Find a, b, c from time specifications

Compare the system with the standard second order form. The final value fixes $c$; the overshoot fixes $\zeta$; the peak time then fixes $\omega_n$.

Interpretation: "$e_{ss} = 0.5$" is taken as the steady state value of the output, $\theta(\infty) = 0.5$ rad, for the 10 Nm step (the usual reading of this problem, since $\theta$ and $T$ have different units).

$$\frac{\theta(s)}{T(s)} = \frac{1}{as^2+bs+c} = \frac{1/a}{s^2 + \frac{b}{a}s + \frac{c}{a}} \;\Rightarrow\; \omega_n^2 = \frac{c}{a},\quad 2\zeta\omega_n = \frac{b}{a}$$

Step 1: Find $c$ from the steady state

$T(s) = 10/s$. By the final value theorem:

$$\theta(\infty) = \lim_{s\to 0} s\cdot\frac{10}{s}\cdot\frac{1}{as^2+bs+c} = \frac{10}{c} = 0.5 \;\Rightarrow\; c = 20\ \text{Nm/rad}$$

Step 2: Damping ratio from 6% overshoot

$$\zeta = \frac{-\ln M_p}{\sqrt{\pi^2 + (\ln M_p)^2}} = \frac{-\ln 0.06}{\sqrt{\pi^2 + (\ln 0.06)^2}} = \frac{2.8134}{\sqrt{9.8696 + 7.9153}} = 0.6671$$

Step 3: Natural frequency from $t_p = 1$ s

$$t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}} \;\Rightarrow\; \omega_n = \frac{\pi}{1\times\sqrt{1 - 0.6671^2}} = \frac{3.1416}{0.7449} = 4.217\ \text{rad/s}$$

Step 4: Find $a$ and $b$

$$\begin{aligned} a &= \frac{c}{\omega_n^2} = \frac{20}{4.217^2} = \frac{20}{17.785} = 1.125\ \text{kg-m}^2 \\ b &= 2\zeta\omega_n a = 2(0.6671)(4.217)(1.125) = 6.328\ \text{Nm-s/rad} \end{aligned}$$

Answer: $a \approx 1.125$ kg-m², $b \approx 6.33$ Nm-s/rad, $c = 20$ Nm/rad.

(If instead $e_{ss} = 10 - \theta(\infty) = 0.5$ is used, then $c = 10/9.5 = 1.053$, $a = 0.0592$, $b = 0.333$; $\zeta$ and $\omega_n$ are unchanged.)