$ω_r=\frac{π}{2}$

$H(s)= \frac{(s+0.1)}{((s+0.1)^2+16)}$ $H(s)= \frac{(s+0.1)}{((s+0.1)^2+(4)^2 )}$ From above equation we can say that, Ω=4. And The value of $ω_r$ is given as $ω_r=\frac{π}{2}$ Now we know that $Ω=\frac{2}{T_s} tan⁡(\frac{ω}{2})$ $∴4=\frac{2}{T_s} tan⁡(\frac{π}{4})$ $∴T_s=\frac{2}{4} tan⁡(\frac{π}{4})$ $∴T_s=0.5 sec$ Using bilinear transformation H(z) can be obtained by putting , $s=\frac{2}{T_s} \frac{((z-1)}{(z+1))}$in the equation of H(s) …