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Time-domain specifications of 25/(s²+8s+25)

Short answer

tr = 0.833 s, tp = 1.047 s, ts = 1 s (2%), Mp = 1.52%.

For a second order system given by the transfer function G(s) = 25/(s² + 8s + 25), find the rise time, peak time, settling time and maximum overshoot.

Compare with the standard second-order form ωn2s2+2ζωns+ωn2\frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}:

  • ωn2=25⇒ωn=5\omega_n^2=25 \Rightarrow \omega_n = 5 rad/s
  • 2ζωn=8⇒ζ=0.82\zeta\omega_n = 8 \Rightarrow \zeta = 0.8 (underdamped)
  • ωd=ωn1−ζ2=5×0.6=3\omega_d = \omega_n\sqrt{1-\zeta^2} = 5\times0.6 = 3 rad/s
  • θ=cos⁡−1(0.8)=36.87∘=0.6435\theta = \cos^{-1}(0.8) = 36.87^\circ = 0.6435 rad
tr=π−θωd=3.1416−0.64353=0.833 stp=πωd=3.14163=1.047 sts=4ζωn=40.8×5=1 s(2% criterion)Mp=e−ζπ/1−ζ2=e−0.8π/0.6=e−4.189=0.0152\begin{aligned} t_r &= \frac{\pi-\theta}{\omega_d} = \frac{3.1416-0.6435}{3} = 0.833\ \text{s} \\ t_p &= \frac{\pi}{\omega_d} = \frac{3.1416}{3} = 1.047\ \text{s} \\ t_s &= \frac{4}{\zeta\omega_n} = \frac{4}{0.8\times5} = 1\ \text{s}\quad(2\%\ \text{criterion}) \\ M_p &= e^{-\zeta\pi/\sqrt{1-\zeta^2}} = e^{-0.8\pi/0.6} = e^{-4.189} = 0.0152 \end{aligned}

With the 5% criterion, ts=3ζωn=0.75t_s = \frac{3}{\zeta\omega_n} = 0.75 s.

Answer: tr=0.833t_r = 0.833 s, tp=1.047t_p = 1.047 s, ts=1t_s = 1 s (2%), Mp=1.52%M_p = 1.52\%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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