24 Butterworth Filters Solutions to Recommended Problems S24.1 (a) For N = 5 and w = ( 2 7)1 kHz, IB(jw)1 2 is given by IB(jw)1 2 1 10 = 1 + (2000 r) (b) The denominator of B(s)B(-s) is set to zero. Thus 0 = 1 + (2000)1O, or s = (-1) 1 j20001r Expressing -1 as e' andj as e j/ 2, we find that the poles of B(s)B(-s) are 2 s = 2000ei[(x/10 + 0 for H(s) to be stable and causal. (c) Figure S24.3 contains a plot of (1 - a)/(1 + a), the pole location of H(z), ver sus a. 1- a 1+ a -1- Figure S24.3 We see that for a > 0, (1 of H(z) occurs at z = (1 a)/(1 + a) is between -- 1 and 1. Since the only pole - a)/(1 + a), H(z) must be stable whenever H(s) is stable, assuming that H(z) represents a causal h[n]. S24.4 (a) For T = 1 and the impulse invariance method, B(jw) must satisfy 1 IB(jw)|1 0.8 for 0 0.2 IB(jw)|1 0 for w 4, 37 fCV Therefore, if we ignore aliasing, B (j B (ij) =_ 1)2N 1 + j/4 4 _ 2N= 1j3r/4 1+ .'W (0. 2) , Signals and Systems S24-6 (b) For T = 1 and the bilinear transformation, B(jw) must satisfy 1 0.2 1B(jo) I 0.8, 0 : o : 2 tan, 1B(ji) 0, 2 tan 8 8) O Therefore, 1+ j2 tan (w/8) 1 (0.8)2, 2= SJwc 1 2= 1+ (0.2)2 j2 tan (3-/8) S24.5 (a) The relation between 0 and w is given by Q = wT, where T = 1/15000. Thus, - IH(e 0 )| 1 0.9 for 0 : - 2wr 5 - IH(e'")|1 0 for 3 -x s7 1 0.1 5 Note that while Hd(jw) was restricted to be between 0.1 and 0 for all o larger than 2r( 4 500), we can specify H(eja) only up to Q = x. For values higher than w, we rely on some anti-aliasing filter to do the attenuation for us. (b) Assuming no aliasing, H(e) = G j Therefore, 3 0.3 1|G(jw)|1 1 IG(jw) 1 2.7, 0, 0 s o 27 7w T 5 s o 15, < 3 (c) The relation between o and Qis given by Q = 2 arctan (w). Thus, 1 0.1 IG(jo)I - 0.9, 0 s - IG(jw)I 0, tan s tan 37r -5 10 T 5) , o < oo (d) If T changes, then the specifications for G(jw) will change for either the impulse variance method or the bilinear transformation. However, they will change in such a way that the resulting discrete-time filter H(e'") will not change. Thus, He(jo) will also not change. Butterworth Filters / Solutions S24-7 Solutions to Optional Problems S24.6 (a) We first assume that a B(s) exists such that the filter specifications are met exactly. Since 1 2 IB(jw)1 1+ we require that IB(j27r)1 2 1 = = (10-0.05)2 = 10-15 = 10-0, 1 IB(j3ir)1 2 = Substituting N = 5.88 and co, = 7.047, we see that the preceding equations are satisfied. (b) Since we know that N = 6, we use the first equation to solve for we: 1 = 10- 1 1 + Solving for co,, we find that w, = 7.032. The frequency response at ( = 0.31 is given by _1 IB(j3)1 2 - 12 1+ 20logo0 B(j3w)| = = 0.02890, 7.032 -15.4dB (c) If we picked N = 5, there would be no value of we that would lead to a Butterworth filter that would meet the filter specifications. S24.7 We require an Hd(z) such that 0 -20 dB 20 logio|Hd(eO)j 20 logio 1 Hd(e)|), -0.75, 0 s Q s 0.2613w, 0.4018w : Q ! w We will for the moment assume that the specifications can be met exactly. Let 9, be the frequency where 20 logI|Hd(eja,)| = -0.75, or Similarly, we define Q, as the frequency where or = -20, 20 log1oHd(eS')| 0 |Hd(es0 ")1 2 = 10-0075 |Hd(e "f)|2 = 10-2 Signals and Systems S24-8 Using T = 1, we find the specifications for the continuous-time filter Ha(jo) as IHa(jo,)| 2 = 10 IH(jW,)| 2 = 10 -.075 2, where o, = 2 tan = 2 tan 22 o, = 2 tan -= 2 =2 0.8703, 2 tan (0418w= 1.4617 2 For the specification to be met exactly, we need N and co such that 1+ j.732N /jO.8703 . and 1 + = 1000" (jl.4617 . 2N 2=10 =102 Solving for N, we find that N = 6.04. Since N is so close to 6 we may relax the specifications slightly and choose N = 6. Alternatively, we pick N = 7. Meeting the passband specification exactly, we choose w, such that 1 + (O.= 03 )14 100.075, we = 0.9805 or jWc The continuous-time filter H,(s) is then specified by H(s)H(-s) 1 = 1+ ( s The poles are drawn in Figure S24.7. We associate with Ha(s) the poles that are on the left half-plane, as follows: s, = -0.9805, S4 = 0.9805e' 0 S2 = 0.9805e"/ 14, S3 = s*, "'/ /" S6 = 0.9805e 2' 14 S5 = S*, Ha(s) is given by Ha(S) = (0.9805)7 7 f i= 1 (s - s) S 7 = s* Butterworth Filters / Solutions S24-9 Hd(z) can be obtained by the substitution ) Hd(Z) = H.(S)I|,= 2 1( _z-1)_(1+z S24.8 (a) Assuming no aliasing, Hd(e'u) is related to Hdtej") = by T = 2 , j $o Hb(jw) Thus, the specifications for Hb(jw) are given by IHb(jw)I 2 2b 2a, 0, I |Hb(jw)| 0 - w - 0.2r/2, 0.37r/2 w (b) Substituting H3(jw) = 2 .2c H 33 for o = 0.27r/ 2 , we have But 3a =.2r H Thus ft(3 0.27r 23a=2 33a=2a 2 Similarly, (.3r) j =2b Thus, A,(s) satisfies the filter specifications for Hb(jw) exactly. (c) H(ej") is given by f(e") = - 1 But ft,(jo) = [j (11 r) .j2 (Q 2x7k H, (jp,). Therefore, ft(e ) ju = 1 02 E2(" = 1 - H0 3 00 k= [ 3 ( 2 2 21rk) 3 ej 3 S24.9 (a) Using properties of the Laplace transform, we have sY(s) = X(s), or H(s) 1 s Signals and Systems S24-10 (b) Here h is given by T, a is given by x[(n - 1)T], and b is given by x(nT). There fore, the area is given by (a + b h = T[x((n 1)T) + x(nT)] = - An (c) From the definition of 9[n], we find that n-i Z P[n - 11 = A, k=-o Subtracting f[n - 1] from P[n], we find n 9[n) - n-1 [n - 11= A= A. Ak-- o k= -ok= Therefore, P[n] = y[n -1] + An. (d) From the answer to part (a), we substitute for An, yielding 9[n] = P[n - 11 + T [x((n 2 {[n = P[n - 1] + 1)T) + x(nT)] - - 1] + &[n]} (e) Using z-transforms, we find f(z) H(z) = z -f(z) =k X(z) + T [z'i(z) + =zT 2 - (1 1+ z zj I(z)], = H(s) s=2 t-21/1 +z-1>

Problem Set Solution 24: Butterworth Filters

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