《信号与系统分析基础》第二章部分习题参考答案

1第二章部分习题参考答案2-6试求下列各函数1()ft与2()ft之卷积。121212(-)001(1)()()()()(0)()()()(-)()(-)11(1)0(2)()tttttttftutfteutftftfftdueutdeedeeetft，解：，2121212()()cos(45)()()()cos[()45]cos(45)(3)()(1)[()(1)]()(1)(2)()()tfttftfttdtfttututftututftft，解：，解：f2(τ)τ021f1(τ)τ01122222221211211()(-1)(-1)-2(-2)(-2)(-1)(-1)-(-2)(-2)2211-(-2)(-2)(-3)(-3)-(-2)(-2)(-3)(-3)22()*()()1,()0123,(1-)(1)21(1)--(12ttfttuttuttuttuttuttuttuttutftftfttftttdttftttt222-112222212111)-222123,(1-)(1)-221()2(1)-2(1-)(-1)211121---152223,()*()0.ttttttdtftttttttttttftft121221--(4)cos,(1)-(-1)()*()()(-)[(1)-(-1)][cos(-)]cos[(1)]-cos[(-1)]fttftttftftfftdtttdtt-212-212--2-2200(5)()(),()sin()()()*()()sin(-)(-)sin(-)sinttttttfteutfttutftftfteututdetdeed3-12-(-)--0022-(-)-33-2-3(6)()2[()-(-3)],()4()-(-2)0,()0.02,()2488-825,88()8(-)5,()0.tttttttttttfteututftututtfttftedeeetftedefteeetft2-8求阶跃响应为32()(21)()ttsteeut的LTI（线性时不变）系统对输入()()txteut的响应。解：()()*()rthtxt3232032323()[()],()[()]()(),LTI()()[][()]()()(-21)()(34)()(34)()()()*(34)()(3tttttttttthtTtstTutdtutdtdstdutTTthtdtdthteeteeuteeutrteuteeute而由于而对于系统有2-32043430043324)()()(34)34(34)[]433434734()[],043431243ttttttttttttteueutdeeeedeeedeeerteeeeeet2-9线性时不变系统输入()xt与零状态响应()yt之间关系为()()(2)ttytexd（1）求系统的()ht。解：-()()*()(-)()tythtxthtxd，（零状态响应）由已知条件，-(-)-(-)--()(-2)(-2)()tttytexdexutd令2，则24则-(--2)-()()(2)tytexutd比较，可得(2)()(2)thteut（2）求当()(1)(2)xtutut时的零状态响应。解：设激励为()ut的零状态响应为()gt，有dtuhtuthtg)()()((2)()()(2)ttgthdeud(2)(1)(2)221,2tttedeet(2)()[1](2)tgteut由线性性质（叠加性）得(1)(4)2()[(1)(2)][1](1)[1](4)ttsytgtgteuteut（3）用简便方法求题图2-9所示系统的响应()yt。图中()ht用（1）之结果，()xt与（2）相同。h(t)δ(t-1)h(t)Σy(t)x(t)+-题图2-9解：由时不变及叠加，有()()*()1()*()()*()(1)*()ythtxtthtxthtxthtxt由（2）的结果知，()*()().zshtxtyt(1)(4)(2)(5)()()(-1)[1](1)[1](4)[1](2)[1](5)]zszsttttytytyteuteuteuteut这里利用了卷积的延迟性：if12()*()()ftftft，then112212()*()()fttfttfttt。52-13用图解的方法粗略画出1()ft与2()ft卷积的波形，并且计算卷积积分12()()ftft。（a）tf1(t)01atf2(t)02b解：12()[()(1)]()[()(2)](2)2bftaututfttututbt，，此表达方式有误在本题中，bffaff2,02,1,012211+1212()()()()()ftftftftft对进行翻转平移处理，不动有：ττ012batt-12200222112222110()001()24412()[(21)]2441()2223()[421)]244tttttttttftabababtftdtabababtftdtttabtabababtftdtt，，，，2(32)43()0abtttft，6(1)(-1)1212(-1)122(-1)22020()()()()()[()(1)]()2()[()(2)](2)2()()[()(2)](2)2[()(2)4tttttftftftftdftbattftddtbfttututbtbftfddututbdbutut也可用有；2(-1)112222](2)[()(2)](2)4()()()()[()(1)][()(2)](2)4[()(2)](2)44butbtututbutdftftftftdtbatttututbutababtututabut222(1)[(1)(3)](3)000141()1222(32)23403tututabuttabttabttabtttt71222112212121()[()(1)]()[()(2)]2()[()(2)-2)2(2)]2()()()[()]()0.5()()()()(()()())()(btftaututftututbfttuttututututtuttututtutfttfttftttftftftftft也可以这样解（利用卷积性质），（利用公式：；；22222222)()[()(1)][()(2)(2)2(2)]2(1)11()(1)(2)(2)(3)(3)2(2)(2)2(3)(3)2222211()(1)(1)[2224](2)[(442222abftututtuttututabttututtuttuttuttutababababtuttuttttutt222222223)2(3)](3)0()0101()41112()(21)(12)44423()(12)(4)(32)4443()(32694124tuttfttftabtabtftabtabtttabababtftttttabtftttttt当，，，，，)0（b）tf2(t)0btf1(t)01a1-1解：2222()()()()()(1)babftaftftft的相对于的，有因此可利用（a）的结论。1212122222()()()()()()(1)(1)()[()(2)](2)(1)[(1)(3)](3)44()(1)(1)[(1)(1)](1)[()(2)](44aabaaabaftftftftftftftftababfttututabuttututabutababftfttututabuttututabut设则而2)8（c）tf2(t)0b-2-2.5tf1(t)0123a解：12()[(2)(3)]()[(2.5)(2)]ftaututftbutut，1122121212()()()()()()()()(22.5)(22.5)(22)()(32.5(0.5)(32)(1)()()()[(0.5)(0.5)()(0ututtutfttfttftttftftabtutabtutabtutabtutftftftabtuttutt，且.5)(0.5)(1)(1)]0.5()00.50()(0.5)00.5()0.50.51()[(0.5)(0.5)](uttuttfttftabttftabtftabtttab当，，，，0.50.5)(1)1()[(0.5)(0.5)(1)]0tabttftabtttt，（d）tf1(t)0tf2(t)11-1011解：1()(1)(1)2()(1)(1)fttuttuttut2()()(1)ftutut利用：①()()()ututtut卷积性质：②2[()]()()2ttututut9③112212()()()fttfttfttt1222222222()()()(1)(1)2()(1)(1)()(1)111(1)(1)()(1)(1)()2221(1)(1)(2)(2)213(1)(1)22ftftfttuttuttutututtuttuttuttuttuttuttuttu2222222222231()(1)(1)(2)(2)22-1()01110()(1)(21)2213101()(1)222133112()(1)(1)2222222()ttuttuttfttftttttfttttttftttttttft当，，，，，22221331(1)(1)(2)02222tttt（e）tf1(t)0tf2(t)21-101-2δ(t-2)δ(t+2)23-2-3解：设上题（d）中的1()(1)(1)2()(1)(1)dfttuttuttut101112121111111()(2)(2)()(2)(2)()()()(4)()()