1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with 2d convolution formula

  1. Nov 23, 2012 #1

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
    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].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook