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// Exa 7.4
clc;
clear;
// Given data
n=2; // Second order Butterworth filter
fH=1000; // Lower cut off frequency(Hz)
// Solution
printf('Let C = 0.1 μF. \n');
C=0.1*10^-6; // Farads
// Since fH = 1/(2 * %pi * R*C);
// Therefore;
R = 1/(2*%pi*fH*C);
printf(' The calculated value of R = %.1f kΩ. \n',R/1000);
printf(' From Table 7.1, for n=2, the damping factor alpha = 1.414.');
alpha=1.414;
A0 = 3-alpha;
printf('\n Then the pass band gain A0 = %.3f. \n',A0);
printf('\n');
printf(' The transfer function of the normalized second order low-pass Butterworth filter is 1.586 ');
printf('\n ----------------');
printf('\n Sn^2+1.414*Sn+1');
// Since Af= 1 + Rf/Ri = 1 + 0.586;
printf('\n Since A0= 1.586 so Let Rf = 5.86 kΩ and Ri = 10 kΩ to make A0 = 1.586.' );
printf(' \n The circiuit realized is as shown in Fig. 7.4 with component value as mentioned above.');
printf('\n By considering minimum DC offset condition, the modified value of R and C comes out to be R = 1.85 kΩ and C=0.086 μF.');
printf('\n\n\n Frequency, f in Hz Gain magnitude in dB 20 log(vo/vi)\n');
// Frequency Response
x=[0.1*fH,0.2*fH,0.5*fH,1*fH,5*fH,10*fH]
for i = 1:1:6
response(i) = 20*log10(A0/(sqrt(1+(x(i)/fH)^4)));
printf(' %d %.2f \n',x(i),response(i));
end
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