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+// Exa 2.7
+
+clc;
+clear;
+
+// Given data
+
+// Fig. 2.11(a) shows the basic differential amplifier
+Rc = 2*10^3; // Ω
+Re = 4.3*10^3; // Ω
+Vcc = 5 ; // Vcc = |VEE|
+Bo = 200;
+Vbe = 0.7; // Volts
+Vt=25*10^-3; // Volts
+
+// Solution
+
+printf(' For V1 = V2 = 0, applying KVL for the base emitter loop, we may write,');
+printf('\n  Vbe+2*(1+Bo)*Ibq*Re-Vee = 0.\n  From this we get Ibq as ');
+Ibq = (Vcc-Vbe)/(2*(1+Bo)*Re);
+printf(' %.2f μA. \n ',Ibq*10^6);
+Icq = Bo*Ibq;
+printf(' The value of Icq = %.3f mA. \n ',Icq*10^3);
+Vo1 = Vcc - Rc*Icq; 
+printf(' The value of Vo1 = Vo2(due to symmetry) = %.3f V. \n ',Vo1);
+Vceq = Vo1-(-Vbe);
+printf(' The value of Vceq = %.3f V. \n ',Vceq);
+gm = Icq/Vt;
+r_pi = Bo/gm;
+// using wq. 2.50 ADM = -gm*Rc;
+ADM = -gm*Rc;
+// using equation 2.53(a) Acm can be given as
+ACM = (-Bo*Rc)/(r_pi+2*(1+Bo)*Re);
+
+CMRR = ADM/ACM;
+CMRR_db = 10*log(CMRR);
+printf(' The remaining values are as follows: \n  ADM = %.2f. \n  ACM = %.2f. \n  CMRR = %.1f = %.1f dB.\n',ADM,ACM,CMRR,CMRR_db);