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+clc
+//Chapter 2:Small Signal Amplifiers
+//example 2.8 page no 42
+//given
+disp('Assuming Vi (input voltage)=1')
+V1=(5+10^6)/(5+2*10^6)//voltage on the positive terminal
+V2=10^6/(5+2*10^6)//the voltage on the inverting terminal
+ed=V1-V2//differential voltage
+ec=(V1+V2)/2//common-mode voltage
+Ad=2*10^3//differentail gain
+Ac=2*10^-3//common mode gain (here 20% of differentail gain)
+Vo=Ad*ed+Ac*ec//actual amplifier output
+mprintf('the voltage gain is %3.2e Volts',Vo)