杨晓非信号习题答案

信号与系统习题解答1120201LIM|LILI|LIPFTDDTEFTDTFT总FT解为功率信号。FT2T1解是矩形脉冲信号，故为能量信号。63解书中已作证明斜坡信号为非公非能信号。02222225|1PLIM|LI5LIM|LIM5JTTTTTFTEFTDTEFTDDTFT总4解为功率信号222240024240SINL|LISIN1LIMLIM1LI4TTTJTJTTJTJTJTJTETFDEDEEED总5解E242401LIM|41124465LIM0SIN2JTJTTEJJEPFET总为能量信号22161ELILIM1LIM02TFDTDTPFT总总解为能量信号123COSSFTTTTFT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。解是无理数改组合正弦信号是非周期信号245|COS|3JTFTE。显然为周期信号为周期信号121214SCOS363/46350FTTTTSMSTFT为周期信号，周期为60S31010210223SIN3IMCOS324RECOSTTJTTJJTJTTFEEETF2COS24IN68672471362COS24JJTTTSFKFKNFTETA5为周期信号，周期为为周期序列，14（波形略）15，是确定下列个信号的零值时间区间。30TF设112TTFTF223304TTFTF1156216试绘出题图16所示各连续信号波形的表达式。A211TTTTFB42CSIN53TTTFD212124TTTTTF2020LIMLI18SINSIN1112074443SISITTTTFTTTTTFTTT17试证明COS4NNTT2202219SISI074N553134|2TTTDTTSAETDETTTTD3201102556527TTDTTD0221011447SIN5SIN82195504TTTTTTTDDTDTD113,1FK12KF1FA2112KFKFA3K41211KFKF52KA1181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119解（1）整理得2532SSIIU（2）2112STCCCTTUDIUDID22224CSCCSSSUIRIIDUITIUIUII整理得2532SUU120解由题意YKYK1YK1YK1FKYK1YK1FK121解由题意Y1F1Y1Y2F2Y1Y1第K个月的全部本利为YK，第K1个月初的全部本利为YK1，则第K个月初存入银行的款数为YK1YK1FK122解由题意YKYK132YKYK10123解由题意1YEX0YXTFDFTSIN0X0X0EX0X0EX0EX0YY满足零输入线性12T12T1T212XFFSINFFDSINFDSINFDYY满足T0T0T01F2零状态线性为线性系统（2）YTSINX0TFT2X0X0SINX0X0TSINX0TSINX0T不满足零输入线性112123不满足分解性，所以是非线性系统；0TFYTDF4是非线性系统；LGX5不满足零线性输入，所以是非线性系统；TY|TF6YT不满足零输入线性DTX0满足零状态线性，故为非线性系统；YFT21210（7）YKKFXK满足零YXXXKKK21010102221输入线性22112121KFFKKKKYYYY不满足零状态线性，因而是非线性系统；（8）KNFXKY000212121KXKYXX因而为线性系统；020102121NNKKKNFFFF124（1）为线性系统；DFTYT因而是时不变系统；DXFXFXDTDT线性02TYF时变DTTDDFTXFX3|YFT非线性121212|FF|DDDTTYT非线性非时变4FTYE非线性非时变52F线性时变6SINY非线性非时变27TTF线性时变8Y线性时变91KYKF非线性非时变02K1251DTT1222FTTFDYTTETE非时变02TRD312220011|TTTTFFYETEE126解由题意，TTXY321TTXY3242TTFY32FT521EETTTTTT33361064TT7127解由题意（1），2132YT（2），FFXTY21ETTXYY3221810321，ETTF2。TYTYETTF3128解KKFX1YYFX122，KKX121KXY2YF221。KKF1KKYYFX214242KK2141291非因果非线性非时变03FTY有225YFT0T当FT非线性非因果时变TF有3非线性非时变因果FYT4线性时变因果COSF5线性非时变非因果FYTT6线性时变因果2FKFF7线性时变因果0FNYF0000KKMNNFFFYK8线性非时变非因果1FYFK010FYF130161285YF2YK3YK2YK1FK1FK3YKYK23FK1FK213113FFY32YK22YK13YK4FK25FK16FK3YK22YK14YKFK1FK或YK2YK14YK2FK1FK2132解有题图可得，YFY0101所以，YFYFY01101整理得，与给定微分方程可得，10101,BA2312112223560,130TTHHTTHYTCECYYTE1、Y解特征方程特征根代入初始状态有解之212120000,COSIN,0,32HJYTCTTCFYFTYTDYTTD，解代入初始状态得、对微分方程两端关于从到作积分有000001,2,68TYFFTYDTTDTT得得，1Y020Y3），4F1,0,YFT上式可写为3YT时微分方程左端只有含冲激，其余均为有限值，故有0T000004YDTYDTTTD得1,02YY4）25,01,TFYFE2TFTE原方程可写为245TYTE000002TDTTYDTTED1,Y03Y314,01,YFYFT解求XT3XX243012,3TTYTCE1021_CX解之TTXETY30求F321TYEPTFTFF设带如原微分方程有即0PTYFP10故321TFTFFEC对原微分方程两端从到关于T积分有_00004DTYDTTYFFFFFFFFF有031021FFFFFFCY解之1F6F31621TETYTTF求全响应。316523316213TTTTTTXEEEYYF0T（2），FY4,10,_TEFY解。02,1TXXECTY2010101222221111,4Y4433XXXXTFTFFFFPTFPTTTTTFCTETCYYEEEYT得求设并代入原微分方程，有得即故010200001DT4DT3Y1Y2TTFFTTFFFFFFFFFFYEDEDC由有01222,3310FFTTFTTFXCYTETTYET解之求21,222112,0,0,COSINCOSIN0,0SIN2XTXXXTXXXXTXFYFYFTJYECTTECTTYYYET解求代入初始状态求FF0000FFFFFFDTT2DT1,0120COSIN1FFFFFFFFFTFYTYYTTYYEATB首先确定与可得则当时，代入初始条件F,0I32SINTFTXFYAYTET求全响应24（1）YK23YK12YK0,1,20XXY解特征方程R23R1R20特征根,12YKKXKXKXKXCRC2121代入初始条件解得21X3,521XXKKXY3502YK22YK12YK01,XXY解1100221212,12XXXXKXKXCJJYJJCKRRKEJJJKYKJKKXX43SIN221,321K03YK22YK1YK01XXY解012R212RR12121XXKXXCYKK2XK0KY1422XY解0KXC故K0XXCYXY502412KK21,XXY解即0232特征根J12,KXKXXJCCKY1231JCX312JCX故JJKYKKXK0KJEKKJKJK32SIN233160167YYY,0X1X2X解即0232313,2KXKXXCY210带入初始条件有38492123101010CYXXXXX3812105CXX解之得，,CX2X故K0KKX25112,01,13YFYY解0222,121KXKXXCK即120221YXXX02421CX解之得故42CX4KKYK（2）121FKY2,1YY解0212KKYYXXX012,21CKXXK321YXXX21CX故0KK（3）1,1,YFY解012J2,1KBAKYXSINCO12Y04632COS52SINKYX2612,1,FYKFYK解2,02KPC,0PKYP,1,Y令20CY所以0,KK,2YKXX其中2,12XXC24KYCKKXFPKFF121212122123,0,4240KKXXXXXXXKKFPYYFFYKCCY解令0121213,612020643FFFFFFFFFFFFPPYYYYCCY则有由得解之得42361411236180236KKFKKKXFKKKY27A解31095TETRHTDTGTETSIUSTIISIB解由图知SLRC其中2DTILCTUILCDTIRLUILL故有SLSIIIIRC51L即故SLLII5241522PPH2SIN5TETHTIL1DTILUL2SIN5TETTDTILUTETETTTT2SIN1COSCOI2THIU28（1）1233321212020TETDEDHTGTEPPPHFYTOT（2）YFT22442PPHP2THTTE20000224TTTTTTGDDETETEA29，求H128YF22184HPP1SIN4HTT2YF2222113331444PPHPP22313COSSINTTHETET3YF22211PHPTTHE462YYFHP6123P2131PPHT231TPEETPTT210求HK1YK2YK1FK1解HE1221KKHE2YK23YK12YKFK1FKHE211232EKKH3YKYK1YK2FK4解HE221EHKHEEEDE211K2ED1K2EK1E21KK4YK4YK18YK2FK解HK4YK18YK2KK0时,有HK4HK18HK202J220842124HCC4K18HK2H004H18H21PH114H08H14222PQ故P1,Q1HK2SINSINK2K42SINSINKK5YK22YK12YKFK12FK解HK22HK12HKK00HKPCOSQSINO2K43KH22PCOSQSIN2Q1所以Q21H1PCOSQSINPO243KP0P1所以HK（）SINCOSK1OK2143KSINK1HKHK12HKO0SINK1K2SINK1K431K43SINK1K2SINK12COSK1K001121,KYFKFHC由图得移序得设有0101222240510561KKKKHCYYFFKEEH又由图可得差分方程1200164200102|4KKTTEHKTFTFD图示法求解下列卷积124400TTTFFTTFTFT124240TT1211112TTTTTTTTTTTTTFEFDEDEFDEDE11TTT1122TTTTEETFTE30T2HZ1KFC1WC,即2KHZ3WHZ3KZFC33594COS2CS511TFTFTF其中。1KZS80解（1）22201111WWWJWF（11112255125,05822YTCOS04TFZ36FF010CSFHZTSFYJFHTJWTT解122MINMIN21552444SF4W,F2010SJSA