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 17:17, 19 January 2010

Author

John Hawkins

Problem Statement

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


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)=∑−∞∞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, 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