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Fourier Transform of a Free Induction Decay Signal

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data

    S(t) = S(0)[itex]e^{-i \pi f_{o}t}[/itex] [itex]e^{-t/T^{*}_{2}}[/itex], 0 [itex]\leq[/itex] t < [itex]\infty[/itex]
    S(t) = 0, t < 0

    Show that the spectrum G(f) corresponding to this signal is given by:

    G(f) = S(0) { [itex]\frac{T^{*}_{2}}{ 1 + [2 \pi (f- f_{o} )T^{*}_{2}]^{2}} + \frac{i2 \pi (f- f_{o} ) (T^{*}_{2})^{2}}{ 1 + [2 \pi (f- f_{o} )T^{*}_{2}]^{2}}[/itex] }

    2. Relevant equations
    [itex]G(f) = \int^{\infty}_{- \infty} S(t) e^{i 2 \pi f t} dt[/itex]
    3. The attempt at a solution

    [itex]G(f) = \int^{\infty}_{- \infty} S(t) e^{i 2 \pi f t} dt[/itex]

    [itex] = \int^{\infty}_{- \infty} S(0) e^{-i 2 \pi f_{o}t} e^{-t/T^{*}_{2}}e^{i 2 \pi f t} dt[/itex]

    [itex] = S(0) \int^{\infty}_{0} e^{i 2 \pi (f - f_{o})t} e^{-t/T^{*}_{2}} dt[/itex] * S(t) is 0 when t is less than 0, so took integral from infity to 0. Took out S(0) and combined the exponents.

    [itex] = S(0) \int^{\infty}_{0} e^{i 2 \pi (f - f_{o})t - t/T^{*}_{2}} dt[/itex] *Combined the exponents

    [itex] = S(0) \int^{\infty}_{0} e^{t[i 2 \pi (f - f_{o}) - 1/T^{*}_{2}]} dt[/itex] *factored out t. Now all the stuff in the square brackets are basically a constant.

    [itex] = S(0) [ \frac{e^{t[i 2 \pi (f - f_{o}) - 1/T^{*}_{2}]}}{i 2 \pi (f - f_{o}) - 1/T^{*}_{2}} ]^{\infty}_{0}[/itex] *This is what I get after integrating. But as t goes to infinity, so does the fraction. Which I think makes sense since its a continuous spectrum. So maybe integrating from 0 to T2*? That way it looks like I am getting closer to the answer. Is my integration wrong?
     
  2. jcsd
  3. Oct 6, 2012 #2
    And I apologize if this is not in the right section.
     
  4. Oct 6, 2012 #3
    Nevermind, I got it.
     
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