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Homework Help: Deriving of the constants in Fourier Analysis

  1. Aug 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Wow LaTex fail.... anyone know how i make this look like not ****? I had it looking all pretty on http://www.codecogs.com/latex/eqneditor.php and it gave me these codes as my latex markup but it didn't come out right... why do they look like source code and not the pretty equations they are supposed to?

    Fourier Analysis equation:
    [tex]$\displaystyle v(t)=a_0 + \sum_{n=1}^N a_ncos(nwt) + \sum_{m=1}^M a_msin(mwt) [/tex]
    [tex]\displaystyle a_n= \frac{2}{T}\int\limits_{\frac{-T}{2}}^{\frac{T}{2}}v\(t)cos(nwt)\, dt [/tex]
    [tex]\displaystyle b_n= \frac{2}{T}\int\limits_{\frac{-T}{2}}^{\frac{T}{2}}v\(t)sin(mwt)\, dt[/tex]

    To show that the coefficients a and b are given by the equations above, we multiply v(t) by cos(2∏kt/T) and integrate with respect to time from t = -T/2 to t = T/2

    [tex]\displaystyle \int\limits_{\frac{-T}{2}}^{\frac{T}{2}}\ ((a_0+\sum_{n=1}^N a_n cos(\frac{2\pi nt}{T}) +\sum_{m=1}^M b_m sin(\frac{2\pi mt}{T})) [/tex]

    which will give us three types of integrals
    [tex]\displaystyle a_0\int\limits_{\frac{-T}{2}}^{\frac{T}{2}}cos\frac{2\pi kt}{T}\,\ dt=0[/tex]
    [tex]\displaystyle \sum_{m=1}^M b_m\int\limits_{\frac{-T}{2}}^{\frac{T}{2}}sin\frac{2\pi mt}{T}cos\frac{2\pi kt}{T}dt=0[/tex]
    [tex]\displaystyle \sum_{n=1}^N a_n\int\limits_{\frac{-T}{2}}^{\frac{T}{2}}cos\frac{2\pi nt}{T}cos\frac{2\pi kt}{T}dt= T/2 (n=k)[/tex]

    My question is why do the first two of these three integrals equal zero?
    (first time using LaTex holy crap took forever
    Last edited: Aug 24, 2013
  2. jcsd
  3. Aug 24, 2013 #2
    Assuming I'm deciphering the equations correctly, then the first integrals equal zero because the integral will be a sin function, and between the limits you will always have integer multiples of pi for any k (sin k*pi = sin(-k*pi) = 0)
    The second integral is zero because it is an odd function integrated between symmetric limits. (an odd function multiplied by an even function is an odd function).
    Last edited: Aug 24, 2013
  4. Aug 24, 2013 #3


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    Use $$ instead of $. More here:https://www.physicsforums.com/showpost.php?p=3977517&postcount=3
  5. Aug 24, 2013 #4
    So sin is an odd function then? Why is cos * sin an odd function? I can't find any good explanations on google.

    P.S thanks Tsny
  6. Aug 24, 2013 #5
    It is kinda simple, just like multiplying numbers, an odd times an even number results in an odd number.

    If you want a more rigorous definition then you can go back to the definition of odd and even. Assume g(x) is odd and f(x) is even.

    Then the product h(x)= f(x)*g(x). To check the parity of h(x) we find h(-x)= g(-x)*f(-x).

    Remembering what it means to be an odd or even function we get g(-x)*f(-x)= -g(x)*f(x)= -h(x).

    Leaving you with h(x)= -h(-x); otherwise known as an odd function.
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