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Fourier Transform (basic table lookup)

  1. Mar 27, 2007 #1
    I have a practice exam I'm going through, and I am stumped on one of the basic problems.

    How is this a transform pair?

    [tex] 10 X(jt) [/tex] <-----> [tex] 20 \pi x (-\omega) [/tex]

    I don't see how one can make this relation. What is the [tex] 10 X (jt) [/tex].

    thanks in advance
  2. jcsd
  3. Mar 28, 2007 #2
    Divide both sides by 10 to simplify!

    X(jt)<-----> 2pi x(-w)

    Still seems wrong to me. The 2pi is normally included in the defn. of the FT.
    What's j ?
  4. Mar 28, 2007 #3
    I'll post the exact question. I don't understand the X(jt) as I posted. Note: [tex] j = \sqrt{-1} [/tex]

  5. Mar 28, 2007 #4
    OK, I got it- but I don't like the j notation. Everyone else writes

    x(t) <-----> X(omega)

    Define FT[x(t)]= Int[e^-iwt x(t)] dt
    FT-1[X(w)]=(1/2pi)Int[e^iwt X(w)] dw

    (you may use a different convention, but the prefactors of FT and FT-1 must multiply to (1/2pi) no matter what convention you use)

    Start with FT[x(t)]=X(w)

    Int [ e^-iwt x(t)] dt = X(w)

    substitute w=t', t=w'

    Int [ e^-iw't' x(w') ] dw' =X (t')

    which is the same as writing

    Int [ e^-iwt x(w) ] dw =X (t)

    Substitute w -> -w

    -(1/2pi) Int [ e^iwt x(-w) ] dw =1/(2pi) X (t)

    The LHS is the inverse FT, which I'll call FT-1

    -FT-1 [x(-w)]=1/2pi X(t)

    Take 2pi to other side
    X(t)=-2pi FT-1 [x(-w)]

    FT both sides

    FT[ X(t) ] =-2pi x(-w)

    X(t) <----> - 2pi x(-w)

    I seem to have an extra minus sign- but who cares? You can have fun checking if it's me or the teacher who messed it up.

    Edit- the minus sign comes from changing the limits from +ininity,-infinity to -infinity,+infinity in the integral, when w-> -w

    You can then turn the integral 'the right way up' with the introduction of a minus sign.
    That always trips me up.
    Last edited: Mar 28, 2007
  6. Mar 28, 2007 #5


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    The conjugation rule is :

    g\left( t \right) = \overline {f\left( t \right)} \Rightarrow G\left( \omega \right) = \overline {F\left( { - \omega } \right)}

    - Warren
  7. Mar 28, 2007 #6
    Is that a proof?
  8. Mar 28, 2007 #7


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    Oops, actually, I read this too quickly. The example given is a form of the inversion rule:

    g\left( t \right) = F\left( t \right) \Rightarrow G\left( \omega \right) = f\left( { - \omega } \right)

    And no, it's not a "proof," it's an rule of the so-called Fourier transform calculus.

    - Warren
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