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+// Exa 3.32
+format('v',4);
+clc;
+clear;
+close;
+// Given data
+bita= 200;
+alpha= bita/(1+bita);
+R_C= 100;// in Ω
+R_B= 10;// in kΩ
+Rsig= 1;// in kΩ
+Rsig= Rsig*10^3;// in Ω
+R_B= R_B*10^3;// in Ω
+V_T= 25*10^-3;
+V=1.5;// in V
+I_E= 10;// in mA
+I_E= I_E*10^-3;// in A
+I_C= alpha*I_E;// in A
+V_C= I_C*R_C;// in V
+I_B= I_C/bita;// in A
+V_B= V-(R_B*I_B)
+gm= I_C/V_T;// in A/V
+rpi= bita/gm;// in Ω
+Rib= rpi;// in Ω
+disp(Rib,"The value of Rib in Ω is : ")
+Rin= R_B*rpi/(R_B+rpi);// in Ω
+disp(Rin,"The value of Rin in Ω is : ")
+// vbe= v_sig*Rin/(Rsig+Rin);
+vbe_by_vsig= Rin/(Rsig+Rin);
+// vo= -gm*vbe*R_C and = -gm*v_sig*Rin/(Rsig+Rin)
+vo_by_vsig= -gm*R_C*vbe_by_vsig;// in V/V
+disp(vo_by_vsig,"Overall voltage gain in V/V is : ")
+// if 
+vo= 0.4;//(±) in V
+vs= vo/abs(vo_by_vsig);// in V
+vbe= vbe_by_vsig*vs;// in V
+disp(vs*10^3,"The value of v_sig in mV is : ")
+disp(vbe*10^3,"The value of v_be in mV is : ")
+
+// Note: There is some difference between in this coding and book solution. But Coding is correct.
+
+
+
+