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Homework Help: Fourier Series of sin(6t)

  1. Sep 12, 2010 #1
    1. The problem statement, all variables and given/known data
    f(t)=sin(|6t|), −pi<t<pi

    with f(t) = f(t+2pi)

    2. Relevant equations
    Show that the Fourier series for f(t) can be written as (24/pi) time the sum, from n=0 to infinity, of ( 1/( 36 - (2k+1)^2 ) )cos(2k+1)t.

    3. The attempt at a solution

    I have an answer of a0 being 0, and a guess that an is also zero, however i cannot onfirm this since i don't know how to do the integral of [sin(nt)6cos(6t)].

    I run into a similar problem with bn, getting stuck at the integral of [sin(6t)sin(nt)]

    any help would be much appreciated
     
  2. jcsd
  3. Sep 12, 2010 #2
    Hi integral of [sin(nt)6cos(6t)] will be zero because integral [sinA.cosB] is zero over a period 2pi.
    Note use the trigonometric formula sinAcosB=sin(A+B)-sin(A-B)
     
  4. Sep 12, 2010 #3
    Interesting problem hmm, can you show how you got a0 and explain hoe you got upto where you are?

    From there I think some trigonometric identities will be useful. But I will put pen to paper once I see where and how you got there!

    Seems like a tricky question!
     
  5. Sep 12, 2010 #4
    well a0 is just a straightforward integration of sin(6t) between -pi and pi, which comes out as -(1/6)cos(6t), which comes out as (-1/6) - (-1/6), which = 0.


    for an, i started with the equation (1/L)int[ f(t)*cos(nt)].dt, for n>=1. (L is given as pi, and the region is -pi < t < pi.
    i am currently working through this but its looking ugly after the sin(nt)cos(6t) integral.


    for bn, i'm starting with the same equation as for an, except with a sin term instead of the cos term.
    this one i am stuck at the integral of sin(6t)sin(nt)
     
  6. Sep 12, 2010 #5
    For an just substituting in I get:

    (1/L) int [ sin(6t) * cos(nt) ] .dt

    Not sure if you started with this....

    If you did I am not sure how this led to:

    the sin(nt)cos(6t) integral.
     
  7. Sep 12, 2010 #6
    yes, that is what i started with, i wasn't too sure how to proceed with it so i used integration by parts (u=sin(6t), dv/dt = cos(nt) ) and that's how it happened.
     
  8. Sep 12, 2010 #7
    Usually for integration of something like sinxcosx i use the identity:

    sin2x = 2sinxcosx

    But not sure how that applies in this case. I am very curious, hopefully someone can answer!
     
  9. Sep 12, 2010 #8

    vela

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    You're finding a0 for sin 6t, not sin |6t|, which is what the problem asked for.
    You can probably get it to work out with integration by parts, but it's simpler using the product-to-sum trig identities.

    http://en.wikipedia.org/wiki/List_o...#Product-to-sum_and_sum-to-product_identities
     
  10. Sep 12, 2010 #9
    thanks for that vela, i looked at the identities prior to you post but I didnt know how to apply them to this case. Would you have any pointers for which one it is? :)
     
  11. Sep 12, 2010 #10
    You're finding a0 for sin 6t, not sin |6t|, which is what the problem asked for.

    For sin |6t| we ignore the - values and we are left with:

    1/2pi [-1/6] = -1/12pi

    Hope that is right!
     
  12. Sep 12, 2010 #11
    You can verify these yourself:
    [tex]
    \int_{-L}^{L} \cos \frac{n\pi x}{L} \cos \frac{m\pi x}{L} = \left\{\begin{array}{cc}0,&\mbox{if } n \neq m\\L, & \mbox {if } n=m>0\\2L, & \mbox {if } n=m=0 \end{array}\right [/tex]
    [tex]
    \int_{-L}^{L} \sin \frac{n\pi x}{L} \sin \frac{m\pi x}{L} = \left\{\begin{array}{cc}0,&\mbox{if } n \neq m\\L, & \mbox {if } n=m>0\end{array}\right [/tex]
    [tex]
    \int_{-L}^{L} \sin \frac{n\pi x}{L} \cos \frac{m\pi x}{L} = 0[/tex]
    (m and n nonnegative integers)
     
  13. Sep 12, 2010 #12
    how does sin(|6t|) change the answer? which are the values you discount. After doing all the integration i'm left with the sum of 2 negative answers
     
  14. Sep 12, 2010 #13

    vela

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    Which one looks like (sin 6t)(cos nt)?
     
  15. Sep 12, 2010 #14

    vela

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    No, it's not. sin |6t| is non-negative, so the area under the curve from -pi to pi is positive.

    Note f(t)=sin |6t| is an even function, so

    [tex]\int_{-\pi}^\pi \sin |6t| \cos nt \, dt = 2\int_0^\pi \sin |6t| \cos nt \, dt = 2\int_0^\pi \sin 6t \cos nt \, dt[/tex]

    You can get rid of the absolute value in the integral over [0,pi] because t will be positive.
     
  16. Sep 12, 2010 #15
    Thanks for that, but from here I am struggling to find the right identity to simplify equation!

    This is not my problem but I cant help be intrigued on how to do it!
     
  17. Sep 12, 2010 #16
    yes i've tried a coupld of times now and i keep ending up at this horrendous thing :

    (2/pi){cos(pi(n+6))[-1/(n(n+6)) - 1/(n(n-6)) ] }


    ummm, yeh, i'm lost
     
  18. Sep 12, 2010 #17

    vela

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    You're not integrating correctly. Show your work.
     
  19. Sep 12, 2010 #18
    Yeah that didnt seem right. The problem I'm having is that very first step! LOL
     
  20. Sep 12, 2010 #19
    i have redone my a0, using a restricted domain of 0 to pi/6, then multiplying it to get the total area under the curve for the actual range.
    This came out to be 8/pi.

    correct?
     
  21. Sep 12, 2010 #20

    vela

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    Oops, I led you astray. The function isn't non-negative over the entire interval, and the constant term should indeed be a0=0.

    It would really help if you would show your work instead of just posting an answer and asking if it's correct. I know you got a0=0 earlier, but I could also tell from what else you had written, you got that result by accident.
     
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