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

$x(t)=\sum _{n=0}^{\infty }\left[a_{n}\cos {\frac {2\pi nt}{T}}+b_{n}\sin {\frac {2\pi nt}{T}}\right]$

where

${\begin{cases}a_{n}={\frac {2}{T}}\int _{c}^{c+T}x(t)\cos {\frac {2\pi nt}{T}}dt\\b_{n}={\frac {2}{T}}\int _{c}^{c+T}x(t)\sin {\frac {2\pi nt}{T}}dt\end{cases}}\ \ \ \ n=0,1,2,3\dots$

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 $n=1,2,3\dots$:

$b_{n}={\frac {2}{1}}\int _{0}^{1}t\sin 2\pi nt\ dt$

which is solved easiest with integration by parts, letting

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

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

so

$b_{n}=2\left[t\left(-{\frac {1}{2\pi n}}\right)\cos 2\pi nt{\bigg |}_{0}^{1}+{\frac {1}{2\pi n}}\int _{0}^{1}\cos 2\pi nt\ dt\right]$

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

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

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

Now, for

Author

John Hawkins