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Fourier evaluation of Series HELP

  1. May 26, 2013 #1
    Fourier evaluation of sum HELP

    1. The problem statement, all variables and given/known data

    Consider the signal:
    f(t) = |sint|, -pi/2 < t < pi/2 where f(t) = f(t+pi)

    2. Relevant equations

    Fourier.

    3. The attempt at a solution

    I determined the General Fourier Series representation for f(t) below:

    2/pi +4/pi + Ʃ(-1/(4n^2))*cos(2nt)

    The question then asks to evaluate the sum: (using the general Fourier representation that was just solved)

    Ʃ (-1^n)/((4n^2)-1)

    I don't know how to approach or even figure this out.

    (all sigmas are from n=1 to infinity).

    Any help would be much appreciated. I've attached an image to help visualize a bit better if needed. IMAGE
     
    Last edited: May 26, 2013
  2. jcsd
  3. May 26, 2013 #2

    Simon Bridge

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  4. May 26, 2013 #3

    mfb

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    Evaluate the series where? And where does the -1 in the denominator come from?
    $$\frac{1}{4n^2-1} = \frac{1}{(2n-1)(2n+1)} = \dots$$
     
  5. May 26, 2013 #4
    Not sure. The coursework doesn't mention using representations to solve series'. That's why I'm a little confused.

    I reworded it all to eliminate confusion (hopefully). Thanks for responding.
     
  6. May 26, 2013 #5

    Simon Bridge

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    Have I got this right?
    You have to evaluate the sum: $$\sum_{n=0}^\infty\frac{(-1)^n}{4n^2-1}$$

    using: $$|\sin t| = \frac{6}{\pi}-\sum_{n=0}^\infty \frac{\cos 2nt}{4n^2}$$
     
  7. May 26, 2013 #6
    You have got that right, granted I solved the general representation of the Series correctly.

    One thing is that n=1, you have n=0 there.
     
  8. May 26, 2013 #7
    Double check you cos series.

    Hint: This is close but wrong.


    The term inside the summation should look like

    [itex] \sum \frac{-1}{4n^2-1} cos \alpha_n t [/itex]

    After you figure out what the correct series is, you'll have to figure out how to relate [itex] \sum \frac{-1}{4n^2-1} cos \alpha_n t [/itex] and [itex] \sum \frac{(-1)^n}{4n^2-1} [/itex]...

    At what value of t are the two sums equal?
    What is |sin t| at this t?
     
  9. May 26, 2013 #8
    I can't seem to pinpoint where the error is. I'm sure the cos reads cos(2t).

    I'm probably making the simplest of errors.

    I'm still unsure of what I would do with the next sum though.

    Are we talking pi/4 for t where they're equal or?
     
  10. May 26, 2013 #9

    vela

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    If you show us your calculations for the coefficients ##a_n##, we can help find where you went wrong.
     
  11. May 26, 2013 #10
    Yeah don't worry about it.

    I think posting here has just confused me more with everyone saying differing things.


    Ill ask a friend or something. Thank you for the effort though.
     
  12. May 27, 2013 #11
    A question, does anyone here actually know how to solve it? Just out of curiosity guys and girls.
     
  13. May 27, 2013 #12

    vela

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    Yes.
     
  14. May 27, 2013 #13
    i've noticed you haven't posted in the thread.

    any ideas? i've asked around from a few guys I know and I can't seem to figure this out.
     
  15. May 27, 2013 #14

    vela

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    The Wolfman essentially told you how to do it. The first thing you need to do is get the correct Fourier series. Until you do that, you can't really go on.

    I did post earlier, suggesting you show your calculations to find the Fourier coefficients, but you didn't seem interested in doing that.
     
    Last edited: May 27, 2013
  16. May 27, 2013 #15

    Simon Bridge

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    We've all solved it already ... but the point is to get you to solve it.

    When you throw a question open it is not terribly unusual to get a variety of answers - part of the skill is to sort through them. However, in this case, the replies are far from contradictory or, even, varied. We all agree that you have the incorrect Fourier series - granted it's close... we all agree you need to fix that before you can continue.

    You seem to be having trouble finding where you went wrong.

    Is there some reason you don't want to show us your working?
    Unless you show us your working - we cannot help you.
     
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