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Peak Time (2nd-Order System)

Calculates the time at which a second-order underdamped system's response reaches its maximum (peak) value.

Written by the Vorzep Team

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Peak Time Calculator2nd-Order System

tp = π / ( ωn · √( 1 − ζ² ) )
tp = peak time (s)  ·  ωn = natural frequency (rad/s)  ·  ζ = damping ratio (0 < ζ < 1)
⟹ Solve⟷tp, ωn, ζ
s
rad/s
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π ≈ 3.14159265359  ·  For underdamped systems: 0 < ζ < 1
Please fix the errors above.
Solve for:
Presets:
Peak Time—
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tp: —ωn: —ζ: —
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Peak Time Magnitude—
Fast (< 0.1 s) Medium (0.1–1.0 s) Slow (1.0–5.0 s) Very Slow (> 5.0 s)
tp = π / (ωn · √(1 − ζ²))  ·  Valid for underdamped systems (0 < ζ < 1).

Interpretation

t_p = π/(ω_n√(1−ζ²)). Time at which the peak overshoot occurs. Used to characterise response speed.

t_p = pi / (omega_n * sqrt(1 - zeta^2))
Peak Time (2nd-Order System)

Variables

SymbolQuantityUnit
t_pTime to reach peak responses
omega_nNatural (undamped) frequency of the systemrad/s
zetaDamping ratio of the system (0 < zeta < 1)

What it means

Peak time is the time it takes for the response to reach its first maximum. It is inversely proportional to the damped natural frequency. It is a performance metric for transient response. Understanding it helps in setting expectations for response times.

Worked example

Peak Time (2nd‑Order) – Two Detailed Examples

Real‑World
Scenario: A second‑order system has damping ratio ζ = 0.5 and natural frequency ω_n = 2 rad/s. The peak time t_p = π / (ω_n × √(1‑ζ²)) = π / (2 × √(1‑0.25)) = 3.1416 / (2 × 0.866) = 3.1416 / 1.732 = 1.814 seconds. This is when the response reaches its maximum overshoot. The engineer uses this to predict when the maximum displacement will occur in a mechanical system, which is important for avoiding collisions.
ParameterValue
ζ (damping ratio)0.5
ω_n (rad/s)2
1t_p = π / (2 × √(1-0.25)) = 3.1416 / (2 × 0.866) = 1.814 s
Result 1.814 s ✓ Peak time
Scenario: A lightly damped system with ζ = 0.1 and ω_n = 1 rad/s has t_p = π / (1 × √(1‑0.01)) = 3.1416 / 0.995 = 3.157 seconds. The long peak time indicates the system oscillates for a long time before settling. The engineer uses this to adjust the damping to achieve a faster response, perhaps by adding a dashpot or modifying the controller gains.
ParameterValue
ζ0.1
ω_n1
1t_p = 3.1416 / (1 × √(0.99)) = 3.1416 / 0.995 = 3.157 s
Result 3.157 s ✓ Long peak time
Insight: Peak time is the time at which the response reaches its maximum value. It is inversely proportional to the damped natural frequency. Systems with low damping have longer peak times and higher overshoot.

Common mistakes

  • Peak time: t_p = π / (ω_n·√(1−ζ²)) – the time of the first peak (overshoot).
  • Units: ω_n in rad/s, t_p in seconds.
  • Damped natural frequency: ω_d = ω_n·√(1−ζ²) – this is the frequency of oscillation.
  • Applies to: Underdamped second‑order systems (0 < ζ < 1).
  • Relation: t_p occurs at the first maximum of the step response.

Applications

Peak time for a second‑order system, t_p = π/(ω_n·√(1−ζ²)), is the time at which the response reaches its first peak. This is used to characterise the speed of response and to tune controllers. Engineers use it to ensure that the system meets response time constraints. A smaller peak time indicates a faster but potentially more oscillatory response. By designing for a specific peak time, they can meet dynamic performance requirements. This formula is widely used in control system design and analysis.

  • Specification of transient response speed
  • Controller tuning for desired peak time
  • Optimisation of response characteristics in servos
  • Analysis of mechanical and electrical systems
  • Educational application of time‑domain metrics

Frequently Asked Questions

Q01What is the peak time for a second‑order system?
A01

The peak time t_p is the time at which the response reaches its first maximum (overshoot). It is given by t_p = π / (ω_n · √(1 – ζ²)), where ω_n is the natural frequency and ζ is the damping ratio.

Q02What is the common mistake when using this formula?
A02

Applying the formula to an overdamped or critically damped system, which has no overshoot and therefore no defined peak time.

Q03How does the peak time depend on ω_n?
A03

It is inversely proportional to ω_n. Higher natural frequency leads to a shorter peak time (faster response).

Q04How does ζ affect the peak time?
A04

As ζ increases (towards 1), the denominator √(1–ζ²) decreases, so t_p increases. The response becomes slower and less oscillatory.

Q05What is the relationship between peak time and settling time?
A05

Both are measures of speed of response. Peak time is the first peak; settling time is when the response stays within a band. For a given system, t_p < t_s.

Q06What is the peak time for ζ = 0.7 and ω_n = 1 rad/s?
A06

t_p = π / (1 × √(1 – 0.49)) = π / √0.51 ≈ 3.1416 / 0.714 = 4.40 seconds.

Q07How do you reduce peak time?
A07

Increase ω_n (e.g., higher gain) or decrease ζ (but this increases overshoot).

Q08What is the peak time for a critically damped system?
A08

Undefined; there is no overshoot.

Q09What are the applications of peak time?
A09

Specifying the desired speed of response in control system design, often along with overshoot.

Q10What is the relationship between peak time and the damped natural frequency?
A10

t_p = π / ω_d, where ω_d = ω_n√(1–ζ²) is the damped natural frequency. The peak time is the time for half a cycle of the damped oscillation.