Fourier series

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Question statement

Consider the following periodic function

$f(t)=\left\{\matrix{\simplify{{p}*t}&\text{if }0\leq t<\pi\\\var{q}&\text{if }\pi\leq t<2\pi}\right.$

where $f(t+2\pi)=f(t)$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

a0	number	0.0707963268
a1	number	-1.2732395447
a2	integer	0
a3	number	-0.1414710605

			Name	Type	Generated Value

b1	number	3.9098593171
b2	rational	-1
b3	decimal	dec("1.303286439034247666666666666666666666667")

			Name	Type	Generated Value

			p	integer	2
			q	integer	-3

			Name	Type	Generated Value

			t1	number	0.0901406829
			t2	decimal	dec("1.363380227632419")

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
(p*pi+2*q)/4

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `number`

0.0707963268

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Advice
"Unnamed part" → "Unnamed gap" - Correct answer

Parts

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1. Gap Mathematical expression
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Determine the Fourier series (up to the third harmonic) of $f(t)$ .

$f(t)=$ Unnamed gap

Using graphing software (e.g. GeoGebra or Desmos), plot the graphs of both the original function and your Fourier series on the same axes.

Advice

First calculate $a_0$ as the mean value of $f(t)$ over one period:

$a_0=\frac{1}{2}\times\var{p}\times\frac{\pi}{2}+\frac{1}{2}\times\var{q}=\var{a0}$

Now calculate the coefficients $a_n$ :

$a_n=\frac{1}{\pi}\int_0^{2\pi}f(t)\cos(nt)\;dt=\frac{\var{p}}{\pi}\int_0^\pi t\cos(nt)\;dt+\frac{\var{q}}{\pi}\int_\pi^{2\pi}\cos(nt)\;dt$

The first integral is performed by parts:

$\frac{\var{p}}{\pi}\int_0^\pi t\cos(nt)\;dt=\frac{\var{p}}{n\pi}\left[t\sin(nt)\right]_0^\pi-\frac{\var{p}}{n\pi}\int_0^\pi\sin(nt)\;dt=\frac{\var{p}}{n\pi}\left[t\sin(nt)\right]_0^\pi+\frac{\var{p}}{n^2\pi}\left[\cos(nt)\right]_0^\pi$

Therefore

$a_n=\frac{\var{p}}{n\pi}\left[t\sin(nt)\right]_0^\pi+\frac{\var{p}}{n^2\pi}\left[\cos(nt)\right]_0^\pi+\frac{\var{q}}{n\pi}\left[\sin(nt)\right]_\pi^{2\pi}$

When $n$ is odd,

$a_n=\frac{\var{p}}{n\pi}\left(0-0\right)+\frac{\var{p}}{n^2\pi}\left(-1-1\right)+\frac{\var{q}}{n\pi}\left(0-0\right)=\simplify{-2*{p}/(n^2*pi)}$

And when $n$ is even,

$a_n=\frac{\var{p}}{n\pi}\left(0-0\right)+\frac{\var{p}}{n^2\pi}\left(1-1\right)+\frac{\var{q}}{n\pi}\left(0-0\right)=0$

Therefore we obtain the following coefficients:

$a_1=\simplify{-2*{p}/pi}=\var{a1}\qquad a_2=0\qquad a_3=\simplify{-2*{p}/(3^2*pi)}=\var{a3}$

Now calculate the coefficients $b_n$ :

$b_n=\frac{1}{\pi}\int_0^{2\pi}f(t)\sin(nt)\;dt=\frac{\var{p}}{\pi}\int_0^\pi t\sin(nt)\;dt+\frac{\var{q}}{\pi}\int_\pi^{2\pi}\sin(nt)\;dt$

The first integral is performed by parts:

$\frac{\var{p}}{\pi}\int_0^\pi t\sin(nt)\;dt=-\frac{\var{p}}{n\pi}\left[t\cos(nt)\right]_0^\pi+\frac{\var{p}}{n\pi}\int_0^\pi \cos(nt)\;dt=-\frac{\var{p}}{n\pi}\left[t\cos(nt)\right]_0^\pi+\frac{\var{p}}{n^2\pi}\left[\sin(nt)\right]_0^\pi$

Therefore

$b_n=-\frac{\var{p}}{n\pi}\left[t\cos(nt)\right]_0^\pi+\frac{\var{p}}{n^2\pi}\left[\sin(nt)\right]_0^\pi+\frac{\var{-q}}{n\pi}\left[\cos(nt)\right]_\pi^{2\pi}$

When $n$ is odd,

$b_n=-\frac{\var{p}}{n\pi}\left(-\pi-0\right)+\frac{\var{p}}{n^2\pi}\left(0-0\right)+\frac{\var{-q}}{n\pi}\left(1--1\right)=\simplify{{p}/n-2*{q}/(n*pi)}$

And when $n$ is even,

$b_n=-\frac{\var{p}}{n\pi}\left(\pi-0\right)+\frac{\var{p}}{n^2\pi}\left(0-0\right)+\frac{\var{-q}}{n\pi}\left(1-1\right)=\simplify{-{p}/n}$

Therefore we obtain the following coefficients:

$b_1=\simplify{{p}-2*{q}/pi}=\var{b1}\qquad b_2=\simplify{-{p}/2}=\var{b2}\qquad b_3=\simplify{{p}/3-2*{q}/(3*pi)}=\var{b3}$

The Fourier series for $f(t)$ (to the third harmonic) is:

$f(t)=a_0+a_1\cos(t)+a_2\cos(2t)+a_3\cos(3t)+b_1\sin(t)+b_2\sin(2t)+b_3\sin(3t)$

On substituting the values of the coefficients $a_n$ and $b_n$ this becomes:

$f(t)=\simplify{{a0}+{a1}*cos(t)+{a2}*cos(2t)+{a3}*cos(3t)+{b1}*sin(t)+{b2}*sin(2t)+{b3}*sin(3t)}$

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Fourier series

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Taxonomy: mathcentre

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Martin Jones 2 years, 5 months ago

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