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Determine the Fourier series

  1. Sep 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Determine the Fourier series for the periodic function of period 2∏ defined by:



    -2 when (-∏ ) ∠ x ∠ (-∏/2)
    f(x)= 2 when ( -∏/2) ∠ x ∠ (∏ /2)
    -2 when (∏/2) ∠ x ∠ (∏)

    how to start i?. I have already drawn it but what next.
    thank you

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 9, 2013 #2
    The function runs from -π to +π. As yourself: is the function even in x or odd in x? If it is odd in x, then you only need to include the sin terms in the Fourier Series. if it is odd in x, then you only need to include the constant term and the cosine terms in the Fourier Series. What is the argument of sin or cosine going to be if the function runs from -π to +π?
     
  4. Sep 9, 2013 #3
    yes, the function runs from -∏ to ∏. its from (-∏ to ∏/2) ( -∏/2 to ∏/2) and ( ∏/2 to ∏). There is no period in it, it doesnt repeat. Than it has to be even in. even in is cosinus than yes?
     
  5. Sep 9, 2013 #4
    dont know what you mean by argument?
     
  6. Sep 9, 2013 #5

    rude man

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    Oh yes it does repeat. That's what 'periodic' means.
     
  7. Sep 9, 2013 #6

    rude man

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    f(x) the argument is x.
     
  8. Sep 9, 2013 #7

    rude man

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    Yes, it's an even function in the argument x.
     
  9. Sep 9, 2013 #8
    I was thinking that the terms in the Fourier Series should be Ancos(nx) where x runs from -π to +π, and n is the sequence of integers 1,2,3,... So, the argument is nx.
     
  10. Sep 9, 2013 #9

    rude man

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    You're right, the argument of each of the harmonics is nx, not x.

    I wanted to say that his function was an even function of x, x being the argument of f(x).

    Sorry Chet!
     
    Last edited: Sep 9, 2013
  11. Sep 10, 2013 #10
    No apology necessary RM. You're the man!

    Chet
     
  12. Sep 10, 2013 #11
    Is this function is odd or even, cause im confused. When it repeats itself its called odd when its not its even.
    This one repeats it self then its odd yes?

    And using co- efficient formulae with 2L= 2π, than L= π.
    after calculating
    [a][/0] =0
    [a][/ 1] = 0

    dont know how to calculate [/0] ????
     
  13. Sep 10, 2013 #12

    No. A function is odd if f(-x)=-f(x), and it is even if f(-x)=+f(x). Your function is even. So, it will be described by a cosine series. Do you have a formula for calculating the coefficients of a Fourier cosine series? If so, please show us, and show us in detail how you applied it to the case of arbitrary n.

    Chet
     
  14. Sep 10, 2013 #13
    ∞ n∏x n∏x
    Sf(x) = a0 +Ʃ [ an cos L + bn sin L ]
    n=1

    1 ∏
    a0= 2∏ ∫ f(x) dx =O
    -∏
    1 ∏
    a1= ∏ ∫ f(x) cos nx dx= 0,
    -∏

    what happenes if even times even = is it odd? I know when odd times odd = even.

    Im sorry i dont have a clue how to use latex, i tried so many times and still doesnt work. Its hard
     
  15. Sep 10, 2013 #14

    vanhees71

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    Do you know, how to calculate the Fourier coefficients? If so, just calculate them ;-)).
     
  16. Sep 10, 2013 #15
    dont know how to calculate b1
     
  17. Sep 10, 2013 #16
    Your result for a1 is incorrect. The integral in the right hand side is not zero. Also, the value of n in the cosine term should be n=1. Try doing the integral for n =1 again, and please be more careful.
     
  18. Sep 14, 2013 #17
    I dont know how to do that, could you please help me?
     
  19. Sep 14, 2013 #18
    [tex]\int_{-\pi}^{+\pi}{f(x)\cos{x}dx}=\int_{-\pi}^{-\pi/2}{(-2)\cos{x}dx}+\int_{-\pi/2}^{+\pi/2}{(2)\cos{x}dx}+\int_{\pi/2}^{\pi}{(-2)\cos{x}dx}[/tex]

    Are you able to evaluate the integrals on the right hand side of this equation?

    Chet
     
  20. Sep 15, 2013 #19
    It looks complicated, wouldnt know where to start!
     
  21. Sep 15, 2013 #20
    Before you can start working with Fourier Series, you will first have to learn how to integrate simple functions.

    Chet
     
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