As we know, both the DTFS (discrete-time Fourier series) and the DFT (discrete Fourier transform) are used to represent discrete-time periodic signals for all time (or the periodic extension of aperiodic signals in the time interval where the DTFS or DFT is computed) using a finite number of different harmonically related real discrete sinusoids or complex exponentials. The only difference between the DTFS and the DFT is a real factor. The FFT is just a faster way to compute the DFT.

If we compare the formulas for the CTFS (continuous-time Fourier series) and for the DTFS, we see they are analogous, the only main difference being that the DTFS requires a finite number of harmonics (as opposed to an infinite number as in the CTFS). On the other hand, if we compare the formulas for the CTFS and the DFT, besides the differences between the CTFS and the DTFS, we additionally see there is a scaling factor difference. Thus, the DTFS is more analogous to the CTFS than the DFT (if you disagree on this, please explain why). But why do we use more often the DFT than the DTFS? Equivalently, why was the FFT algorithm built for the DFT instead of the DTFS?

  • $\begingroup$ "The only difference between the DTFS and the DFT is a real factor". Since that's true, how should we make sense of the question "why was the FFT algorithm built for the DFT instead of the DTFS?" I mean, what algorithmic difference would you expect? $\endgroup$
    – Matt L.
    Jun 20 at 12:41
  • 1
    $\begingroup$ Said another way: why was both the DTFS and DFT invented? Why wasn't just one invented? $\endgroup$ Jun 20 at 20:51
  • $\begingroup$ What book are you reading that calls the Discrete Fourier Series "DTFS"? $\endgroup$ Jun 21 at 1:47
  • 1
    $\begingroup$ This question has been asked and answered before. For some reason, there is a false controversy when the mathematics are perfectly clear. $\endgroup$ Jun 21 at 1:53
  • $\begingroup$ @robertbristow-johnson First book is Signals and Systems (2nd ed.) by Alan V. Oppenheim, Alan S. Willsky & S. Hamid Nawab, on subsection 3.6.2, pages 212-213. Second book is Signals and Systems: Analysis Using Transform Methods and MATLAB® (3rd ed.) by Michael J. Roberts, on section 7.2, pages 307, 310, 311. Both have seem like good books so far for me. You can check the names in these screenshots. $\endgroup$ Jun 21 at 5:17

They are pretty much the same thing and there is no material difference.

The difference in scale factor just comes from the initial approach. The DFT starts with "what's the frequency content of a discrete periodic time signal?" whereas the DTFS starts with "how do I make a discrete periodic time signal out of complex exponentials?"

If done consistently, scaling is somewhat arbitrary and the best method of scaling depends on your signals and application. Interestingly enough, neither DFT or DTFS get Parceval's theorem correctly. That requires using $1/\sqrt{N}$ for both forward and inverse transforms and also makes the transform matrices unitary.

  • $\begingroup$ Hil, do you know who else, in the lit, is calling the Discrete Fourier Series "DTFS"? $\endgroup$ Jun 21 at 4:09
  • $\begingroup$ But then, why aren't there two different versions of the CTFS, one coming from the question "what's the frequency content of a continuous periodic time signal?" and another from "how do I make a continuous periodic time signal out of complex exponentials?"? Or would both approaches yield the exact same formulas? $\endgroup$ Jun 21 at 6:03
  • $\begingroup$ @robertbristow-johnson: I just went with the few hits I saw on Google. $\endgroup$
    – Hilmar
    Jun 21 at 13:09
  • 1
    $\begingroup$ @AlejandroNava: Apparently since we people working on it were a little more thorough and didn't re-invent the wheel. A lot of this is historical. Personally I find both the current naming and scaling conventions sub-optimal, but they are what they are. I suggest you forget about the DTFS (as the term is rarely used) and stick with the DFT instead. Ideally the four FTs would be named consistently (FT, FS, DFT, DFS) but that ship has sailed. $\endgroup$
    – Hilmar
    Jun 21 at 13:17

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.