Fourier transform of a function that is scaled and translated in either order

Consider a function $f(x)$ with Fourier transform $\tilde{f}(k)$ where

$\tilde{f}(k) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} f(x) e^{-ikx } dx \quad \quad \quad \quad (1)$

Then it is straightforward to show that the Fourier transform of the scaled function $f(x/\alpha)$ is

$\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} f(x/\alpha) e^{-ikx } dx = |\alpha| \tilde{f}(\alpha k)$

and the Fourier transform of the $x$ -translated function $f(x - c)$ is

$\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} f(x - c) e^{-ikx } dx = e^{-ikc} \tilde{f}(k)$

What is a little less straightforward is to deduce from these what the Fourier transform must be of a function that has been both scaled and translated in the form

$f\big(\frac{x - c}{\alpha}\big) \quad \quad \quad \quad (2)$

In fact, the Fourier transform of this particular type of scaled and translated function is a simple combination of the two individual adjustments as if they had occurred independently of each other:

$\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} f\big(\frac{x - c}{\alpha}\big) e^{-ikx } dx = e^{-ikc} |\alpha| \tilde{f}(\alpha k) \quad \quad \quad \quad (3)$

We should be able to deduce this by performing the calculations for the scaling and translation sequentially, and we should get the same answer irrespective of the order in which the transformations are performed. However, there are subtleties involved and these are what I am interested in recording in the present note.

Consider performing the scaling first and then the translation. Let $g(x) = f(x/\alpha)$ so that $\tilde{g}(k) = |\alpha| \tilde{f}(\alpha k)$ . Then the combined transformation in (2) is obtained as $g(x - c)$ which has Fourier transform

$e^{-ikc} \tilde{g}(k) = e^{-ikc} |\alpha| \tilde{f}(\alpha k)$

This agrees with (3). However, when going the other way and performing the translation first before the scaling, it is necessary to remember that the translation in (2) is by an amount $c/\alpha$ . Therefore, in this case, we have to let $g(x) = f \big(x - \frac{c}{\alpha} \big)$ so that $\tilde{g}(k) = e^{-ik(c/\alpha)} \tilde{f}(k)$ . Then the combined transformation in (2) is obtained as $g(x/\alpha) = f \big(\frac{x}{\alpha} - \frac{c}{\alpha} \big)$ which has Fourier transform

$|\alpha| \tilde{g}(\alpha k) = e^{-i(\alpha k)(c/\alpha)} |\alpha| \tilde{f}(\alpha k) = e^{-ikc} |\alpha| \tilde{f}(\alpha k)$

This again agrees with (3), but this way round we see that there has had to be a process of cancellation of the $\alpha$ parameter in the complex exponential term. This complication is avoided by performing the scaling first before the translation, making this particular ordering of the transformations a little easier to work with in the case of the combined transformation in (2).

Fourier transform of a function that is scaled and translated in either order

Published by Dr Christian P. H. Salas

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