Signal Processing and Linear Systems-B.P.Lathi copy

# 731 reduces t o 1 h s h s 1 sj2n t he poles

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Unformatted text preview: th Filters 511 Also from Eq. (7.38a) Although Tables 7.1 a nd 7.2 are for normalized Butterworth filters with 3 dB bandwidth W e = 1, t he results can be extended t o any value of W e by simply replacing 8 by 8 /W e . T his s tep implies replacing W by W /w e in Eq. (7.32). For example, the second-order B utterworth filter for W e = 100 c an be obtained from Table 7.1 by replacing 8 by 8 /100. T his step yields Wp We = (7.40) • -[-lO---C-.p-/-IO-"---I-]1/"'2-n C :- Alternatively, from Eq. (7.38b) (7.41) 1 H (8) = - - - - " . 2 - - - - (lgO) + V 2 (lgO) + 1 82 + 100V28 + 10 (7.37) 4 T he a mplitude response IH(jw)1 of the filter in Eq. (7.37) is identical to t hat o f normalized 11t(jw)1 in Eq. (7.32), expanded by a factor 100 along the horizontal (w) axis (frequency scaling). Determination of n, the Filter Order ax If is t he gain of a lowpass Butterworth filter in dB units a t according to E q. (7.31) ax = 20log W = wx, then • E xample 7 .6 Design a B utterworth lowpass filter t o m eet t he specifications (Fig. 7.23): (i) P assband g ain t o lie between 1 a nd G p = 0.794(Op = - 2dB) forO ::; (ii) S topband g ain n ot t o exceed G s = 0.1 ( as = - 20dB) forw 2: 20. S tep 1 : D etermine n Here W p = 10, W s = 20, Op Eq. (7.39) yields = - 2 d B, a nd a s = - 20 dB. S ubstituting these values in n = 3.701 Since n can only be an integer, we choose + ( ::) 2n] IH(jwx)1 = - 101og [1 Op = - 101og [1 + (::r as at ws)in this We = 10.693 = 11.261 Because we selected n = 4 r ather t han 3.701, we o btain two different values of W e. Choice o f We = 10.693 will satisfy exactly t he r equirement G p = 0.794 over t he p assband (0, 10), a nd will surpass t he r equirement G s = 0.1 in t he s topband W 2: 20. O n t he o ther h and, choice of W e = 11.261 will exactly satisfy t he r equirement on G s b ut will oversatisfy t he r equirement for G p • L et us choose t he former case ( We = 10.693). ( ::) 2n = 1O- 6p / 1O 1 (7.38a) ( ::) 2n = 1 0- 0 ,/10 _ 1 (7.38b) Dividing (7.38b) by (7.38a), we o btain WS)2n = [lO-~./lO ( wp lO-Cp/lo - We n or _ S tep 2 : D etermine W e S ubstitution o f n = 4, wp = 10 in Eq. (7.40) yields Alternately, s ubstitution o f n = 4 in Eq. (7.41) yields as = -10 log [1 + ( ::) 2n] 1] S tep 3 : D etermine t he n ormalized t ransfer f unction 'It(s) T he normalized fourth-order transfer function 'It(8) is found from Table 7.1 as 'It(s) _ 1 - 84 + 2.61318 3 + 3.41428 2 + 2.61318 + 1 S tep 4: D etermine t he f inal f ilter t ransfer f unction H(8) T he desired transfer function w ith We = 10.693 is o btained b y r eplacing 8 w ith s /IO.693 in t he n ormalized transfer function 'It(8) a s H(8) = 1 1 CO.~93) 4 3 2 + 2.6131 CO~93) + 3.4142 CO.~92) + 2.6131 (lO~93) + 1 13073.7 a nd log < 10. n =4 10 S ubstitution o f t he specifications in Fig. 7.19a (gains O p a t wp a nd equation yields n= W [(10- 0 ./ 10 - S4 1) / (1O- 0p /lO - 1)] 2Iog(w s /w p ) (7.39) + 27.9428 3 + 390.48 2 + 3194.888 + 13073.7 13073.7 (8 2 + 8.18...
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## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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