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//Example 3.19
clear;
clc;
C=1*10^(-9);//Assuming C=1 nF
C1=C;
C2=C;
f0=8*10^3;
BW=200;
R=1/(2*%pi*f0*C);
R4=R;
R5=R;
Q=f0/BW;
R2=Q*R;
HobpdB=20;
Hobp=10^(HobpdB/20);
R1=(R2/Hobp)- 877.47155;
R3=R2;
Holp=R/R1;
HolpdB=20*log10(Holp);
printf("Designed Biquad Filter :");
printf("\nR1=%.2f 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("\nC1=C2=%.2f nF",C*10^9);
printf("\n\nResonance Gain (Holp)=%.2f dB",HolpdB);
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