1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook