $\endgroup$
    – AChem
    Nov 21, 2020 at 15:24
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    $\begingroup$ @M.Farooq: Yes, the last vertical line in both plots is clearly not at index $N$ but at $N-1$. $\endgroup$
    – Matt L.
    Nov 21, 2020 at 15:26
  • $\begingroup$ and the "first" vertical line is clearly at index $0$. some of us like to call that the "zeroth" vertical line. $\endgroup$ Nov 22, 2020 at 19:56

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