First- and second-order systems

Here is the step response of an RC circuit, its voltage after a battery is switched on, with a time constant of 0.2 s. Watch the two times the markers measure: from 10 % to 90 % of the final value, and until the curve stays within 2 % of it.

One number sets a first-order system

In Differential equations and analog systems (6.2) you met the RC circuit. Switch a battery on and its voltage climbs as $1-e^{-t/\tau}$, where the time constant $\tau$ is $RC$. A system whose equation holds at most a first derivative is called first order, and this curve is the shape of its step response.

I’ll write that step response as $y(t)$. The impulse response (5.1) called it $s[n]$, but on this page $s$ is the exponent in $e^{st}$, so it needs another name.

Here is a question a thermostat has to answer. A room heater switches on, and you want to know how long the room takes to get within a degree of its new temperature. You can get that answer from $\tau$ alone, because every first-order step response has the same shape. Only its horizontal scale changes.

In the picture at the top, the bracket shows that going from 10 % to 90 % takes 0.44 s, which is 2.2 τ. This time is the rise time. The curve is inside the 2 % band by 0.78 s, about 3.9 τ, and it stays there. The time from the start until the curve is inside the band for good is the settling time.

Both numbers come from one step. The curve reaches the fraction $p$ of its final value when $1-e^{-t/\tau}=p$, so $e^{-t/\tau}=1-p$. The natural logarithm $\ln$ undoes $e$, as 6.2 showed, so

For the 10 % point, $p=0.1$ gives $t=\tau\ln(1/0.9)=0.2\times0.10536=0.0211$ s. For the 90 % point, $t=\tau\ln10=0.4605$ s. The difference is the rise time. Because $\ln10-\ln(10/9)=\ln9$, it is $\tau\ln9=0.2\times2.1972=0.4394$ s, which is $2.2\,\tau$.

The 2 % band is the strip within 2 % of the final value. The curve is inside it once $e^{-t/\tau}\le0.02$, which is $t=\tau\ln50=0.2\times3.9120=0.7824$ s, or $3.9\,\tau$. The curve only rises, so it stays inside from then on.

Here is the thermostat answer. If the room follows a first-order response with $\tau=10$ minutes, it takes about 22 minutes to go from 10 % to 90 % of the temperature change, and 39 minutes to settle within 2 % of the change. Double the time constant and both times double.

A mass with three amounts of friction

A first-order system has one number, so it can only creep up to its target. Add a second stored quantity, the speed of the mass as well as its position, and the system can overshoot and ring. The standard example is a mass on a spring with a damper, the rig from the last instrument of 6.2. The damper is the friction.

Think of a car’s shock absorbers: worn, correct, or over-stiff. Push each with a steady force toward a new rest position, which I call 1. The mass, the spring and the time axis are the same for all three. The axis is in plain seconds. Only the friction differs.

Before you watch, answer this: which of these gets the car level first? Watch the flag each curve gets when it enters the 2 % band for good.

The correct one is the first to settle, at 5.8 s. The worn one settles at 11.2 s and the stiff one at 14.9 s. Of these three, the correct one wins: more friction is not faster.

The worn one is the first to touch 1, but touching is not staying level. It overshoots and rings, and takes the longest of the ringing rigs to stay in the band. The stiff one never overshoots, but it creeps. At 10 s it is at $0.926$, while the worn one is already at $1.051$.

With exactly enough friction the mass arrives as fast as it can without ever overshooting. That is called critical damping. Less friction is under-damped, and more is over-damped. The next section gives these a number.

Two numbers set the whole motion

Two numbers describe every rig on that plot. The first, $\omega_n$ (rad/s), is the natural frequency: how fast the mass rings when there is no friction at all. The second, $\zeta$ (zeta), measures the friction against the critical amount. The worn absorber has $\zeta=0.3$, the correct one $\zeta=1$ and the stiff one $\zeta=2$. All three had $\omega_n=1$ rad/s, which is why the axis was in seconds. The equation is

where $x$ is the push, and $\dot y$ and $\ddot y$ are the slope of $y$ and the slope of that slope, as in 6.2.

Now remove the push and look at the free motion. 6.2’s trick was to try $e^{st}$, whose slope is $s\,e^{st}$ and whose second slope is $s^2e^{st}$. Putting it in gives $s^2+2\zeta\omega_ns+\omega_n^2=0$, so

Here $s=\sigma+j\omega$, as in Complex exponentials & phasors (3.4). Its real part $\sigma$ is negative for a decaying motion and says how fast it dies. Its imaginary part $\omega$ says how fast it spins. I call these two values of $s$ the poles of the system. The spin rate of the pair is $\omega_d=\omega_n\sqrt{1-\zeta^2}$, the ringing rate with friction (d for damped).

The poles live in a plane, but this plane holds rates, not positions. Complex numbers for signals (3.3) used a plane of positions. This one shows how fast $e^{st}$ dies, across, and how fast it spins, up.

Watch the two poles as the friction falls, with $\omega_n=1$ rad/s and $\zeta$ going from 2 down to 0.1, and watch the step response beside them.

At $\zeta=2$ the two poles sit on the negative part of the decay-rate axis, far apart, and the step response creeps with no overshoot. As the friction falls the poles slide together. At critical damping, $\zeta=1$, they meet at $-1$. Then they leave the axis as a mirror pair and climb, and the response starts to go past 1. Overshoot is how far the response goes past its final value, as a percentage of that value. It is 20.8 % at $\zeta=0.447$, with poles at $-0.447\pm0.894j$, and 72.9 % at $\zeta=0.1$, where the poles sit high and the response rings.

When the clip stops, drag a pole along the circle to choose your own $\zeta$, from 0.1 to 1.

Two things to see. Once the poles leave the axis they stay on a circle of radius $\omega_n$, because $\lvert s\rvert^2=(\zeta\omega_n)^2+\omega_d^2=\omega_n^2$. And the higher the poles sit, the more the response overshoots.

On the axis, for $\zeta>1$, the poles are $-\omega_n\bigl(\zeta\mp\sqrt{\zeta^2-1}\bigr)$. For $\zeta=2$ and $\omega_n=1$ that is $-0.268$ and $-3.732$. The pole near zero is the slow one, and it sets the creep: $e^{-0.268t}$ has a time constant of $1/0.268=3.73$ s. More friction pushes that pole closer to zero, which is why the stiff rig is slow.

This also corrects a tempting conclusion. Critical damping is the fastest response that never overshoots, but it is not the fastest to settle within 2 %. In units where $\omega_n=1$ rad/s, critical damping settles at 5.83 s, and a slightly under-damped $\zeta=0.8$ settles at 3.76 s. Not every reduction in damping helps: $\zeta=0.7$ gives 5.98 s and $\zeta=0.7071$ gives 5.96 s, both slower than critical.

Why the poles come as a pair

Each pole gives one spiral, $e^{st}$, in the way Complex exponentials & phasors (3.4) showed. Why does a real mass need both? The next instrument answers that for the rig with $\zeta=0.1$.

Look at the gap between the response and its target, $1-y(t)$, the bar 6.2 drew. It starts at 1 and starts flat, because the mass starts at rest. Those two start conditions fix everything about the spirals. The first spiral is $e^{s_1t}$ multiplied by $c=0.503\angle-5.7°$. By the rotate-and-scale rule of 3.3, that means the spiral starts 0.503 long and turned back by $5.7°$ ($-0.1002$ rad). Its partner is $c^*e^{s_2t}$, with $s_2$ the mirror image of $s_1$, and $c^*$ the mirror image of $c$ (3.3’s conjugate).

Watch the first spiral draw alone, then its partner, then their sum.

One pole alone gives a complex signal: its spiral leaves the real line. Its mirror partner turns the other way, from the mirror-image start. With both, the sum is real, and it traces the length of the gap bar as the bar shrinks and changes sign.

Two spirals that turn opposite ways have imaginary parts that cancel, as the two arrows of a cosine did in 3.4. What is left is real, and it is the gap. A real system needs both poles of the pair, and a single complex pole alone would give a complex signal.

The gap has a closed form. With $\omega_n=1$,

with $c=\tfrac12\bigl(1-j\,\zeta/\sqrt{1-\zeta^2}\bigr)$. For $\zeta=0.1$ that is $c=\tfrac12(1-0.1005j)$, whose length is $0.5025$ and angle is $-5.739°$. The start conditions say $c+c^*=1$ (the gap starts at 1) and $s_1c+s_2c^*=0$ (it starts flat), and this $c$ satisfies both.

The slope of the gap is $-e^{-\zeta t}\sin(\omega_dt)/\omega_d$, so it is first zero when $\omega_dt=\pi$. That instant is the peak time, $t_p=\pi/\omega_d$. There the gap is $-e^{-\pi\zeta/\sqrt{1-\zeta^2}}$, so the peak is past 1 by

This is the overshoot of the last section. For $\zeta=0.1$ it is $0.729$, so the peak is $1.729$, at $t_p=\pi/0.995=3.16$ s. For other $\omega_n$ the time axis scales by $1/\omega_n$ and $M_p$ does not change.

How long until it settles? The ringing is wrapped in $e^{-\zeta\omega_nt}$. That envelope reaches 2 % at $\ln50/(\zeta\omega_n)$, which is about $4/(\zeta\omega_n)$. I use this only as a rule of thumb for an under-damped system.

Pushing at the right rhythm

So far the push was a steady force. Now push the mass with a sine instead: the push is $F\sin\omega t$, starting from rest. At first there is a start-up wobble, the natural part of 6.2, and it dies away. What is left is the forced part, and Properties of LTI systems (5.4) says what that must be for a system like this: a sine at the push’s own rhythm. The gain is how big that steady swing is compared with the swing a steady push of the same size would give.

Push the same lightly damped system, $\zeta=0.1$, at three rhythms. A child on a swing knows the result. Pushed at half its natural rate the system follows. Pushed at twice its natural rate it barely moves. Pushed at its natural rate, the swings pile up. This is resonance.

Here are three identical rigs with $\zeta=0.1$, pushed with a sine from rest. Each has $\omega_n=2\pi$ rad/s, so its natural rhythm is one swing per second. Positions are measured in units of the swing a steady push of the same size gives. The swing size is the biggest distance from rest in the last second. Watch the middle rig’s swing size grow, then hold.

The middle rig is the biggest, and it keeps growing for a while: every push adds to the swing. From 6.49 s on it is steady, and the three swings are about 1.32, 5 and 0.33 times the swing of a steady push.

The middle rig reaches five times the swing of a steady push. That number is the quality factor $Q=1/(2\zeta)=1/0.2=5$. I state it without deriving it. Frequency response and Bode plots (8.4) derives the gain at every rhythm. A system with less friction has a larger $Q$ and a sharper resonance.

The other two rigs show the gain at the other rhythms: a little above 1 at half the natural rate, and well below 1 at twice it. Why did the middle rig stop growing? Each push adds to the swing, but a bigger swing loses more energy to friction. The swing stops growing when the two balance.

The same ideas in samples

The first-order recursion of Difference equations (6.1), $y[n]=a\,y[n-1]+x[n]$, has $h[n]=a^n$, a single decay. The number $a$ plays the part that $s$ played above, so I call it the pole of the discrete system. It is the discrete twin of $e^{-t/\tau}$. Decay needs $\lvert a\rvert<1$, and 6.2 showed that sampling an RC circuit gives $a=e^{-T_s/RC}$.

For ringing, use two delay boxes, $y[n-2]$ as well as $y[n-1]$. Take

Here $\Omega$ is the digital frequency in radians per sample, from Sinusoids (3.2), and $r$ is a number between 0 and 1. A plucked string simulated in code can look like this. Its impulse response is

Check the first two values. At $n=0$ it is $\sin\Omega/\sin\Omega=1$. At $n=1$ it is $r\sin2\Omega/\sin\Omega=2r\cos\Omega$, which is what the recursion gives.

This is the height of a spiral. Let

Its height is $\mathrm{Im}\,p[n]=h[n]$. It starts at angle $\Omega$, with length $1/\sin\Omega$. At each sample it turns by $\Omega$ and shrinks to $r$ of its length. Turning by $\Omega$ and shrinking by $r$ is multiplication by $re^{j\Omega}$, the rotate-and-scale of 3.3. That number and its mirror image $re^{-j\Omega}$ are the poles of this system.

Take $r=0.9$ and $\Omega=\pi/6$, which is $30°$ per sample. The picture shows only the plane of positions, because the spiral’s points are positions and not rates. Watch each new point’s height become the next stem.

The first point sits at $2\angle30°$, angle $\Omega$, and its height is the first stem, 1. Each sample turns the point by 30° and shrinks it to 0.9 of its length, and its height is the stem. At $n=5$ the stems, to three decimals, are $1,\ 1.559,\ 1.62,\ 1.263,\ 0.656,\ 0$. The last point is on the negative real axis, at length $1.181$, so its height is 0. The stem at $n=6$ is $-0.531$: the height rings and dies away. Look at where the stems cross zero. A turn of $30°$ per sample means a half turn every six samples, so the stems cross zero at $n=5$ and $n=11$, while the size shrinks by 0.9 at every sample.

Worked example

1. First order, $\tau=0.2$ s. The 10 % point is at $0.2\ln(1/0.9)=0.02107$ s and the 90 % point at $0.2\ln10=0.46052$ s. The rise time is $\tau\ln9=0.2\times2.19722=0.43944$ s. The 2 % settling time is $\tau\ln50=0.2\times3.91202=0.78240$ s.
2. Second order, $\omega_n=10$ rad/s and $\zeta=0.5$. The decay rate is $\zeta\omega_n=5$ and $\omega_d=10\sqrt{0.75}=8.66025$, so the poles are $s=-5\pm8.66025j$. The overshoot is $e^{-\pi(0.5)/\sqrt{0.75}}=e^{-1.81380}=0.16303$, which is $16.3\,\%$, so the peak value is $1.1630$. The peak comes at $\pi/\omega_d=0.36276$ s. The 2 % settling time is about $4/(\zeta\omega_n)=0.8$ s. The quality factor is $Q=1/(2\zeta)=1$.
3. Critical damping. The step response is $y=1-(1+x)e^{-x}$ with $x=\omega_nt$. At $x=1,\ 2,\ 5$ it is $0.26424,\ 0.59399,\ 0.95957$. It enters the 2 % band where $(1+x)e^{-x}=0.02$, at $x=5.834$.
4. Three absorbers, $\omega_n=1$ rad/s. For $\zeta=0.3$ the overshoot is $e^{-\pi(0.3)/\sqrt{0.91}}=0.3723$. The response is $y(4)=1.2944$, $y(5.834)=0.9051$, $y(10)=1.0513$ and $y(16)=1.0063$, and it settles for good at 11.23 s. For $\zeta=1$ it settles at 5.834 s. For $\zeta=2$ the poles are $-0.26795$ and $-3.73205$, and $y=1-1.07735e^{-0.26795t}+0.07735e^{-3.73205t}$ gives $y(4)=0.6311$, $y(10)=0.9261$ and $y(16)=0.9852$, settling at 14.88 s.
5. Mirror spirals, $\zeta=0.1$. The poles are $s=-0.1\pm0.99499j$. The gap is $e^{-0.1t}\bigl[\cos\omega_dt+0.100504\sin\omega_dt\bigr]=c\,e^{s_1t}+c^*e^{s_2t}$ with $c=\tfrac12(1-0.100504j)$. Its length is $0.50252$ and its angle is $-5.739°$. At $t=\pi/\omega_d$ the response is $1+e^{-0.1\pi/0.99499}=1.7292$.
6. Resonance, $\zeta=0.1$. The steady gain is $1/\sqrt{(1-(\omega/\omega_n)^2)^2+(2\zeta\,\omega/\omega_n)^2}$, which gives $1.3216$ at $\omega/\omega_n=0.5$, $5$ at $1$ and $0.3304$ at $2$. The page only shows simulated readouts. The swing sizes at 10 s are $1.322$, $4.991$ and $0.332$.
7. Discrete pair, $r=0.9$ and $\Omega=\pi/6$. With $2r\cos\Omega=1.55885$ and $r^2=0.81$, the recursion gives $h[0]=1$, $h[1]=1.55885$, $h[2]=1.62$, $h[3]=1.26267$, $h[4]=0.6561$, $h[5]=0$ and $h[6]=-0.53144$. For example $h[2]=1.55885^2-0.81=1.62$. The spiral point is $p[n]=2\times0.9^n\angle(n+1)\cdot30°$, and $h[n]=\mathrm{Im}\,p[n]$.
8. Overshoot against $\zeta$. For $\zeta=0.1,\ 0.2,\ 0.3,\ 0.447,\ 0.5$ and $0.7071$ the overshoot is $72.9\,\%$, $52.7\,\%$, $37.2\,\%$, $20.8\,\%$, $16.3\,\%$ and $4.32\,\%$.
9. A mirror partner. A real second-order system has a pole at $-2+3j$. Its other pole is the mirror image, $(-2+3j)^*=-2-3j$. Only with both is its free motion real.

Where you’ll meet this

Many physical systems are one of these two families, or a few of them chained: a heating room, a loudspeaker cone, a suspension, a car seat, a pendulum, a tuned circuit in a radio. When a datasheet quotes a rise time, an overshoot or a quality factor, it is quoting the numbers on this page. The next chapters ask for the response to every rhythm at once, not only three, and for a way to find poles without solving the equation: Frequency response and Bode plots (8.4), The Laplace transform (9.1), Poles, zeros and the s-plane (9.3) and, for the discrete twin, Transfer functions, poles and zeros (16.3). Bandwidth, which belongs to the first-order system too, waits for 8.4 because it needs the response at every rhythm.

Reference card