# Exponential Fourier Series

**Compute the complex exponential fourier series coefficient for:**

So what is an exponential Fourier series, and why do we use it?

The Fourier series is a way to change a signal x(t) from the time domain to the frequency domain X(w)--where w stands for omega--using an infinite series as an approximation. This is because if we wanted to find the output response y(t) when x(t) is put through some system h(t) in the time domain, we would have to perform a convolution integral, which can be very difficult in cases where the input x(t) is periodic like this case.

If using X(w) (the frequency domain), the output y(t) is just X(w) multiplied by H(w) (the fourier series of x(t) and h(t)), bypassing the need for a complicated convolution integral and reducing the problem down to multiplication.

So for a complex exponential fourier series:

Cn is the actual exponential series coefficient. j is the square root of -1 (also called i). w0 is the fundamental frequency, also defined as 2 times pi over the period of x(t), and n is the set of all integers and stands for the harmonic. Fourier series are infinite series, so there are an infinite amount of harmonics that you could calculate and add up (usually the more you calculate the closer the fourier series approximation is to the actual signal x(t)); so for instance if you were asked for the second harmonic, you would plug in n = 2 for Cn.

Before we start, we should also go over what e^(-jnwt) even is:

Now onto the actual problem.

First thing we should do is determine the period T, and from that, w0.

T is the period, so how long it takes for the graph to repeat itself. So T = 3 in this case, since that is how long it takes for the you to go from the left edge of one of the boxes to the left edge of the other:

We then just use the formula for Cn:

Keep in mind that -jnw0 is just a constant while integrating with respect with t.

This raises a problem however. There is an n in the denominator, so this will be undefined for n=0, so we need to specifically solve for that case. We do that by plugging in n = 0 to the Cn formula before we take the integral.