1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fourier series of sin2x

  1. Oct 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Fx= sin2x for -pi<x<0 and 0 for 0<x<pi. Compute the coefficients of the Fourier series.

    2. Relevant equations



    3. The attempt at a solution

    I found Ao=0 even though I came up with a Fourier cosine series. There must be something wrong. And is that possible that I come up with a fourier cosine series which converges to sin2x?? Before I began doing the problem, I thought An will be equal to zero and Bn will be zero for n≠2 which is intuitively correct. So, anyone has any idea?
     
  2. jcsd
  3. Oct 14, 2012 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    First thing to note is that this function is neither even nor odd. Therefore the Fourier series will contain both cosines and sines. You will not end up with a cosine series or a sine series.
    If you show your calculation for [itex]a_0[/itex], you will get feedback regarding what is wrong, if anything.
    Two things to note: first, your function is not simply sin(2x). It is only defined that way for [itex]-\pi < x < 0[/itex], not for [itex]0 < x < \pi[/itex]. Second, you will only end up with a cosine series if your function is even, which is not the case here.
    Your instinct is correct for the function sin(2x). However, once again, be careful: your function is NOT simply sin(2x).
     
  4. Oct 14, 2012 #3
    Well, for Ao;

    Ao=1/2pi ∫sin2x dx (from -pi to 0)

    When I do the integral, I get cos2x and cos0-cos2pi=0.

    For Bn;

    Bn=1/pi ∫sin2x sinnx dx (from -pi to 0)

    When I do the integral, again I get 0. So, the only coefficient i have is An. So, the fourier series of fx becomes fx~ƩAn cosnx which seems incorrect. I checked my algebra maybe 10 times but there is no mistake.
     
  5. Oct 14, 2012 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    OK, that looks right.

    That is not true for every [itex]n[/itex]. For example, if [itex]n = 2[/itex] then the integrand is [itex]\sin^2(2x)[/itex], which certainly does not integrate to zero.
     
  6. Oct 14, 2012 #5
    I did the integral for n=2 and I got Bn=1/2. I should have realized that since I had an n-2 in the denominator. Obviously what I found for n≠2 would not work for n=2. Thank you for that. However, for n=2, An=0 and since A0=0, I just have 1/2*sin2x which does not converge to sin2x. Is that because f(x)=sin2x for -pi<x<0 rather than -pi<x<pi or is the result incorrect?
     
  7. Oct 14, 2012 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Show us your calculations for ##a_n##. You shouldn't be getting them all zero. Putting n=2 aside, the other a values come out every other one 0 and non-zero.
     
  8. Oct 14, 2012 #7
    Well, yes you are right. An=4/π(n+2)(n-2) for n is odd and An=0 for n is even. However, upon plugging A0=0 A2=0, B2=1/2, and n=2 I have;

    F(x) ~ A0 + ƩAncosnx + ƩBnsinnx

    F(x) = 1/2 sin2x. But F(x) must be sin2x for -pi<x<0. I checked my algebra for An but could not find any mistake at all.
     
  9. Oct 14, 2012 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    How do you know you have a mistake? Have you plotted the graph of your series with, say, 10 non-zero terms to see if it is beginning to look like F(x)?
     
  10. Oct 15, 2012 #9

    Ibix

    User Avatar
    Science Advisor

    A slightly belated observation - A0 is the average value of F(x). Since the left-hand half of your function is a complete cycle of a sine wave and the right hand half is flat zero, the average value has to be zero - as your maths is also showing you.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fourier series of sin2x
  1. Fourier Series (Replies: 1)

  2. Fourier series (Replies: 10)

  3. Fourier Series (Replies: 1)

  4. Fourier Series (Replies: 6)

Loading...