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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