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I intuitively understand what the DFT is doing and what the equation means. We are essentially adding up all the points of the input after projection onto the unit circle and then taking an average (dividing by 1/N). I note that this division of 1/N is often done in the IDFT (inverse).

What I don't understand is why when we consider the continuous fourier transform (integrated from -infinity to infinity), I notice there is no average. So in essence, this is just a sum.

Why do we need this 1/N factor in the DFT equation, but it omitted from both the forward and inverse continuous equations? Without it, isn't that just a sum? How mathematically is this equivalent to an average if we aren't dividing?

So it seems there is something weird about why scaling is necessary in DFT but not in the continuous case.

For example, equations 1 and 2 on the Wikipedia page about fourier transform (under definitions), shown the continuous transform, but there is no division by number of points. But isn't that just a sum?

Further, while I understand the 1/N in the DFT case, it isn't clear why we do it only on the inverse transform and why things are still correct in the forward transform without it.

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There is no convention to scaling a discrete or continuous Fourier transform. Different scaling factors are used for different purposes.

For example, when interested in the amplitude of a particular frequency component, typically you would scale by the length of the DFT ($N$ in your example). As a matter of fact, the correct scaling for this purpose is to scale by the sum of the samples of the window used to get the segment. In the case of a rectangular window, that’s $N$.

In the case of an impulse response and its DFT, the frequency response, you’d likely be interested in the gain at a particular frequency, in which case you wouldn’t scale the DFT.

The reason scaling is applied in the definitions of both forward and inverse transforms, whether continuous or discrete, is to conserve symmetry between them, so that applying both simultaneously to a function gives you the same function back. In that sense, the scaling factor can be $1/2\pi$ ($1/N$ in the discrete case) for the forward and $1$ for the inverse, or $1/\sqrt{2\pi} (1/\sqrt{N}$ in the discrete case) for both. Or $100$ and $1/(100N)$ if you really felt like it.

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    $\begingroup$ That $\frac{1}{2\pi}$ or $\frac{1}{\sqrt{2\pi}}$ factor is only necessary for the continuous Fourier transform that uses angular (radian) frequency, $\omega$. If you use "ordinary frequency" $f$ (that is Hz or kHz, not radians/sec) then there is no $\frac{1}{2\pi}$ factor scaling the amplitude, but there is a $2 \pi$ factor in the exponent of both forward and inverse Fourier transform. $\endgroup$ Sep 7, 2023 at 6:06
  • $\begingroup$ Subtle but important detail, thank you! $\endgroup$
    – Jdip
    Sep 7, 2023 at 6:27
  • $\begingroup$ Thank you, but what about the continuous fourier transform from -infinity to infinity and also its inverse? I notice there is no summation in either? $\endgroup$ Sep 7, 2023 at 13:01
  • $\begingroup$ If you read both my answer and @robertbristow-johnson’s comment, you’ll see that there is indeed a scaling factor when the continuous transform uses angular frequency. $\endgroup$
    – Jdip
    Sep 7, 2023 at 14:16
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That factor of $\frac1N$ can be split up and distributed between the forward DFT and inverse DFT any which way you want as long as the two scaling factors multiply to be $\frac1N$.

It has to do with the fundamental derivation of the DFT.

Maybe this answer will help.

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  • $\begingroup$ Thanks, but what about the continuous integral from -infinity to infinity? I notice there is no scaling factor $\endgroup$ Sep 7, 2023 at 13:02
  • $\begingroup$ The derivation of the continuous Fourier Transform (beginning with Fourier Series) is a bit tougher than the derivation of the DFT. You need to have the concept of the Riemann Integral down well, and you have two limits going to infinity simultaneously. Do you have a good textbook? If I do that here at this SE, it's gonna be long and take some time to be both rigorous enough that no one accuses me of handwaving, yet be basic and clear enough that someone with an understanding of calculus can understand the derivation. $\endgroup$ Sep 7, 2023 at 16:06

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