Error constants of 108/[s²(s+4)(s²+3s+12)], r = 2+5t+8t²

The steady state error of a unity feedback system depends on the type of $G(s)$ and on the static error coefficients $K_p$, $K_v$ and $K_a$.

Type of the system

$$G(s) = \frac{108}{s^2(s+4)(s^2+3s+12)}$$

There are two poles at the origin, so the system is Type 2.

Static error coefficients

$$\begin{aligned} K_p &= \lim_{s\to 0} G(s) = \infty \\ K_v &= \lim_{s\to 0} sG(s) = \lim_{s\to 0}\frac{108}{s(s+4)(s^2+3s+12)} = \infty \\ K_a &= \lim_{s\to 0} s^2G(s) = \frac{108}{(4)(12)} = \frac{108}{48} = 2.25 \end{aligned}$$

Error for each input term

For a unity feedback system:

Input: $r(t) = 2 + 5t + 8t^2$.

• Step part $2$: $e_1 = \dfrac{2}{1+\infty} = 0$
• Ramp part $5t$: $e_2 = \dfrac{5}{\infty} = 0$
• Parabolic part $8t^2 = 16\cdot\dfrac{t^2}{2}$, so $C = 16$: $e_3 = \dfrac{16}{K_a} = \dfrac{16}{2.25} = 7.11$

By superposition:

$$e_{ss} = 0 + 0 + \frac{16}{2.25} = \frac{64}{9} = 7.11$$

Answer: $K_p = \infty$, $K_v = \infty$, $K_a = 2.25$, and $e_{ss} = 7.11$ (units of the output).

Note on stability: the closed-loop characteristic equation is

$$s^5 + 7s^4 + 24s^3 + 48s^2 + 0\cdot s + 108 = 0$$

The $s$ term is missing, so by the R-H necessary condition the closed loop is actually unstable, and strictly a steady state does not exist. The value $7.11$ is the result the error-coefficient method gives, which is what is normally expected for this question.