基于最小方差法低通FIR的设计说明书_第1页
基于最小方差法低通FIR的设计说明书_第2页
基于最小方差法低通FIR的设计说明书_第3页
基于最小方差法低通FIR的设计说明书_第4页
基于最小方差法低通FIR的设计说明书_第5页
已阅读5页,还剩10页未读 继续免费阅读

下载本文档

版权说明:本文档由用户提供并上传,收益归属内容提供方,若内容存在侵权,请进行举报或认领

文档简介

1、基于最小方差低通FIR滤波器设计说明书一设计目标根据所学的数字信号处理和MATLABf关知识,用最小方差法设计一个低通FIR滤波器。从FIR数字滤波器的系统函数可以看出,极点都是在z平面的原点,而零点的分布是任意的。不同的分布将对应不同的频率响应,最优化设计实际上就是调节这些零点的分布,使得实际滤波器的频率响应Hd与理想滤波器的频率响应Hd之间的最大绝对误差最小。二低通FIR滤波器技术指标1Wp0.25通带截止频率pWs0.35阻带截止频率p1dB通带衰减s40dB阻带衰减ap201gH(ej0)H(ejWc通带最大衰减H(ej0)as201gRj)阻带最小衰减三氐通FIR滤波器的设计3.1低

2、通FIR滤波器阶数的估计N-20-s-134614.6(sp)/2由于N为偶数,所以可以设计一个1型的低通FIR滤波器。3.2对于基于最小方差的线性相位FIR滤波器的设计下面式子为误差的简化为k2W(i)H(i)D(i)i1其中H()是低通FIR的振幅响应,D()是要求的振幅响应,W()是权重函数。由于所有四种类型的线性相位FIR滤波器的振幅响应可以表示为iH()Q()kcosk03.3H()Q()kcos式中Q()、k、L的确定k0aQ()的确定由于不同类型Q()也就不尽相同,不同类型时Q()的表达式如下Q()=1对于1型Q()=cos对于2型Q()=sin对于3型Q()=sin对于4型由于

3、我们设计的低通FIR滤波器为1型所以Q()=1bk的确定同样根据不同的类型其k的表达式也不一样_k=bkk=k一k=dk我们选择k=k,对于1型0hM,k2hMk,1kM的最大整数,所以M=22,L=22。23.4k2W(i)H(i)D(i)中D()、W()的确定i1根据最小方差的相关要求可知W( ) 1在通带中W( ) 0在阻带中D( ) 1在通带中D( ) 0在阻带中3.5根据上面式子可以确定Q(),L的值和k的表达式,由于最小方差是滤波器参数k的一个函数。为了得到的最小值,令由它可生成L+1个等式的线性方程组,用来求解我们考虑1型线性相位fir滤波器的设计。在这种情况下,Q()=1,k=

4、k且L=22。则均方误差的表达式为 TOC o 1-5 h z kM HYPERLINK l bookmark13 o Current Document W(i)kcos(ik)D(i)i1k0M2W(i)kcos(ik)W(i)D(i)k0若有W( 1) W( 1)COS( 1)W( 1)COS(M 1)W(2)W(2)COS(2).W(2)COS(M2)HW(k)W(k)COS(k).W(k)COS(Mk)a0a1.aMTdW(1)D(1)W(2)D(2).W(k)D(k)teTe,式中eHad。计算如下1求H,1-15时川()取1,16-22时取0将0到0.35上取均匀的22点最后求的H

5、=10.99870.99500.98880.9800.9680.9550.9390.9210.9000.8770.8530.8250.7960.7650.7320.6970.6600.6220.5820.5400.498;10.9950.9800.9550.9210.8770.8250.7650.6970.6220.5400.4540.3630.2680.1700.070.0280.1280.2260.3220.4150.588;10.9880.9550.9000.8250.7320.6220.4980.3630.2190.0710.0780.2260.3690.5040.6270.7370.

6、8290.9040.9570.9890.999;10.9800.9210.8250.6970.5410.3630.1700.0280.2260.4150.5880.7370.8560.9420.9890.9980.9670.8970.7920.6540.492;10.9680.8770.7320.5400.3150.0710.1770.4150.6270.8000.9240.9890.990.9370.8210.6550.3120.2120.0350.2820.510;10.9550.8250.6220.3630.0710.2260.5040.7370.9040.9890.9880.8970.

7、7270.4920.2120.00850.3760.6330.8330.9590.999;10.9390.7650.4980.1700.1770.5040.7690.9420.9990.9370.7600.2770.1630.1840.5100.7740.9440.9990.9340.7550.485;10.9210.6970.3630.0280.4150.7360.9420.9980.8970.6550.3090.0080.4670.7740.9590.9930.8700.6100.2540.1420.516;10.9000.6220.2190.2260.6270.9030.9990.897

8、0.6160.2120.2330.6330.9060.9990.8940.6100.2050.2400.6380.9090.999;1 TOC o 1-5 h z 0.8770.5410.0710.4150.8000.9890.9370.6550.2120.2820.7070.9590.9770.7550.3490.1420.5990.9090.9970.8410.479;10.8520.4540.0780.5870.9230.9870.7600.3090.2330.7070.9720.9510.6490.1560.3820.8090.9970.8910.52200.522;10.8250.3

9、630.2260.7360.9890.8970.4920.0850.6330.9590.9510.6100.0570.5160.9090.9850.7170.1990.3890.8410.999;10.7960.2680.3690.8560.9940.7270.1630.4670.9070.9770.6490.0570.5580.9460.9490.5640.0490.6440.9750.9090.473;10.7650.1700.5040.9400.9370.4910.1840.7740.9990.7550.1560.5160.9460.9320.4790.1980.7830.9990.74

10、60.1420.528;10.7320.0710.6270.9890.8210.2120.5100.9590.8940.3490.349-0.9090.9480.4780.2470.8410.9840.5990.1070.7550.999;0000000000000000000000;0000000000000000000000;0000000000000000000000;0000000000000000000000;0000000000000000000000;0000000000000000000000;0000000000000000000000算这个H时算出来的负值取了它的绝对值2求

11、k令d=1111111111111110000000然后最小均方解可以根据求解线性方程HTHaHTd得到。令rHTHnHTdTa=ka=n*pinv用MATLAB1以计算出ac=ac=1.0e+005*Columns1through190.95392.5572-1.4336-0.37351.9754-3.75723.5858-2.6165-2.31652.7508-3.48832.3659- TOC o 1-5 h z 3.81731.90290.59010000Columns20through22|000k=0.95392.5572-1.4336-0.37351.9754-3.75723.5

12、858-2.6165-2.31652.7508-3.48832.3659-3.81731.90290.590100000003最终H()的结果由于N的限制所以所设计的滤波器为1型所以Q()=1k=k=cL=M=22i根据表达式H()Q()kcosk0H()=0.9539cos()+2.5572cos(2)-1.4336cos(3)-0.3735cos(4)+1.9754cos(5)-3.7572cos(6)+3.5858cos(7)-2.6165cos(8)-2.3165cos(9)+2.7508cos(10)-3.4883cos(11)+2.3659cos(12)-3.8173cos(13)

13、+1.9029cos(14)+0.5901cos(15)16至ij22项为0这里我们只是求出了幅频特性,但由于其相频特性是确定的所以在设计中不考虑其相频特性。 TOC o 1-5 h z j122,33、ze求H(Z)=0.0916Vzz+0.1171Vzz-0.6515(zz)-0.1867(z4z4)+0.9877(z5z5)-1.8786(z6z6)+1.7929(z7z7)-1.3082(z8z8)-1.199101011111212582(zz)+1.3754(zz)-1.7441(zz)+1.1829(zz)-1.9086,1313、/1414、/1515、(zz)+0.9514(

14、zz)+0.2950(zz)4计算误差由eTe和eHad可求出误差用MATLAB算误差如下h=10.99870.99500.98880.9800.9680.9550.9390.9210.9000.8770.8530.8250.7960.7650.7320.6970.6600.6220.5820.5400.498;10.9950.9800.9550.9210.8770.8250.7650.6970.6220.5400.4540.3630.2680.1700.070.0280.1280.2260.3220.4150.588;10.9880.9550.9000.8250.7320.6220.4980

15、.3630.2190.0710.0780.2260.3690.5040.6270.7370.8290.9040.9570.9890.999;10.9800.9210.8250.6970.5410.3630.1700.0280.2260.4150.5880.7370.8560.9420.9890.9980.9670.8970.7920.6540.492;10.9680.8770.7320.5400.3150.0710.1770.4150.6270.8000.9240.9890.990.9370.8210.6550.3120.2120.0350.2820.510;10.9550.8250.6220

16、.3630.0710.2260.5040.7370.9040.9890.9880.8970.7270.4920.2120.00850.3760.6330.8330.9590.999;10.9390.7650.4980.1700.1770.5040.7690.9420.9990.9370.7600.2770.1630.1840.5100.7740.9440.9990.9340.7550.485;10.9210.6970.3630.0280.4150.7360.9420.9980.8970.6550.3090.0080.4670.7740.9590.9930.8700.6100.2540.1420

17、.516;10.9000.6220.2190.2260.6270.9030.9990.8970.6160.2120.2330.6330.9060.9990.8940.6100.2050.2400.6380.9090.999;10.8770.5410.0710.4150.8000.9890.9370.6550.2120.2820.7070.9590.9770.7550.3490.1420.5990.9090.9970.8410.479;10.8520.4540.0780.5870.9230.9870.7600.3090.2330.7070.9720.9510.6490.1560.3820.809

18、0.9970.8910.52200.522;1 TOC o 1-5 h z 0.8250.3630.2260.7360.9890.8970.4920.0850.6330.9590.9510.6100.0570.5160.9090.9850.7170.1990.3890.8410.999;10.7960.2680.3690.8560.9940.7270.1630.4670.9070.9770.6490.0570.5580.9460.9490.5640.0490.6440.9750.9090.473;10.7650.1700.5040.9400.9370.4910.1840.7740.9990.7

19、550.1560.5160.9460.9320.4790.1980.7830.9990.7460.1420.528;10.7320.0710.6270.9890.8210.2120.5100.9590.8940.3490.349-0.9090.9480.4780.2470.8410.9840.5990.1070.7550.999;0000000000000000000000;0000000000000000000000;0000000000000000000000;0000000000000000000000;0000000000000000000000;0000000000000000000

20、000;0000000000000000000000a=0.95392.5572-1.4336-0.37351.9754-3.75723.5858-2.6165-2.31652.7508-3.48832.3659-3.81731.90290.59010000000;s=h*ad=1111111111111110000000;c=s-de=c*ce=Columns1through193.62352.4192-0.7942-1.72131.15432.20881.91280.4541 TOC o 1-5 h z 1.29032.47103.52100.6483-4.8485-0.3134-9.87

21、2700002.41921.6152-0.5303-1.14920.77071.47471.27710.30320.86151.64982.35080.4329-3.2372-0.2093-6.59160000-0.7942-0.53030.17410.3773-0.2530-0.4841-0.4193-0.0995-0.2828-0.5416-0.7718-0.14211.06280.06872.16400000-1.7213-1.14920.37730.8177-0.5483-1.0493-0.9086-0.2157-0.6129-1.1738-1.6726-0.30802.30320.1

22、4894.689901.15430.7707-0.2530-0.54830.36770.70360.60930.41100.78711.12160.2065-1.5445-0.0998-3.1450 TOC o 1-5 h z 0002.20881.4747-0.4841-1.04930.70361.34641.16600.78651.50632.14630.3952-2.9556-0.1911-6.01820001.91281.2771-0.4193-0.90860.60931.16601.00970.68111.30441.85860.3422-2.5594-0.1655-5.211600

23、00.45410.3032-0.0995-0.21570.14470.27680.23970.16170.30970.44130.0813-0.6076-0.0393-1.23730001.29030.8615-0.2828-0.61290.41100.78650.68110.45950.87991.25380.2309-1.7265-0.1116-3.51560002.47101.6498-0.5416-1.17380.78711.50631.30440.87991.68502.40110.4421-3.3064-0.2137-6.73250003.52102.3508-0.7718-1.6

24、7261.12162.14631.85861.25382.40113.42140.6300-4.7114-0.3046-9.59340000.64830.4329-0.1421-0.30800.20650.39520.34220.23090.44210.63000.1160-0.8675-0.0561-1.7665000-4.8485-3.23721.06282.3032-1.5445-2.9556-2.5594-1.7265-3.3064-4.7114-0.86756.48780.419413.2106000-0.3134-0.20930.06870.1489-0.0998-0.1911-0

25、.16550.144700.276800.239700.056900.161700.309700.441300.08130-0.60760-0.03930-0.1116-0.2137-0.3046-0.05610.41940.02710.8540-9.8727-6.59162.16404.6899-3.1450-6.0182-5.2116-1.2373 TOC o 1-5 h z -3.5156-6.7325-9.5934-1.766513.21060.854026.8998000000000000000000000000000000000000000000000000000000000000

26、00000000000000000000000000000000000000000000000000000000000000000000000000000Columns20through22000000000000000000000000000000000000000000000000000000000000四用直接型和级联型这两种结构实现1直接型直接型是利用输入信号x(n)和滤波器单位脉冲响应h(n)的线性卷积来描述输出信号y(n)。虽然有22阶但16到22阶H()为0所以直接型结构如下五用FDATOOL析设定相关参数后但到其幅频特性如下0510152。Frequency tkHz)OO相频特性曲线如下01015Frequericy (kHz)20零极图如下11后1.11i一:01/1%I1,11、n*-1於1I1I11115。)111n*!,1二o111111*t#J1d1:A1111一.J.11-1I*1_产一士外11,2-1.5-1-Q,5C0.511石2RealPart六误差分析误差产生的原因因为在数字信号处理中,需要将输入的离散信号和系统的参数进行量化,而量化的结果必然与原来的数值之间存在误差,误差的大小要依据计算机的字长而定。

温馨提示

  • 1. 本站所有资源如无特殊说明,都需要本地电脑安装OFFICE2007和PDF阅读器。图纸软件为CAD,CAXA,PROE,UG,SolidWorks等.压缩文件请下载最新的WinRAR软件解压。
  • 2. 本站的文档不包含任何第三方提供的附件图纸等,如果需要附件,请联系上传者。文件的所有权益归上传用户所有。
  • 3. 本站RAR压缩包中若带图纸,网页内容里面会有图纸预览,若没有图纸预览就没有图纸。
  • 4. 未经权益所有人同意不得将文件中的内容挪作商业或盈利用途。
  • 5. 人人文库网仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对用户上传分享的文档内容本身不做任何修改或编辑,并不能对任何下载内容负责。
  • 6. 下载文件中如有侵权或不适当内容,请与我们联系,我们立即纠正。
  • 7. 本站不保证下载资源的准确性、安全性和完整性, 同时也不承担用户因使用这些下载资源对自己和他人造成任何形式的伤害或损失。

评论

0/150

提交评论