# fourier transform normalization constant

I'm confused about the two Fourier transform formulas that crop up. One where the constant of $\frac{1}{\sqrt{2\pi}}$ is used in both the forward and reverse transform and the other where just $\frac{1}{{2\pi}}$ is used for the inverse. I understand that it's a normalization constant or something and so both are equivalent somehow, but if the Fourier transform of a signal tells you what the amplitude of the signal is at various frequencies how is one of these not wrong. If the first formula tells you that there's a peak and 5Hz and the other says something else then surely only one of them is right. Or will they both give the same result, and if so how is that possible if they are multiplied by different constants. I hope that makes sense, thanks in advance

• Yes, the normalization constant can affect if the $L^2$ norm / signal energy / power stays the same in Parcevals / Plancherel theorems or if it is scaled by a constant. – mathreadler Aug 25 '17 at 20:57
• In contrary to the Fourier series and the discrete Fourier transform, you can't interpret directly a single value of $\hat{f}(\omega_0)$ as the amplitude of a sine component. But with the Fourier transform $\hat{f}(\omega) = \int_{-\infty}^\infty f(t) e^{-i \omega t}dt$ you can look at the energy in the frequency band $[a,b]$ : $E_{[a,b]} =\frac{1}{2\pi} \int_a^b |\hat{f}(\omega)|^2 d\omega$ and Parceval theorem is that $E_{(-\infty,\infty)} = \int_{-\infty}^\infty |f(t)|^2dt$. – reuns Aug 25 '17 at 21:02
• The factor $\frac{1}{2\pi}$ won't affect at what frequency the amplitudes have a peak. It will only affect the size of the amplitude. – md2perpe Aug 25 '17 at 21:33
• Thanks for all the speedy replies, I can finally go to sleep satisfied. – richard davies Aug 25 '17 at 23:16

You are right that the purely formulaic presentation of Fourier transform and Fourier inversion give the impression that Plancherel and/or other aspects of the situation depend on the normalizations. And, of course, in a way, they do, because one can choose partly-correct normalizations and then make incompatible choices in the rest. Also, coordinate-dependent interpretations of formulas are doomed... Rather, I'd recommend thinking that "Fourier (-Plancherel) transform" is an isometry (or "isometric isomorphism") from one Hilbert space to another, and Fourier inversion is the inverse isometry.

The several differing choices of locations of the $2\pi$ can easily be accommodated by altering the measure for the $L^2$ space that is the target of Fourier (-Plancherel), and similarly for the inverse.

That is, the way to think about this (in my opinion) is that there cannot possibly be any genuine issue about mathematical facts, only one of correct interpretation of formulaic things in terms of the underlying reality.

• Awesome, It took me a while to fully understand the response but i get it now. Thanks a million – richard davies Aug 25 '17 at 23:16
• @richarddavies ... :) ... – paul garrett Aug 25 '17 at 23:29