1
$\begingroup$

The Fourier transform of a function $f(t)$ is defined as: $$F(\omega)=\int_{-\infty}^{\infty} f(t)e^{-2\pi i\omega t} dt$$ Express the Fourier transform of $g(t)=f(at)$ in terms of $F,a$ and $\omega$.

Now, for $a>0$ and $a<0$, I know that the Fourier transform is $g(t)=f(at)$ $$G(\omega)=\frac{1}{|a|}F\left(\frac{\omega}{a}\right)$$

Since the question does not specify anything about $a$, I was wondering what the Fourier transform of $g(t)=f(at)$ would be if $a$ is complex, $a\in \mathbb{C}$, i.e. $$a=\text{Re}(a)+i\text{ Im}(a)$$

This is what I have tried so far, I'm not sure if this is the right approach $$G(\omega)=\int_{-\infty}^{\infty} f(at)e^{-2\pi i\omega t} dt$$ Now using the substitution $t'=at=(\text{Re}(a)+i\text{ Im}(a))t$, we have $$G(\omega)=\frac{1}{a} \int_{-\infty}^{\infty} f(t')e^{-2\pi i(\omega/a) t'} dt'$$ $$\frac{1}{a}=\frac{1}{\text{Re}(a)+i\text{ Im}(a)}=\frac{\text{Re}(a)-i\text{ Im}(a)}{\text{Re}(a)^2+\text{ Im}(a)^2}$$ Looking at it, I don't think that this will work.

$\endgroup$
2
$\begingroup$

First notice a small detail. When you do the transformation $ t = a t' $, it converts to $ dt= | a | d t' $, right? Infinitesimals only make sense as sort of real numbers, remember this.

Now, with that correction, write $x = \frac{\omega }{a } $ in the formula $$ G ( \omega ) = \frac{1}{|a|} \int_{-\infty }^{\infty } f(t' ) e^{-2 \pi i x t' } = \frac{1}{|a| } F(x) \, . $$

The substitution $x = \frac{\omega }{a } $ is only to make clear that your right hand side is in fact the Fourier transform of $f$. You should try and train yourself to make these natural observations even before writing up the substitution.

$\endgroup$
  • $\begingroup$ I just realised that you also have to transform the limits too i.e. $t=\infty$ transforms to $t'=\infty+i\infty$ and $t=-\infty$ transforms to $t'=-\infty-i\infty$. How would this change the result? $\endgroup$ – Hogg Oct 3 '16 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.