Exercise: Sawtooth Redone With Exponential Basis Functions

Author

John Hawkins

Problem Statement

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

$x(t)=t-\lfloor t\rfloor$

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 $a_n$ such that

$x(t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n{e}^{j2\pi nt/T}$

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

$a_n=\frac{1}{T}\underset{c}{\overset{c+T}{\int }}x(t)e^{-j2\pi nt/T}dt$

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

$a_n=\frac{1}{1}\underset{0}{\overset{1}{\int }}t{e}^{-j2\pi nt/1}dt$

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

$$\begin{gathered} a_0=\underset{0}{\overset{1}{\int }}t \\ dt=\frac{1}{2} \end{gathered}$$

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

$u=t\Rightarrow du=dt$

$dv=e^{-j2\pi nt}dt\Rightarrow v=\frac{1}{-j2\pi n}{e}^{-j2\pi nt}$

we have

$a_n=t\left(\frac{1}{-j2\pi n}\right)e^{-j2\pi nt}{|}_0^1-\underset{0}{\overset{1}{\int }}\left(\frac{1}{-j2\pi n}\right)e^{-j2\pi nt}dt$

$=[\frac{1}{-j2\pi n}{e}^{-j2\pi n}-0]-{\left(\frac{1}{-j2\pi n}\right)}^2{e}^{-j2\pi nt}{|}_0^1$

$=\frac{1}{-j2\pi n}{e}^{-j2\pi n}-{\left(\frac{1}{-j2\pi n}\right)}^2(e^{-j2\pi n}-1)$

But

${e}^{-j2\pi n}=\cos (-2\pi n)+j\sin (-2\pi n)=1+j0=1$

So

$a_n=\frac{1}{-j2\pi n}(1)-{\left(\frac{1}{-j2\pi n}\right)}^2(1-1)=\frac{1}{-j2\pi n}$

Therefore,

$x(t)=\frac{1}{2}-\sum _{n=\pm 1}^{\pm \mathrm{\infty }}\frac{1}{j2\pi n}{e}^{j2\pi nt}$

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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