Exercise: Sawtooth Wave Fourier Transform

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Problem Statement

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

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


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

 x(t)=\frac{a_0}{2}+\sum^\infty_{n=1} \left[a_n\cos\frac{2\pi nt}{T}+b_n\sin\frac{2\pi nt}{T}\right]



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 since this allows us to reduce the equation to x(t)=t. It remains, then, only to find the expression for a_n and b_n. We proceed first to find b_n.

b_n=\frac{2}{1}\int_0^1t\sin 2\pi nt\ dt

which is solved easiest with integration by parts, letting

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

dv=\sin2\pi nt\ dt\qquad\Rightarrow\qquad v=-\frac{1}{2\pi n}\cos 2\pi nt


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 a_n we must consider the case when n=0 separately.

a_0=\frac{2}{1}\int_0^1t\ dt=t^2\bigg|_0^1=1

For n=1,2,3\dots, we have

a_n=\frac{2}{1}\int_0^1t\cos 2\pi nt\ dt

which again is best solved using integration by parts, this time with

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

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


a_n=2\left[t\left(\frac{1}{2\pi n}\right)\sin 2\pi nt\bigg|_0^1-\int_0^1\frac{1}{2\pi n}\sin 2\pi nt\ dt\right]

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

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

\ =0

Therefore, the Fourier Transform representation of the sawtooth wave given is:

x(t)=\frac{1}{2}-\sum_{n=1}^\infty \frac{1}{\pi n}\sin 2\pi nt

Solution Graphs

The figures below graph the first few iterations of the above solution. The first graph shows the solution truncated after the first 100 terms of the infinite sum, as well as each of the contributing sine waves with offset. The second figure shows the function truncated after 1, 3, 5, 10, 50, and 100 terms. The last figure shows the Error between the Fourier Series truncated after the first 100 terms and the function itself. These figures were constructed using the following matlab code: SawToothFourier.

First 100 Terms.jpg

First n Terms.jpg


Solution as number of terms varies from 1 to 30.


John Hawkins

Read By

Christopher Garrison Lau I

Reviewed By

Colby Fullerton