Fourier series coefficients

A wave that is 1 for half of each period and 0 for the rest, wound around a circle. Watch the dot: the average position of the wound-up curve.

From arrows to a formula

On the page Signals as sums of sinusoids (7.1) you saw that a repeating wave is a sum of spinning arrows. You also saw that one arrow can be measured alone: multiply the wave by that arrow’s harmonic, average, and double. This page turns that trick into a formula that works for any repeating wave.

The formula comes in three forms, and an exam may ask for any of them. I will show that they describe the same wave. Then I will draw the answer as a row of lines, and use symmetry to know in advance which lines are empty.

The wave I use throughout is a 0-to-1 square wave. It is 1 for the first half of each period $T$ and 0 for the second half. I chose it, and not 7.1’s wave between $-1$ and $+1$, because it has a constant part, and I want that part to have something to show. It equals $\tfrac12$ plus half of 7.1’s wave, so you already know most of its arrows.

There is one change of reading from 7.1. There I followed the height of an arrow’s tip, so a sine started at height 0. From here on I follow the horizontal position of the tip, as on the page Complex exponentials & phasors (3.4).

A cosine is then an arrow that starts pointing right. A sine is a cosine delayed by a quarter turn, as in Sinusoids (3.2), so it is an arrow that starts pointing down.

Winding the wave up

Here is 7.1’s trick, rewritten with complex numbers. To measure harmonic $k$, multiply the wave by the arrow $e^{-jk\omega_0t}$ and average. Here $\omega_0=2\pi/T$ is the fundamental, and $\theta=\omega_0t$ is the angle it has turned by time $t$.

Multiplying by a point on the unit circle only rotates, as on the page Complex numbers for signals (3.3). So at every instant this turns the wave’s value back by $k\theta$. I call that winding the wave at setting $k$.

The wave’s value is a point on the real axis of the complex plane. Where the wave is 1, the wound-up point rides the unit circle. Where the wave is 0, it stays at the origin. The average of where this point has been is a complex number, and I call it $c_k$.

Go back to the picture at the top of the page. While the curve is drawn, the dot is the average position of the part drawn so far, so it moves. When the curve is complete, the dot is $c_k$, the curve’s balance point. At setting 1 the curve leans to one side, and the dot settles $0.318$ from the centre. At setting 2 the curve closes into a full circle, and the dot sits at the centre.

Setting 0 is no winding at all, so the dot is the average of the wave itself, which is 0.5. At the end the clip lifts the wave by 0.5, which raises every value of the wave by 0.5. The dot slides from 0.5 to 1.0, and only the bar in the “average” slot moves.

When the clip has finished, drag along the bars to pick any setting from 0 to 5. Settings 2 and 4 put the dot at the centre. Odd settings lean it to one side, and the dot is nearer the centre for larger settings.

The strip says “half an arrow’s length” for a reason. In 7.1 a real wave needs one arrow for each harmonic. On the page Complex exponentials & phasors (3.4) that arrow is really two, spinning in opposite directions at $k$ and $-k$, and each is half as long. The winding at setting $k$ measures one of the two. For the 0-to-1 wave, 7.1’s arrow 1 is $0.637$ long and the bar for setting 1 is $0.318$. The next section draws the pair.

Now the formula. Averaging a complex quantity means averaging its real parts and its imaginary parts separately, as when you add points in Complex numbers for signals (3.3). Written with the area from $0$ to $T$ divided by $T$, as in 7.1,

This is the analysis equation. Any one period gives the same answer for a repeating wave, so I use $0$ to $T$ unless another period is easier, as in step 4 of the worked example.

Why does the winding find harmonic $k$ and nothing else? Suppose the wave is a sum of arrows, $x(t)=\sum_m c_m e^{jm\omega_0t}$, with $m$ running over all whole numbers, positive and negative. Multiply by $e^{-jk\omega_0t}$. The term for harmonic $m$ becomes $c_m\,e^{j(m-k)\omega_0t}$, an arrow that turns $m-k$ laps per period.

For $m\ne k$, that arrow keeps turning, and its average is zero. For $m=k$ the arrow stands still and survives with the value $c_k$. The zero comes from one line of calculus: for a constant $a$ that may be complex, $\int e^{at}\,dt=e^{at}/a$. With $a=jn\omega_0$ and $n\neq0$,

because $\omega_0T=2\pi$, so $e^{jn\omega_0T}=e^{j2\pi n}=1$. This is the same cancellation as in 7.1, now for any pair of harmonics. It also gives the other half of the pair of equations, the synthesis equation:

Let’s use the analysis equation on the running wave, which is 1 on $(0,T/2)$ and 0 after that. For $k=0$ nothing winds, and $c_0$ is the average of the wave, which is $\tfrac12$. For $k\neq0$,

For even $k$, $e^{-jk\pi}=1$ and $c_k=0$, which is the circle that closes. For odd $k$, $e^{-jk\pi}=-1$, so $c_k=\dfrac{1}{jk\pi}=\dfrac{-j}{k\pi}$, since $1/j=-j$. That gives $c_1=-0.3183j$, $c_3=-0.1061j$ and $c_5=-0.0637j$, the bars of the instrument.

One harmonic, three ways to write it

Each harmonic with $k\ge1$ is a real wave at $k\omega_0$. You can write it in three ways, and the picture below shows that they are one wave. I use the example from the page Complex exponentials & phasors (3.4): $3\cos\theta+4\sin\theta=5\cos(\theta-53.13°)$.

Watch the traced wave. It never changes, while the description of it does.

The two arrows, 3 and 4, join into one arrow of length 5, and that arrow splits into a mirror pair of arrows 2.5 long, spinning in opposite directions.

The three forms are these. In the trigonometric form, a harmonic is $a_k\cos k\theta+b_k\sin k\theta$. In the compact form, it is one shifted cosine, $A_k\cos(k\theta+\phi_k)$. In the complex form, it is a mirror pair of arrows, $c_ke^{jk\theta}+c_{-k}e^{-jk\theta}$. On this page $a_k$ and $b_k$ are Fourier coefficients; on the page Difference equations (6.1) the same letters name filter coefficients, which are a different thing.

The two arrows of the complex form are mirror images of each other. For a real wave, the analysis equation gives

since $x(t)$ is real and conjugating the integrand turns $e^{-jk\omega_0t}$ into $e^{+jk\omega_0t}$. Here $c_k^*$ is the mirror image of $c_k$ from Complex numbers for signals (3.3). The pair adds to $2\,\mathrm{Re}\,(c_ke^{jk\theta})$, which is real.

Now put $c_k=(a_k-jb_k)/2$ into that pair. Multiplying out $2\,\mathrm{Re}\left[\tfrac{a_k-jb_k}{2}(\cos k\theta+j\sin k\theta)\right]$ gives $a_k\cos k\theta+b_k\sin k\theta$. So

Here $\mathrm{atan2}(b_k,a_k)$ is the angle of the point $a_k+jb_k$, in the range $(-180°,180°]$, as the readouts use and as on the page Complex numbers for signals (3.3). The minus sign appears because the arrow that starts pointing down is a sine with a positive $b_k$. The constant is $a_0=A_0=c_0$. Some books write $a_0/2$ for the constant, with $a_0=2c_0$.

For $3\cos\theta+4\sin\theta$ we have $a_1=3$ and $b_1=4$. Then $c_1=1.5-2j$, so $\lvert c_1\rvert=2.5$ and $\angle c_1=-53.13°$ ($-0.9273$ rad). The arrow lengths of 7.1 are $A_k=2\lvert c_k\rvert$, which is $5$ here. That is why the bars of the first instrument were half an arrow: each real arrow is the sum of two halves.

For the 0-to-1 wave, $c_k=-j/(k\pi)$ is purely imaginary. So $a_k=0$ and $b_k=2/(k\pi)$ for odd $k$, which is 0.6366, 0.2122 and 0.1273 for $k=1,3,5$. These are half of 7.1’s lengths $4/(\pi k)$ for the wave between $-1$ and $+1$, as they should be.

The line spectrum, and what a delay does

Plot the length $\lvert c_k\rvert$ and the angle $\angle c_k$ against $k$, for negative and positive $k$ alike. The result is a row of lines, the line spectrum of the wave. For the 0-to-1 wave the line at $k=0$ has length $0.5$. The odd lines have length $1/(\lvert k\rvert\pi)$, and their angles are $-90°$ for $k>0$ and $+90°$ for $k<0$, as the mirror rule requires.

Now delay the wave by a time $t_0$, as on the page Shifting, reversing and scaling time (2.1). The notes of a recording played a moment later are the same notes at the same loudness. So what happens to the lines?

Put $s=t-t_0$ in the analysis equation. The delayed wave $x(t-t_0)$ has the coefficient

The length of every line is unchanged. Line $k$ turns by the angle $-k\omega_0t_0$, which is $k$ times as far as line 1. Before you watch, decide: if the whole wave moves later by $T/4$, does every line turn by the same angle?

Press play and watch the dial under each line as the wave moves later, first by $T/8$, then by $T/4$.

At first every angle reads $-90°$. The dial for line 3 turns three times as far as the dial for line 1, and line 5 turns five times as far. The answer to the question is no: one time shift is a different angle on every line. The readouts give angles between $-180°$ and $180°$.

At a delay of $T/8$ the angle step is $k\omega_0t_0=k\cdot\tfrac{2\pi}{T}\cdot\tfrac T8=k\cdot45°$. So line 1 reads $-90°-45°=-135°$. Line 3 is at $-90°-135°=-225°$, which the readout shows as $+135°$. Line 5 is at $-90°-225°=-315°$, shown as $+45°$. At $T/4$ the turns are $-90°$, $-270°$ and $-450°$, and the lines read $180°$, $0°$ and $180°$.

Line 0 never moves. Its length stays $0.5$ and its angle stays $0°$. But the wave near $t=0$ does change. At first it jumps from 0 to 1 at $t=0$, so just after $t=0$ it is 1. After the delay of $T/4$ it is 0 there. So the line at $0$ is the average, not the value of the wave at $t=0$.

Symmetry tells you what is zero

You can often tell that whole families of coefficients are zero without computing any integral. The reason is this: a cosine is even and a sine is odd. These words are from Kinds of signals (1.2): a wave is even if its mirror image about $t=0$ lies exactly on it, and odd if the mirror image is the wave turned upside down.

Take the mirrored copy of a wave, $x(-t)$. It leaves every cosine term as it was and flips every sine term. So if the copy equals the wave, no sine term can be present, because each would have to equal its own negative. If the copy is the wave turned upside down, no cosine term can be present, including the constant.

There is a third kind of symmetry. A wave has half-wave symmetry if its second half is its first half turned upside down: $x(t+T/2)=-x(t)$.

Advance the wave by $T/2$ and use the delay rule from the last section with $t_0=-T/2$. Each coefficient is multiplied by $e^{jk\pi}=(-1)^k$, so odd $k$ flips and even $k$ stays. Half-wave symmetry says the advanced wave is $-x(t)$, so $(-1)^kc_k=-c_k$, and every even $c_k$ is zero.

Even and odd depend on where you place $t=0$, as in Decomposing signals (2.3).

Here are three tests, one after another. The bars are signed arrow lengths, as in 7.1: each bar for $k\ge1$ is $\pm2\lvert c_k\rvert$, and $a_0$ is $c_0$. A dashed bar marked ”?” could be anything. Press play and watch the dashed bars.

First, an even triangle is mirrored across the vertical axis and fits onto itself, so every sine bar is “0”. The even-$k$ bars keep their ”?”, because mirror symmetry says nothing about them.

Next, the 0-to-1 wave first loses its constant part, and what remains is an odd wave between $-0.5$ and $+0.5$. A copy turned over both axes fits onto it, so no cosine bars remain.

Last, the triangle is slid by half a period and turned upside down, and it fits onto itself. Now the even bars change from ”?” to “0”. For the triangle only the odd cosine bars survive. The odd wave of the second part passes the same test, so only its odd sine bars survive.

Moving the origin matters too. The delay of $T/4$ in the last section turned lines 1, 3 and 5 to $180°$, $0°$ and $180°$. Those lines point along the real axis, so those $c_k$ are real numbers, and $b_k=-2\,\mathrm{Im}\,c_k=0$. The delayed wave has no sine terms. It is even, although the original was not.

Where the power goes

The last quantity is power. As on the page How big is a signal (1.3), the power of a wave is the average of $\lvert x\rvert^2$.

Write $\lvert x\rvert^2=x\,x^*$ with $x=\sum_m c_me^{jm\omega_0t}$, and average. The cross terms between different harmonics spin and average to zero, as before. Only the terms with the same harmonic survive, and each contributes $\lvert c_k\rvert^2$:

This is Parseval’s relation. For a real wave, lines $k$ and $-k$ carry the same power. Together they give $2\lvert c_k\rvert^2=A_k^2/2$, which is 7.1’s fact that $\sin^2$ averages to $\tfrac12$ for an arrow of length $A_k$. The total is $a_0^2+\sum_{k\ge1}A_k^2/2$.

I will check it on the wave between $-1$ and $+1$, because its square is 1 at every instant and so its power is exactly 1. Fill a tank with one layer per arrow: the arrow for harmonic $k$ adds $A_k^2/2=\dfrac{8}{\pi^2k^2}$, and only odd $k$ have one. Watch “Power so far” climb towards 1.

The first arrow alone fills $0.8106$, which is $81\,\%$ of the total. Four arrows reach $0.9496$ and eight reach $0.9747$. Each later arrow adds a thinner layer, so the level rises more slowly towards 1.

I do not sum the infinite series here. The total of 1 comes straight from $x^2=1$. The small gap that remains after a given number of arrows is the subject of the page Convergence and the Gibbs phenomenon (7.3). The wave panel of this instrument shows the sum of the arrows overshooting near each jump. That overshoot is explained there too.

For the 0-to-1 wave, $x^2=x$, so its power is its average, $0.5$. The constant part carries $a_0^2=0.25$ of that total, which is half.

Worked example

1. Square wave between $-1$ and $+1$. It is $+1$ on $(0,T/2)$ and $-1$ on $(T/2,T)$. Then $c_k=\dfrac1T\left[\displaystyle\int_0^{T/2}-\int_{T/2}^{T}\right]e^{-jk\omega_0t}\,dt=\dfrac{2}{jk\pi}$ for odd $k$, and $0$ for even $k$ and for $k=0$. So $c_1=-0.6366j$, $c_3=-0.2122j$ and $c_5=-0.1273j$. Then $2\lvert c_k\rvert=4/(\pi k)$, the lengths of 7.1.
2. The 0-to-1 square wave. It equals $\tfrac12+\tfrac12(\text{the wave of step 1})$. So $c_0=0.5$, and $c_k=\dfrac1{jk\pi}$ for odd $k$: $c_1=-0.3183j$, $c_3=-0.1061j$, $c_5=-0.0637j$. The arrow lengths $2\lvert c_k\rvert$ are 0.6366, 0.2122 and 0.1273. Lifting the wave by $0.5$ moves only $c_0$, from $0.5$ to $1.0$.
3. Three forms. For $3\cos\theta+4\sin\theta$: $a=3$, $b=4$, $A=5$, $\phi=-\mathrm{atan2}(4,3)=-0.9273$ rad $=-53.13°$. Also $c_1=(a-jb)/2=1.5-2j$ with $\lvert c_1\rvert=2.5$ and $\angle c_1=-53.13°$, and $c_{-1}=c_1^*=1.5+2j$. Check: $2\,\mathrm{Re}\,(c_1e^{j\theta})=3\cos\theta+4\sin\theta$.
4. A pulse train. Take $T=1$ s and a pulse of half-width $T_1=0.25$ s, so $x=1$ for $\lvert t\rvert<T_1$ and $x=0$ elsewhere in the period. Integrating from $-T/2$ to $T/2$, with $\omega_0T=2\pi$, $$c_k=\frac1T\int_{-T_1}^{T_1}e^{-jk\omega_0t}\,dt=\frac{e^{jk\omega_0T_1}-e^{-jk\omega_0T_1}}{jk\omega_0T}=\frac{2\sin(k\omega_0T_1)}{k\omega_0T}=\frac{\sin(k\omega_0T_1)}{k\pi}.$$ Here $k\omega_0T_1=k\pi/2$, so $c_k=\dfrac{\sin(k\pi/2)}{k\pi}$, and $c_0=2T_1/T=0.5$. Then $c_1=0.3183$, $c_2=0$, $c_3=-0.1061$, $c_4=0$ and $c_5=0.0637$. Its power is $0.5$, because $x^2=x$.
5. A time shift. Delay the wave of step 1 by $T/8$. Each $c_k$ becomes $c_ke^{-jk\pi/4}$, so $\angle c_1=-135°$, $\angle c_3=-225°\equiv+135°$ and $\angle c_5=-315°\equiv+45°$. The magnitudes and $c_0$ are unchanged.
6. Parseval. For the wave of step 1, harmonic $k$ carries $2\lvert c_k\rvert^2=\dfrac{8}{\pi^2k^2}$: 0.8106, 0.0901, 0.0324 and 0.0165 for $k=1,3,5,7$. The running totals are 0.8106, 0.9006, 0.9331 and 0.9496. With eight arrows the total is $0.9747$, and the exact total is $1$.
7. Three quick cases. Lifting a wave by $0.5$ changes only $c_0$, by $+0.5$. For $x=3\cos\omega_0t+4\sin\omega_0t$, $A_1=5$ and $\lvert c_1\rvert=2.5$. A delay of $T/4$ keeps every $\lvert c_k\rvert$ and turns line 3 by $-3\cdot90°=-270°$, not $-90°$.

Where you’ll meet this

The lines of a spectrum are what an audio analyser shows: the length of each line is how much of that frequency a sound contains. The phase dials explain why a sound stays recognisable when it is delayed, while a different delay for each frequency, as in some filters, changes its shape. This is the subject of Fourier series and LTI systems (7.4), where each $c_k$ passes through a system on its own.

Power per harmonic is how harmonic distortion is measured: the power in the harmonics that should not be there, compared with the power in the one that should. The next step is to let the period $T$ grow without limit, so that the lines crowd together into a curve. That is the page From series to transform (8.1). The rule that a delay turns each line by an angle in proportion to its frequency returns on the page Properties of the Fourier transform (8.2).

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