For the closed$-$loop system shown below, with a step input, determine the following:

a$)$ Settle time, Ts

b$)$ Percent overshoot, OS$\%$

c$)$ Step steady state error

d$)$ Sketch the time step response and indicate steady state error, settle time, and overshoot.

$T=\frac{20}{S^2+2S+36}$

${}_n=\sqrt[\phantom{2}]{36}=6$

$S_1=2J{}_n=2$

$,J=\frac{1}{6}=\mathrm{.}1667$

a$)T_s=\frac{4}{j^{{}_n}}=4$

b$)9,05=e^{-\frac{j}{\sqrt[\phantom{2}]{1-J^2}}}100=58\mathrm{.}8\%$

c$)$

$k_p=\underset{s0}{lim}\frac{20}{s^2+2s+10}=\frac{20}{16}=\frac{5}{4}=>e_{\text{s}\text{t}\text{e}\text{p}}=\frac{1}{1+k_p}=\frac{1}{1+5}4$

$=\frac{4}{4+5}=\frac{4}{9}=\mathrm{.}44$

$d$

EXPLAIN IT AND HAVING TROUBLE UNDERSTADING PART C on why it is $20/16$ instead of $20/36$