# Common Functions of The Fourier Series

 Function Fourier Series 1. Square wave $$f(t)=\frac{A \tau}{T}+\frac{2 A}{T} \sum_{n=1}^{\infty} \frac{1}{n} \sin \frac{n \pi \tau}{T} \cos n \omega_0 t$$ 2. Sawtooth wave $$f(t)=\frac{A}{\pi}+\frac{A}{2} \sin \omega_0 t-\frac{2 A}{\pi} \sum_{n=1}^{\infty} \frac{1}{4 n^2-1} \cos 2 n \omega_0 t$$ 3. Triangular wave $$f(t)=\frac{2 A}{\pi}-\frac{4 A}{\pi} \sum_{n=1}^{\infty} \frac{1}{4 n^2-1} \cos n \omega_0 t$$ 4. Rectangular pulse train $$f(t)=\frac{A \tau}{T}+\frac{2 A}{T} \sum_{n=1}^{\infty} \frac{1}{n} \sin \frac{n \pi \tau}{T} \cos n \omega_{0} t$$ 5. Half-wave rectified sine $$f(t)=\frac{A}{\pi}+\frac{A}{2} \sin \omega_{0} t-\frac{2 A}{\pi} \sum_{n=1}^{\infty} \frac{1}{4 n^{2}-1} \cos 2 n \omega_{0} t$$ 6. Full-wave rectified sine $$f(t)=\frac{2 A}{\pi}-\frac{4 A}{\pi} \sum_{n=1}^{\infty} \frac{1}{4 n^{2}-1} \cos n \omega_{0} t$$
Example 1: Find the Fourier series expansion of f (t) given in Fig. 1.
Fig. 1:For Example 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.
Fig. 2:For Example 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.
Fig. 3:For Example 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.