中南大学课件

中南大学课件

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时间:2019-06-20

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1、DiscreteLTIsystems:theconvolutionsumConsidersignalx[n]:Anarbitrarysequencecanberepresentedasalinearcombinationofshiftedunitimplulses[n-k],wheretheweightsinthislinearcombinationarex[k].Writeas:……-4-3-2-101234nx[n]DiscreteLTIsystems:theconvolutionsum(s

2、iftingproperty—筛选性)DiscreteLTIsystems:theconvolutionsumLeth[n]denotetheresponseoflinearsystemto[n].i.e.h[n]—theunitimpulseresponse.then,each[n]ofx[n]response:….….DiscreteLTIsystems:theconvolutionsum….….Thisresultisreferredtoastheconvolutionsum,andt

3、heoperationontheright-handsideofEQ.isknownastheconvolutionsumofx[n]andh[n].DiscreteLTIsystems:theconvolutionsumWerepresenttheoperationasy[n]=x[n]h[n](2.7)Thesame,isreferredtoastheconvolutionsumofx[n]and[n].Somenotes:AnLTIsystemiscompletelycharacteri

4、zedbyitsh[n].DiscreteLTIsystems:theconvolutionsumThegraphofconvolutionsum.Transformindependentvariable:x[n],h[n]x[k],h[k]andh[k]h[-k]Shifth[-k]nstepsh[n-k].Foranyn,x[k]multipliedbyh[n-k]andx[k]·h[n-k].DiscreteLTIsystems:theconvolutionsumExample2.2

5、determiney[n]=x[n]h[n].Answer:(a)DiscreteLTIsystems:theconvolutionsumandh[k]h[-k](b)Shifth[-k]totheright(n>0)ortotheleft(n<0).n<0,y[n]=0n=0,y[0]=x[0]h[0]=0.51=0.5n=1,y[1]=x[0]h[1]+x[1]h[0]=0.5+2=2.5DiscreteLTIsystems:theconvolutionsum(c)Foranypart

6、icularvalueofn,wemultiplythesetwosignalsandsumoverallvaluesofk.n=2,y[2]=x[0]h[2]+x[1]h[1]=0.5+2=2.5n=3,y[3]=x[1]h[2]=2n4,y[n]=0DiscreteLTIsystems:theconvolutionsumory[n]={0.5,2.5,2.5,2}n=0,1,2,3,Themainstepsofconvolutionsumgraph:reversalshiftmultiply

7、sum(3)Calculationofconvolutionsumcanbedonebyuprightmultiplication.DiscreteLTIsystems:theconvolutionsumh[n]{0.52}(0)x[n]{111}(0)0.520.520.52y[n]{0.52.52.52}(0)DiscreteLTIsystems:theconvolutionsumExample2.3x[n]andh[n]givenbydeterminey[n].Answer:forn0(S

8、eeproblem1.54)ContiuousLTIsystems:theconvolutionintegralWebeginbyconsideringapulseor“staircase”approximation,x(t)toacontinuoussignalx(t).IfdefineContiuousLTIsystems:theconvolutionintegralThen,,andtheshadepulseis,isAs,writtenas……tkx(k)Contiu

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