Why does the fourier transform invert scales?

I am learning about the Fourier transform and have looked at some popular functions and their transforms and have noticed and wondered why the scale is inverted.

I also have seen this as a general property of the Fourier transform from time/spatial domain to frequency domain. I.e. $$f(ax)$$ in spatial domain is $$\frac{1}{|a|} F({\omega}{a})$$

or how the impulse function with spacing T is also just an impulse but with spacing 1/T. Or the gaussian with its variance inverted, etc.

Would someone have an intuitive answer as to why?

Thank you so much!

• This property you have observed is a manifestation of the uncertainty principle. See for example, Tao's blog post on this topic. terrytao.wordpress.com/2010/06/25/the-uncertainty-principle Oct 24 at 8:07
• @ashay thanks! I am going to pour over this in detail sometime, darn, was really hoping there would be some simpler explanation. Hope I may get the general idea somehow. Oct 24 at 8:22
• There is probably a way to explain it more simply -- I suggested Tao because I find his exposition can be useful even if I don't understand everything he's talking about. Hopefully someone else comes along with an answer that's more helpful! Oct 24 at 8:28

• Have the fundamental result in mind (from the theory fo distributions): the Fourier transform of a Dirac "pulse" is $$1$$. From an engineering point of view, a "Dirac" pulse, which has an infinitely small time support, equally excites all the frequencies. That is the reason why when you want to study the response of a mechanical system (to study vibrations or acoustic for instance) you use a hammer or a shot noise that mimics the Dirac pulse to excite all frequencies.
Now concerning your formula : $$\mathcal{F}\left(t\mapsto f(at)\right)=\frac{1}{\mid a \mid}\hat{f}\left(\frac{\xi}{a} \right)$$
• When $$a$$ becomes bigger and bigger you are more and more localized in the time domain due to the factor $$at$$. The signal time support is shortened and you get "closer" to the Dirac pulse. By consequence, you spread more and more in the frequency domain (the $$\frac{\xi}{a}$$ factor), to get closer to $$1$$. Now you also have to take into account energy conservation (the Parseval's theorem). In the time domain, like you reduce the time support without changing the amplitude of the signal, its energy is diminished (energy is roughly $$support \times amplitude^2$$). In the frequency domain, on the contrary, the support increases. By consequence to conserve energy you must add the $$\frac{1}{\mid a\mid}$$ factor that diminishes amplitude of the response in the frequency domain.
• @oliver I am glad it helps. Concerning your question, imagine a function $f$ that is one if $x\in [0,1]$ and zero elsewhere. Then the function $x->f(100x)$ is one if $x\in[0,1/100]$, zero elsewhere: bigger $a$ gives shorter $f$ support, $f$ is more "localized". That was the picture I tried to gives. Oct 24 at 14:44