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 17:45, 19 January 2010

Author

John Hawkins

Problem Statement

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


x(t)=tt


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 an such that


x(t)=n=anej2πnt/T


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


an=1Tcc+Tx(t)ej2πnt/Tdt


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


an=1101tej2πnt/1dt


Again, the case when n=0 needs to be considered separately. In this case,


a0=01tdt=12


For n0, the above integral is solved easiest using integration by parts. So letting

u=tdu=dt


dv=ej2πntdtv=1j2πnej2πnt


we have

an=t(1j2πn)ej2πnt|0101(1j2πn)ej2πntdt

=[1j2πnej2πn0](1j2πn)2ej2πnt|01


=1j2πnej2πn(1j2πn)2(ej2πn1)


But


ej2πn=cos(2πn)+jsin(2πn)=1+j0=1


So


an=1j2πn(1)(1j2πn)2(11)=1j2πn


Therefore,


x(t)=12n=±1±1j2πnej2πnt


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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