数字信号处理邵曦lecture22.ppt

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1、11.6 High-Order Filters High-order filter can implement sharp cutoff specifications.In this lessen,N-order Butterworth lowpass AF serves as the prototype(原 型原 型)in the bilinear transformation method.Specifications of a typical lowpass filter:LPDF：,stoppassstoppassAAff（Fig.11.6.1）passf：passband frequ

2、ency 通带（临界）频率通带（临界）频率 211pass:attenuation of 2|)(|passfH relative to 2|)0(|H(DC)passA：corresponding attenuation in decibels at passf Normalizing|H(0)|=1,)1lg(10)11lg(1022passpasspassA (11.6.4)passA is also the maximum attenuation required in the passband passAfA)(0，for passff 0 （11.6.1）The meaning i

3、s similar for:stopf(stopband frequency),stopA(minimum attenuation achieved in the stopband).By appropriate choice of,stoppassstoppassAAff,the filter can be made as close to an ideal lowpass filter as desired.Higher-order filter required Specification prewarpping:When using binlinear transformation m

4、ethod,the prewarped specification for LPAF should be:,stoppassstoppassAA where)2tan(passpass，)2tan(stopstop (11.6.5)11.6.1 Analog Lowpass Butterworth Filters Procedures to design an analog lowpass filter:(summarized from the section)a)给出一种低通形状的模平方幅度函数给出一种低通形状的模平方幅度函数2|)(|H （H(s)是是 s 的有理分式的有理分式2|)(|H

5、应是应是2的有理分式的有理分式）b)确定技术指标与确定技术指标与2|)(|H中待定参数的关系中待定参数的关系 c)求求 H(s)，使之具有如上，使之具有如上2|)(|H a)Magnitude response squared for Butterworth lowpass filters:NH202)(11|)(|(11.6.7)Two parameters:0:frequency of 3dB attenuation （2|)0(|H1，20|)(|H1/2）N:order of the filter.N sharper response,but 3dB frequency remains

6、 0 b)Determine N、0by specifications,stoppassstoppassAA Attenuation at frequency:)(1lg10|)(|lg10)(202NHA (11.6.8)Thus,the given specifications requires:stopstoppasspassAAAA)()(（p.606）1 2 Solving the equations(1)and(2),)/ln()/ln(passstoppassstopexactN （11.6.10）exactN may not be an integer,we chooseexa

7、ctNN (11.6.12)Taking N back to equation(1),we get 0(11.6.13)which causes stopstopAA)(.Taking N back to equation(2),we get a slightly different 0 which causes passpassAA)(Setting H(s)1/D(s),we have NNNsjssDsD2020*)()1(1)(1)()(11.6.16)2N roots of)()(*sDsD distributed uniformly on a circle of radius 0

8、in conjugate pairs (Fig.11.6.2)21(2,0iNNesijii（i=12N）（11.6.17）To make H(s)causal and stable,we should choose N roots Nss,.,1 in the left-hand s-plane to obtain D(s).Write D(s)in cascade form of SOS:)12()()()()2()()()()(1021KNsDsDsDKNsDsDsDsDKK (11.6.18)001)(ssD (first-order section，zero:0)2020*cos21

9、)1)(1()(sssssssDiiii（i=1K）(SOS，conjugate roots:is、*is)Replacing 0s by s,we get Table 11.6.1 of)(sD(Butterworth polynomials)under various order N.)()()()(10sHsHsHsHK（11.6.21）)(1)(sDsHii 11.6.2 Digital Lowpass Filters Design lowpass DF from Butterworth lowpass AF prototype:(p.612)step1：Specification p

10、rewarping)2tan(passpass,)2tan(stopstop step2：Determine Butterworth filter N、0from,stoppassstoppassAA.Look up Table 11.6.1 to obtain)(sH.step3：1111|)()(zzssHzH)()()(|)(|)(|)(1011111110111111zHzHzHsHsHsHKzzsKzzszzs Alternatively,one can obtain final H(s)in SOS form from calculating(11.6.23)(11.6.26)Ex

11、ample 11.6.3 passf4kHz，passA0.5dB，stopf5kHz，stopA10dB，sf20kHz Solution：step 1：DF specifications:spasspassff2,sstopstopff2 prewarping：)2tan(passpass，)2tan(stopstop step 2：Calculate N、0 for Butterworth lowpass AF.(stoppass,are introduced for convenience)N=7.We get 1412,1410,148,321 from Table 11.6.1 e

12、xcept the unlisted 0 corresponding to the root 0.step 3：Take 321,、0 into（11.6.23）（）（11.6.24）and 0 into（11.6.25）（）（11.6.26）to calculate the SOS coefficients,resulting in the SOS cascade form of H(z).11.6.3 Digital Highpass Filters Starting from Butterworth lowpass AF prototype,the designing of highpass DF should use bilinear transformation 1111zzs The steps are similar to that of designing lowpass DF,except that the equation in Step1 is（11.6.31）and equations in Step3 are（11.6.32）(11.6.36)