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//Example 4.2(a)
clear;
clc;
n=6;
fc=13*10^3;
//For a 1dB ripple Chebyshev low pass filter with n=6 requires 3 second order stages with :
//f01=0.995*fc, Q1=8
//f02=0.747*fc, Q2=2.20
//f03=0.353*fc, Q3=0.761
f03=0.995*fc;
Q1=0.761;
f02=0.747*fc;
Q2=2.20;
f01=0.353*fc;
Q3=8.00;
n1=(4*Q1^(2))+0.0016978;
C1=2.2*10^(-9);
C11=n1*C1;
C21=C1;
k1=(n1/(2*(Q1^(2))))-1;
m1=k1+(((k1^2)-1)^(0.5));
k11=(m1*n1)^(0.5);
R1=1/(k11*2*%pi*f01*C1);
R11=m1*R1;
R21=R1;
n2=(4*Q2^(2))+0.2478431;
C2=510*10^(-12);
C12=n2*C2;
C22=C2;
k2=(n2/(2*(Q2^(2))))-1;
m2=k2+(((k2^2)-1)^(0.5));
k12=(m2*n2)^(0.5);
R2=1/(k12*2*%pi*f02*C2);
R12=m2*R2;
R22=R2;
n3=(4*Q3^(2))+25.818182;
C3=220*10^(-12);
C13=n3*C3;
C23=C3;
k3=(n3/(2*(Q3^(2))))-1;
m3=k3+(((k3^2)-1)^(0.5));
k13=(m3*n3)^(0.5);
R3=1/(k13*2*%pi*f03*C3);
R13=m3*R3;
R23=R3;
printf("Designed Chebyshev Filter :");
printf("\nSection I :")
printf("\nR1=%.2f kohms",R11*10^(-3));
printf("\nR2=%.2f kohms",R21*10^(-3));
printf("\nC1=%.2f nF",C11*10^9);
printf("\nC2=%.2f nF",C21*10^9);
printf("\n\nSection II :")
printf("\nR1=%.2f kohms",R12*10^(-3));
printf("\nR2=%.2f kohms",R22*10^(-3));
printf("\nC1=%.2f nF",C12*10^9);
printf("\nC2=%.2f pF",C22*10^12);
printf("\n\nSection III :")
printf("\nR1=%.2f kohms",R13*10^(-3));
printf("\nR2=%.2f kohms",R23*10^(-3));
printf("\nC1=%.2f nF",C13*10^9);
printf("\nC2=%.2f pF",C23*10^12);
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