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时间:2019-03-07
《信号与系统、第四章习题解答new》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、4-2解题过程:(1)由ft()=−sinwtututT⎡⎤⎣⎦()(−/2)=+sinwtut()sin[wtT(−−/2)]utT()/2得LL⎡⎤⎡⎣⎦⎣ft()=+sinwtut()⎤⎦L{sin[wtT(−/2)]utT(−/2)}2sTωω−=+e22222ss++ωωsTω⎛⎞−=+⎜⎟1e222s+ω⎝⎠(2)由sin()wt+=ϕsinwtcosϕϕ+coswtsin得LLL⎡⎤⎣⎦sin()(wt+=ϕsinwtcosϕϕ)+(coswtsin)ωϕcosssinϕ=+2222ss++ωωωϕcos+ssinϕ=22s+ω4-3解题过程:−−(t2)−2(1)由
2、f()teute=−()2⋅得⎡⎤=−−2⎡⎤−−()t2LL⎣⎦ft()e⎣⎦eut()2−−221s=⋅⋅ees+11−+2(s1)=es+11−2s(2)L⎡⎤⎣⎦f()te=s+1−t2(3)由f()teute=⋅()得22−teLL⎡⎤⎣⎦ft()==e⎣⎦⎡⎤eut()s+1(4)由ft()=−sin2⎡⎤⎣⎦(t1)+−2ut(1)=−cos2sin2⎡⎤⎣⎦()tu1(t−1)+−sin2cos2⎡⎤⎣⎦(tu1)(t−1)得1LL⎡⎤⎣⎦ft()=−cos2{sin2⎡⎤⎣⎦(t1)ut(−1)}+−sin2L{sin2⎡⎤⎣⎦(t1)()ut−1}2cos2−−
3、ssssin2=+ee22ss++442cos2+ssin2−s=e2s+4(5)由ft()(=−t11222)(ut−−−)(t)ut(−−−)ut()得LL⎡⎤⎡⎣⎦⎣ft()=−−(t11)(ut)⎤⎡⎦⎣−−−L(t22)ut()(⎤⎡⎦⎣−−Lut2)⎤⎦111−−−ss2s=−−eee22sss1−s=−⎡⎤11()+se2⎣⎦s4-4解题过程:−−1⎛⎞1t(1)L⎜⎟=e⎝⎠s+1⎡⎤342−−11⎡⎤42⎢⎥−2t(2)由=得LL==⎢⎥2e23s+3⎢⎥⎣⎦23s+3s+⎢⎥s+2⎣⎦2⎛⎞⎡⎤44⎜⎟11−−11⎡⎤441⎡⎤41−1⎢⎥(3)由=−⎜⎟得LL
4、⎢⎥=−⎢⎥L⎢⎥ss()233+⎜⎟s+3⎣⎦ss()233+⎣⎦s3⎢⎥+3ss⎝⎠2⎣⎦211⎛⎞1s(4)由=−⎜⎟得22ss(5++)5⎝⎠ss5−−11⎡⎤111⎡⎤1−1⎡s⎤1LL⎢⎥=−L=()1cos5−t22⎢⎥⎢⎥⎣⎦ss(5++)5⎣⎦s5⎣s55⎦33⎛⎞11(5)由=−⎜⎟得()ss++4(2)2⎝⎠ss++24−−11⎡⎤331⎡⎤31−1⎡⎤3−2tt−4LL⎢⎥=−=⎢⎥L⎢⎥()ee−⎣⎦()ss++4(2)22⎣⎦s+24⎣⎦s+236s3(6)由=−得()ssss++++4(2)422−−11⎡⎤36s⎡⎤⎡⎤−13−4tt−2LL⎢⎥=−
5、=L63ee−⎢⎥⎢⎥⎣⎦()ss++4(2)⎣⎦⎣⎦s+4s+2−1⎡⎤1(7)L+=+1sin(ttδ)⎢⎥2⎣⎦s+1111(8)由=−得2ssss−+32−21−−−11⎡⎤11⎛⎞−1⎛1⎞2ttLL=−=⎜⎟⎜⎟Lee−⎢⎥2⎣⎦ss−+32⎝s−2⎠⎝s−1⎠⎡⎤t111−11−(9)由=−得L⎢⎥=−1eRCsRCs()+1s1⎣⎦sRCs()+1s+RC⎡⎤−t11−RCs2−11RCs−(10)由=−得L⎢⎥=−12eRCsRCs()+1s1⎣⎦sRCs()+1s+RC⎛⎞1⋅wwR11Cws⎜⎟RCw(11)由⋅=⎜⎟−+得221121122swR++()Cs
6、+()RCw⎜⎟ss++sw+⎝⎠RCRC⎡⎤⎡t⎤−1wR11Cw−L⎢⎥⋅=⎢ewRC−costw+sint⎥222⎣⎦swR++()Cs11+()RCw⎣RCw⎦45s+73−13⎡⎤45s+−−tt2(12)由=−得L=−73ee2⎢⎥2ssss++56+3+2⎣⎦ss++56100()ss++50100(50)(13)由=得2ss++201200()ss++1(200)⎡⎤100()s+50100−−1tt−200L⎢⎥=+()49ee1502⎣⎦ss++201200199s+3kkkk1234(14)令=+++332()ss++12()ss++21()ss++11()s
7、+3s+3则k==−1,k==2,132()s+1s+1s=−1s=−22ds⎛⎞+3ds⎛⎞+3k==⎜⎟−1,k=⎜⎟=1342dss⎝⎠+1ds⎝⎠s+1s=−1s=−13s+−31211从而=+−+332()ss++12()ss++21()ss++11()⎡⎤s+3−−12tt−所以L⎢⎥=−ette+()−+13⎢⎥⎣⎦()ss++12()AAK−1⎡⎤AA(15)由=⋅得L=sinKt2222⎢⎥22sKKsK++⎣⎦sKK+⎡⎤(16)由于L−1⎢⎥s=
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