// Exa 7.1

clc;
clear;

// Given data

n=2;  // Second order Butterworth filter
fL=1000; // Higher cut off frequency(Hz)

// Solution

printf('Let C = 0.1 μF. \n');
C=0.1*10^-6; // Farads

//    Since fL = 1/(2 * %pi * R*C);
//    Therefore;
R = 1/(2*%pi*fL*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 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\n\n Frequency, f in Hz          Gain magnitude in dB 20 log(vo/vi)\n');
// Frequency Response
x=[0.1*fL,0.2*fL,0.5*fL,1*fL,5*fL,10*fL]
for i = 1:1:6 
      response(i) = 20*log10(A0/(sqrt(1+(fL/x(i))^4)));
      printf(' %d                                %.2f      \n',x(i),response(i));
end