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Difference between Forward and Backward Fourier Transforms?

  1. Jun 16, 2010 #1
    What is the difference between forward and backward Fourier transforms? I'm look:

    [tex]
    F(k) = \int_{-\infty}^{\infty} f(x)\ e^{- i 2\pi k x }\,dx
    [/tex]

    [tex]
    f(x) = \int_{-\infty}^{\infty} F(k)\ e^{ i 2\pi k x }\,dk
    [/tex]

    If I swap the x and the k in the second equation, the transforms are then:

    [tex]
    F(k) = \int_{-\infty}^{\infty} f(x)\ e^{- i 2\pi k x }\,dx
    [/tex]

    [tex]
    F(k) = \int_{-\infty}^{\infty} f(x)\ e^{ i 2\pi x k }\,dx
    [/tex]

    and the only difference is the minus sign in the exponent. What gives? Why aren't the forward and backwards transforms identical?
     
  2. jcsd
  3. Jun 17, 2010 #2
    Actually minus sign is introduced for generality.

    Fourier transforms have reciprocal (reversible) property i.e., if you take Fourier transform on a signal two times you will end up with original signal. So if [tex] f(x) [/itex] is even function then both forward and inverse transforms can have the same sign and reciprocal property is valid. But if [tex] f(x) [/itex] is odd or neither even nor odd, then reciprocal property is lost (i.e, if you take FT two times you will end up with minus of original function) if you use same sign in forward and inverse transforms.

    In order to generalize Fourier transform for any signal (even, odd, neither of two) and to preserve reciprocal property minus sign is introduced.
     
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