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Fourier Transform (Another question):

  1. Mar 8, 2008 #1
    I need to find the fourier transform of f(x) which is given by the equation:
    [tex]-\frac{d^2f(x)}{dx^2}+\frac{1}{a^3}\int_{-\infty}^{\infty}dx'exp(-\lambda|x-x'|)f(x')=\frac{b}{a^2}exp(-\lambda|x|)[/tex]

    ofcourse Iv'e taken the fourier tarnsform of both sides, but I don't see how to calcualte the fourier tranform of the integral in the above equation, I feel I need to use the definition of dirac's delta function, but don't see how to do this, any ideas, hints?

    thanks in advance.
     
    Last edited: Mar 8, 2008
  2. jcsd
  3. Mar 8, 2008 #2
    Without doing the fourier transform, it looks to me like you'll need to use the property that [itex]\delta(x'-x)f(x')=f(x)[/itex]. Just note which of x and x' is actually a variable for integration and which is the "constant" inside the integration. Use Parseval's Theorem.
     
    Last edited: Mar 8, 2008
  4. Mar 8, 2008 #3
    I'm not sure it's correct I got that the F.T of the integral without the constant 1/a^3 is:
    2pi*f(0)*e^(a-\lambda|x|), is this correct?
     
  5. Mar 8, 2008 #4
    Well actually disregard my previous advice, sorry! Note that the integral is equivalent to [itex]e^{-\lambda |x|} \ast f(x)[/itex] where the operator is convolution. In the frequency domain, this becomes multiplication. Now you simply need to know [itex]e^{-\lambda |x|} \Leftrightarrow \frac{2\lambda}{\lambda^2 + \omega^2} [/itex] which is actually the same answer as you'd arrive by from what I previously said, but in much less time.
     
    Last edited: Mar 8, 2008
  6. Mar 8, 2008 #5
    so using convolution you say, ok I'll try it, thanks.
     
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