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Show that the frequency response of an Nth-order low-pass Butterworth filter is maximallyflat at R = 0 in the sense that the first 2N - I derivatives of J H , ( J R )are equal to zero at R = 0. I~

Fig. 6-58

Without any loss in generality, we may assume that Q, = 1 and evaluate the derivative of the function G(Q) = at Q = 0. Multiplying both sides of this equation by (1

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b 1 f (x) dx b a a Let m and M be the minimum and maximum values of f on [a, b] (The existence of m and M is guaranteed by Theorem 14.2.) Thus, m f (x) M for all x in [a, b], so that Problem 30.3(c) gives

Again setting R = 0, we have G"(Q)],=, = 0 If we continue to differentiate k times, where k 5 2N - I, we have an equation of the form

X 150 and ma X X (b) mjA if mX 200 and mb c X 120 and ma X (c) mjA if mb X 73 (d) mjA if mX ma c X X c (e) mjA if ma :mb :mX 1:4:7

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where G(')(R) is the ith derivative of G(R), and F(Q) is a polynomial in 52. Given that Gt')(Q) is equal to zero at Q = O f o r i = I , . .. , k - I , itfollowsthat G("(R)ln=a = 0 Differentiating 2N times, however, we have

60 110 70

Therefore,

(6.14) 40 30 45 50

G ' ~ ~ ) ( Q ) ~ ,= -G(Q)[,, ,

m(b a)

(6.14)

. (2N)! = -(2N)!

Design a low-pass Butterworth filter that has a 3-dB cutoff frequency of 1.5 kHz and an attenuation of 40 dB at 3.0 kHz.

Fig. 6-59

Given the 3-dB cutoff frequency of the Butterworth filter, all that is needed is to find the filter order, N, that will At give 40 dB of attenuation at 3 kHz, or Q, = 2 n .3000. the stopband cutoff frequency a,, the magnitude of the frequency response squared is

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X 200 X 95 (b) mjA if ma X x (c) mjA if ma X (90 x) (d) mjA if ma X 3 ma X (e) mjA if mb X ma 50 X (f) mjA if mb X X 84 (g) mjA if mb ma X X 5:1 (h) mjA if mb :ma

f (x) dx M(b a)

Therefore, if we want the magnitude of the frequency response to be down 40 dB at R, = 2n .3000, the magnitude squared must be no larger than or

X X 7:3 X 5ma X (j) mjA if mb X (k) ma if mjA 35 X (l) ma if mjA y X (m) mb if mjA 60 X (n) mb if mjA x X (o) mb if AP ' AQ

or N = 7. For a seventh-order Butterworth filter, the 14 poles of Ha(s)Ha(-s) = lie on a circle of radius Q, = 2 n ,3000, at angles of

Find x and y in each part of Fig. 6-60 (t and t9 are tangents). (6.14)

k = 0 , 1 , . . . , 13

But then, by the intermediate-value theorem (Theorem 17.4), the value V is assumed by f somewhere in [a, b].

Fig. 6-60

as illustrated in the following figure:

The poles of Ha@)are the seven poles of Ha(s)H,(-s) that lie in the left-half s-plane, that is,

(6.15)

Except for the isolated pole at s = -Q,, the remaining six poles occur in complex conjugate pairs. The conjugate pairs may be combined to form second-order factors with real coefficients to yield factors of the form

Fig. 6-61

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