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Fourier Analysis - sawtooth

  1. Sep 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Sawtooth signal with To = 1, at T=0, x = 0, at T=1, x =1

    verify:
    [itex]

    a_{k} = \left\{\begin{matrix}
    \frac{1}{2}, for k=0; & \\\frac{j}{2\pi k}, for k \neq 0;
    &
    \end{matrix}\right.

    [/itex]


    2. Relevant equations

    [itex]\frac{1}{T_{0}} \int_{0}^{T_{0}} te^{-j(2\pi/T_{0}))kt}dt[/itex]

    3. The attempt at a solution

    for k = 0

    [itex] a_{0} = \int_{0}^{1} t dt[/itex]

    [itex]a_{0} = \frac{1}{2} t^{2} [/itex] from 0 to 1 = 1/2

    for k != 0

    [itex]\int_{0}^{1} te^{-j(2\pi) kt}dt[/itex]

    u = t
    du = dt
    dv = [itex]e^(-j2\pi kt)[/itex]

    [itex] v = \frac{-1}{j2\pi k}e^{-j2\pi kt} [/itex]


    [itex] t * \frac{-1}{j2\pi k}e^{-j2\pi kt} - \int \frac{-1}{j2\pi k}e^{-j2\pi kt} dt[/itex]

    [itex] t * \frac{-1}{j2\pi k}e^{-j2\pi kt} - \frac{e^{-j2\pi kt}}{4\pi^2k^2} [/itex]

    -1/j = j

    [itex]t * \frac{j}{2\pi k}e^{-j2\pi kt} - \frac{e^{-j2\pi kt}}{4\pi^2k^2}[/itex]

    [itex]e^{-j2\pi kt} (t \frac{j}{2\pi k} - \frac{1}{4\pi^2 k^2})[/itex]

    getting close but not seeing where to go from here.
     
    Last edited: Sep 30, 2013
  2. jcsd
  3. Oct 1, 2013 #2

    Päällikkö

    User Avatar
    Homework Helper

    Check the integration by parts rules: You seem to have forgotten to evaluate the first part at the boundaries (in particular, if you integrate over t from 0 to 1, there is no way t should remain in the final expression)
    [itex] \int_a^b u(x)v'(x)\,dx = \left[u(x)v(x)\right]_a^b - \int_a^b u'(x)v(x)\,dx[/itex],
    first term on the right hand side.
     
  4. Oct 1, 2013 #3


    [itex] \frac{je^{-j2\pi k} }{2\pi k} - \frac{e^{-j2\pi k} }{4\pi^2 k^2} - \frac{1}{4\pi^2 k^2}[/itex]
     
  5. Oct 1, 2013 #4

    Päällikkö

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    Homework Helper

    You seem to have a sign error. Also, remember that k is an integer (a periodic function is mapped into a series in fourier space), and you should be able to arrive at the result.
     
  6. Oct 1, 2013 #5
    Okay, see the sign error but still not seeing how that is going to get
    the other terms to fall out leaving just j/(2pik).
     
  7. Oct 1, 2013 #6

    Päällikkö

    User Avatar
    Homework Helper

    k is an integer. What is [itex]\exp(-j2\pi k)[/itex] for k integer?
     
  8. Oct 1, 2013 #7
    Thank you Paallikko, I did not have that one in my notes.
     
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