# Indexing in DFT (from an old paper)

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.

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 $$\Delta T$$).
• @M.Farooq: Yes, the last vertical line in both plots is clearly not at index $N$ but at $N-1$. – Matt L. Nov 21 '20 at 15:26
• and the "first" vertical line is clearly at index $0$. some of us like to call that the "zeroth" vertical line. – robert bristow-johnson Nov 22 '20 at 19:56