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Property of the convolution product

  1. Feb 25, 2014 #1
    It is known that:[tex]\mathcal{F}\{f\ast g\}=\mathcal{F}\{f\}\mathcal{F}\{g\}[/tex][tex]\mathcal{F}\{f g\}=\mathcal{F}\{f\}\mathcal\ast{F}\{g\}[/tex]

    But this property is valid for inverse tranform too?[tex]\mathcal{F}^{-1}\{F\ast G\}=\mathcal{F}^{-1}\{F\} \mathcal{F}^{-1}\{G\}[/tex][tex]\mathcal{F}^{-1}\{F G\}=\mathcal{F}^{-1}\{F\}\ast \mathcal{F}^{-1}\{G\}[/tex]
     
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  3. Feb 25, 2014 #2

    Ben Niehoff

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    Think about this for a moment...
     
  4. Feb 25, 2014 #3
    The only difference between ##\mathcal{F}## and ##\mathcal{F}^{-1}## is the sign in the exponential in the kernel, thus, appears there isn't reason for the property of the convolution not be valid for inverse transform, I think. Although I never saw this affirmation/property for inverse transform in anywhere.
     
  5. Feb 25, 2014 #4

    Ben Niehoff

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    You don't need to worry about the integral definitions. Just use this:

    and apply ##\mathcal F^{-1}## to each line.
     
  6. Feb 25, 2014 #5
    Oh yeah! Thank you!
     
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