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Fourier transform problem

  1. Mar 4, 2015 #1

    joshmccraney

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    hi pf!

    My book presents a problem and has it boiled down to $$S(u) = -S(f(x)) \exp(- \omega y) / \omega$$ where ##S(u)## is the sine fourier transform of the function ##u##. However, we cannot directly take the transform back since the singularity at ##\omega = 0##. Thus the book then takes $$\frac{\partial}{\partial y} S(u) = S(f(x)) \exp(- \omega y)$$ and now performs an inverse COSINE transform on the exponential (also they use convolution). My question is, why are they using a cosine transform instead of a sine transform?

    Thanks so much for your help!
     
  2. jcsd
  3. Mar 6, 2015 #2

    joshmccraney

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    nevermind, i got it! when using convolution on sine/cosine we use one of each!
     
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