This preview shows page 1. Sign up to view the full content.
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
secondorder 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) • [lOC.p/IO"I]1/"'2n
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
lOCp/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 fourthorder 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...
View
Full
Document
This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.
 Spring '13
 Bayliss
 Signal Processing, The Land

Click to edit the document details