2

$$f\left(x,y\right)=g\left(\frac{c_2 \cdot \left(x-y\right)}{c_1}+y\right)$$

$$\mathcal{F}\left(X,Y\right)=\frac{c_1}{c_2} \cdot \delta\left(Y-\left(\frac{c_1}{c_2}-1\right) \cdot X\right) \cdot \mathcal{G}\left(\frac{c_1}{c_2} \cdot X\right)$$

Can someone explain me how you obtain the following Fourier transform of the above 2D function $f$ in terms of the 1D function $g$?

Matthias
  • 163

1 Answers1

1

Let $\mathcal{F}\{u(x)\}(X)$ be the Fourier transform of $u(x)$. Then it's straightforward to show the inverse Fourier transform matches the original function in question.

\begin{align*} &\mathcal{F}^{-1}\{F(X,Y)\}(x,y) \\ &= \frac{1}{(2\pi)^2}\iint_\limits{\mathbb{R}^2} F(X,Y) \exp(i(xX+yY)) dXdY \\ &= \frac{1}{(2\pi)^2} \iint_\limits{\mathbb{R}^2} \frac{c_1}{c_2} \cdot \delta\left(Y-\left(\frac{c_1}{c_2}-1\right) \cdot X\right) \cdot \mathcal{G}\left(\frac{c_1}{c_2} \cdot X\right) \exp(i(xX+yY)) dXdY \\ &= \frac{c_1}{c_2} \frac{1}{(2\pi)^2}\int_{\mathbb{R}} \mathcal{G}\left(\frac{c_1}{c_2} \cdot X\right) \exp\left(i\left(xX+y\left(\frac{c_1}{c_2}-1\right) \cdot X\right)\right) dX \end{align*}

Letting $X' = \frac{c_1}{c_2}X$ yields equality to $f(x,y)$.

Chester
  • 1,409
  • Thanks your solution also refreshed some of my Fourier derivation skills. The coefficients in the fraction of the mapping still need to be swapped (but is less than 6 characters so I cannot change that ;) ) – Matthias Aug 26 '15 at 19:33
  • whoops. fixed. thanks. – Chester Aug 26 '15 at 20:33