Ejercicion 4 (lugar de las raices)

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Lugar de las raices

Calcular los polos y ceros
Vamos a calcular las raices de $s^{4}+1$ para saber los polos de la funcion.

$\begin{displaymath}s^{4}+1=0\end{displaymath}$

$\begin{displaymath}s^4=-1; s^2=\pm j\end{displaymath}$

$\begin{displaymath}(s+a+jb)(s+c+jd)=(s+a)(s+c)+jb\cdot(s+c)+jd\cdot(s+a)-bd=s^{2}+(a+c)\cdot s+a\cdot c-bd+j(b\cdot s+b\cdot c+d\cdot s+d\cdot a)\end{displaymath}$

$\begin{displaymath}s^{2}+(a+c)\cdot s+a\cdot c-bd+j(b\cdot s+b\cdot c+d\cdot s+d\cdot a)=(s^{2}+j)\end{displaymath}$

$\begin{displaymath}ac-bd=0;a+c=0;b+d=0;b\cdot c+d\cdot a=1\end{displaymath}$

$\begin{displaymath}ac-bd=0;a=-c;b=-d;b\cdot c+d\cdot a=1\end{displaymath}$

$\begin{displaymath}-c^{2}+d^{2}=0;a=-c;b=-d;-d\cdot c+d\cdot (-c)=1\end{displaymath}$

$\begin{displaymath}c^{2}=d^{2};a=-c;b=-d;d\cdot c=-\frac{1}{2}\end{displaymath}$

$\begin{displaymath}c^{2}=d^{2};a=-c;b=-d;d^{2}\cdot c^{2}=\frac{1}{4}\end{displaymath}$

$\begin{displaymath}c^{2}=d^{2};a=-c;b=-d;d^{4}=\frac{1}{4}\end{displaymath}$

$\begin{displaymath}c^{2}=d^{2};a=-c;b=-d;d=\frac{1}{\sqrt{2}}\end{displaymath}$

$\begin{displaymath}c^{2}=d^{2};a=-c;b=-d;d=\frac{1}{\sqrt{2}};c=-\frac{1}{2\cdot d}=-\frac{1}{\sqrt{2}}\end{displaymath}$

$\begin{displaymath}a=\frac{1}{\sqrt{2}};b=-\frac{1}{\sqrt{2}};c=-\frac{1}{\sqrt{2}};d=\frac{1}{\sqrt{2}}\end{displaymath}$

$\begin{displaymath}(s+\frac{1}{\sqrt{2}}-j\cdot \frac{1}{\sqrt{2}})\cdot (s-\frac{1}{\sqrt{2}}+j\frac{1}{\sqrt{2}})=(s^{2}+j)\end{displaymath}$

$\begin{displaymath}s^{2}+(a+c)\cdot s+a\cdot c-bd+j(b\cdot s+b\cdot c+d\cdot s+d\cdot a)=(s^{2}-j)\end{displaymath}$

$\begin{displaymath}-c^{2}+d^{2}=0;a=-c;b=-d;-d\cdot c+d\cdot (-c)=-1\end{displaymath}$

$\begin{displaymath}c^{2}=d^{2};a=-c;b=-d;d\cdot c=\frac{1}{2}\end{displaymath}$

$\begin{displaymath}c^{2}=d^{2};a=-c;b=-d;d=\frac{1}{\sqrt{2}}\end{displaymath}$

$\begin{displaymath}c^{2}=d^{2};a=-c;b=-d;d=\frac{1}{\sqrt{2}};c=\frac{1}{2\cdot d}=\frac{1}{\sqrt{2}}\end{displaymath}$

$\begin{displaymath}a=-\frac{1}{\sqrt{2}};b=-\frac{1}{\sqrt{2}};c=\frac{1}{\sqrt{2}};d=\frac{1}{\sqrt{2}}\end{displaymath}$

$\begin{displaymath}(s-\frac{1}{\sqrt{2}}-j\frac{1}{\sqrt{2}})(s+\frac{1}{\sqrt{2}}+j\frac{1}{\sqrt{2}})=(s^{2}-j)\end{displaymath}$

$\begin{displaymath}s^{4}+1=(s+\frac{1}{\sqrt{2}}-j\frac{1}{\sqrt{2}})\cdot(s-\fr... ...ac{1}{\sqrt{2}})\cdot(s+\frac{1}{\sqrt{2}}+j\frac{1}{\sqrt{2}})\end{displaymath}$

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