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+// Exa 2.8
+
+clc;
+clear;
+
+// Given data
+
+// With reference to differential amplifier designed in example 2.6
+// 2 applied inputs are
+t = [0 :1:100]; // time in mSec
+ v1=  15*sin(2*%pi*60*t) + 5*sin(2*%pi*1000*t); // in mV
+ v2 = 15*sin(2*%pi*60*t) - 5*sin(2*%pi*1000*t); // in mV
+fi = 60; // frequency of interference signal(Hz)
+fo = 1000; // frequency at which signal is to be processed(Hz)
+
+// Solution
+
+// We know from Example 2.6
+gm=4; // mʊ
+Rc=125 ; // kΩ
+Re= 1.25; // kΩ
+Bo=200;
+r_pi= Bo/gm;  // in kΩ
+
+ADM=-500; // from example 2.6(given)
+// From eq. 2.53(a) we get ACM as
+ACM = (-Bo*Rc)/(r_pi*1000+2*(1+Bo)*Re);
+printf(' The value of ACM = %.2f \n',ACM);
+// from eqns 2.56(a and b)
+vDM = (v1-v2)/2;
+vCM = (v1+v2)/2;
+
+//from Eq. 2.57(a and b)
+vo1 = ADM*vDM+ACM*vCM;
+vo2 = -ADM*vDM + ACM*vCM;
+
+printf('  Therefore final equations are- \n');
+disp("vo1 = -2500*sin(2*%pi*1000*t)-0.75*sin(2*%pi*60*t) mV ");
+disp("vo2 = 2500*sin(2*%pi*1000*t)-0.75*sin(2*%pi*60*t) mV");