Fourier series and LTI systems

Here one spinning arrow goes into an RC circuit, the circuit behind a guitar’s tone knob, and then a faster arrow follows it. Make a prediction: does a faster or a slower arrow come out? Watch the output arrow’s spin rate, its length and the angle it falls behind.

A spinning arrow through an RC circuit

On the page Fourier series coefficients (7.2) you drew a repeating wave as a row of lines, one for each spinning arrow. This page asks what a system does to that row. The answer is simple: a linear time-invariant (LTI) system treats every arrow separately.

The picture I want you to keep is the tone knob on a guitar. Turning it down does not block any note. Every harmonic of the sound is turned down, and the high ones are turned down most. The circuit behind that knob is an RC circuit, so I use the one from Differential equations and analog systems (6.2). It is the continuous-time twin of the leaky integrator in Difference equations (6.1).

That page wrote the circuit as $RC\,\dot v+v=v_s$, where $v_s$ is the source and $v$ is the voltage across the capacitor. Here I rename the source as the input $x$ and the voltage as the output $y$, so the equation is

Now feed in one spinning arrow, $x(t)=e^{j\omega t}$, as on the page Complex exponentials & phasors (3.4). Try an output that is the same arrow times one fixed complex number $H$, so $y(t)=He^{j\omega t}$. The rule from 6.2, $\tfrac{d}{dt}e^{st}=s\,e^{st}$, with $s=j\omega$ gives $\dot y=j\omega\,He^{j\omega t}$.

Put both into the equation and cancel the common factor $e^{j\omega t}$:

The guess works, so the circuit answers an arrow with the same arrow, scaled and turned by the number $H(j\omega)$. That is the property from Properties of LTI systems (5.4), now for a continuous-time system. One condition: this is the answer after the start-up part of 6.2, the natural part that decays, has died away.

What is $1/(1+j\omega RC)$ as an arrow? On the page Complex numbers for signals (3.3), multiplying multiplies the lengths and adds the angles. Dividing undoes that. So $1/z$ has length $1/\lvert z\rvert$ and angle $-\angle z$: you divide the lengths and subtract the angles.

Take $z=1+j$, one step across and one step up. Its length is $\lvert1+j\rvert=\sqrt{1^2+1^2}=\sqrt2$ and its angle is $45°$. So

an arrow $0.707$ as long as the unit arrow, turned back by $45°$.

For $RC=1$ s and an arrow spinning at $\omega$ rad/s, $H=1/(1+j\omega)$. Its length, the gain $\lvert H\rvert$, is $1/\sqrt{1+\omega^2}$. Its angle is $-\arctan\omega$, and I call the amount turned back, $-\angle H$, the lag.

Now go back to the picture at the top of the page. At 1 rad/s the arrow comes out spinning at the same rate, $0.707$ as long and $45.0°$ behind: the example you just divided out. Then the input speeds up to 3 rad/s, and the output speeds up with it. The faster arrow comes out $0.316$ as long and $71.6°$ behind. This is $1/(1+3j)$: $\lvert1+3j\rvert=\sqrt{10}=3.162$, so the length is $0.316$, and the angle $-\arctan3=-71.6°$.

So the answer to the prediction is neither. The output spins at the same rate as the input. Only its length and its angle change, and the faster input is cut more and lags more.

One more fact, so that real waves work. The equation $RC\,\dot y+y=x$ has real coefficients. A cosine is the horizontal position of the arrow, as in 7.2, so the response to $\cos\omega t$ is the horizontal position of the response to the arrow. From here on I read all arrows that way.

A square wave, one arrow at a time

Now a repeating input. Take the square wave between $-1$ and $+1$ with fundamental $\omega_0=1$ rad/s, and keep $RC=1$ s. From Fourier series coefficients (7.2) and 7.1, its first four arrows have lengths $b_k=4/(\pi k)$ for $k=1,3,5,7$. Each starts pointing down, at $-90°$, because the harmonics of this wave are sines.

The system is linear, so the output is the sum of what each arrow produces on its own. Each arrow comes out as $H(jk\omega_0)$ times itself. For $RC=1$ s, $H(jk)=1/(1+jk)$ has gain $1/\sqrt{1+k^2}$ and lag $\arctan k$. For $k=1,3,5,7$ the gains are $0.7071,\ 0.3162,\ 0.1961,\ 0.1414$ and the lags are $45.0°,\ 71.6°,\ 78.7°,\ 81.9°$.

Watch each arrow come out of the box shorter and turned, and then the shape of the output wave.

Arrow 3, for example, goes from $0.4244$ to $0.1342$: it is $0.3162$ as long and held back $71.6°$. The output wave at the end is the sum of every arrow that came out, these four and the smaller ones after them. It is the circuit’s exact response, so it includes every harmonic, not only the four drawn. It is smooth, delayed and has rounded corners. Each harmonic is held back by its own angle, not in proportion to $k$, so the shape changes, not just the size.

Compare this with a plain delay. If the circuit only delayed the wave, Fourier series coefficients (7.2) says line $k$ would turn by $k$ times the first angle. A delay of $0.785$ s turns line 1 by $45°$ at $\omega_0=1$ rad/s, so the lines would turn by $45°,\ 135°,\ 225°,\ 315°$. The RC turns them by $45°,\ 71.6°,\ 78.7°,\ 81.9°$. The harmonics slip against each other, and that is why the corners round off.

Write the whole recipe in one line. If $x(t)=\sum_k c_ke^{jk\omega_0t}$, then

The line $c_k$ becomes $H(jk\omega_0)\,c_k$, and nothing else changes. The arrow at $-k\omega_0$ gets $H(-jk\omega_0)=H^*(jk\omega_0)$, the mirror image of the arrow at $k\omega_0$, because the circuit is real. So the output is still a mirror pair, and still a real wave.

The same line shows something else. Every arrow comes out at the spin rate it went in with, so no arrow is created. An LTI system never adds a frequency that was not in its input. I come back to that in the worked example.

Turning the highs down: the tone knob

The gain $\lvert H(jk)\rvert=1/\sqrt{1+(k\omega_0RC)^2}$ is a smooth slope across the harmonics. The corner frequency $\omega_c=1/RC$ is where it reads $0.707$. You met that number in the first clip: at $\omega=1$ rad/s with $RC=1$ s, the gain was $0.707$ and the lag $45°$.

A large $RC$ puts the corner at a low frequency, so more of the harmonics are above it. The next instrument slides $RC$ upward from $0.05$ s to $3$ s, so the corner slides from $20$ rad/s down to $0.33$ rad/s. Watch the slope pass under the bars, and the output wave lose its corners.

With a tiny $RC$ of $0.05$ s every bar is turned down a little, the high ones a bit more. The gains are $0.999,\ 0.989,\ 0.970,\ 0.944$, and the wave is almost unchanged. At 5 s, $RC=0.39$ s, the gains are $0.933,\ 0.652,\ 0.459,\ 0.346$, and the corners of the wave are rounded: the higher the bar, the more it is turned down.

At 10 s, $RC=3$ s, every bar is far down the slope. The gains are $0.316,\ 0.110,\ 0.067,\ 0.048$. The output is a rounded triangle, close to a single sine of amplitude $0.4026$ lagging $71.6°$ with pointed tops: the third harmonic is $0.0469$, about $11.6\%$ of the first. Nothing is blocked. Every harmonic is turned down, the high ones far more.

Watch the slope, and not a wall. Nothing is blocked at any setting. The first bar at $RC=3$ s is $0.316$ of its input and the seventh is $0.048$, so the seventh is turned down about $6.6$ times as hard. Smoothing a square wave means turning the high harmonics down more than the low ones, until what is left is nearly the first sine.

When the clip has finished, drag the corner dot along the gain curve to set $RC$ yourself, and press Hear the output to hear the result.

Distortion as a number

Use the buttons “Hear the input” and “Hear the output”. The output is duller. They are not controls: they play what is on screen. Now ask what numbers would say that.

Engineers use the total harmonic distortion, THD. It compares the size of everything beyond the first harmonic with the size of the first. A pure sine has no harmonic beyond the first, so its THD is 0.

The page How big is a signal (1.3) defined the RMS, the square root of the power. Powers of different harmonics add, by Parseval’s relation in 7.2, and sizes do not. So I compare powers first, with $P_k=A_k^2/2$ for an arrow of length $A_k$, and take the square root last:

The square root turns the ratio of powers into a ratio of sizes. It is usually quoted as a percentage.

The next instrument stacks the power in the first harmonic under the power in all the others, for the input and for the output. Watch the top block shrink faster than the bottom one as the filter engages.

The square wave has 81 % of its power in the first arrow and the rest in the others. With no filter the input blocks are $0.8106$ and $0.1894$, because the wave’s total power is 1. The ratio is $0.2337$ and its square root is $0.4834$, so the THD is $48.3\%$.

At 3 s, $RC=0.5$ s, the output blocks are $0.6485$ and $0.0344$. The ratio is $0.05308$, the square root $0.2304$, and the THD $23.0\%$. At 6 s, $RC=1$ s, the output blocks are $0.4053$ and $0.0108$. The ratio is $0.02674$, the square root $0.1635$, and the THD is $16.4\%$, against $48.3\%$ for the input. The upper harmonics were turned down far more than the first, so the THD fell.

Why it falls is the gain squared. At $RC=1$ s the first harmonic’s power is multiplied by $\lvert H(j)\rvert^2=0.5$, and the third harmonic’s by $0.1$. The first block halves, while the top block, which holds the third harmonic and everything above it, falls to about $6\%$ of its input size.

Worked example

1. $H(jk)$ for $RC=1$ s and $\omega_0=1$ rad/s. $H(jk)=1/(1+jk)$. Dividing: $\lvert1+jk\rvert=\sqrt{1+k^2}$ and $\angle(1+jk)=\arctan k$, so $H(jk)$ has length $1/\sqrt{1+k^2}$ and angle $-\arctan k$. For $k=1$, $\lvert1+j\rvert=\sqrt2$, so $H=0.7071\angle-45.00°$ ($-\pi/4$ rad). For $k=3,5,7$: $0.3162\angle-71.57°$, $0.1961\angle-78.69°$, $0.1414\angle-81.87°$. A plain delay of $0.785$ s would give $-45°,\ -135°,\ -225°,\ -315°$. The RC’s lags, expressed as times, are $0.785,\ 0.416,\ 0.275,\ 0.204$ s.
2. The output arrows. The lengths are $b_k\lvert H\rvert$ with $b_k=4/(\pi k)$: $0.9003,\ 0.1342,\ 0.0499,\ 0.0257$. So $$y(t)\approx0.9003\sin(t-45°)+0.1342\sin(3t-71.57°)+0.0499\sin(5t-78.69°)+0.0257\sin(7t-81.87°).$$
3. THD. For the input, the fundamental carries $8/\pi^2=0.8106$ and the rest carry $1-0.8106=0.1894$, since the total is $1$ ($x^2=1$). The ratio is $0.2337$ and $\mathrm{THD}=\sqrt{0.2337}=0.4834$. For the output at $RC=1$ s, harmonic $k$ carries $\dfrac{8}{\pi^2k^2(1+k^2)}$. The fundamental carries $4/\pi^2=0.4053$ and the rest carry $0.0108$. The ratio is $0.02674$ and $\mathrm{THD}=0.1635$. At $RC=0.5$ s the fundamental carries $0.8106/1.25=0.6485$, the rest $0.0344$, the ratio is $0.05308$ and $\mathrm{THD}=0.2304$.
4. Smoothing hard, $RC=3$ s. $\lvert H(jk)\rvert=1/\sqrt{1+9k^2}$ gives $0.3162,\ 0.1104,\ 0.0665,\ 0.0476$ for $k=1,3,5,7$. The first harmonic has amplitude $1.2732\times0.3162=0.4026$ at $-\arctan3=-71.57°$. The third has amplitude $0.4244\times0.1104=0.0469$, about $11.6\%$ of the first. At $RC=0.387$ s the gains are $0.9325,\ 0.6523,\ 0.4588,\ 0.3461$, and at $RC=0.05$ s they are $0.9988,\ 0.9889,\ 0.9701,\ 0.9439$.
5. No new frequencies. At $RC=3$ s the first harmonic is multiplied by $0.316$ and the seventh by $0.048$. Both are turned down, the seventh about $6.6$ times as hard, and neither is blocked. A repeating input with lines at 100, 300 and 500 Hz gives an LTI output with lines at 100, 300 and 500 Hz only, possibly smaller and possibly turned. So a clipper that turns a pure sine into a sine plus a third harmonic is not LTI.

Where you’ll meet this

An audio equaliser, a tone knob and a loudspeaker’s crossover all work this way: each harmonic is scaled and turned by the system’s value at its frequency. The distortion figure on an amplifier’s datasheet is the number of the last section, usually quoted as a percentage.

The next step is to look at $H(j\omega)$ for every frequency, not only at the harmonics, which is the subject of Frequency response and Bode plots (8.4). Letting the period grow without limit, so that the lines crowd into a curve, is the subject of the page From series to transform (8.1). Why a real system’s $H$ satisfies $H(-j\omega)=H^*(j\omega)$, and why its poles come in mirror pairs, is on the pages First- and second-order systems (6.3) and Poles, zeros and the s-plane (9.3). Real filters are chosen and built on the pages Filter specifications (18.1) and Analog prototype filters (20.1). The square wave’s sum of arrows, and the overshoot that stays near its edges, are on the page Convergence and the Gibbs phenomenon (7.3).

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