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+//Chapter 7, Problem 5
+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(0.28,-58)
+s12=p2z(0.08,92)
+s21=p2z(2.1,65)
+s22=p2z(0.8,-30)
+f=1e9                               //frequency in hertz
+vce=15                              //collector to emitter voltage
+ic=5e-3                             //collector current in ampere
+Zs=35-%i*60                         //source impedance in ohm
+Zl=50-%i*50                         //load impedance in ohm
+K=1.168                             //Rollett stability factor
+g=7.94                              //desired gain
+R=50                                //resistance in ohm
+
+[s11m,s11a]=z2p(s11)
+[s22m,s22a]=z2p(s22)
+[s21m,s21a]=z2p(s21)
+[s12m,s12a]=z2p(s12)
+
+Ds=s11*s22-s12*s21
+[Dsm,Dsa]=z2p(Ds)
+D2=s22m^2-Dsm^2
+C2=s22-(Ds*conj(s11))
+G=g/s21m^2
+ro=(G*conj(C2))/(1+(D2*G))
+po=sqrt(1-(2*K*s12m*s21m*G)+(s12m*s21m)^2*G^2)/(1+(D2*G))
+
+//The Smith chart construction is shown in Figure 7.5. The transistor’s output network must transform the actual load impedance into a value that falls on the constant gain 9 dB circle. By plotting, we get Arc AB = series C = –j2.0 ohm and Arc BC = shunt L = –j0.41 S
+r=2
+y=0.4
+C1=1/(2*%pi*f*r*R)
+L1=R/(2*%pi*f*y)
+
+//For a conjugate match at the input to the transistor, the desired source reflection coefficient must be calculated as follows
+refl=p2z(0.82,13)                                           //point C in figure 7.5
+refs=conj(s11+((s12*s21*refl)/(1-(s22*refl))))
+[refsm,refsa]=z2p(refs)
+
+//The point is plotted as point D in Figure 7.6. The actual normalised source impedance is plotted at point A (0.7 – j1.2) ohm. The input network must transform the actual impedance at point A to the desired impedance at point D. we get Arc AB = shunt C2 = j0.63 S, Arc BC = series L2 = j1.08 ohm, Arc CD = shunt C3 = j2.15 S
+
+y1=0.63
+r1=1.08
+y2=2.15
+
+C2=y1/(2*%pi*f*R)
+L2=r1*R/(2*%pi*f)
+C3=y2/(2*%pi*f*R)
+
+printf("For output matching network,\n\n")
+printf("C1 = %.2f pF\n",C1*10^12)
+printf("L1 = %.1f nH\n\n",L1*10^9)
+printf("For input matching network,\n\n")
+printf("C2 = %.1f pF\n",C2*10^12)
+printf("L2 = %.1f nH\n\n",L2*10^9)
+printf("C3 = %.1f pF\n",C3*10^12)
+printf("The completed design (minus biasing network) is shown in Figure 7.7")