Fourier series

Error

There was an error loading the page.

Contributors

Martin Jones

Feedback

From users who are members of Martin's workspace :

Martin Jones

said

Ready to use

2 years, 7 months ago

History

○

Martin Jones 2 years, 6 months ago

Published this.

○

Martin Jones 2 years, 7 months ago

Gave some feedback: Ready to use

○

Martin Jones 2 years, 7 months ago

Created this.

There is only one version of this question that you have access to.

There are 2 other versions that do you not have access to.

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	3.6415926536
a1	number	-2.5464790895
a2	integer	0
a3	number	-0.2829421211

			Name	Type	Generated Value

b1	number	3.3633802276
b2	rational	-2
b3	decimal	dec("1.121126742544139333333333333333333333333")

			Name	Type	Generated Value

			p	integer	4
			q	integer	1

			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.

Can an exam override the value of this variable?

Generated value: `number`

3.6415926536

← Depends on:

This variable doesn't seem to be used anywhere.

Parts

Gap-fill
1. Gap Mathematical expression
  Add

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

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:

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

Give a worked solution to the whole question.

Use this tab to check that this question works as expected.

There was an error which means the tests can't run:

Part	Test	Passed?
Gap-fill
Gap-fill		Hasn't run yet
Mathematical expression
Mathematical expression		Hasn't run yet

This question is used in the following exams:

Loading...

Error

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Martin Jones 2 years, 6 months ago

Martin Jones 2 years, 7 months ago

Martin Jones 2 years, 7 months ago

Error in variable testing condition

Error

Warning

Generated value: `number`

Parts

Gap-fill

Loading...

Error

Fourier series

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Martin Jones 2 years, 6 months ago

Martin Jones 2 years, 7 months ago

Martin Jones 2 years, 7 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

Generated value: `number`