Introduction : We shall provide Fourier Series examples here and explain step by step each example.
Fourier Series : Fourier series of a function is a representation in series form. Fourier series is obtained of a periodic function.
Periodic Function:
If a real value function \(w\) satisfies \(w(t+P)=w(t)\) for all real numbers \(t\) then it is called a periodic function with period \(P\).
Examples:
\(sin(t), cos(t), sec(t), cosec(t)\) are periodic functions all with same period \(P = 2\pi\)\(tan(t), cot(t)\) are periodic functions with period \(P = \pi\)Piecewise function \(w_1(t)=\begin{cases}2 & \text{if } 0\le t < \frac{1}{2}\\0 & \text{if }\frac{1}{2}\le t < 1\end{cases}\) is periodic with period \(P =1\)Piecewise function \(w_2(t)=\begin{cases}\frac{sin(2\pi t)}{3} & \text{if } 0\le t < \frac{3}{2}\\0 & \text{if }\frac{3}{2}\le t < 3\end{cases}\) is periodic with period \(P =3\).
Fourier Series:
If \(w(t)\) is a periodic and \(w(t) , \frac{dw}{dt}\) atleast piecewise continuous on an interval \(I=[a, b]\) of length \( P = b – a\) , then the Fourier series of \(w(t)\) is an infinite sum of sine and cosine series, written as follows
\[w(t) = \frac{\alpha_0}{2} + \sum\limits_{m = 1}^\infty {\alpha_m} \cos (\frac{2m\pi t}{P}) + {\beta_m}\sin (\frac{2m\pi t}{P})\hspace{5 cm}(2.1)\]
Where \(\alpha_0, \alpha_m, \beta_m\) are called constant , Fourier \(cos\) and Fourier \(sin\) Coefficient respectively and determined using formulas
\[\alpha_0 = \frac{2}{P}\int\limits_I {w(t)dt}, \hspace{1.5 cm} \alpha_m = \frac{2}{P}\int\limits_I {w(t)\cos (\frac{{2m\pi t}}{P})dt}, \hspace{1.5 cm} \beta_m = \frac{2}{P}\int\limits_I {w(t)} \sin (\frac{{2m\pi t}}{P})dt\]
Example (1):
Find Fourier Series expansion of the function
\[w(t)=\begin{cases}1 & \text{if } -2<t<0\\t & \text{if }0<t<2\end{cases}\]
such that \(w\left(t+4\right)=w\left(t\right)\)
Solution:
length of the interval or period \( I =\left[-2, 2\right]\) is \(P = 4\)
Therefore Fourier Series of \( w\left(t\right)\) in the interval I is given as
\[w(t) = \frac{\alpha_0}{2} + \sum\limits_{m = 1}^\infty {\alpha_m} \cos (\frac{2m\pi t}{P}) + {\beta_m}\sin (\frac{2m\pi t}{P})\]
\[\Rightarrow w(t) = \frac{\alpha_0}{2} + \sum\limits_{m = 1}^\infty {\alpha_m} \cos (\frac{m\pi t}{2}) + {\beta_m}\sin (\frac{m\pi t}{2})\hspace{5 cm}(1)\]
Step 1: To find Fourier Coefficients
\[\alpha_0 = \frac{2}{4}\int\limits_I {w(t)dt}=\frac{1}{2}\int\limits_{-2}^{2}{w(t)dt}\]
\[=\frac{1}{2}\int\limits_{-2}^{0}{1 dt}+\frac{1}{2}\int\limits_{0}^{2}{t dt}=\frac{1}{2}\left[t\right]_{t=-2}^{0}+\frac{1}{2}\left[\frac{t^2}{2}\right]_{t=0}^{2} =\frac{1}{2}\left[ -2+2\right] = 0 \]
\[ \alpha_m = \frac{2}{4}\int\limits_I {w(t)\cos (\frac{{2m\pi t}}{4})dt}=\frac{1}{2}\int\limits_{-2}^{2}{w(t)\cos (\frac{{m\pi t}}{2})dt}\]
\[=\frac{1}{2}\int\limits_{-2}^{0}{1 \cos (\frac{{m\pi t}}{2})dt}+\frac{1}{2}\int\limits_{0}^{2}{t \cos (\frac{{m\pi t}}{2})dt} \]
\[=\frac{1}{2}\left[\frac{2 \sin \left(\frac{\pi m t}{2}\right)}{\pi m}\right]_{t=-2}^{t=0}+\frac{1}{2}\left[\frac{4 \cos \left(\frac{\pi m t}{2}\right)}{\pi ^2 m^2}+\frac{2 t \sin \left(\frac{\pi m t}{2}\right)}{\pi m}\right]_{t=0}^{2}\]
\[=\frac{1}{2}\left[0+\frac{2 \sin (\pi m)}{\pi m}+\frac{4 \cos (\pi m)}{\pi ^2 m^2}+\frac{4 \sin (\pi m)}{\pi m}-\frac{4}{\pi ^2 m^2}\right]\]
\[ =\frac{2 \cos (\pi m)}{\pi ^2 m^2}+3\frac{\sin (\pi m)}{\pi m}-\frac{2}{\pi ^2 m^2} \]
Since \[ \sin (\pi m)=0, \cos (\pi m) =(-1)^m \forall m\ge1\]
\[ \Rightarrow \alpha_m =2\left(\frac{ (-1)^m-1}{\pi ^2 m^2}\right)\]
\[ \beta_m = \frac{2}{4}\int\limits_I {w(t)} \sin (\frac{{2m\pi t}}{4})dt=\frac{1}{2}\int\limits_{-2}^{2}{w(t)\sin (\frac{{m\pi t}}{2})dt}\]
\[=\frac{1}{2}\int\limits_{-2}^{0}{1 \sin (\frac{{m\pi t}}{2})dt}+\frac{1}{2}\int\limits_{0}^{2}{t \sin (\frac{{m\pi t}}{2})dt} \]
\[=\frac{1}{2}\left[-\frac{2 \cos \left(\frac{\pi m t}{2}\right)}{\pi m}\right]_{t=-2}^{t=0}+\frac{1}{2}\left[\frac{4 \sin \left(\frac{\pi m t}{2}\right)}{\pi ^2 m^2}-\frac{2 t \cos \left(\frac{\pi m t}{2}\right)}{\pi m}\right]_{t=0}^{2}\]
\[=\frac{1}{2}\left[-\frac{2}{\pi m}+\frac{2 \cos (\pi m)}{\pi m}\right]+\frac{1}{2}\left[ \frac{4 \sin (\pi m)}{\pi ^2 m^2}-\frac{4 \cos (\pi m)}{\pi m}-0 \right]\]
\[ =\frac{2 \sin (\pi m)}{\pi ^2 m^2}-\frac{\cos (\pi m)}{\pi m}-\frac{1}{\pi m} \]
Since \[ \sin (\pi m)=0, \cos (\pi m) =(-1)^m \forall m\ge1\]
\[ \Rightarrow \beta_m =-\frac{(-1)^m+1}{\pi m}\]
Step 2: writing Fourier Series
substituing \(\alpha_0 ,\alpha_m, \beta_m \) into eqn (1)
\[w(t) = 0 + \sum\limits_{m = 1}^\infty {2\left(\frac{ (-1)^m-1}{\pi ^2 m^2}\right)} \cos (\frac{m\pi t}{2}) -\left(\frac{(-1)^n+1}{\pi m}\right)\sin (\frac{m\pi t}{2})\]
is the required Fourier series of \(w(t)\)
Case: When length of interval I is \(P = 2\pi\) for examples \(I = [-\pi, \pi]\) , \([0, 2\pi]\) etc.
In this case, Fourier series of \(w(t)\) can be obtained by Substituting \(P = 2\pi\) in eqn. (2.1) and takes the form
\[w(t) = \frac{\alpha_0}{2} + \sum\limits_{m = 1}^\infty {\alpha_m} \cos (m t) + {\beta_m}\sin (m t)\hspace{5 cm}(2.2)\]
Where \(\alpha_0 = \frac{1}{\pi}\int\limits_I {w(t)dt}, \hspace{1.5 cm} \alpha_m = \frac{1}{\pi}\int\limits_I {w(t)\cos (m t)dt}, \hspace{1.5 cm} \beta_m = \frac{1}{\pi}\int\limits_I {w(t)} \sin (m t)dt\)
Example (2) :
Determine Fourier Coefficients of the periodic function
\[w(t)= t \hspace{0.5cm} \text{if } -\pi<t<\pi\]
with period \( 2\pi\)
Solution:
For the interval \(I =[-\pi, \pi]\) , Fourier Coefficients are determined using formulas
\[\alpha_0 = \frac{1}{\pi}\int\limits_I {w(t)dt}\]
\[=\frac{1}{\pi}\int\limits_{-\pi}^{\pi} {(t)dt}\]
\[=\frac{1}{\pi}\left[\frac{t^2}{2}\right]_{t=-\pi}^{\pi}=\frac{1}{\pi}\left[\frac{\pi^2}{2}-\frac{\pi^2}{2}\right]\]
\[\Rightarrow \alpha_0=0\]
\[\alpha_m = \frac{1}{\pi}\int\limits_I {w(t)\cos (m t)dt}\]
\[=\frac{1}{\pi}\int\limits_{-\pi}^{\pi} {t\cos(mt)dt}\]
\[=\frac{1}{\pi}\left[\frac{\cos (m t)}{m^2}+\frac{t \sin (m t)}{m}\right]_{t=-\pi}^{\pi}\]
\[\Rightarrow \alpha_m=0\]
\[\beta_m = \frac{1}{\pi}\int\limits_I {w(t)} \sin (m t)dt\]
\[=\frac{1}{\pi}\int\limits_{-\pi}^{\pi} {t\sin(mt)dt}\]
\[=\frac{1}{\pi}\left[\frac{\sin (m t)}{m^2}-\frac{t \cos (m t)}{m}\right]_{t=-\pi}^{\pi}\]
\[=\frac{1}{\pi}\left(\frac{2 \sin (\pi m)-2 \pi m \cos (\pi m)}{m^2}\right)\]
Since \(\sin (\pi m)=0, \cos (\pi m)=(-1)^m\) for all \(m =1,2,3,….\)
\[\Rightarrow \beta_m=-\frac{2(-1)^m}{m}=2\frac{(-1)^{m+1}}{m}\]
substituting into the formula (2.2), Fourier series of \(w(t)\) is
\[w(t) =2\sum\limits_{m = 1}^\infty \frac{(-1)^{m+1}}{m}\sin(m t)\]
Odd and Even Function:
If \(w(t)\) satisfies \( w(-t) =- w(t)\) then it is called an odd function.
If \(w(t)\) satisfies \( w(-t) = w(t)\) then it is called an even function.
Fourier Series of an odd function:
Example (3):
Find Fourier series expansion of the function
\[w(t)= \frac{t^2}{2} \hspace{0.5cm} \text{if } -\pi<t<\pi\]
Solution:
Fourier series of \(w(t)\) on the interval \(I =[-\pi, \pi]\) is given as
\[w(t) = \frac{\alpha_0}{2} + \sum\limits_{m = 1}^\infty {\alpha_m} \cos (m t) + {\beta_m}\sin (m t)\hspace{5 cm}(1)\]
where
\[\alpha_0 = \frac{1}{\pi}\int\limits_I {w(t)dt}\]
\[=\frac{1}{\pi}\int\limits_{-\pi}^{\pi} {\frac{t^2}{2}dt}\]
\[=\frac{1}{\pi}\left[\frac{t^3}{6}\right]_{t=-\pi}^{\pi}=\frac{1}{\pi}\left[-\frac{\pi ^3}{6}-\frac{\pi ^3}{6}\right]\]
\[\Rightarrow \alpha_0=-\frac{\pi ^2}{3}\]
\[\alpha_m = \frac{1}{\pi}\int\limits_I {w(t)\cos (m t)dt}\]
\[=\frac{1}{\pi}\int\limits_{-\pi}^{\pi} {\frac{t^2}{2}\cos(mt)dt}\]
\[=\frac{1}{\pi}\left[\frac{1}{2} \left(\frac{2 t \cos (m t)}{m^2}+\frac{\left(m^2 t^2-2\right) \sin (m t)}{m^3}\right)\right]_{t=-\pi}^{\pi}\]
\[=\frac{1}{\pi}\left[\frac{1}{2} \left(\frac{2 \pi \cos (\pi m)}{m^2}+\frac{\left(\pi ^2 m^2-2\right) \sin (\pi m)}{m^3}\right)-\frac{1}{2} \left(-\frac{2 \pi \cos (\pi m)}{m^2}-\frac{\left(\pi ^2 m^2-2\right) \sin (\pi m)}{m^3}\right)\right]\]
\[=\frac{1}{\pi}\left[\frac{2 \pi \cos (\pi m)}{m^2}+\frac{\left(\pi ^2 m^2-2\right) \sin (\pi m)}{m^3}\right]\]
\[\Rightarrow \alpha_m=\frac{2 (-1)^m}{m^2}\]
\[\beta_m = \frac{1}{\pi}\int\limits_I {w(t)} \sin (m t)dt\]
\[=\frac{1}{\pi}\int\limits_{-\pi}^{\pi} {\frac{t^2}{2}\sin(mt)dt}\]
\[=\frac{1}{\pi}\left[\frac{1}{2} \left(\frac{2 t \sin (m t)}{m^2}-\frac{\left(m^2 t^2-2\right) \cos (m t)}{m^3}\right)\right]_{t=-\pi}^{\pi}\]
\[=\frac{1}{\pi}\left[\frac{1}{2} \left(\frac{2 \pi \sin (\pi m)}{m^2}-\frac{\left(\pi ^2 m^2-2\right) \cos (\pi m)}{m^3}\right)-\frac{1}{2} \left(\frac{2 \pi \sin (\pi m)}{m^2}-\frac{\left(\pi ^2 m^2-2\right) \cos (\pi m)}{m^3}\right)\right]\]
Since \(\sin (\pi m)=0, \cos (\pi m)=(-1)^m\) for all \(m =1,2,3,….\)
\[\Rightarrow \beta_m=0\]
substituting \(\alpha_0, \alpha_m,beta_m\) in to formula (1)
\[w(t)=-\frac{\pi ^2}{6}+\sum\limits_{m = 1}^\infty {\frac{2 (-1)^m}{m^2}} \cos (m t)\]
Fourier Series of an Even function:
Example (4):
Find Fourier Series representation of the following function
\[w(t)=\begin{cases}-t-π & \text{if } -π<t<0\\π & \text{if }0<t<π\end{cases}\]
such that \(w(t+2\pi)=w(t)\)
Solution:
Given function \(w(t)\) is periodic with period \(P=2\pi =\) length of \([-\pi, \pi]=\pi-(-\pi)=2\pi\).
step (1): To determine Fourier Constant coefficient
\[\alpha_0=\frac{2}{P}\int\limits_I {w(t)dt}\]
\[=\frac{2}{2\pi}\int\limits_{-\pi}^{0} {(-t-π)dt}+\frac{2} {2\pi}\int\limits_{0}^{\pi} {(π)dt}\]
\[=\frac{1}{\pi}\left [-\frac{t^2}{2}-\pi t \right ]_{t=-\pi}^{0}+\frac{1}{\pi}\left[\pi t\right]_{t=0}^{\pi}\]
\[ =\frac{1}{\pi}\left [\frac{\pi^2}{2}-\pi^2 \right ]+\frac{1}{\pi}\left[\pi^2 \right]\]
\[=\frac{1}{\pi}\left [-\frac{\pi^2}{2} \right]+\pi\]
\[=-\frac{\pi}{2}+\pi=\frac{\pi}{2} \]
\[\Rightarrow \alpha_0=\frac{\pi}{2}\hspace{10 cm} (1)\]
step (2): To determine Fourier \(Cos\) coefficient
\[\alpha_m = \frac{2}{P}\int\limits_I w(t)\cos (\frac{2m\pi t}{P})dt\]
\[=\frac{2}{2\pi}\int\limits_{-\pi}^{0} {(-t-π)\cos (\frac{2m\pi t}{2\pi})dt}+\frac{2}{2\pi}\int\limits_{0}^{\pi} {(π)\cos (\frac{2m\pi t}{2\pi})dt}\]
\[=\frac{1}{\pi}\left[-\frac{\cos (m t)}{m^2}-\frac{t \sin (m t)}{m}-\frac{\pi \sin (m t)}{m}\right]_{t=-\pi}^{0}+\frac{1}{\pi}\left[\frac{\pi \sin (m t)}{m}\right]_{t=0}^{\pi}\]
\[=\frac{1}{\pi}\frac{\cos (\pi m)-1}{m^2}+\frac{1}{\pi}\frac{\pi \sin (\pi m)}{m}\]
Since \(\sin (\pi m) = 0\) and \(\cos (\pi m) = (-1)^m\) for all \(m = 1, 2, 3, 4, ….\)
\[\alpha_m=\frac{1}{\pi}\frac{(-1)^m-1}{m^2}\]
<br>\[\Rightarrow \alpha_m=\begin{cases}-\frac{2}{\pi m^2} & \text{if } m \hspace{0.2cm}is \hspace{0.2cm}odd\\0 & \text{if }m \hspace{0.2cm}is \hspace{0.2cm}even\end{cases}\hspace{10 cm} (2)\]
step (3): To determine Fourier \(sin\) coefficient
\[\beta_m = \frac{2}{P}\int\limits_I w(t)\sin (\frac{2m\pi t}{P})dt\]
\[=\frac{2}{2\pi}\int\limits_{-\pi}^{0} {(-t-π)\sin (\frac{2m\pi t}{2\pi})dt}+\frac{2}{2\pi}\int\limits_{0}^{\pi} {(π)\sin(\frac{2m\pi t}{2\pi})dt}\]
\[=\frac{1}{\pi}\left[-\frac{\sin (m t)}{m^2}+\frac{t \cos (m t)}{m}+\frac{\pi \cos (m t)}{m}\right]_{t=-\pi}^{0}+\frac{1}{\pi}\left[-\frac{\pi \cos (m t)}{m}\right]_{t=0}^{\pi}\]
\[=\frac{1}{\pi}\frac{\pi m-\sin (\pi m)}{m^2}+\frac{1}{\pi}\frac{\pi -\pi \cos (\pi m)}{m}\]
substituting \(\sin (\pi m) = 0\) and \(\cos (\pi m) = (-1)^m\) for all \(m = 1, 2, 3, 4, ….\)
\[\beta_m=\frac{1}{m}+\frac{1-(-1)^m}{m}\]
\[\Rightarrow \beta_m=\begin{cases}\frac{3}{m} & \text{if } m \hspace{0.2cm}is \hspace{0.2cm}odd\\\frac{1}{m} & \text{if }m \hspace{0.2cm}is \hspace{0.2cm}even\end{cases}\hspace{10 cm} (3)\]
step (4): writing Fourier Series
Thus \(w(t)\) can be written as
\[w(t) = \frac{\alpha_0}{2} + \sum\limits_{m = 1}^\infty {\alpha_m} \cos (mt) + {\beta_m}\sin (m t)\]
where Fourier Coefficients are given in eqns. (1), (2), (3).