第7章傅立叶变换与滤波器形状.ppt

第 7 章 傅立叶变换与滤波器形状CH7 FOURIER TRANSFORMS AND FILTER SHAPE,7.1 傅立叶变换基础(FOURIER TRANSFORM BASICS)7.2 频率响应及其他形式(FREQUENCY RESPONSES AND OTHER FORMS)7.3 频率响应和滤波器形状(FREQUENCY RESPONSE AND FILTER SHAPE),返回,专业词汇,傅立叶变换：Fourier Transform 滤波器形状：filter shape频率响应:frequency response 频率特性：frequency characteristics离散时间傅立叶变换：Discrete Time Fourier Transform幅度响应：magnitude response 相位响应：phase response传输函数：transfer function 相位差：phase difference采样频率：sampling frequency,7.1 傅立叶变换基础7.1 FOURIER TRANSFORM BASICS,离散时间傅立叶变换(DTFT)将信号或滤波器由时域 频域 研究其频率特性 frequency characteristics magnitude response phase response,对于滤波器DTFT得到的信息称为滤波器的频率响应 frequency response,幅度响应相位响应,信号xn的离散(discrete)时间(time)傅立叶变换(Fourier transform)定义为 Fxn=x()=xn e-jn=xn(cos(n)jsin(n):数字频率，=2 f/fs 弧度 不同，变换x()不同，当 xn 以接近频率 变化时，x()较大，离散时间傅立叶变换 x()反应了信号的频率。,N=-,N=-,FIGURE 7-1 Signal resonance for the discrete time Fourier transform.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-1 Signal resonance for the discrete time Fourier transform.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-1 Signal resonance for the discrete time Fourier transform.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-1 Signal resonance for the discrete time Fourier transform.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-1 Signal resonance for the discrete time Fourier transform.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,例 7.1 求图 7.2 所示信号的离散时间傅立叶变换。,图 7.2,解：只有 4 个非零采样值(n=0，1，2，4)对变换有贡献，因而：X()=xn e-jn=2-e-j+3 e-j2+e-j4 一般情况下，DTFT 是复值。,N=-,例 7.2 求信号 xn=4(un un-3)的 DTFT。,离散时间傅立叶变换的两个重要特性,周期性(periodicity)延时性(time delay),延时性：假设 xn 的DTFT 存在，为X()，则 xn n0 的 DTFT 为 xn n0 e-jn。,N=-,周期性：X(+2)=xn e-jn(+2)=xn e-jn e-jn2,N=-,N=-,欧拉公式：e-jn2n=cos(2n)jsin(2n)=1 X(+2)=xn e-jn=X()DTFT 是周期的，周期为2，也就是 DTFT对于所有的，每2 重复一次。,N=-,返回,7.2 频率响应及其他形式7.2 FREQUENCY RESPONSES AND OTHER FORMS,7.2.1 频率响应和差分方程a0yn+a1yn 1+a2yn 2+aNyn N=b0 xn+b1xn 1+bMxn M每一项进行DTFTa0Y()+a1e-jY()+a2e-j2Y()+aNe-jNY()=b0X()+b1e-jX()+bMe-jMX(),H()=,Y()X(),b0+b1e-j+bMe-jMa0+a1e-j+aNe-jN,例 7.4 求出与如下差分方程相对应的频率响应：yn+0.1yn 1+0.85yn 2=xn 0.3xn 1,7.2.2 频率响应和传输函数 对照式 6.2 传输函数，频率响应是把传输函数中所有 z-1 换为 e-j。,例 7.5 求滤波器 的频率响应，它的传输函数(transfer function)是：,H(z)=,1 0.2z-21+0.5z-1+0.9z-2,7.2.3 频率响应和脉冲响应(impulse response),图 7.3 描述滤波器方法,滤波器的传输函数 H(z)是脉冲响应 hn 的z变换 xn n X()=n e-jn=1 yn hn Y()=hn e-jn H()=Y()频率响应 H()与脉冲响应 hn 的 DTFT 一样。,N=-,N=-,Y()X(),例 7.6 数字滤波器的脉冲响应为：hn=5 n n 1+0.2n 2 0.4n 3 求滤波器的频率响应的表达式。,返回,7.3 频率响应和滤波器的形状7.3 FREQUENCY RESPONSE AND FILTER SHAPE,7.3.1 滤波器对正弦输入的作用。Filter Effects on Sine Wave Inputs Y()=H()X()yn=F-1Y()对于 DTFT 方法一般仅求取正弦输入时的输出。频率响应 H()是复数，可用极坐标(polar form)H()=|H()|e-j()表示（附录A）,|H()|是数字滤波器在数字频率 处的增益(gain)。（无单位，或 dB 20log|H()|)()是数字滤波器在数字频率 处的相位差(phase difference)。（弧度 或 度）在每给定一个频率，增益和相位差可用来预测滤波器的响应。增益是对输入的放大量(amplification)相位差决定了输入的相位变化。,Y()=|Y()|e-jy()=|X()|e-jx()|H()|e-j()=|X()|H()|e-j(x()+(),幅值计算不能用分贝，都要转成线形计算。对于任一给定频率，输出的幅度是滤波器的增益和输入幅度的积，输出的相位是滤波器的相位差和输入相位的和。,例：数字频率为1.5弧度的余弦波通过滤波器，在此频率下，滤波器增益为-21dB，相位差为86，如果输入幅度为20，相位为12，则输出幅度和相位是多少？解：输入简式为，这是余弦信号 的缩写，在1.5弧度处，滤波器增益为-21 dB，但这个值不能用于计算，必须用 转换为线性值。因为相位差为86，频率响应的简式为。输出是频率响应和输入信号在傅里叶变换域的乘积：,7.3.2 幅度响应和相位响应,数字频率 处的频率响应 H()用极坐标形式 H()=|H()|e-j()所有数字频率处的增益的集合称为滤波器的幅度响应所有数字频率处的相位差的集合称为滤波器的相位响应 数字滤波器的频率响应。,例：一系统的频率响应为 求该系统的幅度响应和相位响应，并画出图。幅度响应是增益对数字频率（弧度)的关系图.相位响应是相位(弧度)对数字频率(弧度)的关系图.数字频率范围是,解:弧度间隔选择是任意的,要获得更高的准确度可选用更小的间隔.对于 弧度,采用非极坐标计算:,幅度响应是周期性的,每2 弧度重复一次,幅度响应是偶函数,相位响应是周期性的,每2 弧度重复一次,幅度响应是奇函数,所以,一般没有必要记录幅度响应和相位响应在 左边部分,而且,考虑 的值没有实际意义,所以实际数字滤波器的频率响应一般画出 部分.幅度响应:线性增益 对数字频率 的曲线画出;或者对数形式 对数字频率 的曲线.后者的优点是在增益变化范围非常大时,可以方便地画在一个图上.改变了图的形状相位响应:相位差可用弧度或角度表示.,FIGURE 7-7 Frequency response for Example 7.10.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-8 Frequency response for Example 7.11.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-8 frequency response for Example 7.11.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-9 Frequency response for Example 7.12.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-10 Frequency responses for common filters.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-10 Frequency responses for common filters.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-10 Frequency responses for common filters.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-10 Frequency responses for common filters.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-11 Frequency response for Example 7.13.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-11 Frequency response for Example 7.13.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-12 Frequency response for Example 7.14.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-13 Frequency response for Example 7.15.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-13 Frequency response for Example 7.15.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-14 Magnitude response of comb filter for Example 7.16.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-15 Pulse passed through comb filter.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-15 Pulse passed through comb filter.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-16“Hello”passed through comb filter.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-16“Hello”passed through comb filter.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-17 Magnitude response of comb filter for Example 7.17.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,7.3.3 模拟频率 f 与数字频率,数字滤波器的形状|H()|设计可不依赖采样频率(sampling frequency)，但所选的采样频率将影响滤波器输入频率的范围。当采样频率 fs 已知，可用模拟 f(Hz)代替数字频率(弧度)。=2 f/fs f=fs/2 以数字频率 表示的数字频率特性，只有当采样频率选定后才能确定。根据上式，可将 0 弧度的数字频率用 0 fs/2 Hz的模拟频率代替。,图 7.18 20log|H()|和()的表示的 幅度响应和相位响应。转成 20log|H(f)|f 和(f)f 的表示的幅度 响应和相位响应,FIGURE 7-19 Frequency response plotted against frequency in Hz.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-20 Magnitude response for Example 7.19.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-21 Magnitude response for fs=4 kHz for Example 7.19.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-22 Magnitude response for fs=10 kHz for Example 7.19.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,7.3.4由极零点确定滤波器形状,考虑如下传输函数：该滤波器的频率响应为,幅度响应（或滤波器形状）为：对于特定的，与p之间的距离越小，其幅度响应越大。当 沿单位圆移动，最靠近极点p时，幅度响应为最大值，即 和极点p的相位相符时,可获得最大幅度.而且极点位置越靠近单位圆,这个最大值就越大.,FIGURE 7-25 Graphical view of filter shape.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,由上可扩展到具有多个极零点的滤波器对于 弧度的频率,离滤波器极点越近,离零点越远,则幅度就越大.同样,靠近单位圆的极点,将导致滤波器形状在某一频率上有非常大的幅值,而靠近单位圆的零点将导致滤波器形状在某一频率上有非常小的幅值.这个幅值大小的剧烈变化可增强滤波器的选择性.,例:推断滤波器的形状,滤波器的传输函数为,FIGURE 7-26 Pole-zero plot for Example 7.21.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-27 Filter shape for Example 7.21.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,例:推断滤波器的形状,滤波器的差分方程为yn=xn-1+xn-3,FIGURE 7-28 Pole-zero plot for Example 7.22.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-29 Filter shape for Example 7.22.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-30 Pole-zero plot for Example 7.23.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-31 Filter shapes for Example 7.23.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,7.3.5一阶滤波器 可正可负,其符号对特性有很大影响.,FIGURE 7-32 Frequency response for Example 7.24.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,FIGURE 7-33 Frequency response for Example 7.25.,Joyce Van de VegteFundamentals of Digital Signal Processing,Copyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.,7.3.6二阶滤波器,CHAPTER SUMMARY,The discrete time Fourier transform(DTFT)of a signal xn is given by x()=xn e-jn.It reports the frequencies present in a signal.The DTFT of a signal xn gives the signals spectrum X().The DTFT of a system hn gives the systems frequency response H().The DTFT is periodic with period 2.A difference equation can be expressed as a frequency response.A transfer function can be expressed as a frequency response.The frequency response H()is the DTFT of the impulse response hn.A frequency response H()is a complex number and may be expressed in polar form in terms of a gain|H()|and a phase difference q(W)as H()=|H()|e j q(W)The frequency response can be used to find a filters output for a sinusoidal input.The output is a sinusoid with the same frequency as the input,but with an amplitude multiplied by the gain of the filter and a phase shifted by the phase difference of the filter.For the input xn=Acos(n 0+q x),with a digital frequency 0,the output is yn=H Acos(n 0+q+q x),where the gain of the filter is H()=|H(0)|and the phase difference is q=q(0).,The gains applied by the filter at each frequency form the magnitude response,plotted as|H()|versus,or 20log|H()|versus.The phase differences applied by the filter at each frequency form the phase response,plotted as q(W),in degrees or radians,versus W.It is sufficient to plot a frequency response for a system for the digital frequencies between 0 and radians,as W=radians corresponds to the Nyquist frequency or fs/2.The magnitude response shows the shape of a filter:low pass,high pass,band pass or band stop.Some filters,such as comb filters,have more complicated shapes.The magnitude and phase responses,|H()|and q(W),are normally plotted against digital frequency in radians.They can be plotted instead against analog frequency f in Hz using the equation f=fs/2 With this substitution,the magnitude response may be plotted as|H(f)|versus f,and the phase response as q(f)versus f.The shape of a filter can be deduced from its pole-zero plot.As increases,the complex number e j W moves around the unit circle.Proximity to a zero tends to reduce the magnitude of the filter,while proximity to a pole tends to increase it.The closer the poles and zeros are to the unit circle,the more selective the filter is.,