I'm reading through Paganin's Coherent X-Ray optics textbook and could use a hand with Fourier Transform (FT) manipulation. For the duration of this post, $\mathcal{F}$ is the forward FT and $\mathcal{F}^{-1}$ is the inverse transform. I define the conventions at the bottom of this post.

In the text, Paganin defines the angular-spectrum representation of a propagated wave-field as

\begin{equation} \psi_\omega (x, y, z=\Delta) \approx \exp(ik\Delta) \mathcal{F}^{-1} \exp\bigg[\dfrac{-i\Delta (k_x^2 + k_y^2)}{2k} \bigg]\mathcal{F} \psi_\omega (x, y, z=0) \end{equation}

Here, $\Delta$ is a spatial coordinate and $\Delta > 0$. In the discussion of the convolution formulation of the Fresnel Diffraction section, he defines the following form of convolution: \begin{equation} f(x,y) *g(x,y) = 2\pi \mathcal{F}^{-1}\big( \{ \mathcal{F} [f(x, y)]\} \times \{ \mathcal{F}[g(x,y)] \} \big). \end{equation}

I'm a little confused with what Paganin does after this. He states that we can rewrite the angular-spectrum representation of the wavefield in the following suggestive form: \begin{align} \psi_\omega (x, y, z=\Delta) &= 2\pi \mathcal{F}^{-1} \bigg( \dfrac{\exp(ik\Delta)}{2\pi} \exp\bigg[ \dfrac{-ik\Delta(k_x^2 + k_y^2)}{2k} \bigg] \times \{ \mathcal{F} [\psi_\omega (x, y, z=0] \} \bigg) \\[.5em] &= 2\pi \mathcal{F}^{-1} \bigg( \bigg\{ \mathcal{F}\mathcal{F}^{-1} \dfrac{\exp(ik\Delta)}{2\pi} \exp\bigg[\dfrac{-i\Delta(k_x^2 + k_y^2)}{2k} \bigg] \bigg\} \times \{ \mathcal{F} [\psi_\omega (x, y, z=0)] \} \bigg) \end{align}

I've tried to work out how he does this, but I haven't had any luck. Does anyone have any insight into how we go from the first equation listed to the one just before this sentence?

Since there are a million ways to write the FT, I'll write the definition Paginin uses. \begin{align} f(x,y) &= \frac{1}{2\pi} \iint_{-\infty}^\infty \tilde{f}(k_x,k_y)\exp[i(k_x x + k_y y)] \, dk_x \, dk_y \\[.5em] \tilde{f}(k_x,k_y) &= \frac{1}{2\pi} \iint_{-\infty}^\infty f(x,y)\exp[-i(k_x x + k_y y)] \, dx \, dy \end{align} Here, $\tilde{f}$ is the FT of $f$. i.e., $\tilde f=\mathcal{F}[f]$ and ${f}=\mathcal{F}^{-1}[\tilde f]$.

  • 2
    $\begingroup$ The first equality seems to just be moving $e^{i k \Delta}$ inside the inverse Fourier transform, which is allowed since it's just a scalar. The second equality just seems to be writing $f=\mathcal{F}[\tilde{f}]=\mathcal{F}\mathcal{F}^{-1}[f]$. $\endgroup$ Commented Jan 15, 2021 at 4:53

1 Answer 1


$\def\F{\mathcal{F}} \def\Fi{\F^{-1}} \def\p{\pi} \def\D{\Delta}$Schematically, we have \begin{align*} \psi &= 2\p A\Fi(B\F C) = 2\p\Fi(AB\F C) = 2\p\Fi(\F(\Fi AB)\F C) \\ &= (\Fi AB)\star C = (A\Fi B)\star C. \end{align*} Note that $A = e^{i k\D}/(2\p)$ is not a function of $x,y$.


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