Question

Using Butterworth responses, design active filters to meet the following specifications:

Answer 1

2. Given that dp = 3dB at fp=2kHz. 2 2.2008 at fs=1k the e p = pau bond attenuation ds = stop band at tenutation to - pass band frequency- ts - Stop band frequency cap alfp = 211 (2000) = 12566.37 radlec as = al fs = 27 (1000) = 6283. 185 rad/sec Now order of. Ilie butterworth filter by Nz log looks 10014_1 Tog, (1881890) 1 10011 (3) - - - 0.9988 log (up/rs) = 0.301

2. Given that dp = 3dB at fp=2kHz. 2 2.2008 at fs=1k the e p = pau bond attenuation ds = stop band at tenutation to - pass band frequency- ts - Stop band frequency cap alfp = 211 (2000) = 12566.37 radlec as = al fs = 27 (1000) = 6283. 185 rad/sec Now order of. Ilie butterworth filter by Nz log looks 10014_1 Tog, (1881890) 1 10011 (3) - - - 0.9988 log (up/rs) = 0.301

N 0.9988 -3.315 [ N = 9 : Tu ade a Bulte voorth, filter is 'q! List at Butter worth Polynomials INI Polynial of 14 (6) sti 5 +825+1 (S+ 1) (² + 5+1) 4. (5 +0.765375 + 1) (5 + 1.84755 +1) a see zeven for N=4 HIS) = (8 +0.76595+ (5 +1.84785+1) for the filter te How cut off frequency govern by i rep (1 68 kp-1)/N

и лh CORP-1) 12566-37 (100 1x3 1) 18 de = 12573.83 radise Now substitute - - in equation 2 re HAS) = (C) +076596.) 17 6 5 [5+0.76545-620)+#(5+1 <45pesta 158. 1x106 HA.PS) =- (6 +962957155.484] 8+23365+156.1518)