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Math Help - Fourier transform - odd funktion

  1. #1
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    Fourier transform - odd funktion

    I'm transforming f(x) = 1-x trying to do so in an even and odd fashion... I think I get it right for the even, but I get stuck while tinkering with the odd one.

    The period is 2, odd distribution.

    \dispalystyle b_n = \frac{2}{T} \int_0^2 (1-t) \sin (n \Omega t) dt ; \Omega = \frac{2 \pi}{T} = \pi &  T = 2

    \dispalystyle b_n = \int_0^2 (1-t) \sin (n \pi t) dt = [integrate by parts] = [1-1] - \int_0^2 \cos (n \pi t) dt

    Here I get stuck because  sin(2\pi n ) = sin (0) = 0.

    If anyone could point out what I'm overseeing..?
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  2. #2
    Grand Panjandrum
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    Re: Fourier transform - odd funktion

    Quote Originally Posted by liquidFuzz View Post
    I'm transforming f(x) = 1-x trying to do so in an even and odd fashion... I think I get it right for the even, but I get stuck while tinkering with the odd one.

    The period is 2, odd distribution.

    \dispalystyle b_n = \frac{2}{T} \int_0^2 (1-t) \sin (n \Omega t) dt ; \Omega = \frac{2 \pi}{T} = \pi &  T = 2

    \dispalystyle b_n = \int_0^2 (1-t) \sin (n \pi t) dt = [integrate by parts] = [1-1] - \int_0^2 \cos (n \pi t) dt

    Here I get stuck because  sin(2\pi n ) = sin (0) = 0.

    If anyone could point out what I'm overseeing..?
    Do you integration by parts again more carefully.

    CB
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  3. #3
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    Re: Fourier transform - odd funktion

    No, I can't nail it...

    What if I change the interval over wish integrate to [-1,0] + [0,1]. Maybe that would make it easier, but will my series convert to the odd distribution of 1-t?

    Edit: Oh crap, I just realised something. I think my series could be right. My conclusion on the contrary... I was so preoccupied with the fact that sin (n \pi) = 0 that I sorta forgot the variable t.

    Does this compute..?
     \sum_{n=1}^{\infinity}  \frac{2 \sin (n \pi t)}{n\pi}
    Last edited by liquidFuzz; November 11th 2011 at 12:51 PM.
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  4. #4
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    Re: Fourier transform - odd funktion

    Quote Originally Posted by liquidFuzz View Post
    No, I can't nail it...

    What if I change the interval over wish integrate to [-1,0] + [0,1]. Maybe that would make it easier, but will my series convert to the odd distribution of 1-t?

    Edit: Oh crap, I just realised something. I think my series could be right. My conclusion on the contrary... I was so preoccupied with the fact that sin (n \pi) = 0 that I sorta forgot the variable t.

    Does this compute..?
     \sum_{n=1}^{\infinity}  \frac{2 \sin (n \pi t)}{n\pi}
    If I recall correctly yes

    CB
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  5. #5
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    Re: Fourier transform - odd funktion

    Quote Originally Posted by CaptainBlack View Post
    If I recall correctly yes

    CB
    I plotted the series in MatLab and got a result that points in an other direction. :-(
    Here's the plot I got and the MatLab code I wrote.
    MatLab
    Code:
    N = 20;
    x = (-50:50)/20;   %interval : [-2.5,2.5]
    f = ones(1,101);
    %Odd fourier, no a_0 and a_i included.
    for i = 1:1:N       %from 1 to N 
        f = f + (2 * sin(pi*i*x))  /  (i * pi);  %sum all terms over N iterations.
    end
    plot(x,f)
    The function I was aiming for is 1-t and it should hold the points [0,1] and [2,-1] and then jump to next 'slope'. Eh, right?

    I guess my question is, is it the series or is the MatLab tinkering that's wrong..?
    Attached Thumbnails Attached Thumbnails Fourier transform - odd funktion-fourier_test.png  
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  6. #6
    Grand Panjandrum
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    Re: Fourier transform - odd funktion

    Quote Originally Posted by liquidFuzz View Post
    I plotted the series in MatLab and got a result that points in an other direction. :-(
    Here's the plot I got and the MatLab code I wrote.
    MatLab
    Code:
    N = 20;
    x = (-50:50)/20;   %interval : [-2.5,2.5]
    f = ones(1,101);
    %Odd fourier, no a_0 and a_i included.
    for i = 1:1:N       %from 1 to N 
        f = f + (2 * sin(pi*i*x))  /  (i * pi);  %sum all terms over N iterations.
    end
    plot(x,f)
    The function I was aiming for is 1-t and it should hold the points [0,1] and [2,-1] and then jump to next 'slope'. Eh, right?

    I guess my question is, is it the series or is the MatLab tinkering that's wrong..?
    Why did you initialise f to one?

    Try:

    f = zeros(1,101)

    CB
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  7. #7
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    Re: Fourier transform - odd funktion

    Eh, good question..?

    Thanks for your help and patience dude!
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