Question
1. Design a low-pass Chebyshev filter with the following specifications: (7pts) • Passband edge frequency of,...
1. Design a low-pass Chebyshev filter with the following specifications: (7pts) • Passband edge frequency of, Wp = 2 rads' Passband ripple of 3dB Cut-off frequency is at mid-point of the transition band • Stopband attenuation of 20dB or greater beyond ws=2.5 rads! • Find the filter transfer function H(S)
Answers
Design of low pass clelyshev filter o ② Determine the order of files (N) Determine the normalized chebyster low pass filles Transfer function low pass the From analog cloi pass to stea transformation, find tranfes function, Hels) liven pass band dipple kp= 3dB. Normal verle is up s 10 kp/20 - 10-3/20 Kp = 0.707 stop band attentation. Ks- 2008 Ks=10-20/20 Ks = d.lo pass band edge Shot on réaline and edge Requency, wp=2 healsec frequency, ws=2.5sadlec By kumarrorder of filter s = (1/6.3093 -) 12 id I lilu E = 1 N - ca llup -1) از مردم ا-اع - Cash (101) 2 - 1/2 Il (1070) 2 - 1 Cosh (25 --..* 14.-*--)) (1414 tists - (1.404 + s] to = '{ [1.192 - 0,838) -0123 Cosh! Cosh -119.9) Cash (1.25) Cash (1.25) Co-Yo = a177, K-2, be=2x y sin EXN 2.98 -= 4.2999 35 shos 0.69 nito ng when po is odd BO T - stco K S2+bustck = (د) - k=1; b = 240.177 x 51 2x5 50. 10g k=2 bz = 2x0.177 x sin fe = 0.286 Z na5 - - - = 2 &- (1/160) 17²parametes where Bu wis odd value of are evaluated wsing = 2k-1) 20 4 + cos² BOB, B2 7 HSN) = Kal; c= (6.177) 2 + cos 2x5 0177x0,93 Xo37 17 Cos 27 - DO 3 + I → Bo B, B2= 0.06 e = 0.93 2 Bo=B, =B2 » Bo3 = 0.06 Bo = 0,39 Cos2o = 2 costo- 0. 39 It couro 0.39 039;s H(SN) = = cos20 2 (S+ 0.177) (52+0.1095 +0.93) 0.22 k=2 = (0.177) + cos2 ( + 1 0.39 2x5 (52 + 0.2865 +0.37) = 0.03+ / It cos (6/11 0.06 153 +0.1952 +0.01s +0.16) (52+0.2865 +0.37) 4 = 0.37 0.06 Во Hsw =1 S+0,177 st +0.1095+0.93 § +oolast tools3 +0,1652 +0.28654 B2 + 0.00553 +0.00252 +0045 +0.3753 + 0.000452 +0.00045 toros) s2 +0.2865 +0. 372.045 +0.05 5 +0.354 +0.3853 to. 1682 +0,04ste 2 His en wolmalized Transfles Puxtion and op- escal sec Social Hals) = HS is 0.06 (55 +0,0354 +0.3833 +0.1652 +0.045 (pot cpoliott + 0.05) Asts) 0.06 Hss so = - ( 5 + 0.03 ( + +0,38/53 (feat assots) (slotal +0.16 (3) 2 + 0.04 (2) Doc +0.05 0.06 x 25 Hs(s) = LO so +0,0654 + 1.5253 + 1.2852 +0.645 + 1.6
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