I understand how to prove the scaling property of Fourier Transforms, except the use of the absolute value:

If I transform $f(at)$ then I get $F\{f(at)\}(w) = \int f(at) e^{-jwt} dt$ where I can substitute $u = at$ and thus $du = a dt$ (and $\frac{du}{a} = dt$) which gives me:

$ \int f(u) e^{-j\frac{w}{a}u} \frac{du}{a} = \frac{1}{a} \int f(u) e^{-j\frac{w}{a}u} du = \frac{1}{a} F \{f(u)\}(\frac{w}{a}) $

But, according to various references, it should be $ \frac{1}{|a|} F \{f(u)\}(\frac{w}{a}) $ and I don't understand WHY or HOW I get/need the absolute value here?


First, lets convince ourselves that $F\{f(-t)\}(\omega)=F\{f(t)\}(-\omega)$:

$$ F\{f(-t)\}(\omega)=\int_{t=-\infty}^{t=\infty} f(-t)e^{-j\omega t}dt\quad\star $$ Set $u=-t$, so $dt=-du$. Also note that when $t=-\infty,$ $u=\infty$ and when $t=\infty$, $u=-\infty$. So,
$$ \star=-\int_{u=\infty}^{u=-\infty}f(u)e^{j\omega u}du\quad \star\star $$ recall that $$ -\int_{a}^bf(x)dx=\int_b^af(x)dx $$ which explains the flipping of the integration bounds. Hence $$ \star\star=\int_{u=-\infty}^{u=\infty}f(u)e^{j\omega u}du $$ which is exactly $F\{f(t)\}(-\omega)$

Then, if $a<0$ we can simply write $a=-\vert a\vert$, so that $F\{f(at)\}=F\{f(-\vert a\vert t)\}(\omega)=F\{f(\vert a\vert t)\}(-\omega)=\frac{1}{\vert a\vert}F\{f(t)\}(\frac{-\omega}{\vert a\vert})=\frac{1}{\vert a\vert}F\{f(t)\}(\frac{\omega}{a})$

  • $\begingroup$ Ok, I'm trying to understand how to get rid of the - at the time reversal case. When they (fourier.eng.hmc.edu/e101/lectures/handout3/node2.html) substitute t by -t' they also swap the borders of the integral (because of dt = -dt'; which is what I do not understand correctly). And when they swap them back they have to add another - (I understand why, this is a simple rule with integrals). Those two - then eliminate each other ... but why do they swap the borders in the first place? $\endgroup$ – Daniel Jour Dec 17 '12 at 0:27
  • $\begingroup$ See edits - it's standard integration stuff from "calc II", but its easy to get lost in the negative signs. $\endgroup$ – icurays1 Dec 17 '12 at 0:42
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    $\begingroup$ "when t=∞ u=−∞." THAT was the part I missed/didn't think of. Thank you, that solves my problem! :) $\endgroup$ – Daniel Jour Dec 17 '12 at 3:53

Think about the range of the variable $t$ in the integral that gives the transform. How do the 'endpoints' of this improper integral transform under $t\to at$? Can you see how this depends on the sign of $a$?

  • $\begingroup$ Hm .. I'm not sure whether I got it: If $a$ is negative, then I would "change" the "direction" of the integral which is equivalent to exchanging the 'endpoints' ... hm, no, I don't think I got it. Is it possible to ... well .. use a equation that is explaining this problem? $\endgroup$ – Daniel Jour Dec 16 '12 at 23:11

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