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

128a equivalent t o a desired analog continuous time

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: nevitable. However, for frequencies beyond some Wo, if IHa(jw)1 is a negligible fraction, say 1%, of Ha(jw)lma,,, t hen we c an consider 1 Ha(jw) t o b e essentially bandlimited t o Wo, a nd we can select I- h "" ~ "+ " ..., e ~ 00 0 ..., ~ ~ ~ u ..., ~ ' "C :.:: ..., I ..., ~ ~ " I", ::< '" ::< ~ F ig. 1 2.10 A liasing i n d igital filters, a nd a choice o f t he s ampling i nterval T . I u '" h I- .c T=!!.... (12.46) Wo if, I N <.J ~ ~ ~ ~ co '" t- :.:: ...... 1 ' " '" . . . 1"'' ' '" MI~ . . I~ . . 11 "" co + .., + .., <l ~u u ~ " -..: '+ '".., I- • E xample 1 2.4 Design a digital filter t o realize t he first-order analog lowpass B utterworth filter w ith t he t ransfer function Ha(S)=~ (12.47) s +wc t How c an we a pply t he d iscussion i n C hapter 5, which a pplies t o i mpulse s amples o f c ontinuoustime s ignals, t o d iscrete-time s ignals? Recall o ur d iscussion i n Sec. 10.4 (Fig. 10.8), w here w e showed t hat t he s pectrum o f d iscrete-time s ignal is j ust a s caled v ersion o f t he s pectrum o f t he i mpulse s amples o f t he c orresponding c ontinuous-time s ignal. 738 12 Frequency Response a nd Digital Filters y [ k] F or t his filter, we find t he c orresponding H[z] a ccording t o E q. (12.43) ( or p air 5 i n T able 1 2.1) a s H[z] = z _ e -w,T w eTz 739 12.5 Recursive F ilter Design: T he Impulse Invariance method (12.48) (a) N ext, we select t he v alue o f T a ccording t o E q. (12.46). We find t he e ssential filter b andwidth wo, w here t he filter g ain is 1% o f t he m aximum f ilter g ain. H ere we use = I~I w + we IHa(jw)1 I n t his c ase IHa(jw)lmax = 1. Hence, t he e ssential b andwidth Wo is t hat f requency w here t he I H a(jwo)l = 0.01. O bserve t hat w » We (b) H ence, IH[ejOlT]1 / T = (~) = 1 0- 6 7r lOwe a nd w eT = ..:':. 10 ................../ :--.---- IHa(joo)l T hus, for good results, we s hould s elect T = ;!l- = 1 O- 7 7r. H owever, for t he s ake o f d emonstrating a liasing effect o f t he o verlapping cy"c\es, w e s hall d eliberately s elect a lower v alue o f wo ( higher T). L et u s s elect LH (12.49) 5 x lOS 00- S ubstitution o f t his v alue i n Eq. (12.48) yields H[z] = 0.3142z (12.50) z - 0.7304 (c) A c anonical r ealization o f t his filter is s hown i n F ig. 12.11a b y following t he p rocedure i n Sec. 1 1.4 (see E xample 6.18c, Fig. 6.25b). N ote t he r ecursive n ature o f t he filter. T o find t he f requency r esponse o f t his d igital f ilter, we r ewrite H[z] a s H[z] _ 0 .3142 - 1 - 0.7304z- 1 H[e iwT ] = 0 .3142. 1 - 0.7304e- JwT = F ig. 1 2.11 A n e xample o f f ilter d esign b y t he i mpulse i nvariance m ethod: r ealization ( b) a mplitude r esponse (c) p hase r esponse. 0 .3142 (1 - 0.7304 cos w T) + j O.7304sin w T C onsequently jH[eiWTlI = ......._--- ..... __ . -1<12 T herefore ( a) f ilter Also, a ccording t o E q. (12.47) ( with We = 105) 0 .3142 = 0.3142 V (1 - 0.7304 cos WT)2 + (...
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.

Ask a homework question - tutors are online