第 9 章 有限脉冲响应滤波器.ppt

1、第 9 章 有限脉冲响应滤波器CH9 FINITE IMPULSE RESPONSE FILTERS,9.1 有限脉冲响应滤波器基础 FINITE IMPULSE RESPONSE FILTER BASICS 9.2 再论滑动平均滤波器 MOVING AVERAGE FILTERS REVITED 9.3 相位失真PHASE DISTORTION 9.4 逼近理想低通滤波器 APPROXIMATING AN IDLE LOW PASS FILTER 9.5 窗函数WINDOWS 9.6 低通 FIR 滤波器设计 LOW PASS FIR FILTER DESIGN 9.7 带阻和高通 FIR

2、滤波器 BAND PASS AND HIGH PASS FIR FILTERS,返回,有限脉冲响应滤波器：finite impulse response filter 无限脉冲响应滤波器：infinite impulse response filter 相位失真：phase distortion 理想低通滤波器：idle low pass filter 窗函数：window function 稳定性：stability 通带波纹：pass band ripple 阻带波纹：stop band ripple 通带边缘频率:pass band edge frequency 过渡带宽度：transi

3、tion width 矩形窗：Rectangular Window 汉宁窗：Hanning Window 哈明窗:Hamming Window 布莱克曼窗:Blackman Window 凯塞窗:Kaiser Window 项数:number of terms 衰减:attenuation 增益:gain 采样频率:sampling frequency,9.1 有限脉冲响应滤波器基础9.1 FINITE IMPULSE RESPONSE FILTER BASICS,有限脉冲滤波器（Finite impulse response filters) :(非递归滤波器nonrecursive fil

4、ters)滤波器的输出只与现在和过去的输入有关,跟过去的输出无关。(它总是在有限长个非 0 采样值后，输出为 0 ，脉冲响应有限项) 差分方程: yn= b0+b1xn-1+bMxn-M 脉冲响应：hn= b0+b1 n-1+bM n-M 传输函数: H(z)= b0+b1z-1+bMz-M 频率响应: H()= b0+b1e-j +bMe-jM,FIR 滤波器的设计需要选择系数 bk，要得到理想滤波器 特征，一般需要 100 个以上的系数，系数越多，滤波器 的滚降(roll-off)越陡峭，滤波器的选择性更好。 FIR滤波器的稳定性(stability) H(z)=b0+b1z-1+bMz-

5、M=,b0zM+b1zM-1+bM zM,可以看出它的 M 个极点都在 z=0 处，在单位圆内， 这种形式的滤波器都是稳定的。,返回,9.2 再论滑动平均滤波器9.2 MOVING AVERAGE FILTERS REVITED,它是一个 FIR 滤波器的一个简单例子，它的脉冲响 应有 M 个非 0 项。每个高度为1/M，它可以平滑输入 信号。 M 项滑动平均滤波器 yn=(xn+xn-1+xn-(M-1)/M 脉冲响应 hn=( n+n-1+n-(M-1)/M,5 项滑动平均滤波器频率响应 H()=0.2+0.2e-j+0.2e-j2 +0.2e-j3 +0.2e-j4 频率响应见图 9.2

6、,滑动平均滤波器的特点： 1) 第一个零点出现在 2/M 弧度 2) 截止频率（0.707对应的频率)约为第一个零 点增益频率的一半 /M 3)项数越多，低通滤波器效果强，滤除高频分量多。,返回,例：设计滑动平均滤波器，要求它的-3dB频率为480Hz，采样频率为10kHz.,9.3 相位失真9.3 PHASE DISTORTION,正弦信号通过线性(linear)滤波器时，它的幅度和相位都要改变。 若滤波器在 = 0 时的增益为|H(0)|，相位差 为 (0)，则输入为 Acos(n0) 的信号，输出为 A|H(0)|cos(n0+(0)。 图9.4 给出了输入和输出信号的图形，输出比输入滞

7、后(lagging)三个采样点，此量可设 n0+(0)=0,n= -,(0) 0,n=3 0= /9， 则 (0)= - /3,图 9.4,不同的频率分量通过滤波器产生的相位延迟不同，从而 产生了相位失真。 解决方法：使相位差为频率的线性系数 ()= - K 则 - ()/ 相位延迟与频率无关，等于 K 个采样点。 输入的所有频率分量同时出现在滤波器的输出端。 设计具有线形相位特点，而没有相位失真的滤波器： 关于零点对称的脉冲响应。,FIGURE 9-6 Noncausal impulse response.,Joyce Van de VegteFundamentals of Digital

8、Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,在实际中，所有滤波器为因果（causal)滤波器，而图 9.6 关于0 点对称的脉冲响应是非因果的，(滤波器在有输入之前就有输出了)。要进行时移(shifted in time)。 图 9.7。,hn=h1n-M 时移后 滤波器频率响应 H()= e-jMH1() 写成极坐标H()= H1()ej(1()-M) 时移对幅值没有影响，在相位响应里引入了 M， 通常 H

9、1() 的相位是 0，则 H() 在通常的相位为 M。（相位随 线形变化）此条件保证滤波时不发 生相位失真，也就是脉冲响应关于中点对称的滤波器没 有失真（奇数项）。,例 9.3 考察九项滑动平均滤波器，脉冲响应如下：,hn =(1/9)n-k,8 K=0,解： 因为脉冲响应关于中点对称，所以它的相位响应在通带 内随频率线形变化。幅度响应见图 9.8。,图 9.8,通带在图中标出，通带在幅度响应至少为直流幅度响应 值的 0.707 的频率范围。注意通带内相位的线形特征， 实际上，相位在许多其他范围也是线形的，只要 H() 变号，相位就移动 。这种因果脉冲响应很容易通 过下列差分方程实现： yn=

10、(1/9) xn-k,8 K=0,图 9.9,返回,9.4 逼近理想低通滤波器9.4 APPROXIMATING AN IDLE LOW PASS FILTER,理想低通滤波器的脉冲响应 h1n= sin(n1) 引入 sinc 函数 sincx=sinx/x x=0 时 sinc(0)=1 x=n 时 (n0) sinc(n)=0,1 n,图 9.11a,h1n= = sinc(n1),n1 sin(n1) 1 n n1 ,1是滤波器的截止频率(cut-off frequency)。,图9.12 理想低通滤波器脉冲响应,从实用来讲，这个脉冲响应要解决两个问题 1) 它的无限长度，两边响应值较

11、小的部分截止,图 9.13,2) 它的非因果性，时移成为因果性系统,图9.14,脉冲响应截断后，滤波器的形状不再是理想矩形,图 9.15,描述非理想滤波器 形状的一些参数,p通带波纹(pass band ripple) 1-p 通带边缘增益通带边缘频率fp1 p 阻带波纹(stop band ripple)， 阻带边缘频率fs1 过渡带宽度(transition width) 带宽(bandwidth),图 9.16,例 9.4 对于图9.17所示的低通滤波器，确定通带波纹、 通带边缘频率、阻带波纹、阻带边缘频率、过 渡带宽度、带宽、-3dB或截止频率。,图9.17,解： 此例中，通带内偏移单

12、位增益的最大量由向上偏移的波纹决定。通带内最大增益为 1.0905，所以 通带波纹为0.0905，于是通带边缘在增益下降到 1- p=1 0.0905=0.9095 的点，通带边缘频率为 1060 Hz（阻带边缘频率）处首次下降到此值。过渡带宽度为通带和阻带之间的频率差，即1 337-1 060=277 Hz。对于此低通滤波器，- 3dB 或截止频率点，即增益下降到 0.707的频率为 1 136 Hz。该频率也决定了滤波器的带宽，滤波器的带宽也是 1 136 Hz。,返回,9.5 窗函数WINDOWS,作用：从理想低通脉冲响应 h1n= 的无限 采样点中截取有限长个采样点。 hn=h1nwn

13、 乘积,sin(n1) n,9.5.1 矩形窗Rectangular window,汉宁窗： wn= 0.5+0.5cos Hanning Window 哈明窗： wn= 0.54+0.46cos Hamming Window 布莱克窗： wn=0.42+0.5cos +0.08cos Blackman Window 凯塞窗： wn= Kaiser Window,2n N 1,2n N 1,2n N 1,4n N 1,I0 1 (2n/(N-1) 1)2 I0,FIGURE 9-19 Impulse response for rectangular window.,Joyce Van de V

14、egteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-20 Magnitude response of rectangular window.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.U

15、pper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-21 Filter shape for filter made with rectangular window.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-22 Impulse

16、response for Hanning window.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-23 Magnitude response for Hanning window.,Joyce Van de VegteFundamentals of Digital Signal Processing,Co

17、pyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-24 Filter shape for filter made with Hanning window.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All r

18、ights reserved.,FIGURE 9-25 Impulse response for Hamming window.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-26 Magnitude response for Hamming window.,Joyce Van de VegteFundamen

19、tals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-27 Filter shape for filter made with Hamming window.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper

20、 Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-28 Impulse response for Blackman window.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-29 Magnitude responses for Blac

21、kman window.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-30 Filter shape for filter made with Blackman window.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyri

22、ght 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-31 Impulse response for Kaiser window, = 8.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserve

23、d.,FIGURE 9-32 Magnitude response for Kaiser window, = 8.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-33 Filter shape for filter made with Kaiser window, = 8.,Joyce Van de Vegte

24、Fundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,I0: 零阶修正第一类内塞尔函数，定义为,I0(x)=1+ 2,(x/2)j j !, j=0,凯塞窗的形状由 参数决定， 可以根据阻带要求计 算得出： =0.110 2A 0.958 7 （期望的阻带衰减值(attenuation) A(dB)50dB）,表 9.2 选择凯塞窗参数 预计阻带 实际滤波器阻 衰减(dB

25、) 带衰减(dB) 5.0 54 56 6.0 63 64 7.0 72 72 8.0 81 81 9.0 90 90 10.0 99 100,返回,9.6 低通滤波器设计9.6 LOW PASS FILTER DESIGN,9.6.1 设计指南(Design Guidelines) 表 9.3 FIR滤波器参数,通带边缘增益 20log(1-p)(dB),FIGURE 9-34 Window shape envelopes (N = 101).,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002

26、 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,低通滤波器通带的边缘频率 1要比设计的 1小(p246),根据经验，使用过渡带宽的中点频率（它处于通带边缘和阻带边缘间的中点），来作为设计的通带边缘频率。,图9.35,9.6.2 低通滤波器的设计步骤9.6.2 Steps for Low Pass FIR Design,加窗低通 FIR 滤波器的设计步骤 1. 在过渡带宽度的中间，选择通带边缘频率(Hz): f1=所要求的通带边缘频率+(过渡带宽度)/2 2. 计算 1=2f1

27、/fs，并将此值代入理想低通滤波器的 脉冲响应 h1n 中： h1n = sin(n1)/n 3. 从表 9.3 中选择满足阻带衰减及其他滤波器要求的窗 函数，用表中 N 的公式计算所需要的非零项数目。选 择奇数项，这样脉冲响应可以完全对称，避免了滤波 器产生相位失真，对于|n|(N-1)/2，计算窗函数 wn。,4. 对于|n|(N-1)/2，从式 hn=h1nwn计算 （有限）脉冲响应，对于其他 n 值hn=0，此脉冲响应是非因果的。 5. 将脉冲响应右移 (N-1)/2，确保第一个非零值在 n=0处，使此低通滤波器为因果的。,返回,例：根据下列指标设计低通滤波器 通带边缘频率2kHz 阻

28、带边缘频率3kHz 阻带衰减 40dB 采样频率 10kHz,FIGURE 9-36 Causal impulse response for Example 9.7.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-37 Pole-Zero diagram and filter shape for Example

29、 9.7.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-37 Pole-Zero diagram and filter shape for Example 9.7.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 20

30、02 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,例：根据下列指标设计低通滤波器 通带边缘频率10kHz 阻带边缘频率22kHz 阻带衰减 75dB 采样频率 50kHz,FIGURE 9-38 Causal impulse response for Example 9.8.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc

31、.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-39 Filter shape for Example 9.8.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-40 Spectrum of signal plus noise

32、for Example 9.9.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-41 Impulse response for Example 9.9.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by P

33、earson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-42 Spectrum of signal pulse noise low pass filtering for Example 9.9.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All r

34、ights reserved.,9.7 带通、高通和带阻滤波器9.7 BAND PASS AND HIGH PASS FIR FILTERS,可先设计低通滤波器，再进行频率移位(shifting in frequency)获得待求的滤波器。 采样会导致双边滤波器响应的频谱镜像出现在采样频率的每个倍数上，这是滤波器形状和一系列脉冲函数在频域里卷积的结果。这种卷积的结果使得每个脉冲函数的位置上都出现滤波器形状的一个副本。这种方法也可用来产生频率移位，将滤波器的低通原型转换为带通或高通滤波器。这样，频域中的单个单位脉冲函数必须位于待求滤波器的中心频率上。脉冲函数与低通滤波器形状的卷积可把双边低通滤波

35、器形状的副本移位到新的位置。,FIGURE 9-43 One-sided magnitude response of low pass filter.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-44 Two-sided magnitude response of low pass filter.,Joyce

36、 Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-45 Construction of a band pass filter.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, In

37、c.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-45 Construction of a band pass filter.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-45 Construction of a band

38、pass filter.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,因为余弦函数的频谱是一个尖峰，所以脉冲可以通过余弦函数提供。当频域卷积时，时域就是相乘。为了将上一节设计的非因果低通滤波器 转换为带通或高通滤波器，低通滤波器的脉冲响应必须与余弦函数相乘： 这里选 等于双边滤波器形状的待求中心频率。,设计步骤：低通滤波器设计

39、过程做两个小的修正： 1、在第三步和第四步之间，数字频率必须按下式计算： f0是待求滤波器的中心频率，对于带通滤波器，这个中心频率介于0到fs/2之间。对于高通滤波器，此频率应等于奈奎斯特界限fs/2，这样，,2、第四步中必须包含因子 ，脉冲响应计算公式为 高通滤波器的脉冲响应计算公式为,例：为采样频率为22kHz的系统设计 FIR带通滤波器，中心频率为4kHz，通带边缘频率在3.5和4.5kHz。过渡带宽度500Hz，阻带衰减50dB.,FIGURE 9-46 Band pass filter and low pass filter equivalent for Example 9.10.,

40、Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-47 Band pass filter impulse response for Example 9.10.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by

41、Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-48 Filter shape of band pass filter response for Example 9.10.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights

42、reserved.,例：需要一个通带边缘频率为8kHz,阻带边缘频率为6kHz 的高通滤波器，阻带增益至少比通带增益低40dB,采样频率为22kHz。设计滤波器并给出它的脉冲响应。,FIGURE 9-49 High pass filter and equivalent low pass filter for Example 9.11.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey

43、 07458All rights reserved.,FIGURE 9-50 Impulse response of high pass filter for Example 9.11.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-51 Pole-zero plot and filter shape of h

44、igh pass filter for Example 9.11.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-51 Pole-zero plot and filter shape of high pass filter for Example 9.11.,Joyce Van de VegteFundamen

45、tals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,带阻FIR滤波器 只要正确选择了通带边缘频率，通过把低通和高通滤波器结合起来，就可以构造出带阻滤波器。低通滤波器为阻带低端规定了通带边缘频率，同时，高通滤波器为阻带高端设定了通带边缘频率。为了形成带阻特性，低通滤波器的通带边缘频率一定要低于高通滤波器的通带边缘频率。,FIGURE 9-54 Constructing a band st

46、op filter.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-55 Summing low and high pass filters to create a band stop filter.,Joyce Van de VegteFundamentals of Digital Signal Proces

47、sing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-56 Cascading low and high pass filters to create a band pass filter.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River

48、, New Jersey 07458All rights reserved.,FIGURE 9-57 Constructing a band pass filter.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,例：图显示了数字电压信号采样值得一部分，采样频率为1kHz。怀疑60Hz的干扰信号对测量结果有影响，设计带阻滤波器以

49、消除这个干扰。,FIGURE 9-58 Voltage signal for Example 9.14.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-59 Band stop filter for Example 9.14.,Joyce Van de VegteFundamentals of Digital

50、Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-60 Low pass filter output for Example 9.14.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey

51、 07458All rights reserved.,FIGURE 9-61 High pass filter output for Example 9.14.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-62 Band stop filter output for Example 9.14.,Joyce V

52、an de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.,FIGURE 9-63 Band pass filter output for Example 9.14.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education, Inc.Upper Saddle