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//Example 3.20
clear;
clc;
f0=1*10^3;
fz=2*10^3;
Q=10;
C=10*10^(-9);//Assume C=10 nF
R=(1/(2*%pi*f0*C))-120;
w0=2*%pi*f0;
wz=2*%pi*fz;
R1=Q*R;
R2=100*10^3;//Assumption
R3=R2;
R4num=R2*(w0^2);
R4den=Q*abs((w0^2)-(wz^2));
R4=R4num/R4den;
R5=R2*((w0/wz)^2);//as fz>f0
Hohp=R5/R2;
HohpdB=20*log10(Hohp);
printf("\nDesigned Biquad Filter for a low pass notch response :");
printf("\nR=%.2f kohms",R*10^(-3));
printf("\nR1=%.f kohms",R1*10^(-3));
printf("\nR2=%.2f kohms",R2*10^(-3));
printf("\nR3=%.2f kohms",R3*10^(-3));
printf("\nR4=%.2f kohms",R4*10^(-3));
printf("\nR5=%.2f kohms",R5*10^(-3));
printf("\nC=%.2f nF",C*10^9);
printf("\n\nHigh Frequency Gain (Hohp)=%.2f dB",HohpdB);
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