Is it possible to define a scaling property for fourier transform when the scale factor is complex? Usually the scaling factor is real. What happen when a scaling factor is complex?


there are issues. given this convention for the continuous Fourier transform (and inverse)

$$ \mathscr{F} \Big\{ x(t) \Big\} \triangleq X(f) \triangleq \int\limits_{-\infty}^{+\infty} x(t) e^{-j 2 \pi f t} \ \mathrm{d}t $$

$$ \mathscr{F}^{-1} \Big\{ X(f) \Big\} \triangleq x(t) = \int\limits_{-\infty}^{+\infty} X(f) e^{+j 2 \pi f t} \ \mathrm{d}f $$

it changes the path of integration from the real axis to something else. this comes up when using this fact:

$$ \mathscr{F} \Big\{ e^{- \pi t^2} \Big\} = e^{- \pi f^2} $$

to get, along with using scaling, this result:

$$ \mathscr{F} \Big\{ e^{j \pi t^2} \Big\} = \sqrt{j} \, e^{-j \pi f^2} $$

which is a linearly-swept "chirp" and it's spectrum.

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  • $\begingroup$ Thanks. If the frequency $f$ and the time variable $t$ are multiplied by a complex quantity $e^{i\theta}$ and not simply $i$, does the fourier transform keeps its meaning again? $\endgroup$ – dart Jul 13 '19 at 1:36
  • $\begingroup$ well they were multiplied by $\sqrt{-i}$ in my example. which is at an angle of $\pi/4$. any complex constant will change the path of integration from the real axis to something at an angle. to be strict about this, you would need to apply contour integration on the complex plane and show that the integration of "$g(z)$" along the "tipped" infinite line is the same as the integral along the real axis. $\endgroup$ – robert bristow-johnson Jul 13 '19 at 1:50
  • $\begingroup$ Thanks a lot! Could you show me an example how to do it? $\endgroup$ – dart Jul 13 '19 at 18:50
  • $\begingroup$ That is very hard. It's been 4 decades since I had my course in complex variables and learned about contour integration and learned that integrating an analytic function over a closed curve results in zero. But those are the steps you need to take to prove the scaling property with a complex scaler for a specific function. $\endgroup$ – robert bristow-johnson Jul 14 '19 at 18:34
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    $\begingroup$ Thanks again for your help. $\endgroup$ – dart Jul 14 '19 at 22:01

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