第04章_连续时间信号的采样

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1、Chapter4samplingofcontinous-timesignals4.1periodicsampling4.2discrete-timeprocessingofcontinuous-timesignals4.3continuous-timeprocessingofdiscrete-timesignal4.4digitalprocessingofanalogsignals4.5changingthesamplingrateusingdiscrete-timeprocessing4.5.1samplingratereductionbyaninteg

2、erfactor(downsampling,decimation)4.5.2increasingthesamplingratebyanintegerfactor(upsampling,interpolation)4.5.3changingthesamplingratebyanonintegerfactor4.5.4applicationofmultiratesignalprocessing4.1periodicsampling1.idealsample（理想采样）[]T：sampleperiodxn=x(t)=x(nT)ct=nTcfs=1/T:sampl

3、eratesamples/secondΩs=2π/T:samplerate,radians/secondWerefertoasystemthatimplementstheoperationoftheaboveasanidealcontinuous-to-discrete-time(C/D)converter（理想连续时间到离散时间的转换器）Figure4.1idealcontinuous-time-to-discrete-time(C/D)converterItisconvenienttorepresentthesamplingprocessmathema

4、ticallyinthetwostages.时间轴归一化tÆt/T=n∞=∑δ(t−nT)n=−∞Figure4.2(a)mathematicmodelforidealC/DLetusnowconsidertheFouriertransformof.x(t)sSinceitistheproductofand,it’sFourierxc(t)s(t)transformistheconvolutionoftheFouriertransformXjc()Ωand.Soitcanbeexpressedas:Sj()Ωc∞1Xs(jΩ)=∑Xc(j(Ω−kΩs))T

5、k=−∞For,itsFouriertransformis:x[n]∞1X(ejω)=X(jΩ)

6、=X(j(−k2)/T)∑ωπsΩ=ω/TcTk=−∞证明见课堂笔记理想采样频域变化X(jΩ)=s∞1Ω−Ω≥Ω∑Xc(j(Ω−kΩs))sNNTk=−∞NoaliasingΩ−Ω<ΩsNNaliasingΩs/22π折叠频率πω=ΩTFigure4.3从时域理解以2π为周期：频率为2.1π和0.1π不可区分从时域理解高频折叠成低频：频率为1.1π和0.9π不可区分2.idealreconstruction（理想重构）Iftheconditionsofthes

7、amplingtheoremaremet(noalias),thenthecontinuous-timesignalcanbereconstructedfromitssample.Theidealbandlimitedsignalreconstructionsystemisreferredtoasidealdiscrete-time-to-continuous(D/C)converter（理想离散时间到连续时间的转换器）.Figure4.10(b)idealD/CconverterPrincipleofidealreconstructioninfreque

8、ncydomain.=Ω/2sTheidealreconstructionfilterisalow-passcontinuous-timefilterwithfrequencyresponse:⎧T

9、Ω

10、≤ΩcHr(jΩ)=⎨⎩0

11、Ω

12、>ΩcFigure4.10(a)mathematicmodelforidealD/C⎧T

13、Ω

14、≤ΩcHr(jΩ)=⎨⎩0

15、Ω

16、>ΩcX(jΩ)=X(jΩ)H(jΩ)=X(jΩ)rsrcTheinput/outputrelationofFigure4.7inthetimedomainis:∞sin[π(t−nT)/T]xr(t

17、)=∑x[n]n=−∞π(t−nT)/TProve:Xj()()(