# Intuition behind the scaling property of Fourier Transforms

I had a course in PDE last year where we used fourier transforms extensively; I understand the rules of manipulation and can prove the scaling theorem directly from the definition using a substitution, but I don't really have any good intuitive argument as to why "compressing" a function causes an expansion of its fourier transform, and vice versa. I have been trying to gain solid intuition behind the various properties of the fourier transform; but have not gotten far with this one. If anyone knows of a website / book or a slick argument that covers this; it'd be greatly appreciated. Thanks!

• What is the scaling theorem? – Qiaochu Yuan Jul 6 '11 at 5:07
• @Qiaochu It is usually called the similarity theorem. If $\mathcal{F}(u)$ is the FT of $f(x)$ then $\frac{1}{a}\mathcal{F}\left(\left|\frac{u}{a}\right|\right)$ is the FT of $f(ax)$ – kuch nahi Jul 6 '11 at 5:59

## 1 Answer

Intuitively, One way to think about it is that compressing a sinusoid, increases its frequency. So compessing a sum of sinusoids will expand the frequency spectrum.

• Ok cool, thanks. I thought about it a bit and I can see that compressing a signal is equivalent to compressing its component frequencies, which would then cause the frequency spectrum to be spread over a larger range. Is there a good way to see why the amplitude of each frequency's contribution is the scaled by the reciprocal of the compression factor? – Steve Jul 6 '11 at 6:28
• the amplitude scaling is due to parsevals. compressing the signal reduces the energy. – Digital Gal Jul 6 '11 at 13:14