0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – M. Farooq Nov 21 at 15:24
  • 1
    $\begingroup$ @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 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. $\endgroup$ – robert bristow-johnson Nov 22 at 19:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.