21 Problem A.5.23 Ogata 4edicion (offset)

Laplace.T	Root-locus	Transient response	Frecuency response

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	21 Problem A.5.23 Ogata 4edicion (offset) We will demonstrate that the system suffers offset to unit-step input $\begin{displaymath}\frac{C(s)}{R(s)}=\frac{\frac{K}{(1+T\cdot s)}}{(1+\frac{K}{(1+T\cdot s)})}=\frac{K}{(K+1+T\cdot s)}\end{displaymath}$ $\begin{displaymath}\frac{E(s)}{R(s)}=\frac{R(s)-C(s)}{R(s)}=1-\frac{C(s)}{R(s)}=... ...cdot s+K+1-K}{(T\cdot s+1+K)}=\frac{T\cdot s+1}{(T\cdot s+1+K)}\end{displaymath}$ To input $R(s)=\frac{r}{s}$ ,we obtain the error's Laplace transform: $\begin{displaymath}E(s)=\frac{T\cdot s+1}{(T\cdot s+1+K)}\cdot R(s)=\frac{T\cdot s+1}{(T\cdot s+1+K)}\cdot \frac{r}{s}\end{displaymath}$ The error e(t) at infinity, we obtain using the final-value theorem: $\begin{displaymath}e(\infty)=\lim_{s \rightarrow 0} s\cdot E(s)=\lim_{s \rightar... ...rac{T\cdot s+1}{(T\cdot s+1+K)}\cdot \frac{r}{s}=\frac{r}{1+K} \end{displaymath}$ The offset: $\begin{displaymath}\frac{r}{(1+K)} \end{displaymath}$