23 Problem A.5.25 Ogata 4edition (steady-state error, torque disturbances)

Laplace.T	Root-locus	Transient response	Frecuency response

23 Problem A.5.25 Ogata 4edition (steady-state error, torque disturbances)

Let's see the system's error with torque disturbances. The close-loop transference function is

$\begin{displaymath}C(s)=(D(s)-K\cdot C(s))\cdot \frac{1}{J \cdot s}=D(s)\cdot \frac{1}{J \cdot s}-\frac{K}{J \cdot s} \cdot C(s) \end{displaymath}$

$\begin{displaymath}(1+\frac{K}{J \cdot s})\cdot C(s)=D(s)\cdot \frac{1}{J \cdot s}\end{displaymath}$

$\begin{displaymath}\frac{C(s)}{D(s)}=\frac{\frac{1}{J \cdot s}}{(1+\frac{K}{J \cdot s})}=\frac{1}{J \cdot s+K}\end{displaymath}$

The Laplace Transform's error is:

$\begin{displaymath}\frac{E(s)}{D(s)}=-\frac{C(s)}{D(s)}=-\frac{1}{J \cdot s+K}\end{displaymath}$

Knowing that the disturbance torque is unit-step will have:

$\begin{displaymath}E(s)=-\frac{1}{J \cdot s+K}\cdot \frac{1}{s}\end{displaymath}$

The steady-state error is:

$\begin{displaymath}e(\infty)=\lim_{s \rightarrow 0} s\cdot E(s)=\lim_{s \rightarrow 0} s\cdot -\frac{1}{J \cdot s+K}\cdot \frac{1}{s}=-\frac{1}{K}\end{displaymath}$