Question K (s+6) The unity feedback system shown in Problem 1 with G(s) = is operating with a (s+2)(s+3)(s+5) dominant-pole damping ratio of 0.707. The dominant poles are located at -2.32 t j2.32 b. Design a PD controller so that the settling time is reduced by a factor of 2. Compare the transient and steady-state performance of the uncompensated and compensated K (s+6) The unity feedback system shown in Problem 1 with G(s) = is operating with a (s+2)(s+3)(s+5) dominant-pole damping ratio of 0.707. The dominant poles are located at -2.32 t j2.32 b. Design a PD controller so that the settling time is reduced by a factor of 2. Compare the transient and steady-state performance of the uncompensated and compensated systems Specifically, compare the settling time, peak time, percent overshoot and on and identify whether there is steady state error in the compensated system. Describe any problems with your design
Sol: The dominant poles ane epsi_(c)=-{omega_(n)+-omega_(d)J:}.Settling Time t_(s)=(4)/(2omega_(n))=(4)/(2.32)=1.7241.The specilictions ane tn t_(s)^(')=(t_(s))/(2)=0.8621quad(4)/(epsiomega_(n))=0.8621=>xiomega_(n)=4.64and fo Sigma=0.707quadomega_(n)=6.5629:. The new dominant poles one S_(d)=-{omega_(n)+-omega_(n)sqrt(1-epsi2):}G(s)=(k(s+6))/((s+2)(s+3)(s+5)). The Ansle dpliceny of ' S_(d) ' to thoopen loop poles of G(s) is:{:[theta_(dy)=180-theta_(1)-theta_(2)-theta_(3)+theta_(4)],[theta_(1)=180-tan^(-1)((4.64)/(2.64))],[theta_(2)=180-tan^(-1)((4.64)/(1.64))],[theta_(3)=Tan^(-1)((4.64)/(0.36))" and "theta_(4)=Tan^(-1)((4.36)/(1.36))],[:.quadtheta_(dy)=-60.9327^(@)],[G_(pp)(s)=k(s+a)]:}The zero ' a ' of the Compensator must be pleed attheta_(dy)0 fim S_(d) i-e{:[a=(4.64)/(Tan(60.9327))+4.64],[=7.219 ... See the full answer