# Coursework check: Fourier series, Coefficients

• Bassalisk

## Homework Statement

[PLAIN]http://pokit.org/get/d9a4a8542c2acdfedbb1038292f6817b.jpg [Broken]

Find the coefficients of this function Ck

## The Attempt at a Solution

\begin{align}\Large C_{k}=\frac{1}{T}\cdot \int_{-T/2}^{T/2} f(t)\cdot e^{\frac{-jk2\pi t}{T}}dt=\frac{1}{T}\cdot \int_{-T/2}^{0} a\cdot e^{\frac{-jk2\pi t}{T}}dt+\frac{1}{T}\cdot \int_{0}^{T/2} b\cdot e^{\frac{-jk2\pi t}{T}}dt = \\ \Large=\frac{1}{T}\cdot a\left[\frac{e^{\frac{-jk2\pi t}{T}}}{\frac{-jk2\pi}{T}}\right]_{-T/2}^{0}+\frac{1}{T}\cdot b\left[\frac{e^{\frac{-jk2\pi t}{T}}}{\frac{-jk2\pi}{T}}\right]_{0}^{T/2}= \frac{j\cdot a}{T}\cdot \frac{e^{0}}{\frac{k2\pi}{T}}-\frac{j\cdot a}{T}\cdot \frac{e^{jk\pi}}{\frac{k2\pi}{T}}+\frac{j\cdot b}{T}\cdot \frac{e^{-jk\pi}}{\frac{k2\pi}{T}}-\frac{j\cdot b}{T}\cdot \frac{e^{0}}{\frac{k2\pi}{T}}= \\ \Large=\frac{j}{T\cdot\frac{k2\pi}{T}}\cdot\left(a-a\cdot e^{jk\pi}+b\cdot e^{-jk\pi}-b\right)=|| e^{jk\pi}=(-1)^{k}; e^{-jk\pi}=\frac{1}{(-1)^{k}}=(-1)^{k}||= \\ \Large\frac{j}{T\cdot\frac{k2\pi}{T}}\cdot\left(a-b-(-1)^{k}(a-b)\right)= \frac{j}{k2\pi}\cdot\left((a-b)(1-(-1)^k)\right)\end{align}

For k=2l;
$\left((a-b)(1-(-1)^k)\right)=0; C_k = 0$

For k=2l+1;

$\left((a-b)(1-(-1)^k)\right)=2(a-b); C_k = \frac{j\cdot 2}{(2l+1)2\pi}\cdot (a-b)=\frac{j}{(2l+1)\pi}\cdot (a-b)$

Somebody please check this.

How do I find bn and an.

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Took me half an hour to write this in LaTex please somebody :/