Question Feedback system finding the transfer function with feedback the input is: . Find and graph . retroalimentación T(s), G(s)=5+2 S+1 9. Del siguiente sistema con retroalimentación encontrar la función de transferencia con H(s)= la entrada es y(t)=1. Encontrar y s+2 graficar x(t) Y(s) Error X(s X(s) G(s) H(s) Feedback

Page-1AnsThe above block diagram seduces to:\begin{array}{l}y(t)=1 \Rightarrow y(s)=\frac{1}{s} \\\therefore x(S)=\frac{G_{\alpha}(S) \cdot Y(s)}{1+G_{x}(S) H(s)} \\\therefore x(s)=\frac{\frac{1}{s+2} \cdot \frac{1}{s}}{1+\left(\frac{1}{s+2}\right)\left(\frac{1}{s+1}\right)} \\\therefore x(s)=\frac{\frac{1}{s(s+1)}}{\frac{(s+2)(s+1)+1}{(s+2)(s+1)}}=\frac{1}{s(s+1)} \times \frac{(s+2)(s+1)}{(s+2)(s+1)+1} \\\end{array}Page-2\begin{array}{l}\therefore x(s)=\frac{s+2}{\left(s^{2}+3 s+2+1\right) s} \\\therefore x(s)=\frac{s+2}{\left(s^{2}+3 s+3\right) s}\end{array}Use partial fraction\begin{array}{l}\frac{S+2}{S\left(S^{2}+3 S+3\right)}=\frac{A}{S}+\frac{B S+C}{S^{2}+3 S+3} \\\therefore S+2=A\left(S^{2}+3 S+3\right)+B S^{2}+C S \\\therefore S+2=A S^{2}+3 A S+3 A+B S^{2}+C S\end{array}compare coefficients of S^{2}, S & contants.\begin{array}{l}=L^{-1}\left\{\frac{2 / 3}{s}\right\}-L^{-1}\left\{\frac{\frac{2 s}{3}+1}{s^{2}+3 s+3}\right\} \\\end{array}Page-3\begin{aligned}& =L^{-1}\left\{\frac{2 / 3}{s}\right\}-L^{-1}\left\{\frac{\frac{2 s+3}{3}}{s^{2}+3 s+3}\right\} \\& =\frac{2}{3}(1)-\frac{1}{3} L^{-1}\left\{\frac{2 s+3}{s^{2}+3 s+\frac{9}{4}+\frac{3}{4}}\right\} \\& =\frac{2}{3}-\frac{1}{3} L^{-1}\left\{\frac{2 s+3}{(s+3 / 2)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\right\} \\& =\frac{2}{3}-\frac{2}{3} L^{-1}\left\{\frac{s+3 / 2}{(s+3 / 2)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\right\} \\& =\frac{2}{3}-\frac{2}{3} e^{-3 t / 2} \cdot \cos \frac{\sqrt{3} t}{2} \\x(t) & \left.=\frac{2}{3}\left[1-e^{-3 t / 2} \cos \frac{\sqrt{3} t}{2}\right]\right]\end{aligned}The Graph of the function x(t) is shown in the figure below. It becomes constant after attaining the value of x(t)= 0.6667 ...