第-7-章-傅立葉變換與濾波器形狀課件

第7章傅立葉變換與濾波器形狀

CH7FOURIERTRANSFORMSANDFILTERSHAPE7.1傅立葉變換基礎(FOURIERTRANSFORMBASICS)7.2頻率響應及其他形式(FREQUENCYRESPONSESANDOTHERFORMS)7.3頻率響應和濾波器形狀(FREQUENCYRESPONSEANDFILTERSHAPE)返回專業(yè)詞匯傅立葉變換：FourierTransform濾波器形狀：filtershape頻率響應:frequencyresponse頻率特性：frequencycharacteristics離散時間傅立葉變換：DiscreteTimeFourierTransform幅度響應：magnituderesponse相位響應：phaseresponse傳輸函數(shù)：transferfunction相位差：phasedifference采樣頻率：samplingfrequency

7.1傅立葉變換基礎

7.1FOURIERTRANSFORMBASICS

離散時間傅立葉變換(DTFT)將信號或濾波器由時域頻域研究其頻率特性

frequencycharacteristics

magnituderesponse

phaseresponse對于濾波器DTFT得到的信息稱為濾波器的頻率響應

frequencyresponse幅度響應相位響應信號x[n]的離散(discrete)時間(time)傅立葉變換(Fouriertransform)定義為

F{x[n]}=x(Ω)=∑

x[n]e-jnΩ=∑

x[n](cos(nΩ)–jsin(nΩ))Ω:數(shù)字頻率，Ω=2π

f/fs

弧度

Ω不同，變換x(Ω)不同，當x[n]以接近頻率Ω變化時，x(Ω)較大，離散時間傅立葉變換x(Ω)反應了信號的頻率。

∞N=-∞

∞N=-∞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.FIGURE7-1SignalresonanceforthediscretetimeFouriertransform.JoyceVandeVegte

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Allrightsreserved.例7.1求圖7.2所示信號的離散時間傅立葉變換。圖7.2解：只有4個非零采樣值(n=0，1，2，4)對變換有貢獻，因而：

X(Ω)=∑x[n]e-jnΩ

=2-e-jΩ+3e-j2Ω+e-j4Ω一般情況下，DTFT是復值。

∞N=-∞例7.2求信號x[n]=4(u[n]–u[n-3])的DTFT。解：在n<0和n≥

時，信號值都是零，所以：

X(Ω)=∑x[n]e-jnΩ

=4+4e-jΩ+4e-j2Ω

∞N=-∞離散時間傅立葉變換的兩個重要特性周期性(periodicity)延時性(timedelay)延時性：假設x[n]的DTFT存在，為X(Ω)，則

x[n–n0]的DTFT為∑

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=-∞時域中延遲n0

在頻率中引入一個復指數(shù)e-jn0Ω周期性：X(Ω+2π)=∑

x[n]e-jn(Ω+2π)

=

∑

x[n]e-jnΩ?

e-jn2π

∞N=-∞

∞N=-∞歐拉公式：e-jn2πn

=cos(2πn)–jsin(2πn)=1

∴X(Ω+2π)=∑x[n]e-jnΩ=X(Ω)

∴DTFT是周期的，周期為2π

，也就是DTFT對于所有的Ω，每2π

重復一次。

∞N=-∞返回7.2頻率響應及其他形式

7.2FREQUENCYRESPONSESANDOTHERFORMS

7.2.1頻率響應和差分方程a0y[n]+a1y[n–1]+a2y[n–2]+…+aNy[n–N]=b0x[n]+b1x[n–1]+…+bMx[n–M]每一項進行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-jNΩ例7.4求出與如下差分方程相對應的頻率響應：

y[n]+0.1y[n–1]+0.85y[n–2]=x[n]–0.3x[n–1]解：容易確定系數(shù)(coefficients)為a0=1，a1=0.1，a2=0.85，b0=1，b1=-0.3。濾波器的頻率響應為：H(Ω)==Y(Ω)X(Ω)b0+b1e-jΩ

+…+bMe-jMΩa0+a1e-jΩ+…+aNe-jNΩ=1–0.3e-jΩ1+0.1e-jΩ+0.85e-j2Ω7.2.2頻率響應和傳輸函數(shù)對照式6.2傳輸函數(shù)，頻率響應是把傳輸函數(shù)中所有z-1

換為e-jΩ

。例7.5求濾波器的頻率響應，它的傳輸函數(shù)(transferfunction)是：H(z)=

1–0.2z-21+0.5z-1+0.9z-2解：頻率響應為：H(Ω)=1–0.2e-j2Ω1+0.5e-jΩ+0.9e-j2Ω7.2.3頻率響應和脈沖響應(impulseresponse)

圖7.3描述濾波器方法濾波器的傳輸函數(shù)H(z)是脈沖響應h[n]的z變換

x[n]δ[n]X(Ω)=∑

δ[n]e-jnΩ=1

y[n]h[n]Y(Ω)=∑

h[n]e-jnΩ

H(Ω)==Y(Ω)

∴頻率響應H(Ω)與脈沖響應h[n]的DTFT一樣。

∞N=-∞

∞N=-∞Y(Ω)X(Ω)例7.6數(shù)字濾波器的脈沖響應為：

h[n]=5δ[n]–δ[n–1]+0.2δ[n–2]–0.4δ[n–3]

求濾波器的頻率響應的表達式。

解：頻率響應是脈沖響應DTFT，見式(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頻率響應和濾波器的形狀

7.3FREQUENCYRESPONSEANDFILTERSHAPE7.3.1濾波器對正弦輸入的作用。

FilterEffectsonSineWaveInputsY(Ω)=H(Ω)?X(Ω)y[n]=F-1{Y(Ω)}對于DTFT方法一般僅求取正弦輸入時的輸出。頻率響應H(Ω)是復數(shù)，可用極坐標(polarform)H(Ω)=|H(Ω)|e-jΘ(Ω)

表示（附錄A）|H(Ω)|是數(shù)字濾波器在數(shù)字頻率

Ω處的增益(gain)。（無單位，或dB20log|H(Ω)|)Θ(Ω)是數(shù)字濾波器在數(shù)字頻率

Ω處的相位差(phasedifference)。（弧度或度）在每給定一個頻率，增益和相位差可用來預測濾波器的響應。增益是對輸入的放大量(amplification)相位差決定了輸入的相位變化。

Y(Ω)=|Y(Ω)|e-jΘy(Ω)

=|X(Ω)|e-jΘx(Ω)|H(Ω)|e-jΘ(Ω)

=|X(Ω)||H(Ω)|e-j(Θx(Ω)+Θ(Ω))幅值計算不能用分貝，都要轉(zhuǎn)成線形計算。對于任一給定頻率Ω，輸出的幅度是濾波器的增益和輸入幅度的積，輸出的相位是濾波器的相位差和輸入相位的和。例：數(shù)字頻率為1.5弧度的余弦波通過濾波器，在此頻率下，濾波器增益為-21dB，相位差為86o，如果輸入幅度為20，相位為12o，則輸出幅度和相位是多少？解：輸入簡式為，這是余弦信號的縮寫，在1.5弧度處，濾波器增益為-21dB，但這個值不能用于計算，必須用轉(zhuǎn)換為線性值。因為相位差為86o，頻率響應的簡式為。輸出是頻率響應和輸入信號在傅里葉變換域的乘積：

7.3.2幅度響應和相位響應

數(shù)字頻率Ω處的頻率響應H(Ω)用極坐標形式

H(Ω)=|H(Ω)|e-jΘ(Ω)

所有數(shù)字頻率處的增益的集合稱為濾波器的幅度響應所有數(shù)字頻率處的相位差的集合稱為濾波器的相位響應數(shù)字濾波器的頻率響應。例：一系統(tǒng)的頻率響應為求該系統(tǒng)的幅度響應和相位響應，并畫出圖。幅度響應是增益對數(shù)字頻率（弧度)的關系圖.相位響應是相位(弧度)對數(shù)字頻率(弧度)的關系圖.數(shù)字頻率范圍是解:弧度間隔選擇是任意的,要獲得更高的準確度可選用更小的間隔.對于弧度,采用非極坐標計算:

幅度響應是周期性的,每2弧度重復一次,幅度響應是偶函數(shù)相位響應是周期性的,每2弧度重復一次,幅度響應是奇函數(shù)所以,一般沒有必要記錄幅度響應和相位響應在左邊部分,而且,考慮的值沒有實際意義,所以實際數(shù)字濾波器的頻率響應一般畫出部分.幅度響應:線性增益對數(shù)字頻率的曲線畫出;或者對數(shù)形式對數(shù)字頻率的曲線.后者的優(yōu)點是在增益變化范圍非常大時,可以方便地畫在一個圖上.改變了圖的形狀相位響應: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模擬頻率f與數(shù)字頻率Ω數(shù)字濾波器的形狀|H(Ω)|設計可不依賴采樣頻率(samplingfrequency)，但所選的采樣頻率將影響濾波器輸入頻率的范圍。當采樣頻率fs

已知，可用模擬f(Hz)代替數(shù)字頻率

Ω(弧度)。

Ω=2π

f/fsf=Ωfs/2π

以數(shù)字頻率Ω表示的數(shù)字頻率特性，只有當采樣頻率選定后才能確定。根據(jù)上式，可將0~π弧度的數(shù)字頻率用0~fs/2Hz的模擬頻率代替。圖7.1820log|H(Ω)|~Ω和Θ(Ω)~Ω的表示的幅度響應和相位響應。轉(zhuǎn)成20log|H(f)|~f和Θ(f)~f的表示的幅度響應和相位響應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由極零點確定濾波器形狀考慮如下傳輸函數(shù)：該濾波器的頻率響應為幅度響應（或濾波器形狀）為：對于特定的，與p之間的距離越小，其幅度響應越大。當沿單位圓移動，最靠近極點p時，幅度響應為最大值，即和極點p的相位相符時,可獲得最大幅度.而且極點位置越靠近單位圓,這個最大值就越大.FIGURE7-25Graphicalviewoffiltershape.JoyceVandeVegte

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Allrightsreserved.由上可擴展到具有多個極零點的濾波器對于弧度的頻率,離濾波器極點越近,離零點越遠,則幅度就越大.同樣,靠近單位圓的極點,將導致濾波器形狀在某一頻率上有非常大的幅值,而靠近單位圓的零點將導致濾波器形狀在某一頻率上有非常小的幅值.這個幅值大小的劇烈變化可增強濾波器的選擇性.例:推斷濾波器的形狀,濾波器的傳輸函數(shù)為FIGURE7-26Pole-zeroplotforExample7.21.JoyceVandeVegte

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Allrightsreserved.FIGURE7-27FiltershapeforExample7.21.JoyceVandeVegte

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Allrightsreserved.例:推斷濾波器的形狀,濾波器的差分方程為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.7.3.5一階濾波器可正可負,其符號對特性有很大影響.

FIGURE7-32FrequencyresponseforExample7.24.JoyceVandeVegte

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Allrightsreserved.FIGURE7-33FrequencyresponseforExample7.25.JoyceVandeVegte

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Allrightsreserved.7.3.6二階濾波器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)

Thefrequencyresponsecan