Error constants of 20/[s(0.5s+1)(s+2)], r = 10+5t

The steady state error follows from the static error constants. $G(s)$ has one pole at the origin, so the system is Type 1.

Static error constants

$$G(s) = \frac{20}{s(0.5s + 1)(s + 2)}$$ $$\begin{aligned} K_p &= \lim_{s\to 0} G(s) = \infty \\ K_v &= \lim_{s\to 0} sG(s) = \frac{20}{(1)(2)} = 10\ \text{s}^{-1} \\ K_a &= \lim_{s\to 0} s^2G(s) = 0 \end{aligned}$$

Steady state error for $r(t) = 10 + 5t$

$$\begin{aligned} e_{ss} &= \frac{10}{1 + K_p} + \frac{5}{K_v} \\ &= \frac{10}{1 + \infty} + \frac{5}{10} = 0 + 0.5 = 0.5 \end{aligned}$$

Answer: $K_p = \infty$, $K_v = 10\ \text{s}^{-1}$, $K_a = 0$; $e_{ss} = 0.5$.

Note on stability: $G(s) = \dfrac{40}{s(s+2)^2}$, so the characteristic equation is $s^3 + 4s^2 + 4s + 40 = 0$. Routh's condition for a cubic needs $4 \times 4 > 40$, which fails. So the closed loop as given is unstable, and the error value above (the usual expected answer) holds only in the formal sense. The gain would have to be below 16 (i.e. $K < 8$ in $K/[s(0.5s+1)(s+2)]$) for the result to apply.