V0.8.9.9

docs/Chapters/5-STATE-FEEDBACK.md

@@ -101,4 +101,89 @@ $$
 $$
 
 and so, you just need to substitute these matrices to the equation of the 
-$G_p(s)$ and you get you controller
+$G_p(s)$ and you get you controller
+
+## Inserting the reeference
+Until now we've worked without taking the reference into consideration, so 
+let's add it like we normally do:
+
+![reference input unitary feedback](../Images/State-Feedback/reference-input-1.png)
+
+Here we can compute the error and our $G(s)$ we have in our system:
+$$
+e = r(t) - y(t) \\
+G(s) = \frac{G_c(s)G_p(s)}{1 + G_c(s)G_p(s)}
+$$
+
+
+Another way to deal with the reference is to put it between our blocks:
+![reference input unitary feedback](../Images/State-Feedback/reference-input-2.png)
+
+
+$$
+e = r(t) - \mathcal{L}^{-1}\left\{Y(s)G_c(s)\right\} \\
+G(s) = \frac{G_p(s)}{1 + G_c(s)G_p(s)}
+$$
+
+Now, let's see the difference in both $G_1(s)$ and $G_2(s)$:
+$$
+\begin{align*}
+G_1(s) &= \frac{G_c(s)G_p(s)}{1 + G_c(s)G_p(s)}  
+
+
+\\
+
+
+&= \frac{
+    \frac{N_c}{D_c}\frac{N_p}{D_p}
+}{
+1 + \frac{N_c}{D_c} \frac{N_p}{D_p}
+} \\
+
+
+&= \frac{
+    \frac{N_c N_p}{D_c D_p}
+}{
+    \frac{D_cD_p +N_cN_p}{D_cD_p} 
+} \\
+
+
+&= \frac{
+    N_c N_p
+}{
+    D_cD_p +N_cN_p
+} \\
+
+\end{align*}
+\;\;\;\;
+\begin{align*}
+G_2(s) &= \frac{G_p(s)}{1 + G_c(s)G_p(s)}  
+
+
+\\
+
+
+&= \frac{
+    \frac{N_p}{D_p}
+}{
+1 + \frac{N_c}{D_c} \frac{N_p}{D_p}
+} \\
+
+
+&= \frac{
+    \frac{N_p}{D_p}
+}{
+    \frac{D_cD_p +N_cN_p}{D_cD_p} 
+} \\
+
+
+&= \frac{
+    N_c D_c
+}{
+    D_cD_p +N_cN_p
+} \\
+
+\end{align*}
+$$
+
+While both have the same `poles`, they differ in their `zeroes`