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(chǔ)(FOURIERTRANSFORMBASICS)7.2頻率響應(yīng)及其他形式(FREQUENCYRESPONSESANDOTHERFORMS)7.3頻率響應(yīng)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(yīng):frequencyresponse頻率特性：frequencycharacteristics離散時(shí)間傅立葉變換：DiscreteTimeFourierTransform幅度響應(yīng)：magnituderesponse相位響應(yīng)：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(chǔ)

7.1FOURIERTRANSFORMBASICS

離散時(shí)間傅立葉變換(DTFT)將信號(hào)或?yàn)V波器由時(shí)域頻域研究其頻率特性

frequencycharacteristics

magnituderesponse

phaseresponse對(duì)于濾波器DTFT得到的信息稱為濾波器的頻率響應(yīng)frequencyresponse幅度響應(yīng)相位響應(yīng)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(hào)的離散時(shí)間傅立葉變換。圖7.2解：只有4個(gè)非零采樣值(n=0，1，2，4)對(duì)變換有貢獻(xiàn)，因而：X(Ω)=∑x[n]e-jnΩ=2-e-jΩ+3e-j2Ω+e-j4Ω一般情況下，DTFT是復(fù)值。

∞N=-∞離散時(shí)間傅立葉變換的兩個(gè)重要特性周期性(periodicity)延時(shí)性(timedelay)延時(shí)性：假設(shè)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(shí)域中延遲n0在頻率中引入一個(gè)復(fù)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(duì)于所有的Ω，每2π

重復(fù)一次。

∞N=-∞返回7.2頻率響應(yīng)及其他形式

7.2FREQUENCYRESPONSESANDOTHERFORMS

7.2.1頻率響應(yīng)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(xiàng)進(jìn)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頻率響應(yīng)和傳輸函數(shù)對(duì)照式6.2傳輸函數(shù)，頻率響應(yīng)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(yīng)，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(yīng)為：H(Ω)=1–0.2e-j2Ω1+0.5e-jΩ+0.9e-j2Ω7.2.3頻率響應(yīng)和脈沖響應(yīng)(impulseresponse)

圖7.3描述濾波器方法例7.6數(shù)字濾波器的脈沖響應(yīng)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(yīng)的表達(dá)式。

解：頻率響應(yīng)是脈沖響應(yīng)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(yīng)和濾波器的形狀

7.3FREQUENCYRESPONSEANDFILTERSHAPE7.3.1濾波器對(duì)正弦輸入的作用。FilterEffectsonSineWaveInputsY(Ω)=H(Ω)?X(Ω)y[n]=F-1{Y(Ω)}對(duì)于DTFT方法一般僅求取正弦輸入時(shí)的輸出。頻率響應(yīng)H(Ω)是復(fù)數(shù)，可用極坐標(biāo)(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(wú)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(gè)頻率，增益和相位差可用來(lái)預(yù)測(cè)濾波器的響應(yīng)。增益是對(duì)輸入的放大量(amplification)相位差決定了輸入的相位變化。例：數(shù)字頻率為1.5弧度的余弦波通過(guò)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(jiǎn)式為，這是余弦信號(hào)的縮寫，在1.5弧度處，濾波器增益為-21dB，但這個(gè)值不能用于計(jì)算，必須用轉(zhuǎn)換為線性值。因?yàn)橄辔徊顬?6o，頻率響應(yīng)的簡(jiǎn)式為。輸出是頻率響應(yīng)和輸入信號(hào)在傅里葉變換域的乘積：

7.3.2幅度響應(yīng)和相位響應(yīng)

數(shù)字頻率Ω處的頻率響應(yīng)H(Ω)用極坐標(biāo)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(yīng)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(yīng)數(shù)字濾波器的頻率響應(yīng)。幅度響應(yīng)是周期性的,每2弧度重復(fù)一次,幅度響應(yīng)是偶函數(shù)相位響應(yīng)是周期性的,每2弧度重復(fù)一次,幅度響應(yīng)是奇函數(shù)所以,一般沒有必要記錄幅度響應(yīng)和相位響應(yīng)在左邊部分,而且,考慮的值沒有實(shí)際意義,所以實(shí)際數(shù)字濾波器的頻率響應(yīng)一般畫出部分.幅度響應(yīng):線性增益對(duì)數(shù)字頻率的曲線畫出;或者對(duì)數(shù)形式對(duì)數(shù)字頻率的曲線.后者的優(yōu)點(diǎn)是在增益變化范圍非常大時(shí),可以方便地畫在一個(gè)圖上.改變了圖的形狀相位響應(yīng):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(shè)計(jì)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(dāng)采樣頻率fs已知，可用模擬f(Hz)代替數(shù)字頻率Ω(弧度)。

Ω=2πf/fsf=Ωfs/2π

以數(shù)字頻率Ω表示的數(shù)字頻率特性，只有當(dāng)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(yīng)和相位響應(yīng)。轉(zhuǎn)成20log|H(f)|~f和Θ(f)~f的表示的幅度響應(yīng)和相位響應(yīng)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(diǎn)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(yīng)為幅度響應(yīng)（或?yàn)V波器形狀）為：對(duì)于特定的，與p之間的距離越小，其幅度響應(yīng)越大。當(dāng)沿單位圓移動(dòng)，最靠近極點(diǎn)p時(shí)，幅度響應(yīng)為最大值，即和極點(diǎn)p的相位相符時(shí),可獲得最大幅度.而且極點(diǎn)位置越靠近單位圓,這個(gè)最大值就越大.FIGURE7-25Graphicalviewoffiltershape.JoyceVandeVegte

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Allrightsreserved.由上可擴(kuò)展到具有多個(gè)極零點(diǎn)的濾波器對(duì)于弧度的頻率,離濾波器極點(diǎn)越近,離零點(diǎn)越遠(yuǎn),則幅度就越大.同樣,靠近單位圓的極點(diǎn),將導(dǎo)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(diǎn)將導(dǎo)致濾波器形狀在某一頻率上有非常小的幅值.這個(gè)幅值大小的劇烈變化可增強(qiáng)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(fù),其符號(hào)對(duì)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)

Thefrequencyresponsecanbeusedtofindafilter’soutputforasinusoidalinput.Theoutputisasinusoidwiththesamefrequencyastheinput,butwithanamplitudemultipliedbythegainofthefiltera