Common Functions of The Fourier Series

Example 1: Find the Fourier series expansion of f (t) given in Fig. 1. Solution: The function f (t) is an odd function. Hence $a_0 = 0 = a_n$. The period is $T = 4$, and $ω_0 = 2π/T = π/2$, so that $$\begin{aligned}b_{n} &=\frac{4}{T} \int_{0}^{T / 2} f(t) \sin n \omega_{0} t d t \\&=\frac{4}{4}\left[\int_{0}^{1} 1 \sin \frac{n \pi}{2} t d t+\int_{1}^{2} 0 \sin \frac{n \pi}{2} t d t\right] \\&=-\left.\frac{2}{n \pi} \cos \frac{n \pi t}{2}\right|_{0} ^{1}=\frac{2}{n \pi}\left(1-\cos \frac{n \pi}{2}\right)\end{aligned}$$ Hence, $$f(t)=\frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1}{n}\left(1-\cos \frac{n \pi}{2}\right) \sin \frac{n \pi}{2} t$$ which is a Fourier sine series.

Example 2: Determine the Fourier series for the half-wave rectified cosine function shown in Fig. 2. Solution: This is an even function so that $ b_{n}=0 $. Also, $ T=4, \omega_{0}=2 \pi / T=\pi / 2 $. Over a period, $$f(t) = \left\{\begin{array}{ll}0, & -2 < t < -1 \\ \cos \frac{\pi}{2} t, & -1 < t < 1 \\0, & 1 < t < 2\end{array}\right.$$ $$ \begin{aligned} a_{0}&=\frac{2}{T} \int_{0}^{T / 2} f(t) d t=\frac{2}{4}\left[\int_{0}^{1} \cos \frac{\pi}{2} t d t+\int_{1}^{2} 0 d t\right] \\ &=\left.\frac{1}{2} \frac{2}{\pi} \sin \frac{\pi}{2} t\right|_{0} ^{1}=\frac{1}{\pi} \\ a_{n}&=\frac{4}{T} \int_{0}^{T / 2} f(t) \cos n \omega_{0} t d t\\ &=\frac{4}{4}\left[\int_{0}^{1} \cos \frac{\pi}{2} t \cos \frac{n \pi t}{2} d t+0\right]\end{aligned}$$ But $$ \cos A \cos B=\frac{1}{2}[\cos (A+B)+\cos (A-B)] $$ Then $$a_{n}=\frac{1}{2} \int_{0}^{1}\left[\cos \frac{\pi}{2}(n+1) t+\cos \frac{\pi}{2}(n-1) t\right] d t$$ For $ n=1 $, $$a_{1}=\frac{1}{2} \int_{0}^{1}[\cos \pi t+1] d t=\left.\frac{1}{2}\left[\frac{\sin \pi t}{\pi}+t\right]\right|_{0} ^{1}=\frac{1}{2}$$ For $ n>1 $, $$a_{n}=\frac{1}{\pi(n+1)} \sin \frac{\pi}{2}(n+1)+\frac{1}{\pi(n-1)} \sin \frac{\pi}{2}(n-1)$$ For $ n=\operatorname{odd}(n=1,3,5, \ldots),(n+1) $ and $ (n-1) $ are both even, so $$\sin \frac{\pi}{2}(n+1)=0=\sin \frac{\pi}{2}(n-1), \quad n=\text { odd }$$ For $ n=\operatorname{even}(n=2,4,6, \ldots),(n+1) $ and $ (n-1) $ are both odd. Also, $$ \sin \frac{\pi}{2}(n+1)=-\sin \frac{\pi}{2}(n-1)=\cos \frac{n \pi}{2}=(-1)^{n / 2}, \quad n=\text { even}$$ Hence, $$a_{n}=\frac{(-1)^{n / 2}}{\pi(n+1)}+\frac{-(-1)^{n / 2}}{\pi(n-1)}=\frac{-2(-1)^{n / 2}}{\pi\left(n^{2}-1\right)}, \quad n=\text { even }$$ Thus, $$f(t)=\frac{1}{\pi}+\frac{1}{2} \cos \frac{\pi}{2} t-\frac{2}{\pi} \sum_{n=\text { even }}^{\infty} \frac{(-1)^{n / 2}}{\left(n^{2}-1\right)} \cos \frac{n \pi}{2} t$$ To avoid using $ n=2,4,6, \ldots $ and also to ease computation, we can replace $ n $ by $ 2 k $, where $ k=1,2,3, \ldots $ and obtain $$f(t)=\frac{1}{\pi}+\frac{1}{2} \cos \frac{\pi}{2} t-\frac{2}{\pi} \sum_{k=1}^{\infty} \frac{(-1)^{k}}{\left(4 k^{2}-1\right)} \cos k \pi t$$ which is a Fourier cosine series.

Example 3: Calculate the Fourier series for the function in Fig. 3. Solution: The function in Fig. $ 3 $ is half-wave odd symmetric, so that $ a_{0}=0= $ $ a_{n} $. It is described over half the period as $$f(t)=t, \quad-1 < t < 1$$ $ T=4, \omega_{0}=2 \pi / T=\pi / 2 $. Hence, $$b_{n}=\frac{4}{T} \int_{0}^{T / 2} f(t) \sin n \omega_{0} t d t$$ Instead of integrating $ f(t) $ from 0 to 2 , it is more convenient to integrate from $ -1 $ to 1 . $$\begin{aligned}b_{n} &=\frac{4}{4} \int_{-1}^{1} t \sin \frac{n \pi t}{2} d t=\left.\left[\frac{\sin n \pi t / 2}{n^{2} \pi^{2} / 4}-\frac{t \cos n \pi t / 2}{n \pi / 2}\right]\right|_{-1} ^{1} \\&=\frac{4}{n^{2} \pi^{2}}\left[\sin \frac{n \pi}{2}-\sin \left(-\frac{n \pi}{2}\right)\right]-\frac{2}{n \pi}\left[\cos \frac{n \pi}{2}+\cos \left(-\frac{n \pi}{2}\right)\right] \\&=\frac{8}{n^{2} \pi^{2}} \sin \frac{n \pi}{2}-\frac{4}{n \pi} \cos \frac{n \pi}{2}\end{aligned}$$ since $ \sin (-x)=-\sin x $ as an odd function, while $ \cos (-x)=\cos x $ as an even function. Using the identities for $ \sin n \pi / 2 $ and $ \cos n \pi / 2 $ in Table $ 16.1 $, $$b_{n}=\left\{\begin{array}{ll}\frac{8}{n^{2} \pi^{2}}(-1)^{(n-1) / 2}, & n=\text { odd }=1,3,5, \ldots \\\frac{4}{n \pi}(-1)^{(n+2) / 2}, & n=\text { even }=2,4,6, \ldots\end{array}\right.$$ Thus, $$f(t)=\sum_{n=1}^{\infty} b_{n} \sin \frac{n \pi}{2} t$$ where $ b_{n} $ is given above.