Evaluate the Fourier Transform of a Damped Sinusoidal Wave

  • #1

Homework Statement


Evaluate the Fourier Transform of the damped sinusoidal wave [itex]g(t)=e^{-t}sin(2\pi f_ct)u(t)[/itex] where u(t) is the unit step function.

Homework Equations


[itex]\omega =2\pi f[/itex]
[itex]G(f)=\int ^{\infty}_{-\infty} g(t)e^{-j2\pi ft}dt[/itex]
[itex]sin(\omega _ct)=\frac{e^{j\omega _ct}-e^{-j\omega _ct}}{2j}[/itex]

The Attempt at a Solution


[itex]G(f)=\int ^{\infty}_{-\infty}e^{-t}sin(2\pi f_ct)u(t)e^{-j2\pi ft}dt[/itex]
[itex]sin(2\pi f_ct)=\frac{e^{j2\pi f_ct}-e^{-j2\pi f_ct}}{2j}[/itex]
[itex]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[/itex]
[itex]G(f)=\frac{1}{2j}\int ^{\infty}_{0}e^{-j2\pi ft-t}(e^{j2\pi f_ct}-e^{-j2\pi f_ct})dt[/itex]
[itex]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[/itex]
[itex]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[/itex]

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

Answers and Replies

  • #2
BvU
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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
New approach:

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

Your book answer must be wrong: it's a real
 

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