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Basic Fourier Series question

  • 03-03-2012 10:02pm
    #1
    Smythe
    Registered Users, Registered Users 2 Posts: 434 ✭✭


    f(t) = [t]
    (modulus of t)

    -∏<t<∏
    T=2∏
    (period of 2∏)

    What is a0?


    The answer I have here is a0 = ∏

    Is that correct? If so, why is a0 = ∏?

    Thank you.


Comments

  • #2
    Kohl
    Closed Accounts Posts: 130 ✭✭Kohl


    Smythe wrote: »
    f(t) = [t]
    (modulus of t)

    -∏<t<∏
    T=2∏
    (period of 2∏)

    What is a0?


    The answer I have here is a0 = ∏

    Is that correct? If so, why is a0 = ∏?

    Thank you.

    Your function, f(t), is an even function of t. The Fourier Series for f(t) is

    a_0 / 2 + sum (n=1 to infinity) a_n cos (n pi t / L)

    To find a_0, you substitute n = 0 into the coefficient of the Fourier Series

    a_n = 2/L (integral from zero to L) f(t) cos (n pi t/ L) dt

    L = pi, n = 0, so

    a_0 = 2/pi (integral from zero to pi) t dt

    a_0 = 2/pi (pi^2 / 2) = pi


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