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!
Intuition behind the scaling property of Fourier Transforms
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fourier-analysis
intuition
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0@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 $f(ax)$ – 2011-07-06
1 Answers
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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.
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0the amplitude scali$n$g is due to parsevals. compressing the signal reduces the energy. – 2011-07-06