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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}