西工大,西电第三章连续信号频域分析--答案

第三章习题3.1图题3.1所示矩形波，试将此函数)(tf用下列正弦函数来近似ntCtCtCtfnsin2sinsin)(21。0t1-1f(t)图题3.1答案任一函数在给定的区间内可以用在此区间的完备正交函数集表示，但若只取函数集中的有限项，或者正交函数集不完备，则只能得到近似的表达式。ntdtntdttfCn2sinsin)(由于分母与分母中的被积函数在区间),(内是偶函数，故有1)1(22sin2121cos1sinsin0020nnnnttntnntdtntdtC故得,0,34,0,44321CCCC3.2求图题3.2（a）所示周期锯齿波)(tf的傅里叶级数。tf(t)．．．．．．-2T-T0T2T3T1(a)图题3.2答案将)(tf求导得)('tf，)(''tf的波形分别如图3.2（b），（c）所示。图题3.2t(t)f．．．．．．(1)(1)(1)(1)(1)1/T(b)-2T-T0T2T(c)t(t)f．．．．．．-2T-T0T2T于是得)(''tf的傅立叶系数为2222'''2)(2)(2)(TTTTtjntjndtetTdtetfTjnA2222'220'0)(2)(2)()(2TTTTTTjnjndttTjndttTdttejnteTTjnTjn220故得)(tf的傅立叶系数为jnjnTjnjnTjnjnjnAAn1222)(2)()(222)0(n112)(2000TTtdtTTdttfTA于是得)(tf的傅立叶级数为000)1(212121221)(nntjnnntjnnntjnnejneAAeAtf1sin1121)3sin312sin21(sin121ntnnttt3.3求图题3.3（a）所示信号)(tf的傅里叶级数。tf(t)．．．．．．-T-T/20T/2T1(a)图题3.3答案：)('tf，)(''tf的波形如图3.3(b),(c)所示。于是得)(''tf的傅立叶系数为图题3.3t．．．t(t)f．．．-T-T/20T/2T(1)(1)(1)2/T(b)(t)f．．．-T-T/20T/2T2/T．．．(c)2222'2''2)2()2(2)(22)(2)(TTTTtjnTtjndteTtTtTtTTdtetfTjnAnnTjnTT)1(2)1(4422故得)(tf的傅立叶系数为0)1(1)1(1)()(2222nnjnjnjnAAnnn又2122)(22000TTtdtTTdttfTA故得)(tf的傅立叶级数为00212)(nntjnneAAtf3.4求图题3.4(a)所示信号)(tf的傅里叶级数，sT1。tf(t)．．．．．．-T0T2T3TE图题3.4(a)答案)('tf，)(''tf的波形如图题3.4（b）,(c)所示。于是得)(''tf的傅立叶系数为图题3.4t(t)f．．．．．．EEtEcos(b)t(t)f．．．．．．-T0T2T)2(E)(2tf(c)dtetftETdtetfTjnAtjnTTTtjn0222''1)()(22)(2)(ntjnTTtjnATEdtetfTdtetET20204)(2)(22其中dtetfTAtjnTn0)(2为)(tf的傅立叶系数。故)(tf的傅立叶nA系数可求得如下：nnATEjnAAjn2224)()(即TEAjnn4)(22今sT1故22T代入上式得EAnn44222故得)14(42nEAn于是得)(tf的傅立叶级数为tjntjntjnnenEenEeAtf2222)14(2)14(42121)(3.5设)(tf为复数函数，可表示为实部)(tfr与虚部)(tfi之和，即)()()(tjftftfir，且设)()(jFtf。证明：)()(21)(jFjFtfFr)()(21)(jFjFjtfFi其中)()(tfFjF答案因(1))()()(tjftftfir故(2))()()(tjftftfir式（1）+式（2）得)()(21)(tftftfr式（1）-式（2）得)()(21)(tftfjtfi故得)()(21)(jFjFtfFr)()(21)(jFjFjtfFi3.6求图题3.6所示信号)(tf的)(jF。0tA-Af(t)图题3.6答案2222)2()2()2()(jjjjeeSaAeSaAeSaAjF)2(2)2sin(22)2sin(2)2sin(2222SaAjjAjA故)2()(22SaAjF3.7求图题3.7所示信号)(tf的频谱函数)(jF。tf(t)102/12/1图题3.7答案方法一用时域积分性质求解。因有)()(1'tGtf故)2()()(1'SajGtf又因有dGdftftt)()()(1'故得jSajjGjGjF)2()()()()()(11方法二用卷积性质求解。因有)()()(1tUtGtf故得jSajSajF)2()(1)()2()(3.8求图题3.8所示信号)(tf的)(jF。tf(t)1-3-2-10123图题3.8答案方法一因),2()2()(22tGtGtf又有)2()(SatG取2故得2故)(2)(2SatG22)(2)2(jeSatG22)(2)2(jeSatG故得)(2cos4)(2)(2)(22SaeSaeSajFjj方法二因有)2()()2()()(22ttGttGtf故)(2cos4)(2)(2)(22SaeSaeSajFjj3.9设)()(jFtf。试证：　　；　　djFfdttfF)(21)0()2()()0()1(答案（1）因有dtetfjFtj)()(取0，则得dttfF)()0(（2）因有dejFtftj)(21)(取0t，则得djFf)(21)0(3.10已知)()(jFtf，求下列信号的傅里叶变换。)()2()2()2()1(tftttf)()4()2()2()3(dttdfttft)1()1((6))1()5(tfttf)((8))52()7(ttUtf答案（1）因有)2(21)2(jFtf又有)2(21)2(21)2('jFjFddtjtf故)2(21)2('jFjttf)(2)()(2)()()2()2('jFjjFtfttftft)2(212)2(21)2(2)2()2()2()3(jFjFddjtfttftft故)2()2(21)2()2('jFjFjtft)4(因有djdFtjtf)()(则有)()(jFjdddttdfjt故)()()('jFjFdttdft)1()1()5(tftf因有1)()1(jejFtf故有jejFtftf)()1()1()1()1()1()1((6)ttftftftjjjejjFejFddjejF)()()(')2(21)25(2)52((7)25jejFtftf1)(1)()((8)2'jjddjttU3.11求图题3.11（a）所示信号)(tf的)(jF。tf(t)E图题3.11(a)答案)21()(sin)(tUtUtEtfT2)('tf，)(''tf的波形如图题3.11（b），（c）所示。图题3.11t(t)f)(E(c))(E0T/22Et(t)fEE(b)故有)2()()()(2''TttEtftf故有2221)()()(TjeEjFjFj故得)1()(222TjeEjF3.12求图题3.12所示信号)(tf的)(jF。f(t)t1021图题3.12(a)答案将)(tf分解为)(1tf与)(2tf的叠加。即)()()(21tftftf如图题3.12（b），（c）所示；)('2tf的波形如图题3.12（d）所示，故得2121)2(1)(3)()()(jeSajjFjFjFf(t)(a)t1021t(t)f101.5(b)图题3.12t(t)f20.5-0.51(c)t(t)f2011(d)3.13求下列各时间函数的傅里叶变换。nttt(3)1)2(1)1(2答案（1）方法一由于t1为奇函数，故0sin121tdttjtF今0202sin0dttt故001jjtF又得01011jjtF即)sgn(1jtF方法二利用傅立叶变换的对称性求解。因已知有jt2)sgn(，故有1)sgn(21tj故tj1)sgn(212故得)sgn(1jt（2）因'2)1(1tt故,,)sgn()sgn(12jjtF00（3）因有)(21根据频域微分性质有)(21)()(nnjt故得)(2)(nnnjt3.14已知图题3.14(a)所示信号)(tf的频谱函数)()()(jbajF,)(a和)(b均为的实函数。试求ttftftx000cos)1()1()(的频谱函数)(jX。)()()(0tftftf，其波形如图题3.14(b)所示。图题3.14tf(t)Etf(t)E-101(a)(b)答案ttfttftx0000cos)1(cos)1()(今)()()()(jbajFtf故)()()()()()(jbajbajFtf)(2)())()(00ajFFjjFtfcos)(4)(2)(2)1()1(00aeaeatftfjj故得)()(2cos)(421)(00ajX)cos()(2)cos()(20000aa3.15已知)(jF的模频谱与相频谱分别为)3()3(2)(UUjF23)(求)(jF的原函数)(tf即0)(