# 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. 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 $E_{(-\infty,\infty)} = \int_{-\infty}^\infty |f(t)|^2dt$. 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. Aug 25 '17 at 21:33
• Thanks for all the speedy replies, I can finally go to sleep satisfied. Aug 25 '17 at 23:16

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.