Inspired by Cliff Pickovers tweet and powered by my fascination about the Fourier transform, I took the time to derive and implement the Fourier series of a two dimensional piecewise linear parametric curve.

By clicking on the empty canvas you can design your custom shape. If you close the path, the Fourier series will be animated similarly to the video mentioned above. Have fun!

Derivation

The following derivation shows a way to obtain the equations of the Fourier series of a two dimensional piecewise linear parametric curve. Please be aware, that this is certainly not the most efficient way to derive it.

We will treat the \(x\) and \(y\) component of the function separately. Therefore, we calculate the Fourier series for both components separately.

Fourier series

The complex Fourier series is defined as

with

Our component functions \(y(t)\) are piecewise linear and can be defined as

Let’s calculate the Fourier coefficients. We start with \(k=0\):

Now we look at the general case \(k>0\):

The first term simplifies to

We now look at the second term

Fortunately, we know the identity

We can use it, by applying the coordinate transformation to our integral

We obtain

Now we have to integrals of the identity above. The first integral is

with \(a = ik\omega (T_{i+1} - T_i) \) and \(n=0\). It follows

The second integral is

with \(a = ik\omega (T_{i+1} - T_i) \) and \(n=1\).

It follows

We insert the results back into our equation

We replace the exponential terms with \(\sin\) and \(\cos\)

We now look again at our first term from above and separate it

into

and

We combine the result of the second term with first equation of the separation

Now we incorporate the second part of the separation.

With the helping variable

follows

The resulting formula is

We note that we can \(\frac{1}{T}\frac{1}{k\omega} = \frac{1}{2\pi k}\) and finally obtain

In the final step, we can express our result as

with coefficients

and