Signals and Systems Part 2: Fourier Series

Lesson 8: The Sine/Cosine Fourier Series

The Full Sine/Cosine Fourier Series

In the last few lessons, we learned that you can create odd periodic signals by adding together a bunch of sinewaves, and even periodic signals by adding together a bunch of cosines, as long as each sine and cosine is multiplied by the right amount, which is given by the Fourier Coefficients, $a_n$ (for the cosines) and $b_n$ for the sines. We can also handle signal offsets using the $a_0$ term of the cosine Fourier Series, as long as we remember to divide that by 2. We learned that any signal can be broken up into even and odd parts, and so we can represent any signal using a combination of sines and cosines.

The Recipe

So, here's the recipe for constructing our signal out of sines and cosines:

Find the Sine Fourier Coefficients $b_n$
Find the Cosine Fourier Coefficients $a_n$
Add up the offset $a_0/2$, all the sinewaves $\sum_{n=1}^{n=\infty}b_n*sin(nx)$, and the cosines $\sum_{n=1}^{n=\infty}a_n*cos(nx)$ and you've got your original signal!

Now, usually we can't add all the way up to infinity, I have a finite amount of patience and my computer has a finite amount of memory. So when we want to plot the signals or deal with them on a computer, we usually truncate the series at some point. But when dealing with them mathematically, we can usually just leave the sums as-is, or evaluate them ananlytically. The more important part of the Fourier Series, for engineers at least, is getting an intuitive feel for what is going on behind the Fourier Transform.

A shifted, offset square wave

To show this, let's bring together everything we have learned so far and attack a signal that is neither even nor odd, a square wave shifted (delayed) by $\pi/4$:

We can't use just a Sine Fourier Series for this, and we can't use just a Cosine Fourier series, we have to use both. Let's do them one at a time. First, let's start with the Sine Fourier Series. Since the sine Fourier series can only represent odd signals, it's going to "pull out", so to speak, the odd part of our signal. We don't actually have to find the odd part of the signal ourselves, the Fourier Series will do that for us. But let's do it anyway, just so we can check that everything makes sense - what is the odd part of our signal? Well, remember, from our previous lesson on even and odd signals we can find the odd part by mirroring our original signal and subtracting it. If we do that, we'll get something that looks like this:

Huh. Cute. What do we actually get if we compute the Fourier Coefficients and add them up? Let's start with just the first Fourier Coefficient $b_1$.

What is the first Sine Fourier Coefficient, $b_1$? (Hint: I recommend integrating from $\pi/4$ to $5\pi/4$, since this is the only part of one period where the function is nonzero)
$b_1 = 4/\pi$
$b_1 =2\sqrt{2}/\pi$
$b_1 = 1$

Now, we could go on to compute the second coefficient, and the third coefficient, and so on, but it's much more convenient if we go straight to the nth coefficient. If you do the integral (again, I recommend splitting it into two integrals, from $\pi/4$ to $5\pi/4$ and $5\pi/4$ to $9\pi/4$), after the dust settles, you should get this:

\begin{equation} b_n=\frac{2}{n\pi}\left( cos\left(\frac{n\pi}{4}\right) - cos\left(\frac{5n\pi}{4}\right)\right) \end{equation}

Now, for the big question - does adding up all the sinewaves scaled by their Sine Fourier Coefficients actually produce the signal we expect? Well, let's plot the expected odd part of our signal with just the first sinewave, $b_1*sin(x)$:

Huh. The sinewave does seem to trace out the odd part of our signal, but let's keep adding more sinewaves. Here's what it looks like after adding up to $b_5 * sin(5x)$:

And up to 25 sinewaves:

Fascinating! It's doing exactly what we expect. The Sine Fourier Series is creating the odd part of our signal! We could keep going, and it would get more and more accurate, but I think you get the point. Now, what about the cosine part?

The Cosine Series

Just like we created the odd part of our signal, we can also create the even part, by mirroring our signal and adding it. If you do that, you should get this:

Notice that this signal has an offset - the even part of the signal contains the offset. This makes sense, because an offset (a constant), is the same everywhere, and so is symmetric about the origin. By now you know the drill, let's just skip straight to finding the nth Cosine Fourier Coefficinet $a_n$ with our square wave signal (let's call it $f(x)$ because we're creative like that).

\begin{equation} a_n = \frac{2}{P}\langle f(x), cos(nx) \rangle = \frac{2}{2\pi}\int_{\pi/4}^{9\pi/4}f(x)*cos(nx)dx \end{equation} As with the Sine Fourier Coefficients, we only actually have to do this integral from $\pi/4$ to $5\pi/4$, since that's the only part of $f(x)$ which is nonzero: \begin{equation} a_n = \frac{1}{\pi}\int_{\pi/4}^{5\pi/4}cos(nx)dx \end{equation}

And we end up getting basically the same answer as with the Sine Fourier Coefficients, but with cosines replaced by negative sines:

\begin{equation} a_n = -\frac{2}{n\pi}\left(sin\left(\frac{n\pi}{4}\right) - sin\left(\frac{5n\pi}{4}\right)\right) \end{equation}

Now, let's start adding stuff up and see if we reconstruct the even part of the signal. Let's start with the offset, $a_0/2$. What is this? Well, if you just try to plug it in above, you'll get $0/0$, not terribly useful. By taking the limit as $n\rightarrow0$, you can find the coefficient $a_0$, and the offset of the signal:

What is the coefficient $a_0$ and the signal offset?
$a_0 = 4, offset=2$
$a_0 = 2, offset = 1$
$a_0 = 1, offset=2$

Let's plot the signal we expect to see (the even part of our original signal, in blue) and the offset (in orange) that we found so far:

Looks reasonable to me. The offset of 1 slices our even signal right in half. Now, what happens when we add $a_1*cos(1x)$?

Let's keep going! Let's add up to $a_5 * cos(5x)$:

Let's add all the terms up to $a_{25} * cos(25x)$:

Exactly what we expected! Again, we could keep adding terms to get better and better accuracy, but let's stop there for now.

Bringing it all together

Finally, if we add everything together: the offset $a_0/2$, all the appropriately-scaled cosines $a_1*cos(1x) + a_2*cos(2x) + a_3*cos(3x) + ...$, and the appropriately scaled sinewaves $b_1*sin(1x) + b_2*sin(2x) + b_3*sin(3x) + ...$ we could fully reconstruct our original signal. Here's the first 25 terms of the sine and cosine series added together with the offset, plotted with our original signal:

Absolutely beautiful. Exactly what we wanted.

But it's so complicated...

At this point, you might be seriously doubting the usefulness of the Sine/Cosine Fourier Series. We have to do two integrals (at least!), make sure to divide $a_0$ by 2, and do two infinite sums just to get back our original signal! That sounds like a lot of work. And indeed, there is a better way. Not only can we do away with having to separate out the offset, but we can unify the Sine/Cosine series into a single series, with a single integral and a single sum, while actually making the integral easier, AND keeping all the same information! Sound too good to be true? It isn't. It's the subject of the next lesson.

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