Signals as sums of sinusoids

Smooth spinning arrows, one riding on another, draw a wave with flat tops. Watch the wave beside the arrows as each new arrow joins.

Smooth arrows, flat tops

I want to start with something that bothered me the first time I saw it. Think of the buzz of a cheap synthesiser, or an old doorbell. Its wave is as unsmooth as a wave gets: it sits flat on top, jumps straight down, sits flat on the bottom, and jumps straight back up. This is called a square wave, and it repeats forever.

The picture at the top of the page builds that wave out of nothing but spinning arrows. You met them in the page Complex exponentials & phasors (3.4): an arrow turning at a steady rate, where the height of its tip, followed over time, is a smooth sine wave. Here, each new arrow rides on the tip of the one before it, and each is shorter and faster than the last.

Every arrow is perfectly smooth, but with each one that joins, the top gets flatter and the corners get steeper. By the eighth arrow the curve is nearly the square wave. The little spikes at the corners do not go away as you add arrows; the page Convergence and the Gibbs phenomenon (7.3) explains why.

That is the whole claim of this chapter: almost any repeating signal you will meet, however angular, can be built from smooth sinusoids. The rest of this page takes the animation apart, one piece at a time.

One arrow, one sine wave

Start with a single arrow. Let it turn at a steady rate, and let the dashed line carry the tip’s height across to the wave.

Watch the dot while the arrow turns. The dashed line carries the height of the tip to the wave, so the dot rises when the tip rises and falls when the tip falls. Then watch for the bracket: when the arrow finishes one full lap, the wave has drawn exactly one repeat, and the bracket marks that repeat as T, the period.

So one lap of the arrow is one period. If one lap takes $T$ seconds, the arrow turns at

and it makes $f_0 = 1/T$ laps every second. I’ll call $\omega_0$ the fundamental angular frequency ($f_0$ is the fundamental frequency in Hz): it belongs to the slowest arrow, and it sets the frequency you hear as the note. With $\theta=\omega_0 t$ for the angle this first arrow has turned by time $t$, the dot’s height is $\sin\theta$.

A second arrow, three times as fast

Now ride a second arrow on the tip of the first. I’ll make it turn three times as fast, so it does three laps for every lap of the first, and I’ll make it a third as long.

Watch the dot while the new arrow grows. The dashed line now carries the height of the tip of the whole chain, so the dot’s height is the first arrow’s height plus the second’s. As the second arrow grows, the top of the wave gets flatter than a plain sine, with a dip in the middle of each top.

You met the word in the page Decomposing signals (2.3): an arrow that turns a whole number of times as fast as the fundamental is a harmonic. This one turns at $3\omega_0$, so it is the 3rd harmonic. The $k$th harmonic turns at $k\omega_0$ and adds a term $b_k\sin(k\omega_0 t)$, where $b_k$ is the arrow’s signed length. A negative $b_k$ means the arrow points the opposite way.

Every harmonic fits a whole number of laps into one period $T$, so after $T$ seconds each arrow is exactly back where it started. The sum is therefore back where it started too. Any sum of harmonics of $\omega_0$ repeats with period $T$, however many arrows you add.

Notice that the first arrow is longer than the height of the flat top. Its length is $4/\pi\approx1.27$, so on its own it rises above the dashed square wave. That is on purpose. The later arrows pull the top back down, and the next section gives all the lengths.

How long each arrow must be

So how long should each arrow be? Here is the recipe; the page Fourier series coefficients (7.2) shows where it comes from. For the square wave, the arrows at $k = 3, 5, 7, \dots$ are $1/3, 1/5, 1/7, \dots$ as long as the first. The even ones, $k=2, 4, 6,\dots$, have length zero. The first arrow has length $4/\pi$. Written out, the length of arrow $k$ is

For now, I want you to see that it works. The wave here swings between $-1$ and $+1$, so the flat top should sit at height 1.

Here each arrow is drawn a new way, as a bar whose height shows its length. Watch the readout called “Height at the cursor”, the height of the curve at the middle of a flat top, as each bar joins.

The height lands above 1, then below, then above, and each new arrow moves it by less than the one before. When the animation has finished, drag the cursor along the wave to read the height anywhere.

With the cursor at the middle of a flat top, the height is

with one term for each arrow. The terms alternate in sign, so the sum overshoots and undershoots. With every term included, the bracket adds up to exactly $\pi/4$. This is a known result, the Leibniz series; take it on trust here. The height then approaches $\tfrac4\pi\cdot\tfrac\pi4=1$.

Before we go on, a limit on what this page covers. This square wave is odd, as in the page Decomposing signals (2.3): its left-to-right mirror image is its upside-down copy. It also averages to zero over a period. For waves like that, sines alone are enough, and that is all I use here. A wave that does not average to zero needs a constant added. A wave without the odd symmetry needs cosines too. The page Fourier series coefficients (7.2) brings in both.

Same frequency, different recipe

If the first arrow decides the frequency, what do the other arrows do? Let’s listen, to a sine, a square, a sawtooth and a triangle. The sawtooth here is a falling slope that snaps back up, going from $+1$ down to $-1$. Press a wave button, then watch the first bar.

The first bar never moves, so the “Frequency” readout stays at 220 Hz whatever you choose. What changes is the other bars. The sine has the first bar only. The square adds the odd bars, and the sound becomes thinner. The sawtooth has every bar, falling off as $1/k$, and the sound becomes sharper, with a buzz. The triangle has only odd bars, but they shrink as $1/k^2$, so the sound stays gentle and close to the sine. In the triangle, bars 3, 7 and 11 hang below zero. Their signed lengths are negative, so those arrows point the opposite way, and that turns a flat top into a pointed one. Press Hear it on each wave in turn and compare.

That is why a flute and a violin can play the same note and still sound different. They share the frequency, which is the first arrow, and they differ in how long the other arrows are. The mix of harmonic lengths is a large part of the timbre, the colour of a sound. How a note starts and fades matters too.

At 220 Hz the period is $T = 1/220 = 4.5455$ ms, and the harmonics sit at $220,\ 660,\ 1100,\ 1540$ Hz for $k = 1, 3, 5, 7$.

Why each arrow can be measured on its own

Here is the question I could not stop asking: if someone hands me a wave, how would I find the length of one arrow in it, with all the others mixed in? The answer is a small trick, and it rests on one fact about multiplying waves.

Take two waves, multiply them point by point, and average the result over one period. You met the average of a signal on the page How big is a signal (1.3) as add up and divide. For a wave that runs continuously, the average over one period is the area from $0$ to $T$ divided by $T$:

Here $(\cdot)$ stands for whatever you are averaging. It is the same add-up-and-divide, done with the running area from the pages Impulse, step and ramp (3.1) and Continuous-time convolution (5.3). When the two waves are the same harmonic, the product is never negative, so the average is positive. When they are different harmonics, the product has equal patches above and below the axis, and they cancel.

Here are four pairs, $\sin\theta\times\sin\theta$, $\sin\theta\times\sin3\theta$, $\sin3\theta\times\sin5\theta$, and finally the square wave times $\sin3\theta$. The first three use odd harmonics only, as the square wave does. Watch the shaded patches and the “average so far”: the area collected so far divided by the time elapsed so far, exactly as on the page How big is a signal (1.3).

At the end of the period the average so far equals the period average. The first pair is the same wave twice, so the product is never negative, and the average ends at $0.5$. In the next two pairs the harmonics differ, and the patches above and below the axis are equal. The average ends at $0$.

The fourth pair is the one we wanted. The square wave is a sum of odd arrows. Multiplying by $\sin3\theta$ makes arrows 1, 5, 7, and so on cancel, as in the pairs before.

Only the third arrow survives, and the average ends at $0.212$. Like the first pair, it averages to half the arrow’s length, so double it: 0.424, the length of the third arrow. The worked example below does the same sum by hand.

The algebra below writes each sine as two spinning arrows, as on the page Complex exponentials & phasors (3.4). Why, then, do the arrow animations at the top of this page use one arrow per harmonic? They read only the tip’s height, and one arrow’s height is already a real sine. The pair is a tool for the algebra only.

Why do the pairs cancel? Write $\sin a=\dfrac{e^{ja}-e^{-ja}}{2j}$, and recall $\cos x=\tfrac12\left(e^{jx}+e^{-jx}\right)$. Multiplying out and collecting pairs into cosines,

If you have met the product-to-sum identity, this is it. So $\sin\theta\sin3\theta=\tfrac12[\cos2\theta-\cos4\theta]$. Both cosines complete whole laps in one period, so each averages to zero. For $a=b$ the identity gives $\tfrac12[1-\cos2a]$, which averages to $\tfrac12$.

In general, for whole numbers $m,k \ge 1$,

Now use the trick. Suppose $x(t)=\sum_k b_k\sin(k\omega_0 t)$, and multiply it by $\sin(m\omega_0 t)$ and average. Every term with $k\neq m$ averages to zero. Only the term $k=m$ survives, and it contributes $\tfrac12 b_m$: the arrow’s signed length $b_m$ times the average of $\sin^2$, which is the $0.5$ from the first pair. That is why we double at the end. So

To measure one arrow, multiply by its harmonic, average, and double. The other arrows contribute nothing, so you can find them one at a time without disturbing each other. Try it on the square wave in the worked example below.

Worked example

1. Arrow lengths. For the $\pm1$ square wave, $b_k=\dfrac{4}{\pi k}$ for odd $k$. So $b_1=1.2732$, $b_3=0.4244$, $b_5=0.2546$, $b_7=0.1819$, $b_9=0.1415$.
2. Four arrows at a flat top. With $k=1,3,5,7$ at $\theta=\pi/2$: $\tfrac4\pi\left(1-\tfrac13+\tfrac15-\tfrac17\right)=1.2732\times0.72381=0.9216$, a little below 1.
3. Measuring the 1st arrow. The square wave is $+1$ on $(0,\pi)$ and $-1$ on $(\pi,2\pi)$. The average of $x\sin\theta$ is $$\frac1{2\pi}\left(\int_0^\pi\sin\theta\,d\theta-\int_\pi^{2\pi}\sin\theta\,d\theta\right)=\frac{2+2}{2\pi}=\frac2\pi=0.6366.$$ Doubling gives $b_1=1.2732$.
4. Pencil and paper: the 3rd arrow. First the area under one hump of $\sin3\theta$: $$\int_0^{\pi}\sin3\theta\,d\theta=\left[-\frac{\cos3\theta}{3}\right]_0^{\pi}=\frac{1-(-1)}{3}=\frac23,$$ since $\cos3\pi=-1$. Over $(\pi,2\pi)$ the wave sits below the axis: $$\int_\pi^{2\pi}\sin3\theta\,d\theta=\left[-\frac{\cos3\theta}{3}\right]_\pi^{2\pi}=\frac{-\cos6\pi+\cos3\pi}{3}=\frac{-1-1}{3}=-\frac23,$$ since $\cos6\pi=1$. Check: the whole period gives $\int_0^{2\pi}\sin3\theta\,d\theta=\tfrac23-\tfrac23=0$, as a full number of laps must.
5. The 3rd arrow, finished. The $-1$ on the second half flips the $-\tfrac23$ to $+\tfrac23$, so the average of $x\sin3\theta$ is $$\frac1{2\pi}\left(\frac23+\frac23\right)=\frac2{3\pi}=0.2122.$$ Doubling gives $b_3=0.4244$, matching the recipe.
6. The 2nd arrow. The average of $x\sin2\theta$ is $0$, so $b_2=0$, as the recipe says.

Where you’ll meet this

Every synthesiser that builds a sound by adding sine waves is using this idea. So is every spectrum display that shows how much of each frequency a sound contains. Sound compression keeps the components you can hear and drops the ones you cannot. The next page, Fourier series coefficients (7.2), turns the multiply-and-average trick into the Fourier series formula, with exact values for any reasonable repeating wave.

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