# Should you scale the FFT bins by $1/N$ where $N$ is the number of points in a transient signal?

I am going over the following tutorial. In section 2.1 the author says

To calculate the $$N$$ point FFT the Matlab algorithm 1 can be used. Here, after taking the FFT, its magnitude is calculated and the bins are scaled by $$1/N$$.

Now the author doesn't explain as to why the bins are scaled by $$1/N$$. At first glance it looks like some kind of normalisation -- but why?

I did some further research and it seems there different ways to define the power in a signal, depending on what you are looking at. See slide 3 of this.

http://www.hep.ucl.ac.uk/~rjn/saltStuff/fftNormalisation.pdf

Yes, some kind of normalization needs to be done and it's a matter of convention which one to use. To explain, let $$\mathbf{F}$$ be a DFT matrix containing $${\rm e}^{-\jmath 2\pi \frac{mn}{N}}$$. Then you can show that $$\mathbf{F} \cdot \mathbf{F}^H = N \cdot \mathbf{I}_N$$. This factor $$N$$ needs to be accounted for. Now, you can do either of the following:

• Define DFT via $$\mathbf{D} = \mathbf{F} \cdot \mathbf{d} \quad \rightarrow\quad$$ IDFT becomes $$\mathbf{d} = \frac 1N \mathbf{F}^H \cdot \mathbf{D}$$.
• Define DFT via $$\mathbf{D} = \frac 1N \mathbf{F} \cdot \mathbf{d} \quad \rightarrow\quad$$ IDFT becomes $$\mathbf{d} = \mathbf{F}^H \cdot \mathbf{D}$$.
• Define DFT via $$\mathbf{D} = \frac{1}{\sqrt{N}} \mathbf{F} \cdot \mathbf{d} \quad \rightarrow\quad$$ IDFT becomes $$\mathbf{d} = \frac{1}{\sqrt{N}} \mathbf{F}^H \cdot \mathbf{D}$$.

All three of them can be used and none of them is more or less correct than the other. Matlab uses the second one, Maple the third (if I remember correctly).

If you need the units to resemble Fourier integrals, normalizing by $$\Delta t$$ or $$\Delta f$$ can make sense as well, as you found in the slide set. Depends on what you want to do with the result really.

• Thanks @Florian. Nice answer -- confirms what I have read elsewhere! – Q.P. Jul 12 '19 at 12:03
• Glad to hear it helped! :) – Florian Jul 12 '19 at 13:07
• Quantum, here is another answer that explicitly is about this whole scaling by $\frac1N$ issue. It says what Florian says above but with the explicit summations. – robert bristow-johnson Jul 12 '19 at 21:52
• Florian, i am unfamiliar with the "$H$" in "$\mathbf{F}^H$". i take it to mean "complex conjugate" from the context, but I cannot say for sure. What does that notation mean? – robert bristow-johnson Jul 12 '19 at 21:54
• Yes it was supposed to mean conjugate (Hermitian) transpose. – Florian Jul 14 '19 at 20:51