This refers to the link where it clearly writes that:

The FrFT argument $u$ is neither a spatial one $x$ nor a frequency $\xi$. We will see why it can be interpreted as linear combination of both coordinates $(x,\xi)$

But I did not see the relation between the domain of Fractional Fourier transform ($u$) and space and frequency $(x,\xi)$. I can understand that domain '$u$' is a linear combination of '$(x,\xi)$' but my question what is the relation? Can anyone point out the exact relation?


It think it is more correct to say that the $u$ axis is a linear combination of the $x$ and $\xi$ axes:

$$(x, \xi) = (1, 0)x + (0, 1)\xi = \left((1, 0)\cos(\alpha) + (0, 1)\sin(\alpha)\right)u,$$

that is:

$$ x = u\cos(\alpha)\\ \xi = u\sin(\alpha).$$

The definition of the fractional Fourier transform from that Wikipedia page is:

$$\mathcal{F}_\alpha[f](u) = \sqrt{1-i\cot(\alpha)} e^{i \pi \cot(\alpha) u^2} \int_{-\infty}^\infty e^{-i2\pi (\csc(\alpha) u x - \frac{\cot(\alpha)}{2} x^2)} f(x)\, \mathrm{d}x.$$

If you use another variable, $y$, for the integral, and substitute $u = \sqrt{x^2 + \xi^2}$ and $\alpha = \text{atan}(\xi, x)$, then the fractional Fourier transform can be written as:

$$\mathcal{F}[f](x, \xi) = e^{i \pi \frac{\xi}{x} (x^2+\xi^2)} \sqrt{1-\frac{i \xi}{x}}\int^\infty_{-\infty}e^{i \pi \frac{y}{x} \left(\xi y-2 x^{\frac{1}{\xi}} \sqrt{\xi^2+x^2}\right)} f(y)\, \text{d}y.$$

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  • $\begingroup$ Can we gain any insight by change of variable? I knew that relation to x but did not realize the relation to frequency. $\endgroup$ – Creator Oct 21 '15 at 23:58
  • $\begingroup$ Dunno, but it's the same as going from polar notation to rectangular notation. $\endgroup$ – Olli Niemitalo Oct 22 '15 at 0:07

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