Fourier Transform of a Free Induction Decay Signal

[wolski888]

Homework Statement

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

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

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

Homework Equations

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

The Attempt at a Solution

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

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

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

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

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

$= S(0) [ \frac{e^{t[i 2 \pi (f - f_{o}) - 1/T^{*}_{2}]}}{i 2 \pi (f - f_{o}) - 1/T^{*}_{2}} ]^{\infty}_{0}$ *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?

 

[wolski888]

And I apologize if this is not in the right section.

 

[wolski888]

Nevermind, I got it.