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

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

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

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

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

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

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

we have

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

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

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

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.