杨晓非信号_习题答案

信号与系统习题解答1.120201lim|()|211lim22lim|()|lim()()PftdtdtEftdtdtftt总(1)f(t)=(t)解为功率信号。()ft(2)f(t)=(t)-(t-1)解是矩形脉冲信号，故为能量信号。()6()fttt(3)解：书中已作证明斜坡信号为非公非能信号。0()2222222222()5|()|51Plim|()|1lim2525lim|()|lim25()jtTTTTfteftftdtTdtTEftdtdtft总(4)解为功率信号22222244200(24)(24)0()sin2()lim|()|lim(sin2)()1lim()lim(2)(2)41()lim[]4tttjtjttjtjtjtjtftettftdtetdteeedteeedtjeedt总(5)解：E(24)(24)01()lim[]|42424111()[1]42424124241()[1]44165lim02()sin2()jtjtteejjjjjjEPftett总为能量信号221(6)()()11Elim()lim(1)1lim()111lim02()ftttftdtdttEPft总总解：为能量信号1221()3cos(2)2cos()2()ftttTTft1.2判断下列信号是否为周期信号，如果是周期信号，试确定其周期。(1)解是无理数改组合正弦信号是非周期信号(245)(2)()|cos(2)|(3)()3jtfttfte。显然为周期信号为周期信号1212'1(4)()cos()cos()cos()23632232/422/631251260()fttttTsTsTmTsTsft为周期信号，周期为60s.(3)(100)(1002)(1002)2222(3)()3sin(3)3Im[]3cos(3)2(4)()Re[()]cos(100)2ttjttjjtjtjtfteteeetftjeeeeeftet2cos(22)4()sin()().68(6)()()78227847()7.1.3.(1)()66(2)()2222cos(2)4jjtfttsfkfkNfteftet(5)为周期信号，周期为为周期序列，1.4（波形略）1.530ttf设，是确定下列个信号的零值时间区间。(1)ttf201(2)ttftf2021(3)2302ttf(4)ttftf1021(5)602ttf1.6试绘出题图1-6所示各连续信号波形的表达式。(a)21121ttttf(b)1242ttf(c)1sin53ttttf(d)22211214224tttttttttf220000220220''lim.()lim()()lim()1.8(1)()sin()()sin()()111(2)()sin()()sin()()0.707()4444(3)()sin()()sin()()ttttttfttttftttttftttt1.7试证明(t)=''cos()()111(4)()sin()()sin()()cos()()44444tfttttt2'20221.9(1)sin()()sin0.70744sin5(2)()5(5)()5(3)()()123(4)(1)()(1)|2|()22tttttdtttdtSattdttettdtetttdttttdt32010102200(5)(2)(5)0(6)(2)(5)(52)(5)27ttdtttdttdt0221022111044(7)sin(5)sin5(5)(8)(1)()(1)2()2()21(9)(25)()[25][41]04(10)(1)()[()()]()()tttttttdtddttttdttttddtt1.13:1()fkkk,12()()1kfkk.(1)112()()()1kfkfkkkak.(2)112()()()1kfkfkkkak.(3)112()()()1kfkfkkak.(4)12(1)(1)(1)1kfkfkkkak.(5)12(1)(1)(1)2kfkfkkak.1.18.(1)偶、偶谐（2）偶、奇谐（3）偶、偶谐奇谐（非唯一）（4）奇、奇谐(5)奇偶谐（6）奇、奇谐偶谐1.19解：（1）整理得：'''''''''25532SSSIIIIUUU（2）''''''2'1()2121211()222()StCCCCCtCtUUdUUUUICUUUUIIIUUUdIIUdUU'''22222'''22'''''''''''''''''''''''''2222(2)2(2)222242CSCCcCCSSSSSUUIIURIIIIIIIduICIIdtUIUIIIUIIUIIIUIIIUII整理得：'''''''25532SUUUUU1.20解：由题意y(k)=y(k-1)+αy(k-1)-βy(k-1)+f(k)∴y(k)-(1+α-β)y(k-1)=f(k)1.21解：由题意y(1)=f(1)+βy(1)Y(2)=f(2)+y(1)+βy(1)第k个月的全部本利为y(k)，第k-1个月初的全部本利为y(k-1)，则第k个月初存入银行的款数为Y(k)-(1-β)y(k-1)=f(k)1.22解：由题意y(k)=32y(k-1)∴y(k)-32y(k-1)=01.23解：由题意(1)yx=etx(0)yf=dft)(sin0x1(0)+x2(0)--et[x1(0)+x2(0)]=etx1(0)+etx2(0)=y1x+y2x满足零输入线性f1+f2--t0sinτ[f1(τ)+f2(τ)]dτ=t0sinτf1(τ)dτ+t0sinτf2(τ)dτ=y1f+y2f满足零状态线性∴为线性系统（2）y(t)=sin[x(0)t]+f2(t)x1(0)+x2(0)--sin{[x1(0)+x2(0)]t}≠sin[x1(0)t]+sin[x2(0)t]不满足零输入线性(3))0()()(xtfty+tdtf0)(不满足分解性，所以是非线性系统；(4))(lg)0()(tfxty是非线性系统；(5))0(lg)(xty+)(|tf不满足零线性输入，所以是非线性系统；(6)y(t)=dtfxtt0)()(0不满足零输入线性yyffdtt21210][满足零状态线性，故为非线性系统；（7）y(k)=)2()()0(12kfkfxkyyxxxxxxxxkkk21)0(1)0(1)]0()0([1)0()0(212121222满足零输入线性)()()]2()2()][()([)()(21212121kfkfkkkkkkyyyyyyyy不满足零状态线性，因而是非线性系统；（8）knnfkxky0)()0()()()()0()0()0()0(212121kxkxkkyyxxxx)()()]()([)()(020102121nnnnkkknknknffffff因而为线性系统；1.24（1）dftyt)()(为线性系统；dxxfxdftfxdtddttt)()()(因而是时不变系统；0(2)()()tytfd线性0()()()()ddtttdddtfttftdxtfxfxdx时变(3)()|()|ytft121212||||||ffffff非线性()|()|()dddfttfttytt()(4)()ftyte非线性非时变(5)'2'2yyff非线性非时变(6)'sin'yyf线性时变2(7)['()]2()()ytytft非线性非时变非时变(8)'()2()()ytyttft线性时变(9)()(1)(1)()ykkykfk线性时变(10)()(1)(2)()ykykykfk非线性非时变1.25()(1)()dttdt1222()()[()]()2()fttfdytdytettetdtdt0(2)()()tRtd3122200011()()()|()(1)()22ttttttffytydedtetet1.26解：由题意eettxy3321，eettxy3242，eettfy322fyxyxyty35221eeeeeetttttt333636102064eett322761.27解：由题意（1）2132yyty，（2）ffyxyxytyyxyxyty3212121}eettxyxyyy322181022321，eettfyyy3212222，tytyeettf32。1.28解：kkykykyfx1kkykykykfx12122kkyyykx212221，kxky21kkkyyykf2122221kkkykf21。kkkykykykkfx214421242kkk21241.29(1)'()0(0)23fttyy有非因果非线性非时变(2)'2()2()(5)yttftft0t当()0ft'()(5)ytf有非线性非因果时变(3)()()fytft非线性非时变因果(4)()()cos()fytft线性时变因果(5)()()fytft线性非时变非因果(6)()(2)()fyKfKfK线性时变因果(7)0()()KfnyKfn线性时变因果0000000()()()()KKKmnKnKfKKfnKfmyKK