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+clear;
+clc;
+//Example 4.5
+b=100;
+Vbe=0.7;
+Rc=2;
+Rs=0.5;
+Icq=2.16;
+V_T=0.026;//(V)
+Vceq=4.8
+//ac solution
+r_pi=b*V_T/Icq;
+printf('\nsmall signal resistance=%.2f KOhm\n',r_pi)
+g_m=Icq/V_T;
+printf('\ntransconductance=%3f mA/V\n',g_m)
+//since Va=infinity,ro=Va/Icq is also infinity
+Re=0.4;
+Rib=r_pi+(1+b)*Re;
+printf('\ninput resistance to the base=%.2f KOhm\n',Rib)
+//Ri=R1||R2||Rib
+Ri=10*Rib/(10+Rib);
+printf('\ninput resistance to the amplifier=%.3f KOhm\n',Ri)
+Av=-(1/(r_pi+(1+b)*Re))*b*Rc*Ri/(Ri+Rs);
+printf('\nsmall signal voltage gain=%.2f \n',Av)
+//by approximate expression
+Av=-Rc/Re;
+printf('\nsmall signal voltage gain=%.3f \n',Av)