Exercise: Sawtooth Redone With Exponential Basis Functions

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Revision as of 17:45, 19 January 2010 by John.hawkins (talk | contribs) (Intermediate progress)
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Author

John Hawkins

Problem Statement

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


x(t)=t−⌊t⌋


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=1T∫cc+Tx(t)e−j2πnt/Tdt


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


an=11∫01te−j2πnt/1dt


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


a0=∫01tdt=12


For n≠0, the above integral is solved easiest using integration by parts. So letting

u=t⇒du=dt


dv=e−j2πntdt⇒v=1−j2πne−j2πnt


we have

an=t(1−j2πn)e−j2πnt|01−∫01(1−j2πn)e−j2πntdt

=[1−j2πne−j2πn−0]−(1−j2πn)2e−j2πnt|01


=1−j2πne−j2πn−(1−j2πn)2(e−j2πn−1)


But


e−j2πn=cos⁡(−2πn)+jsin⁡(−2πn)=1+j0=1


So


an=1−j2πn(1)−(1−j2πn)2(1−1)=1−j2πn


Therefore,


x(t)=12−∑n=±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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