Exercise: Sawtooth Redone With Exponential Basis Functions: Difference between revisions

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(First bit of work)
(Intermediate progress)
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<center><math>x(t)=\sum_{-\infty}^\infty a_n e^{j2\pi nt/T}</math></center>
<center><math>x(t)=\sum_{n=-\infty}^\infty a_n e^{j2\pi nt/T}</math></center>


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Noting again that our period for this function is <math>\ T=1</math>, we proceed:
Noting again that our period for this function is <math>T=1</math> and that an obvious choice for <math>c</math> is zero, we proceed:


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<math>a_n=t\left(\frac{1}{-j2\pi n}\right)e^{-j2\pi nt}\Bigg|_0^1-\int_0^1\left(\frac{1}{-j2\pi n}\right)e^{-j2\pi nt}dt</math>
<math>a_n=t\left(\frac{1}{-j2\pi n}\right)e^{-j2\pi nt}\Bigg|_0^1-\int_0^1\left(\frac{1}{-j2\pi n}\right)e^{-j2\pi nt}dt</math>
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<math>=\left[\frac{1}{-j2\pi n}e^{-j2\pi n}-0\right]-\left(\frac{1}{-j2\pi n}\right)^2e^{-j2\pi nt}\Bigg|_0^1</math>

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<math>=\frac{1}{-j2\pi n}e^{-j2\pi n}-\left(\frac{1}{-j2\pi n}\right)^2\left(e^{-j2\pi n}-1\right)</math>

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But

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<math>\ e^{-j2\pi n}=\cos(-2\pi n)+j\sin(-2\pi n) = 1+j0=1</math>

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So

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<math>a_n=\frac{1}{-j2\pi n}(1)-\left(\frac{1}{-j2\pi n}\right)^2(1-1)=\frac{1}{-j2\pi n}</math>

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Therefore,

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<center><math>x(t)=\frac{1}{2}-\sum_{n=\pm 1}^{\pm \infty}\frac{1}{j2\pi n}e^{j2\pi nt}</math></center>
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==Solution Graphs==
I modified the Matlab code used in the [[Exercise: Sawtooth Wave Fourier Transform]] to generate the solution graphs using the equation found above instead of the previously found solution. This code can be found here: [[Sawtooth2 Matlab Code]]. It generates the following analagous three graphs, which as hoped appear exactly identical to those found using the other method.
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[[Image:Sawtooth2_First_100_Terms.jpg|thumb|800px|center]]

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[[Image:Sawtooth2_First_n_Terms.jpg|thumb|800px|center]]

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[[Image:Sawtooth2_Error.jpg|thumb|800px|center]]

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Revision as of 16:45, 19 January 2010

Author

John Hawkins

Problem Statement

Find the Fourier Tranform with exponential basis functions of the sawtooth wave given by the equation



Note that this is the same function solved in Exercise: Sawtooth Wave Fourier Transform, but solved differently to compare the two methods.

Solution

The goal of this method is to find the coefficients such that



In class we showed not only that this was possible, but also that



Noting again that our period for this function is and that an obvious choice for is zero, we proceed:



Again, the case when needs to be considered separately. In this case,



For , the above integral is solved easiest using integration by parts. So letting



we have




But



So



Therefore,



Solution Graphs

I modified the Matlab code used in the Exercise: Sawtooth Wave Fourier Transform to generate the solution graphs using the equation found above instead of the previously found solution. This code can be found here: Sawtooth2 Matlab Code. It generates the following analagous three graphs, which as hoped appear exactly identical to those found using the other method.

Sawtooth2 First 100 Terms.jpg


Sawtooth2 First n Terms.jpg


Sawtooth2 Error.jpg