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authorpriyanka2015-06-24 15:03:17 +0530
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+//Chapter 7, Problem 10
+clc
+funcprot(0)
+// A = p2z(R,Theta) - Convert from polar to rectangular form.
+// R is a matrix containing the magnitudes
+// Theta is a matrix containing the phase angles (in degrees).
+function [A] = p2z(R,Theta)
+ A = R*exp(%i*%pi*Theta/180);
+endfunction
+
+// [R1, Theta1] = z2p(A1) - Display polar form of complex matrix.
+function [R1, Theta1] = z2p(A1)
+ Theta1 = atan(imag(A1),real(A1))*180/%pi;
+ R1=sqrt(real(A1)^2+imag(A1)^2)
+endfunction
+
+//transistor S-parameter
+s11=p2z(0.35,165)
+s12=p2z(0.035,58)
+s21=p2z(5.9,66)
+s22=p2z(0.46,-31)
+refs=p2z(0.68,142) //source reflection coefficient
+f=300e6 //frequency in hertz
+vce=12 //base to emitter voltage
+ic=4e-3 //collector current in ampere
+[s11m,s11a]=z2p(s11)
+[s22m,s22a]=z2p(s22)
+[s21m,s21a]=z2p(s21)
+[s12m,s12a]=z2p(s12)
+
+Ds=(s11*s22)-(s12*s21)
+[Dmag,Dang]=z2p(Ds)
+K=(1+Dmag^2-s11m^2-s22m^2)/(2*s21m*s12m)
+
+y=1.65
+r=0.85
+y1=0.62
+r1=1.2
+R=50
+
+//The design values of the input matching network are shown in Figures 7.10. By plotting, we get Arc AB = shunt C = j1.65 S and Arc BC = series L = j0.85 ohm
+C1=y/(2*%pi*f*R)
+L1=(r*R)/(2*%pi*f)
+
+refl=conj(s11+((s12*s21*refs)/(1-(s22*refs))))
+
+//The design values of the input matching network are shown in Figures 7.12. By plotting, we get Arc AB = shunt L = –j0.72 S and Arc BC = series C = –j1.07 ohm
+
+C2=1/(2*%pi*f*r1*R)
+L2=R/(2*%pi*f*y1)
+
+printf("For input matching network,\n\n")
+printf("C1 = %.2f pF\n",C1*10^12)
+printf("L1 = %.2f nH\n\n",L1*10^9)
+printf("For output matching network,\n\n")
+printf("C2 = %.2f pF\n",C2*10^12)
+printf("L2 = %.2f nH\n\n",L2*10^9)
+printf("The completed design is shown in Figure 7.12")