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Fourier transform, complex exponential and infinity

  1. Mar 20, 2012 #1
    I'm taking the fourier transform of a signal. This integral has bounds from -∞ to ∞, but since the signal is 0 for negative t, the bounds become 0 to ∞

    doing the integration, the antiderivative I get is et*(-3-jω+2j) where j is sqrt(-1)

    Now I have to evaluate this at t=infinity (since it is a proper integral)....I don't really know how to do this since (-3-jω+2j) is a complex number.

    infinity times a negative number is negative infinity
    infinity times a positive number is positive infinity
    infinity times a complex number?????

    any help will be appreciated,

    thank you very much
     
    Last edited: Mar 20, 2012
  2. jcsd
  3. Mar 20, 2012 #2
    Looks like you have:

    [tex]e^{-3t+it(2-w)}\biggr|_0^{\infty}[/tex]

    where I'm using i cus' I'm not an engineer but same dif. Now assume w is a real number, then we could write:

    [tex]e^{-3t}e^{it(2-w)}=e^{-3t}\left(\cos(t(2-w))+i\sin(t(2-w)\right)\biggr|_0^{\infty}[/tex]

    so that because of the [itex]e^{-3t}[/itex] real part, at the upper limit, that is zero and at the lower limit it's one.
     
  4. Mar 20, 2012 #3
    thank you! got it!
     
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