# Fourier Series in Complex Form

1. May 23, 2012

### JamesEllison

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) = $\sum$ Cn e-int
n=1 to ∞

Where

Cn = $\frac{1}{T}$ ∫ f(t) e-int$\frac{2∏}{T}$

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

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

Cn = $\frac{1}{4}$ 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 = $\frac{8}{4}$0∫2 e-int∏/2 dt + 2∫4 e-int∏/2 dt

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

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

Cn = ( $\frac{4}{in∏}$ ) { e-in∏ + 1 + e-in2∏ - e-in∏ }

Cn = $\frac{4}{in∏}$ ( 2(e-in∏ + e-in2∏ + 1 ))

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

Cn = $\frac{4}{in∏}$ ( 2 (Cosθ - iSinθ) + (Cosθ - iSinθ) + 1 )

Cn = $\frac{4}{in∏}$ ( 2Cos(n∏) - 2iSin(n∏) + Cos (2∏n) - iSin(2∏n) )

Cn = $\frac{4}{in∏}$ ( 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