1. Not finding help here? Sign up for a free 30min 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!

Evaluate the Fourier Transform of a Damped Sinusoidal Wave

  1. Sep 8, 2016 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant 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]

    3. 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.
     
  2. jcsd
  3. Sep 8, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  4. Sep 9, 2016 #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.
     
  5. Sep 10, 2016 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Evaluate the Fourier Transform of a Damped Sinusoidal Wave
  1. Fourier Transform (Replies: 2)

  2. Fourier Transforms (Replies: 5)

  3. Fourier Sinusoids (Replies: 13)

Loading...