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+// Exa 5.7
+clc;
+clear;
+close;
+// Given data
+R1= 50;// in kohm
+// Let us choose
+R3= 15;// in k ohm
+R4= R3;
+// Ad= 1+2*R2/R1             (i)
+// Ad= ((1+2*R2/R1)*(V2-V1))/(V2-V1)= 1+2*R2/R1
+// For minimum differential voltage gain
+Ad_min=5;
+Ad= Ad_min;
+R1_max= R1;// since Ad will be minimum only when R1 will be maximum
+// Putting values of Ad and R1 in eq(i)
+R2= (Ad-1)*R1/2;// in k ohm
+// For maximum differential voltage gain
+Ad_max=200;
+Ad= Ad_min;
+// Putting values of Ad and R2 in eq(i)
+R1= 2*R2/(Ad-1);// in k ohm
+R1=floor(R1);
+// For maximum value of Ad, R1 will have minimum value , therefore
+R1_min= 1;// in kohm
+disp("Value of R1_min is : "+string(R1_min)+" k ohm");
+disp("Value of R1 is     : "+string(R1)+"-50 k ohm");
+disp("Value of R2 is     : "+string(R2)+" k ohm");
+disp("Value of R3 is     : "+string(R3)+" k ohm");
+disp("Value of R4 is     : "+string(R4)+" k ohm");
+