Settling Time and Rise Time Calculator

Find the settling time, peak time, and rise time of an underdamped second-order system from its natural frequency and damping ratio.

📉 Settling Time and Rise Time Calculator
rad/s
0.5100 rad/s
0.010.99
Settling time (ts, 2%)
—
Peak time (tp)
—
Rise time (tr, 0-100%)
—
Damped natural frequency (ωd)
—
Step-by-step working

📉 What are Settling Time and Rise Time?

The settling time and rise time calculator finds the key transient-response specifications of an underdamped second-order control system, settling time (ts), peak time (tp), and rise time (tr), directly from its natural frequency (ωn) and damping ratio (ζ). These three numbers describe how fast a system responds to a step input and how much it oscillates before settling at its final value, the standard vocabulary control engineers use to judge whether a design is fast enough and smooth enough for its application.

Second-order transient specs show up constantly in practice: a motor speed loop's rise time determines how quickly a robot arm accelerates to a commanded speed, a temperature controller's settling time determines how long a process must wait before the next production step, and an aircraft autopilot's damping ratio determines whether a pitch correction feels smooth or jittery to passengers. Because so many higher-order systems can be reasonably approximated by a dominant second-order pair, these formulas remain a practical everyday tool well beyond textbook problems.

A common misconception is that rise time and settling time measure the same thing. They do not. Rise time measures only how fast the response first reaches its target (0% to 100% in the convention used here), while settling time measures how long the oscillations that follow the first crossing take to die down within a tolerance band, 2% of the final value in this calculator's convention. A system can have a very fast rise time yet a slow settling time if it overshoots and rings for several cycles before settling.

This calculator is useful for control systems students verifying textbook problems, and for practicing engineers doing a quick sanity check on a PID-tuned loop's expected transient behavior before running it on real hardware.

📐 Formula

ωd = ωn√(1−ζ²)    ts = 4/(ζωn)    tp = π/ωd    tr = (π−β)/ωd
ωn = natural frequency (rad/s)
ζ = damping ratio, strictly between 0 and 1 for an underdamped system
ωd = damped natural frequency (rad/s)
ts = settling time using the standard 2% criterion
β = arccos(ζ), in radians
tr = 0% to 100% rise time, the exact underdamped formula
Example: ωn = 10 rad/s, ζ = 0.5 → ωd = 8.660 rad/s, ts = 0.8 s, tp = 0.3628 s, tr = 0.2418 s.

📖 How to Use This Calculator

Steps

1
Enter the natural frequency. Type the system's natural frequency ωn, in rad/s.
2
Enter the damping ratio. Type ζ, a value strictly between 0 and 1 for an underdamped system.
3
Read the results. See settling time, peak time, rise time, and the damped natural frequency, plus a labeled step-response chart.

💡 Example Calculations

Example 1 — Moderately Damped System

wn = 10 rad/s, zeta = 0.5

1
ωd = 10 × √(1−0.5²) = 8.6603 rad/s
2
ts = 4 / (0.5 × 10) = 0.8000 s, tp = π / 8.6603 = 0.3628 s
3
β = arccos(0.5) = 1.0472 rad, tr = (π − 1.0472) / 8.6603 = 0.2418 s
ts = 0.8000 s, tp = 0.3628 s, tr = 0.2418 s
Try this example →

Example 2 — Lightly Damped, More Oscillatory System

wn = 10 rad/s, zeta = 0.3

1
ωd = 10 × √(1−0.3²) = 9.5394 rad/s
2
ts = 4 / (0.3 × 10) = 1.3333 s, tp = π / 9.5394 = 0.3293 s
3
β = arccos(0.3) = 1.2661 rad, tr = (π − 1.2661) / 9.5394 = 0.1966 s
ts = 1.3333 s, tp = 0.3293 s, tr = 0.1966 s (faster peak, slower settle, than Example 1)
Try this example →

Example 3 — Heavily Damped, Sluggish System

wn = 10 rad/s, zeta = 0.8

1
ωd = 10 × √(1−0.8²) = 6.0000 rad/s
2
ts = 4 / (0.8 × 10) = 0.5000 s, tp = π / 6.0000 = 0.5236 s
3
β = arccos(0.8) = 0.6435 rad, tr = (π − 0.6435) / 6.0000 = 0.4163 s
ts = 0.5000 s, tp = 0.5236 s, tr = 0.4163 s (settles fastest, but slowest to first reach target)
Try this example →

❓ Frequently Asked Questions

What is settling time in a second-order system?+
Settling time (ts) is the time it takes for a system's step response to enter and stay within a specified error band around its final value, commonly 2%. Using the 2% criterion, ts = 4/(zeta x wn), where zeta is the damping ratio and wn is the natural frequency in rad/s.
What is the formula for settling time?+
ts = 4 / (zeta x wn), using the standard 2% settling criterion. For wn = 10 rad/s and zeta = 0.5, ts = 4 / (0.5 x 10) = 0.8 seconds.
What is peak time and how is it calculated?+
Peak time (tp) is when the step response first reaches its maximum (overshoot) value. tp = pi / wd, where wd = wn x sqrt(1 - zeta^2) is the damped natural frequency. For wn = 10 rad/s and zeta = 0.5, wd = 8.660 rad/s and tp = pi / 8.660 = 0.3628 s.
What is the formula for rise time?+
Using the exact 0% to 100% underdamped formula: tr = (pi - beta) / wd, where beta = arccos(zeta) in radians. For wn = 10 rad/s and zeta = 0.5, beta = arccos(0.5) = 1.0472 rad, and tr = (pi - 1.0472) / 8.660 = 0.2417 s.
Why do these formulas only work for zeta strictly between 0 and 1?+
These are the standard underdamped second-order response formulas, which assume the system oscillates while decaying toward its final value. At zeta = 1 the system is critically damped and at zeta greater than 1 it is overdamped, neither of which oscillates, so peak time and this rise-time formula do not apply. At zeta 0 or below the system is undamped or unstable and never settles.
What happens if I enter a damping ratio of 1 or greater?+
The calculator shows an error explaining the system is critically damped (zeta = 1) or overdamped (zeta > 1). Both step responses rise smoothly to the final value with no oscillation and no true peak time, so this page's underdamped formulas do not apply to them.
What is the difference between settling time and rise time?+
Rise time measures how fast the response first reaches the target value (0% to 100% in this calculator's convention). Settling time measures how long the oscillations that follow take to damp down within a tolerance band (2% here) of the final value. A system can rise quickly but still settle slowly if it overshoots and rings for a while.
Does a higher natural frequency always mean a faster response?+
Yes, for a fixed damping ratio, all three transient specs (ts, tp, tr) scale inversely with wn, doubling wn halves settling time, peak time, and rise time. Natural frequency sets the overall speed scale, while damping ratio shapes how oscillatory the response looks at that speed.
Why does this calculator use the 2% settling time criterion instead of 5%?+
The 2% criterion (ts = 4/(zeta x wn)) is the most common convention in control systems textbooks and is what this calculator uses throughout. A 5% band uses ts = 3/(zeta x wn) instead, which gives a smaller settling time for the same system, always confirm which convention a datasheet or textbook problem is using.
What units does this calculator use?+
Natural frequency wn is entered in radians per second (rad/s), damping ratio zeta is dimensionless (between 0 and 1), and all time results (settling time, peak time, rise time, and the damped natural frequency's period) are shown in seconds or rad/s as appropriate.
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What is settling time in a second-order system?

Settling time (ts) is the time it takes for a system's step response to enter and stay within a specified error band around its final value, commonly 2%. Using the 2% criterion, ts = 4/(zeta x wn), where zeta is the damping ratio and wn is the natural frequency in rad/s.

What is the formula for settling time?

ts = 4 / (zeta x wn), using the standard 2% settling criterion. For wn = 10 rad/s and zeta = 0.5, ts = 4 / (0.5 x 10) = 0.8 seconds.

What is peak time and how is it calculated?

Peak time (tp) is when the step response first reaches its maximum (overshoot) value. tp = pi / wd, where wd = wn x sqrt(1 - zeta^2) is the damped natural frequency. For wn = 10 rad/s and zeta = 0.5, wd = 8.660 rad/s and tp = pi / 8.660 = 0.3628 s.

What is the formula for rise time?

Using the exact 0% to 100% underdamped formula: tr = (pi - beta) / wd, where beta = arccos(zeta) in radians. For wn = 10 rad/s and zeta = 0.5, beta = arccos(0.5) = 1.0472 rad, and tr = (pi - 1.0472) / 8.660 = 0.2418 s.

Why do these formulas only work for zeta strictly between 0 and 1?

These are the standard underdamped second-order response formulas, which assume the system oscillates while decaying toward its final value. At zeta = 1 the system is critically damped and at zeta greater than 1 it is overdamped, neither of which oscillates, so peak time and this rise-time formula do not apply. At zeta 0 or below the system is undamped or unstable and never settles.

What happens if I enter a damping ratio of 1 or greater?

The calculator shows an error explaining the system is critically damped (zeta = 1) or overdamped (zeta > 1). Both step responses rise smoothly to the final value with no oscillation and no true peak time, so this page's underdamped formulas do not apply to them.

What is the difference between settling time and rise time?

Rise time measures how fast the response first reaches the target value (0% to 100% in this calculator's convention). Settling time measures how long the oscillations that follow take to damp down within a tolerance band (2% here) of the final value. A system can rise quickly but still settle slowly if it overshoots and rings for a while.

Does a higher natural frequency always mean a faster response?

Yes, for a fixed damping ratio, all three transient specs (ts, tp, tr) scale inversely with wn, doubling wn halves settling time, peak time, and rise time. Natural frequency sets the overall speed scale, while damping ratio shapes how oscillatory the response looks at that speed.

Why does this calculator use the 2% settling time criterion instead of 5%?

The 2% criterion (ts = 4/(zeta x wn)) is the most common convention in control systems textbooks and is what this calculator uses throughout. A 5% band uses ts = 3/(zeta x wn) instead, which gives a smaller settling time for the same system, always confirm which convention a datasheet or textbook problem is using.

What units does this calculator use?

Natural frequency wn is entered in radians per second (rad/s), damping ratio zeta is dimensionless (between 0 and 1), and all time results (settling time, peak time, rise time, and the damped natural frequency's period) are shown in seconds or rad/s as appropriate.

📌 Quick Tips

💡Settling time and damping ratio trade off directly: a higher zeta settles faster relative to its own oscillation period but also slows the rise time, because it takes longer to first reach the target.
💡The 2% settling-time criterion (ts = 4/(zeta x wn)) is the most common convention. Some references use a 5% band instead (ts = 3/(zeta x wn)), so always check which criterion a spec sheet is using before comparing numbers.
💡Peak time tp depends only on the damped natural frequency wd, not on how far the system overshoots, that is governed separately by zeta through the percent overshoot formula.
💡Rise time here uses the exact 0% to 100% underdamped formula. Some references instead define rise time as 10% to 90%, which gives a smaller, non-comparable number for the same system.
💡As zeta approaches 1, the underdamped formulas on this page break down; the system is transitioning to critically damped, where distinct decaying oscillations no longer occur.