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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 $\frac{1}{2}$