Fourier Series Solved Examples

Example 1: Find the Fourier series expansion of the function f(x) = e^x, within the limits  [– π, π].

Solution:

Using Fourier series expansion.

[Tex]a_0 =  \dfrac{1}{2\pi}\int_{- \pi}^{\pi}e^x dx \\= \dfrac{e^\pi – e^{-\pi}}{2\pi} [/Tex] .

[Tex]a_n = \dfrac{1}{\pi}\int_{- \pi}^{\pi}e^x cos (nx) dx \\= \dfrac{1}{\pi}\dfrac{e^x}{1+n^2}[\cos(nx)+ n\sin(nx)]_{-\pi}^{\pi}\\= \dfrac{1}{\pi(1+n^2)}[e^\pi(-1)^n)-e^{-\pi}(-1)^n)] [/Tex] .

[Tex]b_n = \dfrac{1}{\pi}\int e^x sin(nx) dx \\= \dfrac{e^x}{\pi (1+n^2)}[\sin(nx)- n \cos(nx)]_{-\pi}^{\pi}\\= \dfrac{1}{\pi (1+n^2)}[e^\pi(-n(-1)^n) – e^{-\pi}(-n)(-1)^n] [/Tex] .

The Fourier series for this function is given as, 

[Tex]\dfrac{e^\pi -e^{-\pi}}{2\pi} + \sum_{n=1}^{\infty}\dfrac{(-1)^n(e^\pi – e^{-\pi})}{\pi(1+n^2)}[cos nx -n sin nx] [/Tex] .

Example 2: Find the Fourier series expansion of the function f(x) = x , within the limits  [– 1, 1].

Solution:

From Fourier series expansion. Here,

[Tex]A_{0}=\frac{1}{2 } \cdot \int_{-1}^{1} x d x [/Tex] .

[Tex]A_{n}=\frac{1}{1} \cdot \int_{-1}^{1} x \cos \left(\frac{n \pi x}{1}\right) d x, \quad n>0 [/Tex] .

[Tex]B_{n}=\frac{1}{1} \cdot \int_{-1}^{1} x \sin \left(\frac{n \pi x}{1}\right) d x, \quad n>0 [/Tex] .

[Tex]f(x)=A_{0}+\sum_{n=1}^{\infty} A_{n} \cdot \cos \left(\frac{n \pi x}{L}\right)+\sum_{n=1}^{\infty} B_{n} \cdot \sin \left(\frac{n \pi x}{L}\right) [/Tex] .

[Tex]\mathrm{f}(\mathrm{x})=\frac{1}{2 \cdot 1} \cdot \int_{-1}^{1}\left(x\right) d x+\sum_{n=1}^{\infty} \frac{1}{1} \cdot \int_{-1}^{1}\left(x\right) \cos \left(\frac{n \pi x}{1}\right) d x \cdot \cos \left(\frac{n \pi x}{1}\right)+\sum_{n=1}^{\infty} \frac{1}{1} \cdot \int_{-1}^{1}\left(x\right) \sin \left(\frac{n \pi x}{1}\right) d x \cdot \sin \left(\frac{n \pi x}{1}\right) [/Tex] .

Om solving the integrals we get even functions and one odd function. Therefore,

[Tex]f(x) = \sum _{n=1}^{\infty \:}-\frac{2\left(-1\right)^n\sin \left(\pi nx\right)}{\pi n} [/Tex].

Example 3: Suppose a function f(x) = tanx find its Fourier expansion within the limits [-π, π].

Solution:

[Tex] a_0 =  \dfrac{1}{\pi}\int_{- \pi}^{\pi} tanx dx     [/Tex]

[Tex]a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} tanx cosnxdx    [/Tex] 

[Tex]b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} tanx sinnxdx    [/Tex] 

[Tex]\large f(x)=\frac{1}{2}a_{0}+\sum_{n=1}^{\infty}a_{n}cos\;nx+\sum_{n=1}^{\infty}b_{n}sin\;nx    [/Tex]   

Now the integral of tanx⋅sinnx and tanx⋅cosnx cannot be found.

Therefore the Fourier series for this function f(x) = tanx is undefined.

Example 4: Find the Fourier series of the function f(x) = 1 for limits  [– π, π] .

Solution:

Comparing with general Fourier series expansion we get,

[Tex] a_0 =  \dfrac{1}{\pi}\int_{- \pi}^{\pi}1  dx    [/Tex] 

[Tex]a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} 1 \cdot cosnxdx    [/Tex] 

[Tex]b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} 1 \cdot sinnxdx .\\ \large f(x)=\frac{1}{2}a_{0}+\sum_{n=1}^{\infty}a_{n}cos\;nx+\sum_{n=1}^{\infty}b_{n}sin\;nx    [/Tex]  

f(x) = π + 0 + 0

f(x) = π

Example 5: Consider a function f(x) = x² for the limits  [– π, π]. Find its Fourier series expansion.

Solution:

Comparing with general Fourier series expansion we get,

[Tex] a_0 =  \dfrac{1}{\pi}\int_{- \pi}^{\pi}x^2  dx . \\ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cdot cosnxdx . \\ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cdot sinnxdx . \\ \large f(x)=\frac{1}{2}a_{0}+\sum_{n=1}^{\infty}a_{n}cos\;nx+\sum_{n=1}^{\infty}b_{n}sin\;nx \\ f(x) = \frac{{\pi}^{3}}{3} + \sum_{n=1}^{\infty}a_{n}cos\;nx +0 .\\ f(x) =  \frac{{\pi}^{3}}{3} + \sum_{n=1}^{\infty} \frac{4πcosnπcosnx}{n2}. [/Tex]

Example 6: Find Fourier series expansion of the function f(x) = 4-3x for the limits  [– 1, 1].

Solution:

Comparing with general Fourier series expansion we get,

[Tex]a_0 =  \dfrac{1}{\pi}\int_{- \pi}^{\pi}4-3x  dx . \\ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} 4-3x \cdot cosnxdx . \\ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} 4 -3x\cdot sinnxdx .\\ \large f(x)=\frac{1}{2}a_{0}+\sum_{n=1}^{\infty}a_{n}cos\;nx+\sum_{n=1}^{\infty}b_{n}sin\;nx .\\ f(x) =\frac{1}{2\cdot \:1}\cdot \:8+\sum _{n=1}^{\infty \:}\frac{1}{1}\cdot \:0\cdot \cos \left(\frac{n\pi x}{1}\right)+\sum _{n=1}^{\infty \:}\frac{1}{1}\left(\frac{6\left(-1\right)^n}{\pi n}\right)\sin \left(\frac{n\pi x}{1}\right).\\ f(x) = 4+\sum _{n=1}^{\infty \:}\frac{6\left(-1\right)^n\sin \left(\pi nx\right)}{\pi n} . [/Tex]

Example 7: Find the expansion of the function [Tex]1- \frac{x}{\pi} [/Tex]. For the limits  [– π, π].

Solution:

Comparing with general Fourier series expansion we get,

[Tex]a_0 =  \dfrac{1}{\pi}\int_{- \pi}^{\pi}4-3x  dx .\\ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} 4-3x \cdot cosnxdx .\\ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} 4 -3x\cdot sinnxdx .\\ f(x) = \frac{1}{2\pi }\cdot \int _{-\pi }^{\pi }\left(1-\frac{x}{\pi }\right)dx+\sum _{n=1}^{\infty \:}\frac{1}{\pi }\cdot \int _{-\pi }^{\pi }\left(1-\frac{x}{\pi }\right)\cos \left(\frac{n\pi x}{\pi }\right)dx\cdot \cos \left(\frac{n\pi x}{\pi }\right)+\sum _{n=1}^{\infty \:}\frac{1}{\pi }\cdot \int _{-\pi }^{\pi }\left(1-\frac{x}{\pi }\right)\sin \left(\frac{n\pi x}{\pi }\right)dx\cdot \sin \left(\frac{n\pi x}{\pi }\right) .\\ f(x) = =\frac{1}{2\pi }\cdot \:2\pi +\sum _{n=1}^{\infty \:}\frac{1}{\pi }\cdot \:0\cdot \cos \left(\frac{n\pi x}{\pi }\right)+\sum _{n=1}^{\infty \:}\frac{1}{\pi }\left(\frac{2\left(-1\right)^n}{n}\right)\sin \left(\frac{n\pi x}{\pi }\right).\\ f(x) = =1+\sum _{n=1}^{\infty \:}\frac{2\left(-1\right)^n\sin \left(nx\right)}{\pi n} . [/Tex]

Fourier Series is a sum of sine and cosine waves that represents a periodic function. Each wave in the sum, or harmonic, has a frequency that is an integral multiple of the periodic function’s fundamental frequency. Harmonic analysis may be used to identify the phase and amplitude of each harmonic. A Fourier series might have an unlimited number of harmonics. Summing some, but not all, of the harmonics in a function’s Fourier series, yields an approximation to that function. For example, a square wave can be approximated by utilizing the first few harmonics of the Fourier series.

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