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6.3 — Time Response: First and Second Order
A specification says "settle within 2 seconds with less than 5% overshoot." Chapter 6.2 turned that into a region of the s-plane. This chapter derives the formulas that make the translation exact, and explains where each one comes from.
Almost everything reduces to two cases, because of the dominant-pole argument of Chapter 6.2: a first-order system, and a second-order system. Learn those two completely and you can read the behaviour of nearly any real loop.
The letters this chapter uses, said out loud
Control theory reuses a small alphabet relentlessly, and a page of it is unreadable until you know what each letter is measuring. Here is every symbol in this chapter, with where it came from.
\tau — "tau", from the Greek for the first letter of tachos, speed. It is the time constant: the number of seconds a first-order system takes to cover 63% of the gap to its target. Bigger tau means a lazier system. It has units of seconds and nothing else.
\omega_n — "omega sub n", the natural frequency. Omega has meant angular frequency since Euler, and the subscript n is for natural: the rate this system would oscillate at if nothing damped it. Measured in radians per second, so divide by 2\pi to get hertz.
\zeta — "zeta", the damping ratio, and the single most important number in this chapter. It is a pure number with no units, and it is a ratio: how much damping this system has, divided by how much it would need to just barely stop oscillating. Below 1 it rings; at exactly 1 it does not; above 1 it is sluggish.
\omega_d — "omega sub d", the damped natural frequency: the rate it actually rings at, which is always slightly below \omega_n because damping slows the oscillation as well as shrinking it. The relationship \omega_d = \omega_n\sqrt{1-\zeta^2} is derived in section 2.
\sigma — "sigma", the decay rate, \sigma = \zeta\omega_n. It is how fast the envelope of the ringing shrinks, in reciprocal seconds. On the s-plane of Chapter 4.5 it is the horizontal position of the pole, which is why the two chapters use the same letter.
M_p, t_r, t_s — peak overshoot, rise time, settling time. These are the four measurements marked on the step-response figure below, and sections 2 and 3 derive a formula for each from \zeta and \omega_n.
One relationship ties the whole chapter together, and it is worth having before any algebra: \zeta alone decides the shape of the response, and \omega_n alone decides how fast that shape plays out. Change \omega_n and the curve stretches or squashes horizontally without changing form; change \zeta and the form itself changes.
1. First order
G(s)=\frac{K}{\tau s+1}
One pole at s=-1/\tau. Step response, by the partial fractions of Chapter 4.5:
y(t)=K\left(1-e^{-t/\tau}\right)
The same exponential approach as the RC circuit of Chapter 1.5, and every number carries across:
| Time | Fraction of final |
|---|---|
| \tau | 63.2% |
| 2\tau | 86.5% |
| 3\tau | 95.0% |
| 4\tau | 98.2% |
| 5\tau | 99.3% |
Rise time (10% to 90%):
t_r=\tau\ln9=2.2\tau
Derivation: reaching 10% takes -\tau\ln(0.9)=0.105\tau; reaching 90% takes -\tau\ln(0.1)=2.303\tau. The difference is 2.197\tau.
Settling time to 2%:
t_s=4\tau
Overshoot: none, ever. A first-order system cannot overshoot, because it has no mechanism to store momentum. This is worth remembering as a diagnostic: if your supposedly first-order system overshoots, your model is wrong — there is a hidden second pole or a zero you did not account for.
2. Second order — the standard form
G(s)=\frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}
Two parameters do everything:
- \omega_n — the natural frequency, in rad/s. Sets the timescale. Double it and everything happens twice as fast, with the same shape.
- \zeta — the damping ratio, dimensionless. Sets the shape. Overshoot and ringing depend on it alone.
The separation of concerns is complete and useful: \omega_n scales time, \zeta scales nothing and shapes everything.
Poles:
s=-\zeta\omega_n\pm\omega_n\sqrt{\zeta^2-1}
The four regimes
Undamped, \zeta=0. Poles on the imaginary axis at \pm j\omega_n. Oscillates forever at \omega_n. Never a design target, but it is what a system does when a loop's damping is accidentally cancelled.
Underdamped, 0\lt\zeta\lt1. Complex poles at -\zeta\omega_n\pm j\omega_d where
\omega_d=\omega_n\sqrt{1-\zeta^2}
y(t)=1-\frac{e^{-\zeta\omega_nt}}{\sqrt{1-\zeta^2}}\sin\!\left(\omega_dt+\arccos\zeta\right)
Read the structure: an exponentially decaying envelope e^{-\zeta\omega_n t} multiplying a sine at \omega_d. The system overshoots, rings, and settles.
Critically damped, \zeta=1. Two equal real poles at -\omega_n:
y(t)=1-e^{-\omega_nt}(1+\omega_nt)
The factor of t is the signature of a repeated pole, as Chapter 4.5 noted. Fastest possible arrival with no overshoot.
Overdamped, \zeta\gt1. Two distinct real poles, two decaying exponentials, no overshoot, and slower than critical.
The curve every formula below is measuring
All the numbers in the next section are measurements taken off one shape, so here is the shape.
Walk it left to right, because each feature is one of the formulas about to be derived.
The rise — the output climbs from nothing towards the target. How long that takes is the rise time, and it measures how quick the system is.
The overshoot — it does not stop at the target, it sails past. The height of that first peak above the target, as a percentage, is the overshoot, and it is set entirely by the damping ratio \zeta and by nothing else. That fact is what makes \zeta the number a control engineer reaches for first.
The ringing — it comes back, undershoots, and oscillates with shrinking amplitude. The frequency of that ringing is the damped natural frequency \omega_d, slightly below the undamped \omega_n for exactly the reason Volume II Chapter 6.2 gives.
The settling — eventually it stays within a small band, usually ±2%, of the target. The time to get there for good is the settling time, and it is the number a customer actually cares about.
The whole of control design is trading these four against each other, because you cannot improve them all at once. Speed up the rise and you increase the overshoot. Kill the overshoot and the rise slows. Section 3 turns that trade into equations.
3. The performance formulas, derived
Peak time
The first maximum. Differentiate the step response and set it to zero. After some algebra the derivative comes out proportional to \sin(\omega_dt), which is zero at \omega_dt=\pi:
\boxed{t_p=\frac{\pi}{\omega_d}=\frac{\pi}{\omega_n\sqrt{1-\zeta^2}}}
Half a ringing period, which is what "first peak" means.
Overshoot
Substitute t_p back into the response. The sine term becomes \sin(\pi + \arccos\zeta) = -\sin(\arccos\zeta)=-\sqrt{1-\zeta^2}, so:
y(t_p)=1+e^{-\zeta\omega_n\pi/\omega_d}
\boxed{\%OS=100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}}}
\omega_n has cancelled completely. Overshoot depends only on the damping ratio, which is the single most useful fact in this chapter.
| \zeta | Overshoot |
|---|---|
| 0.1 | 72.9% |
| 0.3 | 37.2% |
| 0.5 | 16.3% |
| 0.6 | 9.5% |
| 0.707 | 4.3% |
| 0.8 | 1.5% |
| 0.9 | 0.15% |
| 1.0 | 0% |
Inverting it, which is the form you actually use in design:
\zeta=\frac{-\ln(\%OS/100)}{\sqrt{\pi^2+\ln^2(\%OS/100)}}
Worked example. Specification says 5% maximum overshoot.
\ln(0.05)=-2.996
\zeta=\frac{2.996}{\sqrt{9.87+8.98}}=\frac{2.996}{4.342}=0.690
So \zeta \ge 0.69, and on the s-plane that is an angle of \arccos(0.69)=46.4° from the negative real axis.
Settling time
The envelope is e^{-\zeta\omega_nt}. It reaches 2% when
\zeta\omega_nt=\ln50=3.91
\boxed{t_s\approx\frac{4}{\zeta\omega_n}}
And \zeta\omega_n is exactly the real part of the pole. So:
t_s\approx\frac{4}{|\text{Re}(pole)|}
Settling time depends only on how far left the poles are. That is why Chapter 6.2's picture had a vertical line for a settling specification.
For 5% settling the constant is 3 rather than 4.
Rise time
No closed form, but a good approximation for 0.3\lt\zeta\lt0.8:
t_r\approx\frac{1.8}{\omega_n}
So \omega_n sets the speed and \zeta sets the shape — the separation stated at the start, now with numbers attached.
4. Designing from a specification
Specification: settling time under 1 s, overshoot under 10%.
Step 1 — overshoot gives \zeta.
\ln(0.1)=-2.303, \qquad \zeta=\frac{2.303}{\sqrt{9.87+5.30}}=\frac{2.303}{3.895}=0.591
Round up for margin: \zeta = 0.6.
Step 2 — settling time gives \zeta\omega_n.
\frac{4}{\zeta\omega_n}\le1 \;\Rightarrow\; \zeta\omega_n\ge4
With \zeta=0.6: \omega_n\ge6.67 rad/s.
Step 3 — the pole locations.
\sigma=\zeta\omega_n=4, \qquad \omega_d=\omega_n\sqrt{1-\zeta^2}=6.67\times0.8=5.33
s=-4\pm j5.33
Step 4 — check the other numbers.
t_p=\frac{\pi}{5.33}=0.59\ \text{s}, \qquad t_r\approx\frac{1.8}{6.67}=0.27\ \text{s}
\%OS=100e^{-\pi(0.6)/0.8}=100e^{-2.356}=9.5\% \ ✓
Step 5 — find the gain that puts the poles there. For the plant G(s)=\dfrac{K}{s(s+8)}:
s^2+8s+K=0
Matching to s^2+2\zeta\omega_ns+\omega_n^2: 2\zeta\omega_n=8 so \zeta\omega_n=4 ✓ automatically satisfied, and \omega_n^2=K.
With \omega_n=6.67: K=44.5.
Check: \zeta = 4/6.67=0.6 ✓.
Both specifications met with one gain value, which is lucky and specific to this plant. Usually the plant's pole positions do not let you hit both targets with gain alone, and then you need a compensator — a controller that adds poles and zeros of its own, which is Chapter 6.6.
5. The effect of extra poles and zeros
The formulas above assume exactly two poles and no zeros. Real systems have more, and here is how much it matters.
A third pole at -p. If p \gt 5\zeta\omega_n, the second-order formulas are accurate to a few percent. Closer than that and the response gets slower and the overshoot drops — the extra pole adds lag that damps the ringing.
A zero at -z in the left half plane. If z \gt 5\zeta\omega_n, negligible. Closer and the overshoot increases significantly.
Worked example. A system with \zeta=0.5, \omega_n=10 has 16.3% overshoot. Add a zero at -5 — which is at z/\zeta\omega_n = 1 — and the overshoot rises to roughly 40%.
Where this bites in practice. A PD controller adds a zero (Chapter 6.6). The zero improves the damping, which reduces overshoot, but it also adds this direct overshoot contribution. The net effect is usually good but not as good as the damping calculation alone predicts, and this is the standard reason a designed loop overshoots more than the maths said.
A right-half-plane zero, from Chapter 6.2. Undershoot first, then a slow approach, and a hard bandwidth ceiling. No formula from this chapter applies, and the second-order approximation is worthless.
Pole-zero cancellation. If a pole and a zero are at nearly the same place, their contributions largely cancel and the system behaves as though neither existed.
Two warnings about relying on it. First, cancellation is never exact — component values drift, and a pole at -2.0 cancelled by a zero at -2.1 leaves a small slow term that will show up as a long tail on the step response. Second, never cancel an unstable pole with a zero. On paper the transfer function is fine; in reality the unstable mode is still there, merely invisible from the input and output, and any disturbance excites it and the system runs away. This has destroyed real hardware.
6. Disturbance response
Setpoint tracking is only half the job. What happens when the world pushes back?
For a disturbance D entering at the plant input:
\frac{Y(s)}{D(s)}=\frac{G(s)}{1+G(s)C(s)}
Compare with the setpoint response \dfrac{GC}{1+GC}. The denominators are identical — the poles are the same — but the numerators differ, so the shape of the disturbance response is not the same as the setpoint response.
At low frequency, if |GC|\gg1:
\frac{Y}{D}\approx\frac{G}{GC}=\frac1C
Read that: the disturbance rejection depends on the controller gain, not the plant. High controller gain at low frequency means good rejection.
And with an integrator in C, the controller gain at DC is infinite, so a constant disturbance is rejected completely. This is the same argument as the steady-state error result of Chapter 6.2, applied to disturbances rather than setpoints, and it is the second reason every practical controller has an integral term.
Worked example. A motor speed loop with a proportional gain of 10. A load torque disturbance appears.
Without integral action, the steady-state speed drop is 1/(1+10) = 9\% of what it would be with no control at all. A 90% improvement, and a permanent 9% error.
With an integral term, the drop is transient and the speed returns exactly to setpoint. The load torque is fully rejected.
7. Reading a step response
Given a measured response, work backwards to the model. This is the most practical skill in the chapter.
Step 1 — measure the overshoot. That gives \zeta directly from the table in section 3.
Step 2 — measure the peak time. Then
\omega_d=\frac{\pi}{t_p}, \qquad \omega_n=\frac{\omega_d}{\sqrt{1-\zeta^2}}
Step 3 — measure the final value. That gives the DC gain.
Step 4 — check consistency. Predict the settling time from 4/\zeta\omega_n and compare with the measurement. If it does not match, the system is not second order and there is something else going on.
Worked example. A measured response settles at 2.0, peaks at 2.36 at t=0.4 s, and stays within 2% after 1.6 s.
\%OS=\frac{2.36-2.0}{2.0}=18\%
From the table, \zeta\approx0.48.
\omega_d=\frac{\pi}{0.4}=7.85, \qquad \omega_n=\frac{7.85}{\sqrt{1-0.23}}=\frac{7.85}{0.877}=8.95\ \text{rad/s}
Predicted settling time: 4/(0.48\times8.95)=0.93 s.
Measured: 1.6 s. They do not match, and the measurement is 70% slower than predicted.
What that tells you. There is a slow pole the second-order model is missing — probably a lag somewhere near -1 that is dragging out the tail. The mismatch is the finding, not an error, and it points directly at where to look for unmodelled dynamics.
A ratio the other way — measured settling faster than predicted — usually means a zero near the dominant poles, which speeds the initial response.
Everything so far has assumed the system is stable. Chapter 6.4 provides the tests that tell you whether it is, without computing the roots.
Every formula above, built from scratch
None of the results in this chapter are worth memorising, because each one can be rebuilt in under a minute from something simpler. What follows is that rebuilding, one result at a time, so the formula and the reason for it sit on the same page as the explanation that needed them.
Steady-state error
E(s)=\frac{R(s)}{1+G(s)} \qquad\text{(unity feedback)}
e_{ss}=\lim_{s\to0}\frac{sR(s)}{1+G(s)}
Error constants:
K_p=\lim_{s\to0}G(s), \qquad K_v=\lim_{s\to0}sG(s), \qquad K_a=\lim_{s\to0}s^2G(s)
| Type | Step | Ramp | Parabola |
|---|---|---|---|
| 0 | \dfrac{1}{1+K_p} | ∞ | ∞ |
| 1 | 0 | \dfrac{1}{K_v} | ∞ |
| 2 | 0 | 0 | \dfrac{1}{K_a} |
Type = number of poles at the origin in G(s). Each integrator eliminates the error for one more level of input.
First-order response
G(s)=\frac{K}{\tau s+1}, \qquad y(t)=K\left(1-e^{-t/\tau}\right)
t_r=\tau\ln9=2.2\tau, \qquad t_s(2\%)=4\tau, \qquad \%OS=0
Where 2.2 comes from: 10% is reached at -\tau\ln0.9=0.105\tau and 90% at -\tau\ln0.1=2.303\tau.
Second-order response
G(s)=\frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}, \qquad s=-\zeta\omega_n\pm j\omega_d
\omega_d=\omega_n\sqrt{1-\zeta^2}
y(t)=1-\frac{e^{-\zeta\omega_nt}}{\sqrt{1-\zeta^2}}\sin(\omega_dt+\arccos\zeta)
Peak time. The derivative of the step response is proportional to \sin\omega_dt, first zero at \omega_dt=\pi:
t_p=\frac{\pi}{\omega_d}
Overshoot. Substituting t_p back and simplifying:
\%OS=100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}}
\omega_n cancels entirely — overshoot depends only on \zeta.
Inverted, for design:
\zeta=\frac{-\ln(\%OS/100)}{\sqrt{\pi^2+\ln^2(\%OS/100)}}
Settling time. The envelope e^{-\zeta\omega_nt} reaches 2% when \zeta\omega_nt=\ln50=3.91:
t_s\approx\frac{4}{\zeta\omega_n}=\frac{4}{|\text{Re}(\text{pole})|}
For 5%: t_s\approx3/\zeta\omega_n.
Rise time, empirical for 0.3\lt\zeta\lt0.8:
t_r\approx\frac{1.8}{\omega_n}
Resonant peak in frequency:
M_p=\frac{1}{2\zeta\sqrt{1-\zeta^2}}, \qquad \omega_r=\omega_n\sqrt{1-2\zeta^2}
No peak exists for \zeta\gt0.707, which is where that value's popularity comes from.
What the next chapter fixes
Every response in this chapter assumed the system settles at all. Feed back too much gain and it does not — it oscillates, and then it grows. Chapter 6.4 gives two ways to know in advance: one that answers yes or no from the coefficients alone, and one that draws how the poles move as you turn the gain up.