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Finding Fourier Coefficient

  1. Apr 12, 2017 #1
    1. The problem statement, all variables and given/known data
    I am looking for help with part (d) of this question
    5bdd37a7f76ed036ac58da9e96f81bbf.png

    2. Relevant equations
    3d897f3c764d9eac7086188e1f12bdb1.png



    3. The attempt at a solution
    I have attempted going through the integral taking L = 4 and t0 = -2. I was able to solve for a0 but I keep having the integrate by parts on this one. I've tried it out twice and messed up. Any help would be appreciated.
     
  2. jcsd
  3. Apr 12, 2017 #2

    BvU

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    Hi TBD,

    Well, it's nice to tell us you have attempted, but the idea is that you post your work and then we can comment, hint, etc.
     
  4. Apr 12, 2017 #3

    LCKurtz

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    You haven't told us what ##L## represents. Half period or full period? And your half range formula needs to be consistent with your choice. Also, in half range expansions, you would normally integrate from ##0## to half a period.
     
  5. Apr 13, 2017 #4

    BvU

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    Can't be right. Show us.
     
  6. Apr 13, 2017 #5
    And you'd be right.

    Apologies for not being more clear. L represents the fundamental period of the function. In this case L = 4 because f(t+4) = f(t).

    Sorry for not posting my workings but as it happens they were pretty much useless (I will do this in future though). I ended up splitting the integral up into 4 parts: $$ \frac{1}{2} \int_{-2}^{-1}f(t) cos(\frac{\pi n t}{2}) dt + \frac{1}{2} \int_{-1}^{0}f(t) cos(\frac{\pi n t}{2}) dt + \frac{1}{2} \int_{0}^{1}f(t) cos(\frac{\pi n t}{2}) dt + \frac{1}{2} \int_{1}^{2}f(t) cos(\frac{\pi n t}{2}) dt $$
    Since everything else goes to #0# we only need to consider the integral between -1 and 1. The integral from -1 to 1 is equivalent to twice the integral from 0 to 1:
    $$ \frac{1}{2} \int_{-1}^{1}f(t) cos(\frac{\pi n t}{2})dt = 2*\frac{1}{2} \int_{0}^{1}f(t) cos(\frac{\pi n t}{2}) dt $$and because the values of x are positive, the function ##1-|x| = 1-x## meaning we only have to integrate that and not a nasty ass modulus. In the end I got an answer of
    ##a_n = \frac{4}{\pi^2 n^2}##
     
  7. Apr 13, 2017 #6

    LCKurtz

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    Your last integral is correct. But you have made a mistake in your integration because it doesn't give you ##a_n = \frac{4}{\pi^2 n^2}##. All I can suggest is for you to recheck your integration since you haven't shown your steps. Also, you will need to calculate ##a_0## separately.
     
    Last edited: Apr 13, 2017
  8. Apr 14, 2017 #7
    Okay, no problem. I am right in saying ##1-|x| = 1-x## though amn't I? It's similar to an example our professor gave us which is how I made the leap. Probably just made some small error somewhere in the integration. I did calculate ##a_0## seperately and got ##\frac{1}{2}##.
     
  9. Apr 14, 2017 #8

    LCKurtz

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    Yes. As I said earlier, your last integral is correct with ##f(t) = 1-t##.

    You should get ##a_0 =\frac 1 4##. And your answer for ##a_n## should have ##\cos(\frac{n\pi} 2)## involved in its formula.
     
    Last edited: Apr 14, 2017
  10. Apr 14, 2017 #9
    Okay must have made some massive integration errors. At least I know I have the integral right and that's half the battle :D! Thanks for your comments
     
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