第-7-章------傅立叶变换与滤波器形状课件

第第 7 章章 傅立叶变换与滤波器形状傅立叶变换与滤波器形状CH7 FOURIER TRANSFORMS AND FILTER SHAPE7.1 傅立叶变换基础傅立叶变换基础(FOURIER TRANSFORM BASICS)7.2 频率响应及其他形式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 傅立叶变换基础傅立叶变换基础7.1 FOURIER TRANSFORM BASICS 离散时间傅立叶变换离散时间傅立叶变换(DTFT)将信号或滤波器由时域将信号或滤波器由时域 频域频域 研究其频率特性研究其频率特性 frequency characteristics magnitude response phase response对于滤波器对于滤波器DTFT得到的信得到的信息称为滤波器的频率响应息称为滤波器的频率响应 frequency response 幅度响应幅度响应相位响应相位响应信号信号xn的离散的离散(discrete)时间时间(time)傅立叶变换傅立叶变换(Fourier transform)定义为定义为 Fxn=x()=xn e-jn =xn(cos(n)jsin(n):数字频率数字频率，=2 f/fs 弧度弧度 不同，变换不同，变换x()不同，当不同，当 xn 以接近频率以接近频率 变变化时，化时，x()较大，离散时间傅立叶变换较大，离散时间傅立叶变换 x()反应了反应了信号的频率。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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.例例 7.1 求图求图 7.2 所示信号的离散时间傅立叶变换。所示信号的离散时间傅立叶变换。图图 7.2解：解：只有只有 4 个非零采样值个非零采样值(n=0，1，2，4)对变换有贡献，对变换有贡献，因而：因而：X()=xn e-jn=2-e-j+3 e-j2+e-j4 一般情况下，一般情况下，DTFT 是复值。是复值。N=-例例 7.2 求信号求信号 xn=4(un un-3)的的 DTFT。解：解：在在 n0 和和 n 时，信号值都是零，所以：时，信号值都是零，所以：X()=xn e-jn=4+4e-j+4 e-j2 N=-离散时间傅立叶变换的两个重要特性离散时间傅立叶变换的两个重要特性周期性周期性(periodicity)延时性延时性(time delay)延时性：假设延时性：假设 xn 的的DTFT 存在，为存在，为X()，则则 xn n0 的的 DTFT 为为 xn n0 e-jn。N=-令令 m=n n0,n=m+n0 xm e-j(m+n0)=e-jn0 xm e-jm=e-jn0 X()N=-N=-时域中延迟时域中延迟 n0 在频率中引入一个复指数在频率中引入一个复指数e-jn0 周期性：周期性：X(+2)=xn e-jn(+2)=xn e-jn e-jn2 N=-N=-欧拉公式：欧拉公式：e-jn2n=cos(2n)jsin(2n)=1 X(+2)=xn e-jn=X()DTFT 是周期的，周期为是周期的，周期为2，也就是也就是 DTFT对于对于所有的所有的，每，每2 重复一次。重复一次。N=-返回返回7.2 频率响应及其他形式频率响应及其他形式7.2 FREQUENCY RESPONSES AND OTHER FORMS7.2.1 频率响应和差分方程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例例 7.4 求出与如下差分方程相对应的频率响应：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 解：解：容易确定系数容易确定系数(coefficients)为为 a0=1，a1=0.1，a2=0.85，b0=1，b1=-0.3。滤波器的频率响应为：滤波器的频率响应为：H()=Y()X()b0+b1e-j+bMe-jMa0+a1e-j+aNe-jN=1 0.3e-j1+0.1e-j+0.85e-j27.2.2 频率响应和传输函数频率响应和传输函数 对照式对照式 6.2 传输函数，频率响应是把传输传输函数，频率响应是把传输函数中所有函数中所有 z-1 换为换为 e-j。例例 7.5 求滤波器求滤波器 的频率响应，它的传输函数的频率响应，它的传输函数(transfer function)是：是：H(z)=1 0.2z-21+0.5z-1+0.9z-2解：解：频率响应为：频率响应为：H()=1 0.2e-j21+0.5e-j+0.9e-j27.2.3 频率响应和脉冲响应频率响应和脉冲响应(impulse response)图图 7.3 描述滤波器方法描述滤波器方法滤波器的传输函数滤波器的传输函数 H(z)是脉冲响应是脉冲响应 hn 的的z变换变换 xn n X()=n e-jn=1 yn hn Y()=hn e-jn H()=Y()频率响应频率响应 H()与脉冲响应与脉冲响应 hn 的的 DTFT 一样。一样。N=-N=-Y()X()例例 7.6 数字滤波器的脉冲响应为：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。求滤波器的频率响应的表达式。解：解：频率响应是脉冲响应频率响应是脉冲响应 DTFT，见式见式(7.3)，因而得：，因而得：H()=hn e-jn=5 e-j+0.2e-j2 0.04 e-j3 N=-返回返回7.3 频率响应和滤波器的形状频率响应和滤波器的形状7.3 FREQUENCY RESPONSE AND FILTER SHAPE7.3.1 滤波器对正弦输入的作用。滤波器对正弦输入的作用。Filter Effects on Sine Wave Inputs Y()=H()X()yn=F-1Y()对于对于 DTFT 方法一般仅求取正弦输入时的输出。方法一般仅求取正弦输入时的输出。频率响应频率响应 H()是复数，可用极坐标是复数，可用极坐标(polar form)H()=|H()|e-j()表示表示 （附录（附录A）|H()|是数字滤波器在数字频率是数字滤波器在数字频率 处的增益处的增益(gain)。（无单位，或（无单位，或 dB 20log|H()|)()是数字滤波器在数字频率是数字滤波器在数字频率 处的相位差处的相位差(phase difference)。（。（弧度弧度 或或 度）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。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增益是对输入的放大量(amplification)相位差决定了输入的相位变化。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。幅值计算不能用分贝，都要转成线形计算。对于任一给定频率对于任一给定频率 ，输出的幅度是滤波器的增益和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。相位的和。例：数字频率为例：数字频率为1.5弧度的余弦波通过滤波器，弧度的余弦波通过滤波器，在此频率下，滤波器增益为在此频率下，滤波器增益为-21dB，相位差为相位差为86，如果输入幅度为，如果输入幅度为20，相位为，相位为12，则输出，则输出幅度和相位是多少？幅度和相位是多少？解：输入简式为解：输入简式为 ，这是余弦信号，这是余弦信号 的缩写，在的缩写，在1.5弧度处，弧度处，滤波器增益为滤波器增益为-21 dB，但这个值不能用于计算，但这个值不能用于计算，必须用必须用 转换为线性值。因为相位转换为线性值。因为相位差为差为86，频率响应的简式为，频率响应的简式为 。输出。输出是频率响应和输入信号在傅里叶变换域的乘积：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幅度响应和相位响应 数字频率数字频率 处的频率响应处的频率响应 H()用极坐标形式用极坐标形式 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。数字滤波器的频率响应。例：一系统的频率响应为例：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（弧出图。幅度响应是增益对数字频率（弧度度)的关系图的关系图.相位响应是相位相位响应是相位(弧度弧度)对对数字频率数字频率(弧度弧度)的关系图的关系图.数字频率范围数字频率范围是是解解:弧度间隔选择是任意的弧度间隔选择是任意的,要获得更高要获得更高的准确度可选用更小的间隔的准确度可选用更小的间隔.对于对于 弧度弧度,采用非极坐标计算采用非极坐标计算:幅度响应是周期性幅度响应是周期性的的,每每2 弧度重复弧度重复一次一次,幅度响应是幅度响应是偶函数偶函数相位响应是周期相位响应是周期性的性的,每每2 弧度弧度重复一次重复一次,幅度响幅度响应是奇函数应是奇函数所以所以,一般没有必要记录幅度响应和相位响应在一般没有必要记录幅度响应和相位响应在 左边部分左边部分,而且而且,考虑考虑 的值没有实际意义的值没有实际意义,所以实际数字滤波器的频率响应一般画出所以实际数字滤波器的频率响应一般画出 部分部分.幅度响应幅度响应:线性增益线性增益 对数字频率对数字频率 的曲线画的曲线画出出;或者对数形式或者对数形式 对数字频率对数字频率 的曲的曲线线.后者的优点是在增益变化范围非常大时后者的优点是在增益变化范围非常大时,可可以方便地画在一个图上以方便地画在一个图上.改变了图的形状改变了图的形状相位响应相位响应:相位差可用弧度或角度表示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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.7.3.3 模拟频率模拟频率 f 与数字频率与数字频率 数字滤波器的形状数字滤波器的形状|H()|设计可不依赖采样频率设计可不依赖采样频率(sampling frequency)，但所选的采样频率将影响滤但所选的采样频率将影响滤波器输入频率的范围。波器输入频率的范围。当采样频率当采样频率 fs 已知，可用模拟已知，可用模拟 f(Hz)代替数字频率代替数字频率 (弧度弧度)。=2 f/fs f=fs/2 以数字频率以数字频率 表示的数字频率特性，只有当采样频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，可将率选定后才能确定。根据上式，可将 0 弧度的数弧度的数字频率用字频率用 0 fs/2 Hz的模拟频率代替。的模拟频率代替。图图 7.18 20log|H()|和和()的表示的的表示的 幅度响应和相位响应。幅度响应和相位响应。转成转成 20log|H(f)|f 和和(f)f 的表示的幅度的表示的幅度 响应和相位响应响应和相位响应FIGURE 7-19 Frequency response plotted against frequency in Hz.Joyce Van de VegteFundamentals of Digital Signal ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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由极零点确定滤波器形状考虑如下传输函数：考虑如下传输函数：该滤波器的频率响应为该滤波器的频率响应为幅度响应（或滤波器形状）为：幅度响应（或滤波器形状）为：对于特定的对于特定的 ，与与p之间的距离越小，其幅度之间的距离越小，其幅度响应越大。当响应越大。当 沿单位圆移动，最靠近极点沿单位圆移动，最靠近极点p时，幅度响应为最大值，即时，幅度响应为最大值，即 和极点和极点p的相位相符时的相位相符时,可获得最大幅度可获得最大幅度.而且极而且极点位置越靠近单位圆点位置越靠近单位圆,这个最大值就越大这个最大值就越大.FIGURE 7-25 Graphical view of filter shape.Joyce Van de VegteFundamentals of Digital Signal ProcessingCopyright 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由上可扩展到具有多个极零点的滤波器对于对于 弧度的频率弧度的频率,离滤波器极点越近离滤波器极点越近,离零离零点越远点越远,则幅度就越大则幅度就越大.同样同样,靠近单位圆的极靠近单位圆的极点点,将导致滤波器形状在某一频率上有非常大将导致滤波器形状在某一频率上有非常大的幅值的幅值,而靠近单位圆的零点将导致滤波器形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状在某一频率上有非常小的幅值.这个幅值大这个幅值大小的剧烈变化可增强滤波器的选择性小的剧烈变化可增强滤波器的选择性.例例:推断滤波器的形状推断滤波器的形状,滤波器的传输函数滤波器的传输函数为为FIGURE 7-26 Pole-zero plot for Example 7.21.Joyce Van de VegteFundamentals of Digital Signal ProcessingCopyright 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 ProcessingCopyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.例例:推断滤波器的形状推断滤波器的形状,滤波器的差分方程滤波器的差分方程为为yn=xn-1+xn-3FIGURE 7-28 Pole-zero plot for Example 7.22.Joyce Van de VegteFundamentals of Digital Signal ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 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 ProcessingCopyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.7.3.5一阶滤波器一阶滤波器 可正可负可正可负,其符号对特性有很大影响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 ProcessingCopyright 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 ProcessingCopyright 2002 by Pearson Education,Inc.Upper Saddle River,New Jersey 07458All rights reserved.7.3.6二阶滤波器二阶滤波器CHAPTER SUMMARYThe 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.