Second-Order Systems and Transient Specifications
Add a second pole and the response comes alive: it can now overshoot, ring, and settle in a dozen different ways. Two numbers govern all of it — the natural frequency, which sets the timescale, and the damping ratio, which sets the oscillation. From these two parameters flow every transient specification a designer quotes: peak time, overshoot, rise time, settling time. This is the most-used model in all of control engineering, because almost any real system is read as a second-order one near its dominant poles.
- The standard second-order form and the meaning of natural frequency \(\omega_n\) and damping ratio \(\zeta\).
- The four damping cases — undamped, underdamped, critically damped, overdamped — and their poles.
- The underdamped step response and the pole-plane geometry of \(\sigma\), \(\omega_d\), and \(\beta\).
- The transient specifications: peak time, percent overshoot, rise time, settling time, and delay time.
- Why \(\zeta\) alone fixes the overshoot while \(\omega_n\) scales the time axis.
- How to design a system to meet overshoot and settling-time specifications.
The Standard Second-Order System
A great many systems — an RLC circuit, a mass–spring–damper, a position servo — reduce to the same canonical second-order transfer function. Writing it in standard form exposes the two parameters that control everything: the undamped natural frequency \(\omega_n\) (rad/s), which sets how fast the system oscillates if undamped, and the damping ratio \(\zeta\) (dimensionless), which sets how quickly those oscillations die.
Match any quadratic denominator \(s^2 + as + b\) to this template: \(\omega_n = \sqrt{b}\) and \(\zeta = a/(2\omega_n)\). The two poles, and therefore the entire response, are fixed by \(\zeta\) and \(\omega_n\) alone.
The Four Damping Cases
The discriminant \(\zeta^2-1\) under the square root decides the character of the poles, and with it the entire shape of the response. Four regimes result.
| Damping ratio | Poles | Response |
|---|---|---|
| \(\zeta = 0\) (undamped) | \(\pm j\omega_n\) (on the \(j\omega\)-axis) | Sustained oscillation, never settles |
| \(0 < \zeta < 1\) (underdamped) | \(-\zeta\omega_n \pm j\omega_d\) (complex pair) | Damped oscillation with overshoot |
| \(\zeta = 1\) (critically damped) | \(-\omega_n\) (repeated real) | Fastest response with no overshoot |
| \(\zeta > 1\) (overdamped) | Two distinct real, negative | Sluggish, no overshoot |
The Underdamped Response
The underdamped case \(0<\zeta<1\) is the one that matters most, because it is where designers operate. Its complex poles are \(s = -\sigma \pm j\omega_d\), and the pole-plane geometry below names every quantity in the response.
The response is a sinusoid at the damped frequency \(\omega_d\), wrapped in a decaying envelope \(e^{-\sigma t}\). The envelope's decay rate \(\sigma\) controls settling; the sinusoid's frequency \(\omega_d\) controls the peak time; and the angle \(\beta=\cos^{-1}\zeta\) the pole makes with the imaginary axis fixes the overshoot.
Peak Time and Overshoot
Two specifications describe the overshoot: when the first peak occurs and how high it rises above the final value. Differentiating the step response and setting it to zero gives the first peak at \(\omega_d t = \pi\).
The peak time depends only on the damped frequency \(\omega_d\). The percent overshoot depends only on the damping ratio \(\zeta\) — not on \(\omega_n\). This separation is the key to design: pick \(\zeta\) for the overshoot, then \(\omega_n\) for the speed.
Rise, Settling, and Delay Times
The remaining specifications measure how quickly the response gets to and stays near its final value. The settling time follows from the decay envelope \(e^{-\zeta\omega_n t}\) reaching the tolerance band; the rise time follows from the sine reaching the final value for the first time.
| Specification | Formula (underdamped) | Set by |
|---|---|---|
| Peak time \(t_p\) | \(\dfrac{\pi}{\omega_d}\) | \(\omega_d\) |
| Rise time \(t_r\) | \(\dfrac{\pi - \theta}{\omega_d},\;\; \theta=\cos^{-1}\zeta\) | \(\omega_d,\,\zeta\) |
| Settling time \(t_s\) (2%) | \(\dfrac{4}{\zeta\omega_n}\) | \(\sigma=\zeta\omega_n\) |
| Settling time \(t_s\) (5%) | \(\dfrac{3}{\zeta\omega_n}\) | \(\sigma=\zeta\omega_n\) |
| Delay time \(t_d\) | \(\dfrac{1 + 0.7\zeta}{\omega_n}\) (approx.) | \(\omega_n,\,\zeta\) |
| Maximum overshoot \(M_p\) | \(e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}\) | \(\zeta\) only |
Damping Ratio and Overshoot
Because overshoot is a function of \(\zeta\) alone, it is worth committing a few values to memory. The mapping is steep: a small change in damping makes a large change in overshoot.
| \(\zeta\) | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 | 0.8 |
|---|---|---|---|---|---|---|---|---|
| \(M_p\) | 73% | 53% | 37% | 25% | 16% | 9.5% | 4.6% | 1.5% |
Given a required overshoot, this returns the damping ratio that achieves it. For example, a 10% overshoot needs \(\zeta\approx 0.59\); the popular \(\zeta=0.707\) gives about 4.3% overshoot — fast yet barely overshooting, which is why it is a common design target.
Designing to Specifications
The clean separation of roles — \(\zeta\) for overshoot, \(\omega_n\) for speed — turns specification into a two-step recipe. Given a target overshoot and a target settling (or peak) time, the parameters fall out directly.
| Step | From the specification… | …find |
|---|---|---|
| 1 | Required overshoot \(M_p\) | \(\zeta\) (overshoot relation, Section 8-6) |
| 2 | Required settling time \(t_s\) | \(\zeta\omega_n = 4/t_s\) (2%), hence \(\omega_n\) |
| 3 | \(\zeta\) and \(\omega_n\) | Pole locations \(-\zeta\omega_n \pm j\omega_d\) |
| 4 | Pole locations | Required denominator \(s^2+2\zeta\omega_n s+\omega_n^2\) |
Worked Examples
Problem. For \(\dfrac{C}{R}=\dfrac{25}{s^2+6s+25}\), find \(\omega_n\), \(\zeta\), \(\omega_d\), and the poles. Classify the response.
Solution. Match the standard form: \(\omega_n^2=25\Rightarrow\omega_n=5\); \(2\zeta\omega_n=6\Rightarrow\zeta=0.6\). Then:
Since \(0<\zeta<1\), the system is underdamped.
Problem. For the same system (\(\zeta=0.6,\ \omega_n=5,\ \omega_d=4\)), find \(t_p\), \(M_p\), \(t_s\) (2%), and \(t_r\).
Solution. Apply each formula in turn (\(\theta=\cos^{-1}0.6=0.927\) rad):
Problem. Classify \(\dfrac{9}{s^2+6s+9}\) and \(\dfrac{4}{s^2+5s+4}\).
Solution. First: \(\omega_n=3,\ 2\zeta\omega_n=6\Rightarrow\zeta=1\) — critically damped, double pole at \(-3\). Second: \(\omega_n=2,\ 2\zeta\omega_n=5\Rightarrow\zeta=1.25\) — overdamped:
Problem. A response must not overshoot more than 5%. What minimum damping ratio is required?
Solution. Invert the overshoot relation with \(M_p=0.05\):
Any \(\zeta \ge 0.69\) keeps the overshoot at or below 5%.
Problem. Design a second-order system with at most 5% overshoot and a 2% settling time of 2 s. Find \(\zeta\), \(\omega_n\), and the denominator.
Solution. From Example 4, \(\zeta=0.69\). Then \(\zeta\omega_n=4/t_s=2\Rightarrow\omega_n=2/0.69\approx 2.9\,\text{rad/s}\):
The settling-time spec fixed the real part at \(-2\); the overshoot spec fixed the pole angle. Together they pin the desired pole locations.
Problem. Two systems share \(\zeta=0.5\) but have \(\omega_n=2\) and \(\omega_n=8\). Compare their overshoot and peak time.
Solution. Overshoot depends only on \(\zeta\), so both overshoot \(16\%\). Peak time scales with \(1/\omega_d\), and \(\omega_d=\omega_n\sqrt{1-\zeta^2}=0.866\,\omega_n\):
Quadrupling \(\omega_n\) quarters the peak time while leaving the overshoot untouched — the cleanest illustration of the two parameters' separate roles.
Chapter Summary
\(\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}\): two parameters, \(\omega_n\) and \(\zeta\).
Undamped, underdamped, critically damped, overdamped — set by \(\zeta\).
\(\sigma=\zeta\omega_n\), \(\omega_d=\omega_n\sqrt{1-\zeta^2}\), angle \(\beta=\cos^{-1}\zeta\).
\(t_p=\pi/\omega_d\); \(M_p=e^{-\zeta\pi/\sqrt{1-\zeta^2}}\), set by \(\zeta\) alone.
\(t_s=4/(\zeta\omega_n)\) at 2%; governed by the real part \(\sigma\).
Overshoot → \(\zeta\); settling time → \(\omega_n\); together → desired poles.
Problems
Match each transfer function to the standard form first, then read off \(\zeta\) and \(\omega_n\) before computing specs. Difficulty rises down the list.
- For \(\dfrac{16}{s^2+4s+16}\), find \(\omega_n\), \(\zeta\), \(\omega_d\), and the poles.
- Classify the response of \(\dfrac{100}{s^2+20s+100}\) by its damping ratio.
- Classify \(\dfrac{8}{s^2+6s+8}\) and give its poles.
- For \(\zeta=0.5\) and \(\omega_n=10\), find \(t_p\), \(M_p\), and \(t_s\) (2%).
- A system has \(\zeta=0.707\). Compute its percent overshoot.
- Find the damping ratio that gives exactly 20% overshoot.
- For \(\dfrac{25}{s^2+6s+25}\), find the rise time using \(t_r=(\pi-\theta)/\omega_d\).
- A second-order system must have \(M_p\le 10\%\) and \(t_s\le 1\,\text{s}\) (2%). Find the required \(\zeta\) and \(\omega_n\).
- Write the closed-loop denominator for poles at \(-4\pm j4\). State \(\zeta\) and \(\omega_n\).
- Two systems have the same \(\omega_n=6\) but \(\zeta=0.3\) and \(\zeta=0.6\). Compare overshoot and settling time.
- Show that the peak time is independent of the damping ratio for fixed \(\omega_d\), and explain physically why.
- A position servo must settle within 2% in 0.5 s with no more than 5% overshoot. Determine the desired pole locations.