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

B ut hk m ust have only a finite d uration a nd i t m

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Unformatted text preview: .1326/T T he t erms ± 5wT r epresent t he t ime d elay o f 5 T s econds. N ote t hat b ecause H a(jw) = j w, 3 - 1/2T 0.6821 - 0.3410/T 4 l /T 0.9121 0 .9121/T 5 0 6 - l/T 0.9121 - 0.9121/T 7 1 /2T 0.6821 0.341O/T 8 - 1/3T 0 .3978 - 0.1326/T 9 1 /4T 0 .1678 0 .04195/T - 1/5T 0 .08 - 0.016/T w <O IHa(jw)1 L.Ha(jw) = Iwl = 71"/2 w>0 { -71"/2 w <O (12.91) 0 t Note t hat according t o E q. (12.81a) ha(O) = - 1 J "/2T 2" jwdw = 0 - "/2T 10 Therefore, h[O] = o. q 770 12 F requency R esponse a nd D igital F ilters 12.8 771 N onrecursive F ilter D esign n um2=h2; d en2=denl; H 2=freqz( n um2,den2, W ); m ag2=abs(H2); p hase2=180/pi*unwrap(angle(H2»; s ubplot(2,1,1); p lot(W , magl,W , mag2); s ubplot(2,1,2); p lot(W,phase2,W,phasel); 0 1T ! :; E xercise E12.7 Design a sixth-order nonrecursive filter to realize Ha(jw) Answer: = !:;(WT/7T) Ikl:-:; 3 Ikl > 3 o 00- It T F ig. 1 2.20 An ideal differentiator design by a t enth-order nonrecursive filter using rectangular and Hamming windows. The amplitude response of t he Hamming filter is practically ideal up t o frequency = 27T/3T. I f we wish t o design a digital differentiatorfor audio application, for instance, where t he highest frequency is, say 20 kHz, we should select this frequency t o be less than 27T/3T. T hus, w "" (2/3)(7T/T) 27T X 27T 20,000 :-:; 3 T =? T:-:; 16.67 /los T hus, a choice of T :-:; 16.67 /los would result in a desired differentiator. This example (also F ig. 12.20) shows t hat a Hamming window shrinks the passband. This is generally true of t apered windows. To compensate for t his shrinkage, we s tart w ith a passband somewhat larger (typically 25% larger) t han t he design passband. In t he present case, for instance, selecting T = 16.67 /los would make t he passband W h = 7T I T "" 27T(30, 000), which is 50% h igher t han t he design passband of 20,000 Hz. • 1 2.8-2 Nonrecursive Filter Design by t he Frequency-Domain Criterion: T he Frequency Sampling M ethod T he f requency-domain c riterion is [see E q. (12.36)] I n t his c ase we s hall r ealize t his e quality for r eal f requencies; t hat is, for s = j w: (12.92) F or a n n th-order filter, t here a re o nly No = n + 1 e lements i n h[k], a nd we c an h ope t o force t he t wo f requency s pectra i n E q. (12.92) t o b e e qual o nly a t No p oints. B ecause t he s pectral w idth is we choose t hese f requencies wo = r adls a part; t hat is , '¥' JoT 211" NoT W o=-- 8 C omputer E xample C 12.10 Using MATLAB, find t he frequency response of t he 48th-order digital differentiator i n Example 12.10. P lot t he frequency response for r ectangular and Hamming window differentiators. N O=49; m =(NO-l)/2; k =O:NO-l; T =I; h l=cos( ( k-m)*pi).1 « k-m ) *T) ; hl ( 25)=0; n uml=hl; d enl=[I, z eros(I,NO-l»); W = -pi:pi/l00:pi; H l=freqz(numl,denl,W); m agl=abs(Hl); p hasel= 1 80I pi*unwrap( a ngle(Hl»; h 2=cos( ( k-m ) *pi).1 « k-m) * T). * ( 0.54+0.46*cos(pi* ( k-m) ./m»...
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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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