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Help with 2d convolution formula

  1. Nov 23, 2012 #1
    Hello,

    I consider two functions [itex]f:R^2 \rightarrow R[/itex] and [itex]g:R^2 \rightarrow R[/itex], and the two dimensional convolution [tex](f \ast g)(\mathbf{x}) = \int_{\mathbb{R}^2}f(\mathbf{t})g(\mathbf{x-t})d^2\mathbf{t}[/tex]
    I proved using the Fourier transform and the convolution theorem that the convolution of two "rotated" versions of f and g is equivalent to simply taking the convolution (f*g) and rotating it.

    However I have troubles proving these statement using only the definition of convolution. I will show my attempt. There must be a mistake somewhere.

    I consider an isometry (rotation) [itex]\phi:R^2\rightarrow R^2[/itex] and the two rotated versions of the functions: [itex]f(\phi(\mathbf{x}))[/itex] and [itex]g(\phi(\mathbf{x}))[/itex].
    The convolution would be: [tex]\int_{\mathbb{R}^2}f(\phi(\mathbf{t}))g(\mathbf{x}-\phi(\mathbf{t}))d^2\mathbf{t}[/tex] By setting [itex]\mathbf{y}=\phi(\mathbf{t})[/itex] and observing that [itex]d^2y = d^2t[/itex] we get: [tex]\int_{\mathbb{R}^2}f(\mathbf{y})g(\mathbf{x}-\mathbf{y})d^2\mathbf{y}[/tex] which is not the expected result. Where is the mistake?
     
  2. jcsd
  3. Nov 23, 2012 #2

    haruspex

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    You mean, the convolution of the rotated functions, [tex]f\phi(), g\phi()[/tex]?
    Wouldn't that be [tex]\int_{\mathbb{R}^2}f(\phi(\mathbf{t})) g(\phi(\mathbf{x}-\mathbf{t}))d\mathbf{t}[/tex]
    (And [tex]\phi(\mathbf{x}-\mathbf{t})=\phi(\mathbf{x})-\phi(\mathbf{t})[/tex], right?)
     
  4. Nov 24, 2012 #3
    Thanks!
    yes. Maybe I was tired and didn't notice that mistake! :redface:
    The statement to be proven then follows from the linearity of [itex]\phi[/itex].
     
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