Overshoot, settling time and error for 50/[s(s+2)]

Find $\zeta$ and $\omega_n$ from the closed-loop characteristic equation for the transient part. Then use the error coefficients (the system is Type 1) for the steady state part.

Closed-loop transfer function

$$\frac{C(s)}{R(s)} = \frac{50}{s^2 + 2s + 50}$$ $$\begin{aligned} \omega_n^2 &= 50 \;\Rightarrow\; \omega_n = 7.071\ \text{rad/s} \\ 2\zeta\omega_n &= 2 \;\Rightarrow\; \zeta = \frac{1}{7.071} = 0.1414 \end{aligned}$$

Maximum overshoot

$$M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}} = e^{-\pi(0.1414)/0.9899} = e^{-0.4488} = 0.6384 = 63.84\%$$

Settling time (2% criterion)

$$t_s = \frac{4}{\zeta\omega_n} = \frac{4}{0.1414 \times 7.071} = \frac{4}{1} = 4\ \text{s}$$

(With the 5% criterion, $t_s = 3/\zeta\omega_n = 3$ s.)

Static error coefficients

$G(s)$ has one pole at the origin, so it is Type 1.

$$\begin{aligned} K_p &= \lim_{s\to 0} G(s) = \infty \\ K_v &= \lim_{s\to 0} sG(s) = \frac{50}{2} = 25 \\ K_a &= \lim_{s\to 0} s^2G(s) = 0 \end{aligned}$$

Steady state error for $r(t) = 2 + 4t + 6t^2$

Write $6t^2 = 12\cdot\dfrac{t^2}{2}$.

$$\begin{aligned} e_{ss} &= \frac{2}{1+K_p} + \frac{4}{K_v} + \frac{12}{K_a} \\ &= 0 + \frac{4}{25} + \frac{12}{0} = 0 + 0.16 + \infty = \infty \end{aligned}$$

Answer: $M_p = 63.84\%$, $t_s = 4$ s (2%); $K_p = \infty$, $K_v = 25$, $K_a = 0$; $e_{ss} = \infty$, because a Type 1 system cannot follow a parabolic input.