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Parseval's Relation w/ Fourier Transform

  1. Oct 22, 2011 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img600.imageshack.us/img600/161/parcq.png [Broken]

    2. Relevant equations

    Parseval's Theorem using FT's for this is [itex]∫^{\infty}_{-\infty}[/itex] [itex]|f(t)|^{2}[/itex]dx = [itex]∫^{\infty}_{-\infty}[/itex] [itex]|\tilde{f}(w)|^{2}[/itex]dw

    3. The attempt at a solution

    From what I know, the fourier transform of f(t) = [itex]e^{-a|t|}[/itex] is [itex]\tilde{f} (w) = \frac{2a}{w^{2}+a^{2}}[/itex]

    So for my answer I would simply evaluate [itex]∫^{\infty}_{-\infty}[/itex] [itex]|f(t)|^{2}[/itex]dx for f(t) = [itex]e^{-a|t|}[/itex]

    However in the question there is no "2a" term on the top so I'm confused as how to proceed
    [itex]|\tilde{f}(w)|^{2}[/itex] does not equal [itex]\frac{dw}{w^{2}+a^{2}}[/itex] as given in the question where [itex]\tilde{f} (w) = \frac{2a}{w^{2}+a^{2}}[/itex]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 22, 2011 #2
    So, you need to integrate:
    [tex]
    \int_{-\infty}^{\infty}{e^{-2 a |t|} \, dt}
    [/tex]
    Parseval's Theorem guarantees that this integral (provided that [itex]f(t) = e^{-a |t|}[/itex] is truly the inverse Fourier transform of [itex](\omega^2 + a^2)^{-1}[/itex]) is the same as the integral you are supposed to calculate.

    Then, split the limits of integration so that you get rid of the absolute value sign and do the integrals. They are table.
     
    Last edited: Oct 22, 2011
  4. Oct 22, 2011 #3
    But the Fourier Transform of f(t) = [itex]e^{-a|t|}[/itex] is [itex]\tilde{f} (w) = \frac{2a}{w^{2}+2a^{2}}[/itex] and not [itex]\frac{1}{w^{2}+a^{2}}[/itex] so i don't understand how you can say [itex]∫^{\infty}_{-\infty}[/itex] [itex]|f(t)|^{2}[/itex]dx = [itex]∫^{\infty}_{-\infty}[/itex] [itex]|\tilde{f}(w)|^{2}[/itex]dw when [itex]|f(t)|^{2}[/itex] = [itex]e^{-2a|t|}[/itex] and [itex]|f(w)|^{2}[/itex] = [itex](\frac{2a}{w^{2}+a^{2}})^{2}[/itex]

    Basically it says evaluate [itex]∫^{\infty}_{-\infty}[/itex] [itex]\frac{dw}{w^{2}+a^{2}}[/itex] and not [itex]∫^{\infty}_{-\infty}[/itex] [itex]\frac{2a}{w^{2}+a^{2}}[/itex]dw
     
    Last edited: Oct 22, 2011
  5. Oct 22, 2011 #4
    Can anyone shed some light on this?
     
  6. Oct 22, 2011 #5

    vela

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    Hint:
    [tex]\frac{1}{\omega^2+a^2} = \frac{1}{2a}\left(\frac{2a}{\omega^2+a^2}\right)[/tex]
     
  7. Oct 22, 2011 #6
    Hmm :S , so is it just as simple as evaluating [itex]\frac{1}{2a}∫^{\infty}_{-\infty}[/itex] [itex]e^{-2a|t|}[/itex]dt ?
     
  8. Oct 22, 2011 #7
    no. Because you need [itex]|f(t)|^2[/itex] in the integral.
     
  9. Oct 22, 2011 #8
    oops

    [itex]∫^{\infty}_{-\infty}[/itex] [itex](\frac{e^{-a|t|}}{2a})^{2}[/itex]dt

    =[itex]\frac{1}{4a^{2}}∫^{\infty}_{-\infty}[/itex] [itex]e^{-2a|t|}[/itex]dt
    ?
     
  10. Oct 22, 2011 #9

    vela

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    Yes.
     
  11. Oct 22, 2011 #10
    BTW, methods of contour integration should give the result:
    [tex]
    2 \pi i \lim_{z \rightarrow i a}{\frac{d}{dz} \left( \frac{1}{(z + i a)^2}\right)}
    [/tex]

    and Wolfram Alpha gives the following value for the integral:
    http://www.wolframalpha.com/input/?i=Integrate[1/(x^2+a^2)^2,{x,-Infinity,Infinity}]"

    EDIT:
    Also, your Parseval's Theorem is wrong. It should be:
    [tex]
    \int_{-\infty}^{\infty}{\vert f(t) \vert^2 \, dt} = \int_{-\infty}^{\infty}{\vert \tilde{F}(\omega) \vert^2 \, \frac{d \omega}{2 \pi}}
    [/tex]
    (there's a factor of 2 pi missing in your definition), PROVIDED THAT, your fourier transform is:
    [tex]
    \tilde{F}(\omega) = \int_{-\infty}^{\infty}{f(t) \, e^{i \omega t} \, dt}
    [/tex]
    and the inverse Fourier transform is:
    [tex]
    f(t) = \int_{-\infty}^{\infty}{\tilde{F}(\omega) e^{-i \omega t} \, \frac{d \omega}{2 \pi}}
    [/tex]

    This convention is standard in Physics, but not in EE. See if your function and Fourier transform satisfy these.
     
    Last edited by a moderator: Apr 26, 2017
  12. Oct 22, 2011 #11
    Well I ended up getting [itex]\frac{1}{4a^{3}}[/itex] which is the same as the Wolfram Alpha solution minus the factor of 2[itex]\pi[/itex], so Ill include that in my FT's this time. Thanks for your help!
     
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