Transient response
The overshoot of a second-order system depends only on the damping ratio: 16 % with 0.5 and under 5 % with 0.7. A nearby third pole breaks the rule.
01First order: the time constant
The step response of a first-order system, \( G(s) = K/(\tau s + 1) \), is an exponential that reaches 63 % of the final value in one time constant \( \tau \), 95 % in \( 3\tau \) and 98 % in \( 4\tau \). It has no overshoot. It is the model of the oven and of the simplified motor in the previous units. The solution of the differential equations of circuits was studied in transient response, in Circuit Theory I; here what matters is reading the specifications of a control system from it.
02Second order: ζ and ωn
Many closed loops behave like a second-order system, described by the damping ratio \( \zeta \) and the natural frequency \( \omega_n \):
The step response specifications are related to the poles by closed formulas, valid for \( \zeta < 1 \):
03Dominant poles
A real system has more poles, but if a complex pair is much closer to the imaginary axis than the others, its exponentials are the slowest and dominate the response: the other poles die out quickly. The rule of thumb asks that the other poles be at least five times farther to the left. If they are close, the response becomes slower and has less overshoot than the formulas predict.
The step response simulated with Runge-Kutta, optionally with a third real pole. The data compare the specifications measured on the curve with those of the formulas. The shaded strip is the 2 % band.
04From specifications to poles
The formulas are used in reverse to design: a maximum overshoot fixes a minimum \( \zeta \), and a settling time fixes the minimum real part of the poles. In the s-plane, that defines a region: to the left of a vertical line and inside an angle around the negative real axis. The design consists of bringing the closed-loop poles into that region, with the tools of the root locus or of frequency response.
05In the lab
Measure the step response of a series RLC circuit for three values of R and compute ζ and ωn from the overshoot and the peak time.
Close a proportional position loop around a geared motor and measure how the overshoot and the peak time change with the gain.
Add an RC filter at the output of the RLC circuit and observe how the response changes with its time constant.
06Common mistakes
- Applying the second-order formulas to a system without dominant poles. They can be far off.
- Believing that more damping is always better. With ζ > 1 there is no overshoot, but the response is slow.
- Forgetting the zeros. A zero near the poles increases the overshoot even if ζ is the same.
- Confusing peak time with settling time. One depends on the imaginary part of the poles and the other on the real part.
07Self-assessment
What ζ corresponds to a 10 % overshoot?
\( \zeta = -\ln 0.1/\sqrt{\pi^2 + \ln^2 0.1} \approx 0.59 \).
With ζ = 0.5 and ωn = 4 rad/s, what are tp and ts?
\( t_p = \pi/(4 \cdot 0.866) \approx 0.91 \) s; \( t_s \approx 4/2 = 2 \) s.
What happens to the response if a third pole has the same real part as the complex pair?
It becomes slower and has less overshoot: the pair stops being dominant.
What region of the s-plane satisfies ζ ≥ 0.7?
That of the poles that make an angle of up to 45.6° with the negative real axis.
08Further reading
- Katsuhiko Ogata. Modern Control Engineering. 5th ed., Prentice Hall, 2010 (Spanish edition: “Ingeniería de control moderna”, Pearson). The analysis of transient response with its specifications.
- Norman S. Nise. Control Systems Engineering. 8th ed., Wiley, 2019. Dominant poles and the effect of additional poles and zeros.
- Benjamin C. Kuo and Farid Golnaraghi. Automatic Control Systems. 10th ed., McGraw-Hill, 2017. Time-domain specifications and s-plane regions.