Evaluate the Fourier Transform of a Damped Sinusoidal Wave

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1. Sep 8, 2016

Captain1024

1. The problem statement, all variables and given/known data
Evaluate the Fourier Transform of the damped sinusoidal wave $g(t)=e^{-t}sin(2\pi f_ct)u(t)$ where u(t) is the unit step function.

2. Relevant equations
$\omega =2\pi f$
$G(f)=\int ^{\infty}_{-\infty} g(t)e^{-j2\pi ft}dt$
$sin(\omega _ct)=\frac{e^{j\omega _ct}-e^{-j\omega _ct}}{2j}$

3. The attempt at a solution
$G(f)=\int ^{\infty}_{-\infty}e^{-t}sin(2\pi f_ct)u(t)e^{-j2\pi ft}dt$
$sin(2\pi f_ct)=\frac{e^{j2\pi f_ct}-e^{-j2\pi f_ct}}{2j}$
$G(f)=\frac{1}{2j}\int ^{\infty}_{0}e^{-t}(e^{j2\pi f_ct}-e^{-j2\pi f_ct})e^{-f2\pi ft}dt$
$G(f)=\frac{1}{2j}\int ^{\infty}_{0}e^{-j2\pi ft-t}(e^{j2\pi f_ct}-e^{-j2\pi f_ct})dt$
$G(f)=\frac{1}{2j}\int ^{\infty}_{0}e^{-j2\pi ft-t+j2\pi f_ct}-e^{-j2\pi ft-t-j2\pi f_ct}dt$
$G(f)=\frac{1}{2j}\int ^{\infty}_{0}e^{j2\pi t(f_c-f)-t}-e^{-j2\pi t(f+f_c)-t}dt$

I am not feeling confident on my algebra and I also feel like I should be able to simplify this more before I integrate.

2. Sep 8, 2016

BvU

If you're not allowed to cheat (e.g. google 'Table of Fourier Transform Pairs', or 'convolution theorem' ) then all you can do is grin and carry on ...
A possible simplification is to put ${\bf \alpha} = j2\pi (f_c-f)-1$ etc.

3. Sep 9, 2016

Captain1024

New approach:

$sin(\omega _ct)=\frac{e^{j\omega _ct}-e^{-j\omega _ct}}{2j}$
u(t) changes limits of integration (just as before): $G(f)=\int _0^{\infty}e^{-t}(\frac{e^{j\omega _ct}-e^{-j\omega _ct}}{2j})e^{-j\omega t}dt$
Pulling out constants and combining exponentials outside the parentheses: $=\frac{1}{2j}\int _0^{\infty}e^{-(j\omega t+t)}(e^{j\omega _ct}-e^{-j\omega _ct})dt$
Distributing the integrand: $=\frac{1}{2j}[(\int _0^{\infty}e^{t(-j\omega -1+j\omega _c)}dt)-(\int _0^{\infty}e^{-t(j\omega +1+j\omega _c)}dt)]$
Integrating: $=\frac{1}{2j}([\frac{1}{-j\omega -1+j\omega _c}e^{t(-j\omega -1+j\omega _c)}]_0^{\infty}-[\frac{-1}{j\omega +1+j\omega _c}e^{-t(j\omega +1+j\omega _c)}]_0^{\infty})$
$=-\frac{1}{2j}(\frac{1}{-j\omega -1+j\omega _c}+\frac{1}{j\omega +1+j\omega_c})$
$=(\frac{-1}{2j})(\frac{-j\omega +2+j\omega _c}{j(\omega +1+\omega _c)})$
$=\frac{-j\omega +2+j\omega _c}{2(\omega +1+\omega _c)}$
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Solutions manual gives the following answer: $G(f)=\frac{2\pi f_c}{1+4\pi ^2(f-f_c)^2}$
Which I believe is equivalent to: $G(f)=\frac{\omega _c}{1+\omega ^2-2\omega \omega _c+\omega _c^2}$
Any help would be appreciated.

4. Sep 10, 2016

BvU

I liked your first post and didn't find errors. Here's my bit. Don't have time to do this step by step, so I'll violate PF rules and work it through:
$$G(f)=\frac{1}{2j}\int ^{\infty}_{0}e^{j2\pi t(f_c-f)-t}-e^{-j2\pi t(f+f_c)-t}dt$$ was good. In short: your lack of confidence in your own work was unnecessary!
Use $$\int\limits_0^\infty e^{-\alpha x}dx = {1\over \alpha}$$to get $$G(f) ={1\over 2j}\left [{1\over 1-2\pi j(f_c - f) } - {1\over 1 + 2\pi j(f_c + f)}\right ]$$ so $$G(f) ={1\over 2j}\left [ \left ( 1 + 2\pi j(f_c + f) \right ) - \left ( 1-2\pi j(f_c - f) \right ) \over \left ( 1-2\pi j(f_c - f) \right ) \left ( 1 + 2\pi j(f_c + f) \right ) \right ]$$ and (dividing by $2j$), the numerator becomes $2\pi f_c$, which is good. For the denominator I get $$(2\pi f_c)^2 + (1+j2\pi f)^2$$ which is not in agreement with your book answer. But it does agree with the cheat link I gave (top page 3).