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Convolution theorem; integration

  1. Jan 10, 2014 #1
    1. The problem statement, all variables and given/known data

    Part(a)/(b)/(c): Fourier transform the following:

    Part(d): Prove the convolution theorem

    Part (e): Find total displacement


    2. Relevant equations

    3. The attempt at a solution

    [tex]f = \frac {1}{sqrt{2\pi}} \int_{-\infty}{\infty} F e^{-iωt} dω[/tex]

    [tex]\frac {1}{\sqrt{2\pi}} \int_{-T}{T}[/tex]
    [tex]= \frac{1}{iω\sqrt{2\pi}} [e^{iωT} - e^{-iωT}] [/tex]
    [tex] = \sqrt{\frac{2}{\pi}} \frac {sin (ωt)}{ω}[/tex]

    Part (b)
    [tex]F_2 = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{0}e^{-at}e^{iωt} dt + \frac{1}{\sqrt{2\pi}} \int_{0}^{\infty}e^{-at}e^{iωt} dt [/tex]
    [tex] = -\frac{2a}{\sqrt{2\pi}(a^2 + ω^2)} [/tex]


    Using result from (b):
    [tex] F_2 = -\frac{2a}{\sqrt{2\pi}(a^2 + ω^2)} [/tex]

    Let a = 1,

    [tex]e^{-|t|} = -\frac{1}{2\pi} \int_{-\infty}^{\infty} \frac {2}{1+ω^2} e^{-iωt} dω [/tex]

    [tex] t→-t, [/tex]
    [tex]e^{-|-t|} = \frac {1}{\pi} \int_{-\infty}^{\infty} \frac {1}{1+ω^2} e^{iωt} dω [/tex]

    [tex] t→ω,[/tex]
    [tex] e^{-|ω|} = \frac {1}{\pi} \int_{-\infty}^{\infty} \frac {1}{1+t^2} e^{iωt} dt [/tex]


    [tex] V_{(ω)} = \sqrt{2\pi} F_{(ω)}G_{(ω)} = \frac {2}{\pi} \frac {a sin (ωT)}{a^2 + ω^2}[/tex]

    [tex] sin (ωT) = \frac {1}{2i} (e^{iωT} - e^{-iωT}) ,[/tex]

    Taking inverse fourier transform of Vω:

    [tex] V_{(t)} = \frac {1}{i\pi}(\frac{a}{\sqrt{2\pi}}) \int_{-\infty}^{\infty} \frac {1}{a^2 + ω^2} e^{iω(T-t)} dω + \frac {1}{i\pi}(\frac{a}{\sqrt{2\pi}}) \int_{-\infty}^{\infty} \frac {1}{a^2 + ω^2} e^{-iω(t+T)} dω [/tex]

    My velocity is complex! Strange...

    This is related to inverse fourier transform of F2, f2 but as a function of (t-T) and (t+T):

    [tex]i(\frac{1}{\sqrt{2\pi}}) f_2(t-T) - i(\frac{1}{\sqrt{2\pi}}) f_2(t+T) [/tex]
    [tex]= i\frac{1}{\sqrt{2\pi}}[e^{-a|t-T|} - e^{-a|t+T|}] [/tex]

    To find the displacement, simply integrate the above with respect to t from -∞ to ∞.

    [tex] s = i\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-a|t-T|} dt - i\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-a|t+T|} dt [/tex]
    [tex] = i\sqrt{\frac{2}{\pi a}}(e^{-aT} - e^{aT}) [/tex]

    Again complex displacement.. (and negative)
    Last edited: Jan 10, 2014
  2. jcsd
  3. Jan 11, 2014 #2


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    Homework Helper

    That's some nice work on parts a to c. at the end of part a, it should be ##\tau## (not ##t##), but I'm guessing that is just a miss-type. For part e, there is a mistake in this equation:
    [tex]V_{(ω)} = \sqrt{2\pi} F_{(ω)}G_{(ω)} = \frac {2}{\pi} \frac {a sin (ωT)}{a^2 + ω^2}[/tex]
    there should be another ##\omega## from ##f_1## (which you have got in part a). And it looks like you've missed out the ##\sqrt{2\pi}## which is on the left hand side of the equation.

    Also, there is a simpler way to do part e. You don't need to do the inverse Fourier transform. hint: you have got ##\tilde{V}(\omega )## and you want
    [tex]\int_{-\infty}^{\infty}V(t) dt [/tex]
    So, thinking about the definition of the Fourier transform, what is the easiest way to get this?
  4. Jan 11, 2014 #3
    [tex]\tilde{V}(ω) = \int_{-\infty}^{\infty} V_{(t)} e^{iωt} dt [/tex]

    I've been thinking of a way to 'remove' the exp[iωt] from the integral and everything would be perfect. The only way is if ω = 0. But if ω = 0, LHS = 0..
  5. Jan 12, 2014 #4


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    yes, use ##\omega=0##. The left hand side is not zero, it is ##\tilde{V}(0)## this isn't zero. Even if it was zero, that would still be the correct answer, since the right hand side is exactly what you are looking for.
  6. Jan 12, 2014 #5
    The 0 comes about from the sin (ωT) function and part (a) looks right to me..
  7. Jan 12, 2014 #6


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    you have got the correct answer to part (a), but part (a) evaluated at ##\omega=0## is not equal to zero.

    edit: you have
    [tex]\sqrt{\frac{2}{\pi}} \frac {sin (ω\tau)}{ω}[/tex]
    but this is not zero for ##\omega=0##
  8. Jan 13, 2014 #7
    Ah ok, I get it, thanks so much!
  9. Jan 13, 2014 #8


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    hehe, no problem!
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