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Fourier Series Problem

  1. Nov 6, 2008 #1
    1. The problem statement, all variables and given/known data
    Given that the fourier series of function f1(t)

    = At/π + A , -π < t < 0
    = -At/π +A, 0 < t < π

    is f(t) = 0.5A +4A/π2(cos t + 1/9 cos 3t + 1/25 cos 5t + ...)

    what is the fourier series of function f2(t)
    = -At/π, -π < t < 0
    = At/π, 0 < t < π

    3. The attempt at a solution
    for a function f2(t) = f1(t + π) means the function f2(t) is the function of f1(t) shifted by a distance of π to the left. How do I link this with the above problem?
  2. jcsd
  3. Nov 6, 2008 #2


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    Hint: what is [tex]\frac{A(-t+\pi)}{\pi}[/tex]?
  4. Nov 6, 2008 #3
    Thanks for the reply. The function f1 is a periodic function. The function f2 is a function f1 shifted by a distance π. However, which direction the shifting should be the right answer? I found out that if we shifted f1 either to the right or left by π distance, we still get the same f2. or both answer are acceptable?

    my answer is : f2(t) = A/2 + 4A/π2 {cos(t+π) + 1/9 cos(3(t+π)) + 1/25 cos(5(t+π)) + ...}

    please correct me if i am wrong. thank you.
  5. Nov 6, 2008 #4


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    Shouldn't it be [itex]f_2(t)=f_1(-t+\pi)[/itex] i.e. with a negative sign in front of t :wink:
  6. Nov 6, 2008 #5
    why is it negative t? can you elaborate more on this? from my observation, i can see that the function f1 is just shifting to the left or right by π. since f1 is a periodic function, it looks like sin wave on cosine wave which are π distance from each other.
  7. Nov 7, 2008 #6


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    well, look at the functions....for [itex]0<t<\pi[/itex],


    The same relationship holds for [itex]\pi<t<0[/itex].
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