Error constants of 4/[s(s²+4s+4)], r = 4t+1

Static error constants are the limits $K_p=\lim_{s\to0}G(s)H(s)$, $K_v=\lim_{s\to0}sG(s)H(s)$ and $K_a=\lim_{s\to0}s^2G(s)H(s)$. They give the steady-state error for step, ramp and parabolic inputs.

$$G(s)H(s) = \frac{4}{s(s^2+4s+4)} = \frac{4}{s(s+2)^2}$$

There is one pole at the origin, so the system is type 1.

Stability check (needed before using final value theorem): characteristic equation $s^3+4s^2+4s+4=0$. All coefficients are positive and $4\times4=16>1\times4=4$, so by Routh's condition for a cubic the closed loop is stable.

Static error constants

$$\begin{aligned} K_p &= \lim_{s\to0}\frac{4}{s(s+2)^2} = \infty \\ K_v &= \lim_{s\to0}s\cdot\frac{4}{s(s+2)^2} = \frac{4}{4} = 1\ \text{s}^{-1} \\ K_a &= \lim_{s\to0}s^2\cdot\frac{4}{s(s+2)^2} = 0 \end{aligned}$$

Steady-state error for $r(t)=4t+1$

By superposition, split the input into a step and a ramp:

• Step part: $r_1(t)=1\cdot u(t)$, $A=1$: $e_{ss1} = \frac{A}{1+K_p} = \frac{1}{1+\infty} = 0$
• Ramp part: $r_2(t)=4t$, $A=4$: $e_{ss2} = \frac{A}{K_v} = \frac{4}{1} = 4$

$$e_{ss} = e_{ss1} + e_{ss2} = 0 + 4 = 4$$

Answer: $K_p=\infty$, $K_v=1\ \text{s}^{-1}$, $K_a=0$; steady-state error for $r(t)=4t+1$ is $e_{ss}=4$ (units of the output).