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+// 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
+