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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+//Caption:calculate_Td,tr,Tp,Mp
+//example 11_10_b
+//page 478
+s=%s;
+syms Td
+G=16/(s^2+1.6*s)
+G1=1+s*Td
+H=1;
+a=G*G1
+CL=a/.H
+CL=simple(CL)
+disp(CL,"C(s)/R(s)=");
+zeta=0.8//given
+//since zeta=0.8 so 2*zeta*Wn=1.6+16*Td
+Wn=sqrt(16)
+//so on solving
+Td=0.3
+//so transfer function takes the form:
+CL1=(16+4.8*s)/(s^2+6.4*s+6)
+disp(CL1,"C(s)/R(s)=");
+Wn1=sqrt(16)
+//2*zeta1*Wn1=6.4
+zeta1=6.4/(2*Wn1)//damping ratio
+Wd=Wn1*sqrt(1-zeta1^2)//damped frequency
+Mp=(exp(-(zeta1*%pi)/sqrt(1-zeta1^2)))*100//max. overshoot
+tr=(%pi-atan(sqrt(1-zeta1^2)/zeta1))/(Wn1*sqrt(1-zeta1^2))//rise time
+Tp=%pi/Wd//peak time
+disp(Tp,"peak time=");
+disp(Mp,"max. peak overshoot=")
+disp(tr,"rise time=")