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+//Example 2.8 page 33
+//Determine the cutoff frequency and the order for the anti-image filter
+//given a DSP system with a sampling rate of 16,000 Hz and specifications
+//for the anti-image filter as shown in Figure 2.24.
+//
+//Design requirements:
+//& Maximum allowable gain variation from 0 to 3,000 Hz ¼ 2dB
+//& 33 dB rejection at the frequency of 13,000 Hz
+//& Butterworth filter assumed for the anti-image filter
+clc,clear,close;
+
+fs = 16000;
+T = 1/fs;
+f1 = 3000,f2 = 13000;//Hz
+x1 = f1*T;
+gain1 = sin(x1*%pi)/(x1*%pi);
+gaindb1 = 10*log(gain1);
+disp("For f = 3000Hz, gain = "+string(gain1)+" = "+string(gaindb1));
+
+x2 = f2*T;
+gain2 = sin(x2*%pi)/(x2*%pi);
+gaindb2 = 10*log(gain2);
+disp("For f = 13000Hz, gain = "+string(gain2)+" = "+string(gaindb2));
+
+n = .5*log((10^2-1)/(10^0.154 - 1))/log(f2/f1)
+disp("n = "+string(n));
+n = round(n);
+
+fc2 = f2/(10^2-1)^(1/(2*n));
+fc1 = f1/(10^0.154-1)^(1/2/n);
+
+disp("fc1 = "+string(fc1)+" fc2 = "+string(fc2));
+disp("we choose the smaller one, that is " + string(min(fc1,fc2)));
+\ No newline at end of file