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当前位置:首页 > 行业资料 > 能源与动力工程 > 《信号与系统分析基础》第3章习题解答
第三章习题解答3.2求下列方波形的傅里叶变换。(a)解:110002()()11()2tjttjttjttjtjaFjftedteedtjettSej(b)解:200000022()111()[]1(1)1(1)tjttjtttjtjtttjtjtjtjtjtjttFedteedttdejjjteedtjeeeje(c)解:13112211()()22111()()2211()cos21()21()21112()2()22jtjtjtjtjtjtjtjtFtedteeedteedteejj()()()()22221111[][]2222jjjjeeeejj1tt01()ft1tt02()ft11t03()ft1cos2t2222sin()sin()cos()cos()cos2222()()2222(d)解:24222()()2222()()22()()()()2222()sin1()21()2112()2()sin[(22()2()TjtTTjtjtjtTTjtjtTTTjtjtTTTTTTjjjjFtedteeedtjeedtjeeTeeeejjjj)]sin[()]2()()Tjj3.3依据上题中a,b的结果,利用傅里叶变换的性质,求题图3.3所示各信号的傅里叶变换.(a)解:11111()()()ftftft11()ft就是3.2中(a)的1()ft如果1()()ftF,则1()()ftF111111111222()()()()()sin()42()[]sin()sin()2222jjaftftftFFtSeejj(b)解:2()()()ftgtgt,而()()2agtS2()(3)2()aaFSS如利用3.2中(a)的结论来解,有:211'()(3)(1)ftftft,其中,'2.3211'()()()(3)2()jjaaFeFeFSS(如()()ftF,则00()()jtftteF)12Tt03()ft2Tsint11tt01()ftt113t02()ft321221(c)解:32222()2()2(),1ftftft由3.2(b)知,2221()(1)jtjtFejte32222222222222()2()2(),1112(1)2(1)222222444cos(1cos)jtjtjjjjjjjjFeFeFeejeeejejeje(d)解:设,0()0,ttftelse由3.2知,21()(1)jtjtFejte而本题中,4()(0.5)(0.5)ftftft由傅里叶变换的尺度变换特性有:41()()()jbaftfatbeFa在本题中,a=0.5,b=0.4222222222222()2(2)2(2)21(21)(21)42()()22cos(2)sin(2)jtjtjtjtjtjtjtjtFFFejteejtejeejeejjj(e)解:设1,01()0,tftelse由3.2知,2()()2jaFSe根据5()ft的波形,将5()ft用()ft表示为0t3()ft2110t4()ft2t112t0t5()ft1111sin6t566()[()()]sin(6)1[()()]()2jtjtftftfttftfteej22[()()]()()()()2()cos222sincossin22222()2jjaaaftftFFSeeSS由频移特性.00()()jtfteF52222221()[2(6)2(6)]21sin(6)sin(6)[(6)sin(6)(6)sin(6)][]66(6)(sin6sinsin6sin)12sin3636aaFSSjjjjj(f)解:设3()()ftft2222224()(1cos)1sin[1cos(2)]248()2sin()sin()22FF61()()cos(10)()()2jwtjwtftfttftee利用频移特性有:6222211()(10)(10)22410410sin()sin()(10)2(10)2FFF3.4利用对称性求下列各函数的傅里叶变换.(1)sin2(2)(),(2)tfttt解:()2[2(2)]aftSt0t6()ft12213cos10()tft而()()2agtS或4()4(2)agtS由对称性,4411(2)2()()42aStgg224,22[2(2)]()0,2jjaeStge(2)222(),.fttt解:222te,由对称性,2222et(3)444444444244()[2][2]1[2]()21111()[2][2][()]*[()][()()]22282,()()0.22,()()2;26,()()aaaaaftStStStgftStStggggggggdggd解:而,利用数域卷积特性,得:积分:2444246.6,()()0gg3.82[()](),.1(2);()()(),[(-)()],1()()().11()2(2)22nnnFftFtftdFftFFjtftddFdFtftjjdddFtftjd若已知试求下列函数的频谱()由频域微分特性,如则当(2)(1)ft;1[()]().1,1.(1)()jbajFfatbeFaaabfteF在本题中,(3)()(2)()2()dFtftjFd(4)(1)(1)tft;解:由频域的微分特性,得:()()dFtftjd.由时移特性:()(1)(1)jdFtftjed.(5)()dfttdt;()()().()().()-[][()]()()[()]().nnndftjFdtdftjFdtdftdjtjFdtddftddFtFFdtdd由时域微分特性:而由频域的微分特性,有:(6)(25)ft;由1[()]()jbaFfatbeFaa,2,5,ab2.51(25)()22jfteF3.9计算下列各信号的傅里叶变换.(2)3()2(32)()2[2()]2uttutt3(3)2232()1,1[()]().2,3.112(32)21,()().21()2(32)()jbajjjtFfatbeFaaabteeutjuttej由(7)33(2)63(3)9[(2)(3)](2)(3)ttteututeuteeute33(2)23(3)31.311();(2)331(3)3tttjtjejeuteutejjeutej同理:362931[(2)(3)]()3tjjeututeeeej3.10利用傅里叶变换性质,求题图3.10所示函数的傅里叶逆变换。(a)解:00()0,()()0,jtjAeFFe000000000200000200222010200()(),()2(),222()2()()()()()()()(),()().[()][()]aaaajtjtjtagtSgtSStgStggFAegftFftteFFAegAStt利用对称性,有:在本题中,根据时移特性:有:()()F0A00t00000(b)020()2000,,0()(),00,jjjAeFFeAe0000000022000000()()()22()(),()2()2()()()222jjaaaFAgeAgegtSgStgStg由于为偶函数利用对称性,;00()(),()().jtftFfteF根据频移特性:0000000000000200021002200002222000022()()222()()222()()22()()2222()()()s222tjatjajjttjjjjaattjjaaStegStegFAgeAgeAeSteAeSteAjASteeSt020in()22sin()2tAtt补充题:求图示周期信号f(t)的傅里叶级数系数.解:02()ftt4321543211223,3TT;在一个周期内:2,20()22,01ttfttt()()F0A00t0000022111101220232111()(2)(22)339(1)4()jntjntjntnjnnFftedttedttedtTFen此题也可利用傅里叶级数的微分性质求解。将信号求导两次,则一个周期内:()''()3()31)gtfttt(其傅里叶级数系数为:122313[(1)()](1)jnjntnGttedteT则由关系21()nnGjnF,可得2329(1)4()jnnFen3.13已知阶跃函数和正弦、余弦函数的傅里叶变换如下:0000001[()]()[cos][()()][sin][()()]FutjFtFtj求单边正弦函数和单边余弦函数的傅里叶变换。解:1.单边正弦函数:0()sin()sfttut
本文标题:《信号与系统分析基础》第3章习题解答
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