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Settling Time (2nd-Order System)

Estimates the time for a second-order underdamped control system's response to settle within about 2% of its final value.

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Settling Time Calculator2nd-Order System

ts = 4 / (ζ · ωn)
ts = settling time (2% criterion)  ·  ζ = damping ratio  ·  ωn = natural frequency
⟹ Solvets, ζ, ωn
rad/s
s
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Solve for:
Presets:
Settling time (ts)
ζ: ωn: ts:
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Settling time gauge (s)
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ts = 4 / (ζ·ωn)  ·  2% settling time criterion for underdamped 2nd-order systems

Interpretation

t_s = 4/(ζ·ω_n). Time to settle within 2% of final value for a second‑order system. Used to evaluate speed of response.

t_s = 4 / (zeta * omega_n)
Settling Time (2nd-Order System)

Variables

SymbolQuantityUnit
t_sSettling time (2% criterion)s
zetaDamping ratio of the system
omega_nNatural (undamped) frequency of the systemrad/s

What it means

For a standard second‑order system with damping ratio ζ and natural frequency ω_n, the settling time for a 2% tolerance band is approximately t_s = 4/(ζω_n). This is a key performance metric in control design. A shorter settling time means faster response. It is used to specify system performance and to tune controllers. Understanding this helps in meeting design requirements.

Worked example

Settling 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 settling time t_s ≈ 4 / (ζ × ω_n) = 4 / (0.5 × 2) = 4 / 1 = 4 seconds. The control engineer uses this to ensure the system reaches its final value within an acceptable time. This is crucial for applications like robotic arm positioning, where fast settling is needed to maintain productivity.
ParameterValue
ζ (damping ratio)0.5
ω_n (rad/s)2
1t_s = 4 / (0.5 × 2) = 4 / 1 = 4 s
Result 4 s ✓ Settling time
Scenario: A critically damped system has ζ = 1 and ω_n = 1 rad/s. The settling time is t_s = 4 / (1 × 1) = 4 seconds. A critically damped system is often desired for applications like measuring instruments, where overshoot must be avoided. The engineer uses this to design a system that settles quickly without overshooting.
ParameterValue
ζ1
ω_n1
1t_s = 4 / (1 × 1) = 4 s
Result 4 s ✓ Critically damped
Insight: Settling time is the time required for a system's response to stay within a specified band (typically ±2% of the final value). It depends on the damping ratio and natural frequency. Increasing ω_n or ζ reduces the settling time.

Common mistakes

  • Settling time (2nd‑order): t_s = 4 / (ζ·ω_n) – for a 2% settling criterion.
  • Damping ratio ζ: Dimensionless – between 0 and 1 for underdamped.
  • Natural frequency ω_n: In rad/s – must be consistent with t_s in seconds.
  • Approximation: This is the 2% criterion (within 2% of final value) – for 5%, use 3/(ζ·ω_n).
  • Assumes: Standard second‑order underdamped system with no zeros.

Applications

Settling time for a second‑order system, t_s = 4/(ζ·ω_n), is the time required for the response to stay within a certain percentage (typically 2%) of the final value. This is a key performance metric for control systems. Engineers use it to assess the speed of response and to set controller gains. A shorter settling time indicates a faster system, but may increase overshoot. By adjusting the damping ratio ζ and natural frequency ω_n, they can achieve the desired trade‑off. This formula is used in designing motor drives, robots, and aerospace control systems.

  • Performance specification for servo systems and actuators
  • Controller tuning for desired response speed
  • Trade‑off analysis between speed and stability
  • Design of underdamped and critically damped systems
  • Education on second‑order system dynamics

Frequently Asked Questions

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

The settling time (t_s) is the time required for the response to stay within a certain percentage (usually 2% or 5%) of the final value. For a 2% criterion, it is approximately t_s = 4 / (ζ·ω_n), where ζ is the damping ratio and ω_n is the natural frequency.

Q02What is the common mistake when using this formula?
A02

Using the 2% settling‑time approximation when a different settling criterion was actually required, since the constant changes accordingly (e.g., for 5%: t_s = 3 / (ζ·ω_n)).

Q03What factors affect the settling time?
A03

The settling time is inversely proportional to ζ·ω_n. A higher natural frequency or higher damping ratio reduces the settling time. However, increasing ζ beyond 0.7 may increase the rise time.

Q04What is the settling time for a critically damped system (ζ=1)?
A04

For ζ=1, t_s (2%) ≈ 4/ω_n. The response is the fastest without overshoot.

Q05How does the settling time relate to the pole locations?
A05

The settling time is determined by the real part of the dominant poles: t_s ≈ 4 / (ζ·ω_n) = 4 / |real part of poles|. The poles are at s = –ζω_n ± jω_d.

Q06What is the effect of increasing ω_n on settling time?
A06

Increasing ω_n reduces the settling time (faster response), provided ζ is kept constant. This is why systems with higher natural frequency respond faster.

Q07What is the settling time for an underdamped system (ζ < 1)?
A07

The formula t_s = 4/(ζω_n) is approximate. The exact settling time depends on the damping and the tolerance band. For ζ between 0.5 and 0.8, the approximation is good.

Q08What is the relationship between settling time and rise time?
A08

For a given ω_n, a higher ζ reduces overshoot but increases rise time. There is a trade‑off. Typically, ζ ≈ 0.7 gives a good balance.

Q09What are the applications of settling time?
A09

Control system design, performance specifications, and evaluating the speed of response of a system.

Q10How do you improve settling time?
A10

Increase ω_n (e.g., by increasing the gain or using a faster actuator) or adjust ζ to an optimal value (usually around 0.7).