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//Example 2.10(a)
clear;
clc;
Amin=1;
Amax=10^3;
AI=0.5;
R1=100*10^3;//Tolerance (1%)
R2=AI*R1;//Tolerance (1%)
AImin=Amin/AI;
AImax=Amax/AI;
//AImin<=AI<=AImax
//AImin=1+((2*R3)/(R4+R1)) -> 1+((2*R3)/(R4+R1))-Amin=0 -> (1-AImin)*R4+2*R3+(1-AImin)*R1=0...(i) and AImax=1+((2*R3)/(R4+0)) ->(1-AImax)*R4+2*R3=0....(ii)
//Solving these two equations will give R3 and R4
A=[2 (1-AImin);2 (1-AImax)];
B=[(1-AImin)*R1;0];
R=linsolve(A,B);
R3=R(1,1);//Tolerance (1%)
R4=R(2,1);//Tolerance (1%)
printf("Designed Instrumentation Amplifier :");
printf("\nR1=%.2f kohms",R1*10^(-3));
printf("\nR2=%.2f kohms",R2*10^(-3));
printf("\nR3=%.f kohms",R3*10^(-3));
printf("\nR4=%.f ohms",R4);
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