Question

Problem 1 (25 Pts) Design a low pass multistage Butterworth filter that simultaneously meets the following design requirements: 1. Minimum attenuation of 24 dB at 1000 Hz and 2. Minimum attenuation of 48 dB at frequency of 2000 Hz or higher. Consider equal source and load impedances at 50 2. Part a) 15 pts Solve for both the order of the Butterworth filter and the cut-off frequency required to meet the above design criteria Part b) 10 pts Find the values of the capacitors and inductors required to construct the filter and draw the circuit diagram of the filter with all the required components.

Answer 1

natin (P) Given 24 dB @ w = 1000 H 2 44848@ w7;2000 H2 24 = 20 =502 Alto order of the Bultenoooth filled n=10910 (10 401) We cutof frequency 2.809,6 ( Wilwe) we = is the frequency at atenuation We wc Since W1 - 2000 H2. af A=48.00 cufqf frequency uc=1000H2 to the orderq filter n = fog,0 (1040|10,) =797 4/8 Ans (2000 So for finding the value of inductor of capacitor value find parauneler'g' 96 = zsin 4 90 - intl - 1 2,109,10 CLS 29 96 = Dein ( my ) 1.6529 - 2sina 90=984 - g g = 1 "(T) = 0.3902 92 = 2.sin ( 311|16) = 1. In SIT (16) = 106629 9 7 = 28ìn / 13 IT - billil 98 = zsin (13, 2.8in (ISO) = 0,3902 93 2in 84 2.sin (FrT116) = 1.5615 2010 (977716) 1.9615 of 92 SO Value Lk = 2ook f cx = Ik wc 20. Wc 니 2051 co LOC 118 = 2013 2owc Cu = 94 L5=20.gs 106 = 96 2o. Wc (f = 20,97 Co- 90 WC 20. We wc 20. We uc obole 11-150(0.3902) (P2 19.51 mH & C2 = lill 1000 18-50 (1.6629) 50(1000) 831145mH 1000 = 22.2 eir 464 = 1.9615 15- 50(1.9615) 39.23 elf 50x1000 1000 98. 075 ml ¢ C6 = 1,6629 33.26-41f 50x1000 L 7 =50x1.IIL 55.56MH 50X1000 = 7.8041F :: Resulting Circuit diagram 18 ILS LF 502 Vs 21 cul co f 08 Or 3902 mit 50.2 Scanned with CamScanner

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