2203 Fourier Series Definitions and Coefficients

We will first state Fourier's theorem for periodic functions with period $P = 2\pi$. In words, the theorem says that a function with period $2\pi$ can be written as a sum of cosines and sines which all have period $2\pi$.

Theorem (Fourier)

Suppose $f(t)$ has period $2\pi$ then we have $$\begin{aligned} f(t)&\sim\frac{a_0}{2}+a_1\cos t+a_2\cos 2t+a_3\cos 3t + \ldots+b_1\sin t+b_2\sin 2t+b_3\sin 3t+\ldots\\ &=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos nt+b_n\sin nt\tag{1} \end{aligned}$$ where the coefficients $a_0, a_1, \ldots$ and $b_1, b_2 \ldots$ are computed by $$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)dt\\ a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(nt)dt\\ b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin(nt)dt\tag{2}$$ Some comments are in order.
1. As we saw in the quiz above, each of the functions $\cos(t), \cos(2t), \cos(3t), \ldots$ all have $2\pi$ as a period. The same is clearly true for $\sin(t), \sin(2t), \sin(3t), \ldots$. 2. The series on the right-hand side $(1)$ is called a Fourier series; and the coefficients, $a_0, a_1, \ldots b_1, b_2 \ldots$ in $(2)$ are called the Fourier coefficients of $f(t)$. 3. The letter a is used in $a_0/2$ because we can think of it as the coefficient of $\cos(0\cdot t) = 1$. We don't need a $b_0$ term because $\sin(0\cdot t) = 0$. The term constant term $\frac{a_0}{2}$ is written in this way to make the formula for $a_0$ look just like those of the other cosine coefficients $a_n$. (We will see why we need the factor of $\frac{1}{2}$ in a later note when we prove that these formulas really do give the coefficients.) 4. In $(1)$ we used the symbol $\sim$ instead of an equal sign because the two sides of $(1)$ might differ at those values of $t4 where$f(t)$is discontinuous. For us, this is a minor point and we will allow ourselves to use an equal sign from now on. 5. There is some terminology coming from acoustics and music: the$n = 1$frequency is called the **fundamental**, and the frequencies$n \geq 2$ are called the higher harmonics (or overtones). We will explore the connection between Fourier series and sound in a later session.

Example

Compute the Fourier series of $f(t)$, where $f(t)$ is the square wave with period $2\pi$. which is defined over one period by $$f(t)=\begin{cases} -1&\text{for }-\pi\leq t < 0\\ 1&\text{for }0\leq t <\pi \end{cases}$$ The graph over several periods is shown below.

Solution Computing a Fourier series means computing its Fourier coefficients.
In applying these formulas to the given square wave function, we have to split the integrals into two pieces corresponding to where $f(t)$ is +1 and where it is -1. We find
$$\begin{aligned} a_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(nt)dt\\ &=\frac{1}{\pi}\int_{-\pi}^{0}(-1)\cdot \cos(nt)dt+\frac{1}{\pi}\int_{0}^{\pi}(1)\cdot \cos(nt)dt \end{aligned}$$ Thus, for $n\neq0$
$$a_n=-\frac{\sin(nt)}{n\pi}\bigg|_{-\pi}^0 + \frac{\sin(nt)}{n\pi}\bigg|_0^\pi=0$$ and for $n=0$
$$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)dt=0 \tag{characteristic of odd function}$$ Likewise $$ $$\begin{aligned} b_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin(nt)dt\\ &=\frac{1}{\pi}\int_{-\pi}^{0}(-1)\cdot \sin(nt)dt+\frac{1}{\pi}\int_{0}^{\pi}(1)\cdot \sin(nt)dt\\ &=\frac{\cos(nt)}{n\pi}\bigg|_{-\pi}^0 - \frac{\cos(nt)}{n\pi}\bigg|_0^\pi\\ &=\frac{1-\cos(-n\pi)}{n\pi}-\frac{cos(n\pi)-1}{n\pi}\\ &=\frac{2}{n\pi}(1-\cos(n\pi))\\ &=\frac{2}{n\pi}(1-(-1)^n)\\ &=\begin{cases} \frac{4}{n\pi}&\text{for $n$ odd}\\ 0&\text{for $n$ even} \end{cases} \end{aligned}$$ $$ We have used the simplification $\cos n\pi = (-1)^n$ to get a nice formula for the coefficients $b_n$.

This then gives the Fourier series for $f(t)$: $$f(t)=\sum_{n=1}^\infty b_n\sin(nt)=\frac{4}{n\pi}(\sin t+\frac{1}{3}\sin 3t + \frac{1}{5}\sin 5t + \ldots)$$ seeing the convergence of a Fourier series
However, it is not easy to see that the sum on the right-hand side is in fact converging to the square wave $f(t)$. So let's use a computer to plot the sums of the first $N$ terms of the series. for $N = 1, 3, 9, 33$. We get the following four graphs:

Notice that since a finite sum of sine functions is continuous (in fact smooth), the partial sums cannot jump when $t$ is an integer multiple of $\pi$, the way the square way $f(t)$ does. But they are certainly "trying" to become the square wave $f(t)$! And the more terms you add in, the better the fit, with the theoretical limit as $N \rightarrow \infty$ being exactly equal to $f(t)$ (except actually at the jumps $t = n\pi$, as we'll see).

Note: In this case we don't have any cosine terms, just sine. This turns out to be not an accident: it follows from the fact that $f(t)$ here is an odd function, i.e. $f(-t) = -f(t)$, and such functions have only sines (which are also odd functions) in their Fourier series. Similarly for even functions and cosine series: if $f(t)$ is even ($f(-t) = f(t)$) then all the $b_n$'s vanish and the Fourier series is simply $f(t)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos nt$; while if $f(t)$ is odd then all the $a_n$'s vanish and the Fourier series is $f(t)=\sum_{n=1}^{\infty}b_n\sin nt$.