9 Problem A8.5 OGATA 4ed(Polar plot)

Laplace.T	Root-locus	Transient response	Frecuency response

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	9 Problem A8.5 OGATA 4ed(Polar plot) Let's show that the polar plot is a semicircle. $\begin{displaymath}G(jw)=\frac{j\cdot w \cdot T}{1+j\cdot w \cdot T}\end{displaymath}$ Solution: $\begin{displaymath}G(jw)=\frac{(j w T)\cdot(1-j w T)}{(1+j w T)\cdot(1-j w T)}=\... ...)^{2}}=\frac{(w T)^{2}}{1+(w T)^{2}}+\frac{(j w T}{1+(w T)^{2}}\end{displaymath}$ $\begin{displaymath}X=\frac{(w T)^{2}}{1+(w T)^{2}}\end{displaymath}$ $\begin{displaymath}Y=\frac{(w T}{1+(w T)^{2}}\end{displaymath}$ If is a circle has the form: $\begin{displaymath}(X-a)^{2}+Y^{2}=R^2\end{displaymath}$ We check it: $\begin{displaymath}X^{2}-2\cdot X \cdot a+a^{2}+Y^{2}=R^{2}\end{displaymath}$ $\begin{displaymath}X^{2}-2\cdot X \cdot a+Y^{2}+a^{2}=R^{2}\end{displaymath}$ $\begin{displaymath}\frac{(w T)^{4}}{(1+(w T)^{2})^{2}}-2\cdot a \cdot \frac{(w T... ...}}{1+(w T)^{2}}+\frac{(w T)^{2}}{(1+(w T)^{2})^{2}}+a^{2}=R^{2}\end{displaymath}$ $\begin{displaymath}\frac{(w T)^{4}}{(1+(w T)^{2})^{2}}-2\cdot a \cdot \frac{(w T... ...T)^{2})^{2}}=R^{2}\cdot \frac{(1+(w T)^{2})}{(1+(w T)^{2})^{2}}\end{displaymath}$ $\begin{displaymath}\frac{(w T)^{4}}{(1+(w T)^{2})^{2}}-2\cdot a \cdot \frac{(w T... ...{2}}=(R^{2}-a^{2})\cdot \frac{(1+(w T)^{2})}{(1+(w T)^{2})^{2}}\end{displaymath}$ $\begin{displaymath}(w T)^{4}-2\cdot a \cdot (w T)^{2}\cdot (1+(w T)^{2})+(w T)^{2}=(R^{2}-a^{2})\cdot (1+(w T)^{2})\end{displaymath}$ $\begin{displaymath}(w T)^{4}-2\cdot a \cdot (w T)^{2}-2\cdot a \cdot (w T)^{4}+(w T)^{2}=(R^{2}-a^{2})\cdot (1+(w T)^{2})\end{displaymath}$ $\begin{displaymath}(w T)^{4}\cdot (1-2\cdot a)+(w T)^{2}\cdot (1-2\cdot a)=(R^{2}-a^{2})\cdot (1+(w T)^{2})\end{displaymath}$ $\begin{displaymath}R=a\end{displaymath}$ $\begin{displaymath}a=\frac{1}{2}\end{displaymath}$ The radius circle is $\frac{1}{2}$ and the circle is centered in $\frac{1}{2}$