Time Response and Specifications
Rise time, overshoot, settling time and damping, and how poles set the shape of the step response.
What we specify
A control loop is judged by how its output moves after a step in the setpoint (or a disturbance). The standard measures of that step response are:
| Specification | Meaning | Usual target |
|---|---|---|
| Rise time tr | time from 10 to 90 percent of the final value | as short as the actuator allows |
| Overshoot | the peak above the final value, in percent | 0 to 10 percent for most loops; 0 for a heater that must not exceed the limit |
| Peak time tp | time of the first peak | — |
| Settling time ts | time to stay within 2 (or 5) percent | the time before the next move can start |
| Steady-state error | the final difference from the setpoint | zero for a step (needs an integrator) |
| IAE, ISE | the integral of the absolute or squared error | a single number to compare tunings |
First-order response
A first-order system never overshoots. With time constant τ the 10 to 90 percent rise time is 2.2 τ and the 2 percent settling time is about 4 τ. For the oven (τ = 120 s) in open loop that is a rise of 264 s and a settling time of 480 s (plus the 15 s of dead time). Feedback with a proportional gain makes the time constant shorter by the factor 1 + K Kp (Laplace Transforms and Transfer Functions).
Second-order response
overshoot = exp(−π ζ ÷ √(1 − ζ²)) tp = π ÷ (ωn √(1 − ζ²)) ts ≈ 4 ÷ (ζ ωn) tr ≈ (1 + 1.1 ζ + 1.4 ζ²) ÷ ωn
| Damping ζ | Overshoot | Peak time (ωn = 10) | Settling time (2 %) | Pole angle from the negative real axis |
|---|---|---|---|---|
| 0.2 | 52.7 % | 0.321 s | 2.00 s | 78° |
| 0.3 | 37.2 % | 0.329 s | 1.33 s | 73° |
| 0.5 | 16.3 % | 0.363 s | 0.80 s | 60° |
| 0.7 | 4.6 % | 0.440 s | 0.57 s | 46° |
| 1.0 | 0.0 % | — | 0.40 s | 0° |
Worked example: the servo
Worked example: the mass-spring-damper servo, ωn = 10 rad/s, ζ = 0.5
- Overshoot = exp(−π × 0.5 ÷ √0.75) = 16.3 percent. A numerical check of the step response finds a peak of 16.3 percent at t = 0.363 s.
- Peak time tp = π ÷ (10 × 0.866) = 0.363 s.
- Settling time (2 percent) ≈ 4 ÷ (0.5 × 10) = 0.80 s; the exact curve settles at 0.81 s.
- Rise time (10 to 90 percent): measured 0.164 s (the approximation gives 0.190 s from 0 to 100 percent).
- To go twice as fast (ωn = 20 rad/s at the same damping) the settling time halves to 0.40 s, but the spring needs four times the stiffness (k ∝ ωn²), or the control loop must supply it.
- The specification “overshoot under 10 percent, settling under 1 s” translates to ζ ≥ 0.6 and ζ ωn ≥ 4: the poles must lie to the left of the line s = −4 and inside the 53° wedge about the negative real axis.
Pole locations and the s-plane
The shape of the response follows from where the poles are:
- The real part σ of a pole sets the decay rate: the farther left, the faster the transient dies (settling time ≈ 4 ÷ |σ|). For the servo, σ = −ζ ωn = −5.
- The imaginary part sets the ringing frequency ωd = ωn √(1 − ζ²) = 8.66 rad/s.
- The angle of the pole from the negative real axis is arccos ζ = 60°, so the overshoot is fixed by the angle: a pole at 45° (ζ = 0.707) gives 4.3 percent, one at 60° (ζ = 0.5) 16 percent, one at 78° (ζ = 0.2) 53 percent.
- The distance from the origin is ωn = 10 rad/s.
Control design is largely the art of moving the closed-loop poles to a good place with the controller gains.
Dominant poles, zeros and delay
A real loop has more than two poles, but if one pair is much closer to the imaginary axis than the others (at least 3 to 5 times nearer), it dominates the response and the second-order formulas apply. Zeros in the left half-plane add overshoot and speed (a PI controller’s zero is the reason for its overshoot); a zero in the right half-plane (non-minimum phase, for example a boiler drum level or an aircraft’s elevator) makes the response first go the wrong way and limits the achievable bandwidth. Dead time adds lag but no new poles in the simple picture; it limits how fast the loop can be made (see Stability: Routh, Root Locus and Margins). A rough rule for processes with dead time θ is to keep the closed-loop time constant at least comparable to θ: the oven’s 15 s.
Steady-state error and system type
The system type is the number of pure integrators in the loop (poles at the origin). A type-0 loop has a constant error to a step (r ÷ (1 + Kv)), a type-1 loop has zero error to a step and a constant error to a ramp, and a type-2 loop zero error to both. For example a position loop with a motor and an integrating drive (type 1) with a velocity constant Kv = 4 s⁻¹ follows a ramp of 1 rad/s with a steady lag of 1 ÷ Kv = 0.25 rad. Adding an integral term to a type-0 process makes it type 1 and removes the offset.
Related
- Previous: Laplace Transforms and Transfer Functions. Next: PID Control and Tuning.
- PID Tuner (response simulation).
- Control Systems: the whole guide.
FAQ
Why 4 time constants for settling?
Because e−4 = 0.018: the output is within 2 percent. For 5 percent use 3 τ.
What damping ratio should I aim for?
0.7 is a good compromise: 4 percent overshoot and a fast settle. 1 or more for no overshoot; 0.5 when speed matters more than ringing.
Can I make the response faster without overshoot?
Only by increasing ωn at fixed damping, which needs more actuator power and bandwidth: the limit is saturation, noise and unmodeled dynamics.
Do the formulas hold for a PID loop?
Approximately, when the closed loop has a dominant pole pair. Zeros and delay change the numbers; simulate to be sure.
References
- Ogata, K., Modern Control Engineering, Pearson (transient-response analysis); Nise, N. S., Control Systems Engineering, Wiley.
- Franklin, G. F., et al., Feedback Control of Dynamic Systems, Pearson.
- Åström, K. J., Murray, R. M., Feedback Systems, Princeton University Press (free online).
Last reviewed: October 2026. Rules of thumb for rise time and settling time are approximations.