1. Limited time only! Sign up for a free 30min personal 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!

Obtaining the fourier series

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Please take into account I'm not very good at maths so I would just like to make sure that what I am doing so far is correct

    Obtain the fourier transform for:

    f(x) = x(2∏-x) 0<x<2∏ f(x)=f(x+2∏)


    2. Relevant equations

    f(x) = 1/2 a0 + Ʃ {an cosnx + bn sinnx} = ao/2 + a1cosx +a2cos2x + b1sinx+b2sin2x +...

    3. The attempt at a solution

    an = 1/x ∫x(2∏-x)cos(nωt) dt

    I've attached the rest of my working out below:
     

    Attached Files:

  2. jcsd
  3. Jan 21, 2012 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Your formula for an is incorrect. You should have ##1/\pi## in front, not 1/x. What is ω equal to in this case? What about the bn coefficients?

    I'm not sure exactly what you're doing with the integration. Why did you break it up into two pieces?
    The second piece is definitely wrong. There's no cosine in it, for one thing.
     
    Last edited: Jan 21, 2012
  4. Jan 21, 2012 #3
    Ah that makes more sense!

    The question itself is:

    Obtain the fourier series for:

    f(x)=x(2∏-x) 0<x<2∏ f(x)=f(x+2∏)

    I split the integration because there was an f(x) and an f(x) so thought you were supposed to. There was also no mention of what the values for ω is and I hadn't gotten to the bn bit yet.

    In the examples I have from my lecturer he has them with cosines in so I figured that this one would be the same. Please note I'm not very good at fourier series!
     
    Last edited: Jan 21, 2012
  5. Jan 21, 2012 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Where did you get the formulas for an and bn? The definition of ##\omega## should have been given there. Show us the formula for an exactly as given (before you plug in f). Let's get all these little details straight before you dive into the problem.
     
  6. Jan 21, 2012 #5
    The formulas I got the formulas for an and bn from my lecture notes. The exact formulas are written as:

    Periodic functions as Fourier series

    f(x)= [itex]\frac{1}{2}[/itex][itex]a{0}[/itex]+[itex]Ʃ^{∞}_{n=1}[/itex]{an cosnx + bn sinnx=[itex]\frac{a0}{2}[/itex] + a1 cosx + a2 cos2x+....+b1 sinx + b2 sin2x +...

    a0=[itex]\frac{2}{T}[/itex][itex]\int[/itex][itex]^{T/2}_{-T/2}[/itex] f(x)dx ie. 2xmean value of f(x) over period

    an= [itex]\frac{2}{T}[/itex][itex]\int[/itex][itex]^{T/2}_{-T/2}[/itex] f(x) * cosnx dx ie. 2x mean value of f(x)*cosnx over period

    bn= [itex]\frac{2}{T}[/itex][itex]\int[/itex][itex]^{T/2}_{-T/2}[/itex] f(x) * sinnx dx ie. 2x mean value of f(x)*cosnx over period


    I have also attached a printscreen of the lecturers example that I have been working off of
     

    Attached Files:

  7. Jan 21, 2012 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The above formula is applicable to functions with a period of ##2\pi##.

    These formulas aren't quite right. It's mixing up the case where the period is ##2\pi## and the more general case of a period of T.

    When the period is ##2\pi##, you have
    \begin{align*}
    f(x) &= \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx) \\
    a_n &= \frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos nx\,dx \\
    b_n &= \frac{1}{\pi}\int_{-\pi}^\pi f(x)\sin nx\,dx
    \end{align*}

    When the period is T, you have
    \begin{align*}
    f(x) &= \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos n\omega_0x + b_n \sin n\omega_0x) \\
    a_n &= \frac{2}{T}\int_{-T/2}^{T/2} f(x)\cos n\omega_0x\,dx \\
    b_n &= \frac{2}{T}\int_{-T/2}^{T/2} f(x)\sin n\omega_0x\,dx
    \end{align*}where ##\omega_0 = 2\pi/T##. If you set ##T=2\pi## in these formulas, you'll recover the ones above.

    For your particular f(x), the period is ##2\pi##, so you can use the first set of formulas.
     
  8. Jan 21, 2012 #7
    Thank you so much for giving me those equations!

    I have attempted them using those equations and I've attached when I've done so far.

    The red circle is something that I am unsure about and everything below it is also what I want to check if it is okay?
     

    Attached Files:

  9. Jan 21, 2012 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Sorry, I should have been a little more careful. The equations above assume one period of the function is defined on the interval [-T/2, T/2], but one period of f(x) in this problem is defined on [0, 2pi]. You want to use these equations instead:
    \begin{align*}
    f(x) &= \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx) \\
    a_n &= \frac{1}{\pi}\int_0^{2\pi} f(x)\cos nx\,dx \\
    b_n &= \frac{1}{\pi}\int_0^{2\pi} f(x)\sin nx\,dx
    \end{align*}What the limits of the integrals are don't really matter other than they have to cover exactly one period.
     
  10. Jan 22, 2012 #9
    Hi,

    the answer I got for Ao was 4pi^3 / 3. I've attached a copy of the worknigs just to do a final check that i'm doing it right!

    Thank you so much for helping me as well :)

    Also is fourier series similar to the complex fourier series or would I have to use a whole new set of equations?
     

    Attached Files:

  11. Jan 22, 2012 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Close. It should be ##4\pi^2/3##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Obtaining the fourier series
  1. Fourier series (Replies: 10)

  2. Fourier Series (Replies: 1)

  3. Fourier Series (Replies: 6)

Loading...