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Fourier Series in Complex Form

  1. May 23, 2012 #1
    Hi,
    Just wanted to verify my process to complete the Fourier series in complex form.
    With the Function
    f(t) = 8, 0≤t≤2
    -8, 2≤t≤4

    f(t) = [itex]\sum[/itex] Cn e-int
    n=1 to ∞

    Where

    Cn = [itex]\frac{1}{T}[/itex] ∫ f(t) e-int[itex]\frac{2∏}{T}[/itex]

    Cn = [itex]\frac{1}{T}[/itex] 0∫T f(t) e-int2∏/T dt

    Cn = [itex]\frac{1}{4}[/itex] 0∫4 f(t) e-int2∏/T dt ]

    Cn = [itex]\frac{1}{4}[/itex] 0∫2 8e-int2∏/T dt + 2∫4 -8 e-int2∏/T dt + -∞∫0 0 e-in2∏/t dt + 4∫∞ 0 e-in2∏/t dt

    Cn = [itex]\frac{8}{4}[/itex]0∫2 e-int∏/2 dt + 2∫4 e-int∏/2 dt

    Cn = 2 ( [itex]\frac{2e[itex]\frac{-int∏}{2}[/itex]}{-in∏}[/itex] 0 ->2 + [itex]\frac{2e-int∏/2}{-in∏}[/itex] ) 2 -> 4

    Cn = 2 ( [itex]\frac{2e-in∏}{-in∏}[/itex] + [itex]\frac{2e0}{in∏}[/itex] ] + [ [itex]\frac{2e-in2∏}{in∏}[/itex] - [itex]\frac{2e-in∏/}{in∏}[/itex] ] }


    Cn = ( [itex]\frac{4}{in∏}[/itex] ) { e-in∏ + 1 + e-in2∏ - e-in∏ }

    Cn = [itex]\frac{4}{in∏}[/itex] ( 2(e-in∏ + e-in2∏ + 1 ))

    Use e-iθ = Cosθ - iSinθ and let θ = n∏ and θ = n2∏

    Cn = [itex]\frac{4}{in∏}[/itex] ( 2 (Cosθ - iSinθ) + (Cosθ - iSinθ) + 1 )

    Cn = [itex]\frac{4}{in∏}[/itex] ( 2Cos(n∏) - 2iSin(n∏) + Cos (2∏n) - iSin(2∏n) )

    Cn = [itex]\frac{4}{in∏}[/itex] ( 2 Cos (n∏) + 1)



    Something along those lines??

    PS sorry about my terrible notation within the question, i tried using as many symbols with the code as I could. Mainly for Tiny Tim :D
     
    Last edited: May 23, 2012
  2. jcsd
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