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Fourier series - DC component, integration problem

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the Fourier series representation of:

    f(t)={-t , -∏<t<0
    f(t)={0 , 0<t<∏

    This is a piecewise function.

    T=2∏ (the period)

    2. Relevant equations

    [tex]a_{0}=\frac{2}{T}*\int_0^T f(t),dt[/tex]


    3. The attempt at a solution

    I need help only with calculating the DC component. Once I get that working, I can continue on my own. Tried multiple times, I keep getting pi/2.

    The correct answer should be pi/4.

    This is my graph:
    hzRy6UY.jpg
    Is this graph correct?

    I consider the area of the function from -pi to 0 identical if I change it from:

    f(t)=-t -∏<t<0

    to

    f(t)=t 0<t<∏

    Is this permitted? Surely it's the same result. Seems alright to me

    [tex]a_{0}=\frac{2}{2\pi}*\int_0^∏ -t,dt[/tex]

    [tex]a_{0}=\frac{1}{\pi}*(\frac{\pi^2}{2}-0)[/tex]

    [tex]a_{0}=\frac{\pi}{2}[/tex]

    And the book says it should be pi/4.

    What am I doing wrong? Is my graph correct? Is my integration correct? Is the book's solution wrong?
     
  2. jcsd
  3. Apr 16, 2013 #2
    I think I figured it out.... my a0 seems correct to be ∏/2,

    but when I type out the actual Fourier series, the formula is:

    a0/2, making it ∏/4.

    If anybody wants to confirm my thinking, that would be helpful!
     
  4. Apr 16, 2013 #3

    LCKurtz

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    Textbooks are not consistent on the formula for ##a_0## because some formulas assume the constant term in the series is ##a_0## and some use ##\frac {a_0} 2##, which causes the formula for ##a_0## to differ by ##2##. As long as you are consistent, you should be OK. But there is no need to go to a different interval than that where the function is given. If the constant term of the series is ##a_0## and the period is ##2p## then$$
    a_0 = \frac 1 {2p}\int_{-p}^p f(t)\, dt = \frac 1 {2\pi}\left(\int_{-\pi}^0 -t\, dt
    +\int_{-\pi}^0 0\, dt\right) = \frac \pi 4$$
     
  5. Apr 16, 2013 #4
    Thank you for helping me with this!

    I think perhaps an error has snuck into your maths though, because you write:

    [tex]a_{0}=\frac{1}{2p}.........=\frac{1}{2\pi}......[/tex]

    But p=2∏, so it should be [tex]a_{0}=\frac{1}{4\pi}[/tex] using your formula.

    Can you comment on this?

    I understand how there are different formulas though, as you say, when I created the question I hadn't realized that I had to divide it by 2 again, but when I did I wanted somebody to confirm my logic.

    I also understand, of course, that I don't have to tinker with the range and function, but that saved me having to deal with negative values etc, I should stick with the actual function though, I have to be at a level where that kind of thing doesn't phase me.
     
  6. Apr 16, 2013 #5

    LCKurtz

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    No, I said the period was ##2p## so ##p=\pi## in your case. And when you do the integral you will have a ##t^2## in the numerator with a limit of ##-\pi## which will give a ##\pi^2## in the numerator.
     
  7. Apr 16, 2013 #6
    Yes of course, I was too quick in reading there, thanks again!
     
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