solvedB35 - S 35.1 (P 4.10) _ (i) bar(0:36, b1),...

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Sheet1 Page 1 S 35.1 (P 4.10) ______________ (i) bar(0:36, b1), title('Filter Impulse Response'), xlabel('Time n') The impulse response is symmetric about n=M/2=20. (ii) f = (0:999)/1000 f H = fft(b1,1000) Ha = abs(H) plot(f,Ha), title('Amplitude Response'), xlabel('\Omega/2\pi'), grid Highpass filter. (iii) For a closer inspection of the passband: plot(Ha(411:500)), grid Then A1 = max(Ha) A2 = min(Ha(441:500)) resulting in A1 = 1.0880 A2 = 0.9123 and A1/A2 = 1.193 (iv) For a closer inspection of the stopband: plot(Ha(1:380)), grid Then A3 = max(Ha(1:375)) resulting in A3 = 0.00148 and A2/A3 = 617.5 (iv) The original passband begins roughly at fp = (min(find(Ha>=A2))-1)/1000 = 0.426 (cycles/sample) and ends at 0.5. The conjugate symmetric band starts at 0.5 and ends at 1-fp = 0.574 (see plot of part (i)). The two bands together form a single composite band centered at 0.5. To create a passband centered at fc=0.25 together with its symmetric (centered at fc=0.75), multiply the impulse response of the filter by
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This note was uploaded on 03/21/2010 for the course ENEE 241 taught by Professor Staff during the Spring '08 term at Maryland.

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solvedB35 - S 35.1 (P 4.10) _ (i) bar(0:36, b1),...

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