Library / Control engineering / CTL-04
Control engineering
Peak time of an underdamped second-order system
Time at which the step response of a standard second-order system first reaches its peak.
- Ref
- CTL-04
- Rev
- v1
- Field
- Control engineering
- Result
- tp [s]
tp is the time of the first maximum of the unit-step response of the standard second-order system. It is half of the damped period, π/ωd, where ωd = ωn√(1 − ζ²) is the damped natural frequency.
Time at which the step response of a standard second-order system first reaches its peak. It is half of the damped oscillation period.
Calculator
- Damped natural frequency, ωn√(1 − ζ²)3.464 rad/s
- Damped period, 2tp1.814 s
- Overshoot at the peak16.3 %
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- At the defaults the first peak arrives after 0.9069 s, where the overshoot is 16.3 %. The full damped period is 1.814 s.
- Doubling ωn halves the peak time to 0.4534 s without changing the overshoot, because overshoot depends on ζ only. Changing ζ changes both.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (s) | Local step | Change in result |
|---|---|---|---|---|---|---|
| ωn Natural frequency | 1 to 20 rad/s | 0.1814 | 3.628 | 3.446 | +10% | −9.09 % |
| ζ Damping ratio (0 < ζ < 1) | 0.1 to 0.9 | 0.7894 | 1.802 | 1.012 | +10% | +3.7 % |
Sweep of Natural frequency
| Natural frequency (rad/s) | Peak time (s) |
|---|---|
| 1 | 3.628 |
| 3.375 | 1.075 |
| 5.75 | 0.6309 |
| 8.125 | 0.4465 |
| 10.5 | 0.3455 |
| 12.88 | 0.2818 |
| 15.25 | 0.2379 |
| 17.63 | 0.2058 |
| 20 | 0.1814 |
Sweep one variable
Variables and how each one contributes
Peak time
What the equation gives you. Everything else feeds it.
Damping ratio (0 < ζ < 1)
The damping ratio, 0 < ζ < 1. Peak time lengthens as damping rises, because the damped frequency ωd falls. Raising it by 0.1 (to 0.6) moves the result by +8.25 %, from 0.9069 to 0.9817 s.
Natural frequency
The natural frequency in rad/s: the oscillation frequency if there were no damping. Inversely proportional: a faster system peaks sooner by the same percentage. Raising it by 10 % (to 4.4) moves the result by −9.09 %, from 0.9069 to 0.8245 s.
Worked example with the values above
- Damped frequency: ωd = ωn√(1 − ζ²) = 4 × √(1 − 0.25) = 3.464 rad/s.
- Half a damped period: tp = π / ωd = 3.142 / 3.464.
- tp = 0.9069 s.
Assumptions and limits
- A standard second-order system with no zeros and 0 < ζ < 1.
- A unit step from rest in a linear system.
Common mistakes
- Using ωn instead of ωd. The peak time is π/ωd, not π/ωn.
- Looking for a peak when ζ ≥ 1. An overdamped response rises monotonically.
Statistics and uncertainty
tp is inversely proportional to ωn, so a 5 % error in ωn gives a 5 % error in tp. The dependence on ζ is weak at low damping and strong near 1: at the defaults an error of ±0.05 in ζ moves tp from 0.8795 s to 0.9404 s.
Where it is used
- Estimating when an actuator or attitude loop reaches its first peak.
- Matching a second-order model to a measured step response using the time and size of the first peak.
Maths you need
- Damped frequency: ωd = ωn√(1 − ζ²)
- Period and half period: T = 2π / ω