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

# 725 for n 3 a similar procedure applies t o a ny nj i

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Unformatted text preview: er found in Example 7.6. H( ) = 8 0.3269 (fa + 0.3689)( fa + 0.1844 + jO.9231)( fa + 0.1844 - In t he p resent case, f = 2 d B means t hat [see Eq. (7.47)] (2 S tep 1 : D etermining n A ccording t o Eq. (7.49b), we have = 10°·2 _ 1 - 1 [ 102 _ 1 ] 1/2 cosh --100.2 - 1 1(1.65) = cosh = 2.999 S tep 2 : D etermining ?-(8) We may u se t he Table 7.4 t o d etermine ?-(8). For n = 3 a nd f = 2 dB, we r ead t he coefficients o f t he d enominator polynomial o f ?-(8) as ao = 0.3269, a 1 = 1.0222, a nd a2 = 0.7378. Also in Eq. (7.53), for o dd n , t he n umerator is given by a c onstant K n = ao = 0.3269. Therefore, = 0.3269 83 + 0.737882 + 1.02228 + 0.3269 (7.54) Because t here a re infinite possible combinations of n and f , Table 7.4 (or 7.5) can list values o f t he d enominator coefficients for values of f in q uantum i ncrements only. 1 For t he values of n and f not listed in t he Table, we c an compute pole locations from Eq. (7.51). For t he sake of demonstration, we now recompute ?-(8) using this method. I n t his case, t he value of e is [see Eq. (7.47)] e= VlO f / lO - 1= VlOO. 2 - 1 = 0.7647 1 VI + 0.5849(4w 3- 3W)2 T his is t he normalized filter amplitUde response. T he a ctual filter response I H(jwll is o btained by replacing w w ith ~; t hat is, w ith Tii in ' It(jw) Because n m ust b e a n integer, we select n = 3. Observe t hat even with more stringent requirements, t he Chebyshev filter requires only n = 3. T he p assband behavior of t he B utterworth filter, however, is superior (maximally fiat a t w = 0) c ompared to t hat o f t he Chebyshev, which has rippled passband characteristics. ?-(8) 1 = 0.5849 T he frequency response is [see Eq. (7.42) a nd T able 7.3] I?-{(jw) I = n jO.9231) 326.9 83 + 7.3788 2 + 102.228 + 326.9 IH(jw)1 = --;====~1==== 1 + 0.5849 [4 (Tii) 3 - 3 ( Tii) 2] 103 - V9.3584w 6 - 1403.76w 4 + 52640w 2 + 106 Observe t hat d espite more stringent specifications t han t hose in Example 7.6, t he C hebyshev filter requires n = 3 c ompared t o t he B utterworth filter in Example 7.6, which requires n = 4. F igure 7.26 shows t he a mplitude response. • o C omputer E xample e 7.S Design a lowpass Chebyshev filter for t he specifications in Example 7.7 using functions from Signal P rocessing Toolbox in MATLAB. W p=lO;Ws=16.5;r=2;Gs=-20; [ n,Wp)=cheblord(Wp,Ws,r,-Gs,'s'); [ num,den)=chebyl(n,r,Wp,'s'); 522 7 Frequency Response a nd Analog Filters MATLAB returns n = 3 and num= 0 0 0 326.8901, den= 1 7.3782 102.219 326.8901; t hat is, H(8) _ 326.8901 - 83 + 7.37828 2 + 102.2198 + 326.8901 a result, which agrees with the solution in Example 7.7. To plot amplitude response, we can use the last three functions from Example C7.5. 0 ' " E xercise E 7.4 Determine n , the order of the lowpass Butterworth filter to meet the following specifications: ap = -0.5dB, a" = -20dB, Answer: 5. \ l '" wp = 100, andw. = 200. 7.6 = 2 dB, o C omputer E xample C 7.9 Design a lowpass inverse Chebyshev filter for the specifications in...
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