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!

Find the following fourier series in trigonometric form

  1. Nov 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the following fourier series in trigonometric form.
    6oooi8.png

    2. Relevant equations
    $$y(t)=a_0+\sum\limits_{n=1}^{\infty} a_n cos(n\omega_{0}t)+b_n sin(n\omega_{0}t)$$

    3. The attempt at a solution
    The graph above is represented by the function:
    $$
    x(t) = \left\{
    \begin{array}{ll}
    -t+1 & \quad 0 < t \leq 1 \\
    0 & \quad 1 < t \leq 2
    \end{array}
    \right.
    $$

    ##T_0=2## and ##\omega_{0}=\pi##

    To find ##a_0##:
    $$a_0=\frac{1}{T_{0}}\int_{0}^{T_{0}} x(t)dt = \frac{1}{2}\int_{0}^{1} (-t+1)dt = \frac{1}{4}$$
    The integral of 0 can be ignored since it equals 0.

    To find ##a_n##:
    $$a_n=\frac{2}{T_{0}}\int_{0}^{T_0}x(t) cos(n\omega_{0}t)dt=\frac{2}{2}\int_{0}^{1} (-t+1)cos(\pi nt)dt=\int_{0}^{1} -tcos(\pi nt)dt +cos(\pi nt)dt$$

    To find ##b_n##
    $$b_n=\frac{2}{T_{0}}\int_{0}^{T_{0}} x(t)sin(n\omega_{0}t)dt = \frac{2}{2}\int_{0}^{1} (-t+1)sin(\pi nt)dt=\int_{0}^{1} -tsin(\pi nt)+sin(\pi nt)dt$$

    I am assuming that the ##T_0## in these equations are for the period of each function, since the function x(t)=0 is from 1 to 2, I only use 1 for ##T_0## for the function x(t)=-t+1.

    Now I can find these integrals for ##a_n## and ##b_n## using integration by parts, but I'm wondering if I am going along with this correctly and what I will do once I find it (the summation really confuses me).


    Thank you for any help!
     
    Last edited: Nov 23, 2014
  2. jcsd
  3. Nov 23, 2014 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It looks right, go head. Find an bn, in terms of n. No need to sum, the sum must be the original function:)
     
  4. Nov 23, 2014 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Except at x = 0 and x = 2, where the sum will be 1/2 instead of 1 and 0, respectively.
     
  5. Nov 23, 2014 #4
    The original function is continuous in negative and positive direction, sorry I should of mentioned that.
     
  6. Nov 23, 2014 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Wrong assumption. The periodic extension of that function has period ##T=2##. Even though the integral for the coefficients is zero for half the period, you still have to use the full period in the trig functions. That is, for example, ##sin(\frac {n \pi t}{T}) = sin(\frac {n \pi t}{2})## and similarly for the cosine.
     
  7. Nov 23, 2014 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The function must be periodic to have a Fourier series. So it is not continuous at x=2k (for integer k). And Ray is right, the Fourier series returns 1/2 at x=2k instead of f(2k)=0, as the original function.
     
  8. Nov 23, 2014 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The OP was referring to the upper limit of the integral, which can be set to 1 because x(t) is 0 from t=1 to t=2. What he wrote confused me too. The rest of the calculations properly use ##T_0=2##.
     
  9. Nov 23, 2014 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Not sure what you mean by that. If you actually want the periodic extension to represent a continuous function for all ##t## you would have to take the even extension of your function, use ##T=4##, and do a half range cosine expansion. Unless you were given instructions in the problem you haven't told us, I doubt that is what is wanted.
     
  10. Nov 23, 2014 #9

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I don't think so. I don't see ##\sin(\frac{n\pi t} 2)## in those integrands.
     
  11. Nov 23, 2014 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    If the period is T=2, then ##\omega=2\pi/T = \pi##, so the argument of the trig functions is ##n\pi t##.
     
  12. Nov 23, 2014 #11

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Arghh. You're right. @iRaid: Ignore my posts, except (maybe) post #8.
     
  13. Nov 23, 2014 #12
    What I mean is that the function repeats itself forever, it doesn't just go from 0 to 2. What would I end up doing once I figure out the 2 integration by parts? The summation really confuses me and I never understand how examples of these problems get rid of sine or cosine. The examples will say something like "clearly this equals 0 so the sine term can be excluded" but it is not clear to me.
     
  14. Nov 24, 2014 #13

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You will get an and bn, and then write up the Fourier series as
    ##y(t)=a_0+\sum\limits_{n=1}^{\infty} a_n cos(n\omega_{0}t)+b_n sin(n\omega_{0}t)##
    Perhaps, you find a formula to write up the general terms.
    Some of the integrals turn out to be zero if you substitute the boundaries. If you get sin(nπt), for example, it is zero both at t=0 and t=1..Just do the integrals you obtained in the OP and you will see.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Find the following fourier series in trigonometric form
Loading...