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

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

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

$a_{n}={\frac {1}{T}}\int _{c}^{c+T}x(t)e^{-j2\pi nt/T}dt$

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

$a_{n}={\frac {1}{1}}\int _{0}^{1}te^{-j2\pi nt/1}dt$

For $n\neq 0$, the above integral is solved easiest using integration by parts. When $\ n=0$,however, IBP does not work, so the case when $\ n=0$ needs to be considered separately. In this case,

$a_{0}=\int _{0}^{1}t\ dt={\frac {1}{2}}$

For $n\neq 0$, we continue with IBP, letting

$u=t\quad \Rightarrow \quad du=dt$

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

This gives

$a_{n}=t\left({\frac {1}{-j2\pi n}}\right)e^{-j2\pi nt}{\Bigg |}_{0}^{1}-\int _{0}^{1}\left({\frac {1}{-j2\pi n}}\right)e^{-j2\pi nt}dt$

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

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

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,

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

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. Note that the "terms" mentioned in the titles of the graphs should now be interpreted as the sum of the $n$th and $-n$th terms.

Analytical Comparison of Two Solutions

To convince myself that the two solutions are actually the same, I performed the following analysis. Let $t_{n}$ be the nth term of the solution found on this page. Then for $n\neq 0$,

$t_{n}+t_{-n}={\frac {1}{-j2\pi n}}e^{j2\pi nt}+{\frac {1}{-j2\pi (-n)}}e^{j2\pi (-n)t}$

$={\frac {1}{-j2\pi n}}(\cos 2\pi nt+j\sin 2\pi nt)-{\frac {1}{-j2\pi n}}\left(\cos(-2\pi nt)+j\sin(-2\pi nt)\right)$

$={\frac {1}{-j2\pi n}}\left(\cos 2\pi nt+j\sin 2\pi nt-\cos 2\pi nt+j\sin 2\pi nt\right)$

$={\frac {1}{-j2\pi n}}\left(2j\sin 2\pi nt\right)$

$=-{\frac {1}{\pi n}}\sin 2\pi nt$

Therefore,

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

As we had hoped and expected.

Reviewed By

Christopher Garrison Lau I

Tyler Starr