数字信号处理基础答案第五章sol

《数字信号处理基础答案第五章sol》由会员上传分享，免费在线阅读，更多相关内容在教育资源-天天文库。

1、Chapter5Solutions5.1(a)x[n]=d[n]+0.6065d[n–1]+0.3679d[n–2](b)Theonlynon-zerocomponentsinthesumarek=0,k=1andk=2.5.2Thesystem’sresponsetod[n]ish[n],whiletheresponsetod[n–1]ish[n–1].Therefore,whentheinputx[n]=d[n]+d[n–1]isappliedtothesystem,theoutputisy[n]

2、=h[n]+h[n–1].Thetwosignalsh[n]andh[n–1]canbesummedsamplebysample.h[n]nh[n–1]nh[n]+h[n–1]n5.3(a)h[n–2]h[n–2]n(b)h[n+1]h[n+1]n(c)h[–n]h[–n]n(d)h[3–n]h[3–n]n5.4Startingwithn=0,theinputx[n]hasthesamplevalues0,–1and–2.Startingwithn=0,theimpulseresponseh[n]ha

3、sthesamplevalues1,2,3and4.Thesesamplesareinsertedintoaconvolutiontable.Beyondn=5,alloutputsarezero.Thisisadirectresultofthefinitelengthsoftheinputandimpulseresponsesignals.x[k]0–1–2h[–k]4321y[0]=0h[1–k]4321y[1]=–1h[2–k]4321y[2]=–4h[3–k]4321y[3]=–7h[4–k]

4、4321y[4]=–10h[5–k]4321y[5]=–8h[6–k]4321y[6]=05.5(a)x[k]1234567h[–k]1000y[0]=0h[1–k]1000y[1]=0h[2–k]1000y[2]=0h[3–k]1000y[3]=1h[4–k]1000y[4]=2h[5–k]1000y[5]=3h[6–k]1000y[6]=4(b)Theimpulseresponseh[n]=d[n–k]delaysthesignalbyksteps.5.6(a)Alloutputsamplesbe

5、foren=0andaftern=8arezero.x[k]1111111h[–k]–0.2–0.41?y[0]=1.0h[1–k]–0.2–0.41?y[1]=0.6h[2–k]–0.2–0.41y[2]=0.4h[3–k]–0.2–0.41y[3]=0.4h[4–k]–0.2–0.41y[4]=0.4h[5–k]–0.2–0.41y[5]=0.4h[6–k]–0.2–0.41y[6]=0.4h[7–k]–0.2–0.41?y[7]=–0.6h[8–k]–0.2–0.41?y[8]=–0.2(b)T

6、heoutputsinfluencedbyboundaryeffectsareindicatedbyquestionmarks.5.7Theconvolutionoftheimpulseresponsewiththestepfunctionproducesthestepresponse.Thesamplesinthestepresponseare,ofcourse,cumulativesumsoftheimpulseresponsesamples.n0123456789h[n]1.0000.8500.

7、7230.6140.5220.4440.3770.3210.2730.232s[n]1.0001.8502.5733.1873.7094.1524.5304.8505.1235.354n10111213141516171819h[n]0.1970.1670.1420.1210.1030.0870.0740.0630.0540.046s[n]5.5515.7185.8615.9816.0846.1726.2466.3096.3636.408After17samples,thestepresponsesa

8、mpleschangebylessthan1%.Thispointmaybeconsideredthebeginningofanapproximatesteadystate.5.8(a)x[k]010–1010–10h[–k]1y[0]=0.0h[1–k]–0.61y[1]=1.0h[2–k]0.36–0.61y[2]=–0.6h[3–k]–0.2160.36–0.61y[3]=–0.64h[4–k]0.130–0.2160.36–0.61y[4]=0.