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Math Help - Fourier Series Approximation

  1. #1
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    Fourier Series Approximation

    How do I find the fourier series approximation to the even extension of g(t), where g(t) is given as shown below using the fact that :

    sin(2pir/3)=(-1)^r+1sin(pir/3)
    Attached Thumbnails Attached Thumbnails Fourier Series Approximation-gt.bmp  
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  2. #2
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    Hi

    I am struggling with this too. I can't make sense of the writing down the odd and even extensions. Have you got that far yet?
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  3. #3
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    I was thinking that perhaps since the period is 6 that we need to change the intervals so they are equal to something like 0<t<1/2

    Alternatively, it may have something to do with the range -3<t<3

    In which case the even extensions will be those that have positive y and the odd will have negative y.

    Its really difficult - hope someone on this forum can help out with this one.
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  4. #4
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    I think you are right about the range being -3<t<3, as in part a of the question the function f(t)=f(t + 2pi)

    2pi was the period of the function, so if you look at the formulas for even and odd extension they use the letter L and  2L is defined as the period.

    If L is equal to 3 as the fundamental interval is [-L,L] then 2x3 equals 6 giving us the desired period...... i think.
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  5. #5
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    That makes sense - though when i substitute this into formula

    F(t)=A_0 + \sum_{n=0}^{\infty}A_rcos(2r\pi/T)

    and then try and work our A_0    and    A_n

    I still can see how we should use

    sin(2r\pi/3)=-1^r+1sin(r\pi/3)
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  6. #6
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    Am I right to assume that the use of sin refers to an odd extension?

    Is g(t) an odd function?

    I'm so confused!!
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  7. #7
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    The sin function is odd because of the shape of the curve (i.e. not symmetrical about y axis). Whereas the cos function is even since if can be reflected in y axis. See bottom of page 9 of unit 21 for explanation
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  8. #8
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    have you completed b)i) yet?
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  9. #9
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    Nope - still waiting for some help on that. Have you got any ideas
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  10. #10
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    Yeah

    I think for part i) it is only asking you to write the extensions in function form.

    So I have something like

    G(odd) (t) =
    -1 (-3<= t <-2)
    0 (-2<= t < -1)
    1 (-1<= t < 0)

    then

    G(even) (t) =
    1 (-3<= t <-2)
    0 (-2<= t < -1)
    -1 (-1<= t < 0)

    Then draw each graph...

    Does that make some sense? (sorry i'm still not up to speed on Latex)
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  11. #11
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    That looks good too me - thanks for starting me off on part b
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  12. #12
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    No worries...

    How have you got on for question 2?
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  13. #13
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    Just had a thought - wouldnt the even extensions be positive i.e. range from 0<=t<3
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  14. #14
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    I don't think so. If you look at the odd and even functions on pg 61 of handbook - it shows the range to be -L<t<0
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  15. #15
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    For Q21i have a look at page 27. The answer is virtually the same

    For part ii - I think to have non-trivial solutions it needs to be such that u<0. i.e. negative. There is something in the text about this but I havent got it in from of me
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