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I have to solve an exercise, but if i could use the following theorem, it would be piece of cake

Similarity Theorem

if $ \mathscr{F}\{g(x,y)\}= G( f_x,f_y)$ then

$ \mathscr{F}\{g(ax,by)\}= \frac {1} {| a \cdot b|}G( f_x /a, f_y/b) $

i just need the proof, (You can find the theorem at page 8 here: http://ymk.k-space.org/elective_2DFT.pdf )

can you help? probably needs Jacobian and change of variable?

Cris
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1 Answers1

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$$x' = ax, y' = by$$

Jacobian is $\dfrac{1}{|ab|}$

$$\int e^{i(xf_x + yf_y)} g(ax,by) \,dx\,dy = \dfrac{1}{|ab|}\int e^{i\left(\dfrac1ax'f_x + \dfrac1by'f_y\right)} g(x',y') \,dx'\,dy' \\ = \dfrac{1}{|ab|}\int e^{i\left(x'\dfrac{f_x}{a} + y'\dfrac{f_y}{b}\right)} g(x',y') \,dx'\,dy'\\ = \dfrac{1}{|ab|} G\left(\dfrac{f_x}{a},\dfrac{f_y}{b}\right)$$

BaronVT
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