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

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(New page: ==Problem Statement== Find the Fourier Tranform with exponential basis functions of the sawtooth wave given by the equation <br /> <center><math>x(t)=t-\lfloor t \rfloor</math></cente...)
 
(First bit of work)
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==Author==


John Hawkins


==Problem Statement==
==Problem Statement==
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<br />
<br />

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 <math>a_n</math> such that

<br />

<center><math>x(t)=\sum_{-\infty}^\infty a_n e^{j2\pi nt/T}</math></center>

<br />

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

<br />

<center><math>a_n=\frac{1}{T}\int_c^{c+T}x(t) e^{-j2\pi nt/T}dt</math></center>

<br />

Noting again that our period for this function is <math>\ T=1</math>, we proceed:

<br />
<center>

<math>a_n=\frac{1}{1}\int_0^1te^{-j2\pi nt/1}dt</math>

</center>
<br />

Again, the case when <math>\ n=0</math> needs to be considered separately. In this case,

<br />
<center><math>a_0=\int_0^1t\ dt=\frac{1}{2}</math></center>
<br />

For <math>n\neq 0</math>, the above integral is solved easiest using integration by parts. So letting
<br />

<center>

<math>u=t\quad\Rightarrow\quad du=dt</math>

<br />

<math>dv=e^{-j2\pi nt}dt \quad\Rightarrow\quad v=\frac{1}{-j2\pi n}e^{-j2\pi nt}</math>

</center>

<br />
we have
<br />

<center>

<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>

Revision as of 16:17, 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 , 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