傅立叶变换与滤波器形状课件

第7章傅立叶变换与滤波器形状

CH7FOURIERTRANSFORMSANDFILTERSHAPE7.1傅立叶变换基础(FOURIERTRANSFORMBASICS)7.2频率响应及其他形式(FREQUENCYRESPONSESANDOTHERFORMS)7.3频率响应和滤波器形状(FREQUENCYRESPONSEANDFILTERSHAPE)返回专业词汇傅立叶变换：FourierTransform滤波器形状：filtershape频率响应:frequencyresponse频率特性：frequencycharacteristics离散时间傅立叶变换：DiscreteTimeFourierTransform幅度响应：magnituderesponse相位响应：phaseresponse传输函数：transferfunction相位差：phasedifference采样频率：samplingfrequency

7.1傅立叶变换基础

7.1FOURIERTRANSFORMBASICS

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

frequencycharacteristics

magnituderesponse

phaseresponse对于滤波器DTFT得到的信息称为滤波器的频率响应frequencyresponse幅度响应相位响应FIGURE7-1SignalresonanceforthediscretetimeFouriertransform.JoyceVandeVegte

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Allrightsreserved.FIGURE7-1SignalresonanceforthediscretetimeFouriertransform.JoyceVandeVegte

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Allrightsreserved.FIGURE7-1SignalresonanceforthediscretetimeFouriertransform.JoyceVandeVegte

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Allrightsreserved.FIGURE7-1SignalresonanceforthediscretetimeFouriertransform.JoyceVandeVegte

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Allrightsreserved.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(Ω)=∑x[n]e-jnΩ=2-e-jΩ+3e-j2Ω+e-j4Ω一般情况下，DTFT是复值。

∞N=-∞离散时间傅立叶变换的两个重要特性周期性(periodicity)延时性(timedelay)延时性：假设x[n]的DTFT存在，为X(Ω)，则x[n–n0]的DTFT为∑x[n–n0]e-jnΩ。

∞N=-∞令m=n–n0,n=m+n0∑x[m]e-j(m+n0)Ω=e-jn0Ω∑x[m]e-jmΩ=e-jn0ΩX(Ω)

∞N=-∞

∞N=-∞时域中延迟n0在频率中引入一个复指数e-jn0Ω周期性：X(Ω+2π)=∑x[n]e-jn(Ω+2π)

=

∑x[n]e-jnΩ•

e-jn2π

∞N=-∞

∞N=-∞欧拉公式：e-jn2πn

=cos(2πn)–jsin(2πn)=1

∴X(Ω+2π)=∑x[n]e-jnΩ=X(Ω)

∴DTFT是周期的，周期为2π

，也就是DTFT对于所有的Ω，每2π

重复一次。

∞N=-∞返回7.2频率响应及其他形式

7.2FREQUENCYRESPONSESANDOTHERFORMS

7.2.1频率响应和差分方程a0y[n]+a1y[n–1]+a2y[n–2]+…+aNy[n–N]=b0x[n]+b1x[n–1]+…+bMx[n–M]每一项进行DTFTa0Y(Ω)+a1e-jΩY(Ω)+a2e-j2ΩY(Ω)+…+aNe-jNΩY(Ω)=b0X(Ω)+b1e-jΩX(Ω)+…+bMe-jMΩX(Ω)H(Ω)==Y(Ω)X(Ω)b0+b1e-jΩ

+…+bMe-jMΩa0+a1e-jΩ+…+aNe-jNΩ7.2.2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-jΩ。例7.5求滤波器的频率响应，它的传输函数(transferfunction)是：H(z)=1–0.2z-21+0.5z-1+0.9z-2解：频率响应为：H(Ω)=1–0.2e-j2Ω1+0.5e-jΩ+0.9e-j2Ω7.2.3频率响应和脉冲响应(impulseresponse)

图7.3描述滤波器方法例7.6数字滤波器的脉冲响应为：h[n]=5δ[n]–δ[n–1]+0.2δ[n–2]–0.4δ[n–3]

求滤波器的频率响应的表达式。

解：频率响应是脉冲响应DTFT，见式(7.3)，因而得：

H(Ω)=∑

h[n]e-jnΩ=5–e-jΩ+0.2e-j2Ω–0.04e-j3Ω

∞N=-∞返回7.3频率响应和滤波器的形状

7.3FREQUENCYRESPONSEANDFILTERSHAPE7.3.1滤波器对正弦输入的作用。FilterEffectsonSineWaveInputsY(Ω)=H(Ω)•X(Ω)y[n]=F-1{Y(Ω)}对于DTFT方法一般仅求取正弦输入时的输出。频率响应H(Ω)是复数，可用极坐标(polarform)H(Ω)=|H(Ω)|e-jΘ(Ω)

表示（附录A）|H(Ω)|是数字滤波器在数字频率Ω处的增益(gain)。（无单位，或dB20log|H(Ω)|)Θ(Ω)是数字滤波器在数字频率Ω处的相位差(phasedifference)。（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。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86º，如果输入幅度为20，相位为12º，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-21dB，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86º，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Ω处的频率响应H(Ω)用极坐标形式H(Ω)=|H(Ω)|e-jΘ(Ω)

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。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.FIGURE7-7FrequencyresponseforExample7.10.JoyceVandeVegte

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Allrightsreserved.FIGURE7-8FrequencyresponseforExample7.11.JoyceVandeVegte

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Allrightsreserved.FIGURE7-8frequencyresponseforExample7.11.JoyceVandeVegte

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Allrightsreserved.FIGURE7-9FrequencyresponseforExample7.12.JoyceVandeVegte

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Allrightsreserved.FIGURE7-10Frequencyresponsesforcommonfilters.JoyceVandeVegte

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Allrightsreserved.FIGURE7-10Frequencyresponsesforcommonfilters.JoyceVandeVegte

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Allrightsreserved.FIGURE7-10Frequencyresponsesforcommonfilters.JoyceVandeVegte

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Allrightsreserved.FIGURE7-10Frequencyresponsesforcommonfilters.JoyceVandeVegte

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Allrightsreserved.FIGURE7-11FrequencyresponseforExample7.13.JoyceVandeVegte

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Allrightsreserved.FIGURE7-11FrequencyresponseforExample7.13.JoyceVandeVegte

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Allrightsreserved.FIGURE7-12FrequencyresponseforExample7.14.JoyceVandeVegte

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Allrightsreserved.FIGURE7-13FrequencyresponseforExample7.15.JoyceVandeVegte

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Allrightsreserved.FIGURE7-13FrequencyresponseforExample7.15.JoyceVandeVegte

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Allrightsreserved.FIGURE7-14MagnituderesponseofcombfilterforExample7.16.JoyceVandeVegte

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Allrightsreserved.FIGURE7-15Pulsepassedthroughcombfilter.JoyceVandeVegte

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Allrightsreserved.FIGURE7-15Pulsepassedthroughcombfilter.JoyceVandeVegte

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Allrightsreserved.FIGURE7-16“Hello”passedthroughcombfilter.JoyceVandeVegte

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Allrightsreserved.FIGURE7-16“Hello”passedthroughcombfilter.JoyceVandeVegte

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Allrightsreserved.FIGURE7-17MagnituderesponseofcombfilterforExample7.17.JoyceVandeVegte

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Allrightsreserved.7.3.3模拟频率f与数字频率Ω数字滤波器的形状|H(Ω)|设计可不依赖采样频率(samplingfrequency)，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Ω(弧度)。

Ω=2πf/fsf=Ωfs/2π

以数字频率Ω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~π弧度的数字频率用0~fs/2Hz的模拟频率代替。图7.1820log|H(Ω)|~Ω和Θ(Ω)~Ω的表示的幅度响应和相位响应。转成20log|H(f)|~f和Θ(f)~f的表示的幅度响应和相位响应FIGURE7-19FrequencyresponseplottedagainstfrequencyinHz.JoyceVandeVegte

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Allrightsreserved.FIGURE7-20MagnituderesponseforExample7.19.JoyceVandeVegte

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Allrightsreserved.FIGURE7-21Magnituderesponseforfs=4kHzforExample7.19.JoyceVandeVegte

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Allrightsreserved.FIGURE7-22Magnituderesponseforfs=10kHzforExample7.19.JoyceVandeVegte

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Allrightsreserved.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.FIGURE7-25Graphicalviewoffiltershape.JoyceVandeVegte

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Allrightsreserved.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FIGURE7-26Pole-zeroplotforExample7.21.JoyceVandeVegte

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Allrightsreserved.FIGURE7-27FiltershapeforExample7.21.JoyceVandeVegte

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Allrightsreserved.例:推断滤波器的形状,滤波器的差分方程为y[n]=x[n-1]+x[n-3]FIGURE7-28Pole-zeroplotforExample7.22.JoyceVandeVegte

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Allrightsreserved.FIGURE7-29FiltershapeforExample7.22.JoyceVandeVegte

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Allrightsreserved.FIGURE7-30Pole-zeroplotforExample7.23.JoyceVandeVegte

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Allrightsreserved.FIGURE7-31FiltershapesforExample7.23.JoyceVandeVegte

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Allrightsreserved.7.3.5一阶滤波器可正可负,其符号对特性有很大影响.

FIGURE7-32FrequencyresponseforExample7.24.JoyceVandeVegte

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Allrightsreserved.FIGURE7-33FrequencyresponseforExample7.25.JoyceVandeVegte

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Allrightsreserved.7.3.6二阶滤波器CHAPTERSUMMARYThediscretetimeFouriertransform(DTFT)ofasignalx[n]isgivenbyx(Ω)=∑x[n]e-jnΩ.Itreportsthefrequenciespresentinasignal.TheDTFTofasignalx[n]givesthesignal’sspectrumX(Ω).TheDTFTofasystemh[n]givesthesystem’sfrequencyresponseH(Ω).TheDTFTisperiodicwithperiod2π.Adifferenceequationcanbeexpressedasafrequencyresponse.Atransferfunctioncanbeexpressedasafrequencyresponse.ThefrequencyresponseH(Ω)istheDTFToftheimpulseresponseh[n].AfrequencyresponseH(Ω)isacomplexnumberandmaybeexpressedinpolarformintermsofagain|H(Ω)|andaphasedifferenceq(W)

asH(Ω)=|H(Ω)|ejq(W)

Thefrequencyresponsecanbeusedtofindafilter’soutputforasinusoidalinput.Theoutputisasinusoidwiththesamefrequencyastheinput,butwithanamplitudemultipliedbythegainofthefiltera