Exercise: Sawtooth Wave Fourier Transform

Problem Statement

Find the Fourier Tranform of the sawtooth wave given by the equation

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

Solution

As shown in class, the general equation for the Fourier Transform for a periodic function with period $T$ is given by

$x(t)=\sum _{n=0}^{\mathrm{\infty }}[a_n\cos \frac{2\pi nt}{T}+b_n\sin \frac{2\pi nt}{T}]$

where

$$\begin{gathered} \{\begin{array}{l}{a}_n=\frac{2}{T}\underset{c}{\overset{c+T}{\int }}x(t)\cos \frac{2\pi nt}{T}dt\\ b_n=\frac{2}{T}\underset{c}{\overset{c+T}{\int }}x(t)\sin \frac{2\pi nt}{T}dt\end{array} \\ n=0,1,2,3\dots \end{gathered}$$

For the sawtooth function given, we note that $T=1$, and an obvious choice for $c$ is 0. It remains, then, only to find the expression for $a_n$ and $b_n$. We proceed first to find $b_n$. For $b_n$ we can ignore the case when $n=0$ because $\sin 0=0$. Hence, we proceed for $n=1,2,3\dots$:

$$\begin{gathered} b_n=\frac{2}{1}\underset{0}{\overset{1}{\int }}t\sin 2\pi nt \\ dt \end{gathered}$$

which is solved easiest with integration by parts, letting

$u=t\Rightarrow du=dt$

$$\begin{gathered} dv=\sin 2\pi nt \\ dt\Rightarrow v=-\frac{1}{2\pi n}\cos 2\pi nt \end{gathered}$$

so

$$\begin{gathered} b_n=2[t(-\frac{1}{2\pi n})\cos 2\pi nt{|}_0^1+\frac{1}{2\pi n}\underset{0}{\overset{1}{\int }}\cos 2\pi nt \\ dt] \end{gathered}$$

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

$=2[-\frac{1}{2\pi n}(1)+0]$

$=-\frac{1}{\pi n}$

Now, for

Author

John Hawkins