信号与系统第二版课后答案 西安交大 奥本海姆(可编辑)

信号与系统第二版课后答案西安交大奥本海姆第一章13解T214TAELIMXTDTEDT,P0T4T0TT121BPLIMXTDTLIMDT1TT2T2TTTT22ELIMXTDTXTDTTTT22ELIMXTDTCOSTDT,TTCTT111COS2T12PLIMXTDTLIMDTTT2TT222TTNN1211142NDPLIMXNLIMLIM0NNN2N12N122N13NNN02NN142ELIMXNN23NN0EXN1,E2NN112PLIMXNLIM11NN2NN121NNNNN112FPLIMXNN2N12NNN2ELIMXNNNN2M019A10,TB非周期的C7,N200001052N0DE非周期的N100112解N1K对于N4时,为1K3即N4时,XN为0,其余N值时,XN为1易有XNUN3,MN1,30115解A1YNYNXN2XN3,又,XNYN2XN4XN122221112YN2XN24XN3XN32XN4,XNXN11111YN2XN25XN32XN4其中XN为系统输入。B交换级联次序后YNYN2XN4XN21112XN24XN3XN32XN422222XN25XN32XN4其中XN为系统输入通过比较可知,系统S的输入输出关系不改变116解A不是无记忆的,因为系统在某一时刻的输出还与时刻的输入有关。NN200B输出YNANAN22ANN20C由B可得,不论A为任意实数或者复数,系统的输出均为零,因此系统不可逆。121122和123画图均略126解30A,为有理数,XN具有周期性,且周期N72710B,为无理数,XN无周期性21622C由周期性的定义,如果存在N,使得COSNNCOSN,则函数有周期2811222N2NN16K性,即NN2KN,对全部N成立取88N的最小值N8,即为周期。131DXNCOSNCOSNCOSNCOSN,与A同理,XN具有周2424431期性,对COSN存在N8,对COSN存在N8,基波周期N81244周期N16E与上题同理,N8,N16,N4123127A系统具有线性性与稳定性E系统具有线性性,时不变性与因果性与稳定性128C系统是无记忆的,线性的,因果的E系统是线性的,稳定的G系统是线性的,稳定131解A如图PS217A所示。XTXTXT2YTYTYT2211211BXTXT1XTYTYT1YT如图PS217B所示。311311YTYT232TT4013021122AB1331正确。设XN的周期为N。如果N为偶数,则YN的周期为N/2如果N为1奇数,则必须有,才能保证周期性,此时的周期为。22NNYNNN010N2不正确。设XNGNHN,其中GNSIN,对所有N,4N1,N奇HN显然XN是非周期的,但YN是周期的。310,N偶3正确。若XN的周期为N,则YN的周期为2N。24正确。若的周期为N,则N只能是偶数。XN的周期为N/2。YN2TXTYDT137AXYYXBTT,奇部为零。XXXXCTTT,TTXYXXYYXX142解A结论正确。设两线性时不变系统如下图所示级联。当时,则有XTAXTBXT12WTAWTBWT,于是YTAYTBYT,因此整个系统是线性的。1212若输入为XTT,则由于时不变性可知系统1的输出为WTT,这正是系统200的输入,因此总输出为YTT。即整个系统是时不变的。0YTXTWTHTHT12B结论不对。如系统1为WTXT3T,系统2为YTWT3T。虽然两系统都不是线性的,但它们的级联YTXT却是线性的。C设系统1的输出为WN,系统2的输出为ZN11YNZ2NW2NW2N1W2N22411XNXN1XN224146解AYNN1YN1,N0,YN0,N1,YN1,N2,YN1N1YN1UN1BYNUN1YN1,N0,YN0,N1,YN1,N2,YN0N3,YN1,N4,YN0,N5,YN1147解AYNSXNCLXNC,C为系统的零输入响应。111YNSXNXNYN11LXNXNCYN11LXNLXNCYNLXN11N/2,NEVEN1,YNN,2,YN00CN1/2,NODD3非增量线性系统4YTXTTDXT/DT,非增量线性系统25增量线性系统,YNCOSN第二章21解AYNXNHNX0HNX1HN1X3HN312N14N2N12N22N4图略BYNXN2HNYN2212N34N22N12N2N2图略CYNXNHN2YN图略329N425解YNXKHNK,由Y45可知K0由Y140可知9N114,即N4所以N4211解AT3时,YT0T35TYTUT3HTU3HTD时,33T3T1E3TED3363T51EE53TT5时,YTT3UT5UHTED330,T33T31E因此YT,3T5363T51EE,5T3DXTBTT35DTDXT3TT335GTHTHT3HT5EUT3EUT5DTDYTCGTDTNNN1111213解A将代入式子得HNUNUNAUN1N555N1即UN5AUN1N51从而可得51A,即A51B由A可知HNHN1N51则S的逆系统的单位脉冲响应为HNNN1S1125216解A对。若NNN,即NNN,则XK与HNK没有公共部分,显2112然有XNHN0。B错。YN1XKHN1KXNHN1KC对。YTXRHTRDR,令R,则YTXHTDXHTDXTHTD对。若,则没有公共部分,故时,XTHT0。TTTTTT21121219AYNYN1WN,WNYNYN11将WN代入后经比较可得,1。4B根据书上例题215,利用递推算法,可求得系统S1,S2的脉冲响应为NN11,HNUNHNUN1224NN11则总系统的单位脉冲响应为HNHNHN2UN12244NNN118/91/84,N6221AYNUNCYNN8/91/2,N62512T2T2T2T22T5T1222B当时,YTEDEDE2EE0222512T2T22T22T513T当时,YTEDEDE2EET122512TT252当36T时,YTEDEET12当T6时,YT0EXT是周期信号,由此可推知YTXTHT也是周期的,且周期也为2。因此只需求出YT的一个周期。1T11122当T时,YT1TD1TDTT1T12242224解AHNNN1,2HNHNHNHNHNN2N1N21221HNHN2HN1HN2,根据HN的图形可推出H1N111H101,H113,H123,H141,H150N5,HN0BYNHNHN1N1228解AHNUN55N0HN0当时,因而是因果的。HK,因而是稳定的。4KN1CHNUN2K当N0时,HN0,因而是非因果的。HK2,因而是KK非稳定的。N1NEHNUN101UN12当N0时,HN0,因而是因果的。HK,因而是非稳定的。KN1GHNNUN13K1当N0时,HN0,因而是因果的。HKK3,因3KK1而是稳定的。6T229解BHTEU3T36T0时,HT0,因而是非因果的。HDED,因而是非稳定的。2TDHTEU1T1122T0时,HT0,因而是非因果的。HDEDE,2因而是稳定的。F因果的,稳定的。N3231解系统最初松弛,当时,YN0由YNXN2XN22YN1可递推得出Y2X22X42Y31Y1X12X32Y20,Y0X02X22Y15,Y1X12X12Y04,Y2X22X02Y116,Y3X32X12Y227,Y4X42X22Y358,Y5X52X32Y4114,N5N6时,YN1142TT2TT2T2240解AYTEX2DXEDXTEUT2T2HTEUT2B由图PS37知,当T1时,YTXTHT0T12T114TYTED1E当时,2T12TT41当T4时,YTEDEET2244ATTTBNNN,NNNMMM1312402513YXH247AXT2XT,HTHT00如图A所示。YT2XTHT2YT000BXTXTXT2,HTHT000YTYTYT2如图B所示。00CXTHT1YT1,XT2HT1YT1000000YTXT2HT1YT1如图C所示。000D信号不能确定EYTXTHTXHTDXHTDYT00000FYTXTHTYT如图F所示。000248解A正确。HT为周期性非零函数时,HTDT。B错误。若系统的冲激响应为,则其逆系统的冲激响应为,显然TT,T0TT000是非因果的。C错误。若HNUN,显然HN1但,因此系统不稳定。HNND正确。HN为有限长时,必然有HN。NE错误。若HTUT,显然系统是因果的,但由于HT,因此系统不稳定。0F错误。若系统A的冲激响应HTT3,系统B的冲激响应HTT5系AB统A非因果,系统B因果但它们级联后有HTHTHTT2,显然AB是因果的。TG错误。若某系统的HTEUT,显然该系统稳定,但其阶跃响应TTSTED1EUT并不绝对可积。0N0H正确。UNNK,SNHNK,如果时,SN0,则必有KK00N0时,HN0,从而系统是因果的。反之,若系统因果,则N0时,HN0,从NN0而必有,SNHK0。K250解A系统B是系统A的逆系统,图P312所示的整个系统是恒等系统。系统A对AXTBXT的响应为AYTBYT,因此系统B对输入AYTBYT的响应121212为AXTBXT。12B系统A对的响应是,XTYT11系统B对YT的响应是XT。11第三章35解由于XT只是对XT做了平移变换21所以,12JKJK11而由傅立叶级数的性质有,BBBAEAEKKK2KK1JK1EAAKK38解由1,AAA,A是虚的奇函数KKKK2由2,T2,T由3,XT至多有三个非零傅立叶级数系数,A,A,A0111又AXTDT0,AA011TTJTJTXTAEE1221由4,利用PARSEVAL定理,AA1,即AA1111222AJ,AJ1122XT2SINT311解由1,A是实偶函数K由2,3可知,N10,A5AA511119951222由4,XNA50A50KK10N0K0K5又AA5115,KN1AK0,K取其它值522JNJNNN综上,XNAEAE10COSNKK5KNK5故有,A10,B,C05322解AAT2,XT是实的奇函数,A00K1111J1JKTJKT11JKTATETE|E|,K0K11122JKJKKKK1JJ11JKTJKTXTEEKKKK11T6B,A020KEVENJKWT0KA,XTAEKKJ1KODDKKCT3,A103JJK2/3JK/3AESINK2/32ESINK/3,K0K222KJKT0XTAEKK328解4JK572EKSIN6JKN1177AAN7,AXNEK77N0SINK74JK553JKNJKN1111E33BN6,AXNEEKJK666N0N031EJK22ESINK1231K5A063SINK622N2JKNJKJKJKJK1133333CAXNEE2E12EEK66N21212COSKCOSK,0K563333JNJNN144CCXN1SIN1EE,0N342J11888JKNJNKJNK11122222AEEEK48J8JN0N0N011J2KJ2KJ2K2211E11E11E114JK8JJK8JJK222221E1E1E2J2K11E124JK222COSK21E2132即A112,0441KK1A112COS,K1,2,3K4233111111JKNJNKJNK11166262DAEEEK1224J24JN0N0N033J2KJ2KJ2K2211E11E11E33JKJKJK1224J24J6261E621E1E2J2K11E12JK12662COSK21E6211122即A1061222K2COS11216A,1K11K12122COSK2COSK63J/4330N6,AA1,AA1/2,BBBE/2,011112JJ/4/4CCAB,可求得CCOS/4/2,CCE/2,CCE/2KLKL01122L2JKT0334解设则其中分别是XT和YT的傅里YTBE,BAHKAB、KKK0KKK叶级数系数。84TJTHJEEDT216NCXT1TNT2,0N10,K偶11JKTJK2ATT1EDT1EK1221,K奇20,K偶B1K,K奇4JKXTD由图所示可得T1,2011SINK/2A,A,K1,2,0K22K/20,KK偶,01BB,SINK/20K,K奇8K4J2KJNJJN336解EHEE,将此代入差分方程中可得11JNJJJNJJNJEHEEEHEE,求得HE,14J1E4AN8,信号中的谐波分量为正负3次谐波,可得AA1/2J,33输出信号中的傅立叶级数系数为JJ3/43/4BAHE,BAHE3333BN8,信号中的谐波分量为正负1次谐波与正负2次谐波,可得AA1/2,AA11122输出信号中的傅立叶级数系数为J/4J/4J/2J/2BAHE,BAHE,BAHE,BAHE11112222343解222TJKTT/2JKTTJKT11TTTAAXTEDTXTEDTXTEDTK00T/2TT22TT/2JKT/2JKTT1KTT若XTXT,则AXTEDT1XTEDTK002T2T/2JKT2T当K为奇数时,AXTEDTK0T当K为偶数时,A0K只有奇次谐波TBXTT2,奇谐信号,XTXT2TT,01XT1TT,10XT如下图所示。1111JKTJKTJKTATEDTTEEK0JKJK0,K为奇数122JKKA0,K为偶数K344解2由T6,可得0T3JKTJKT333由条件4可知,AEAEKKK1即,所以K为奇数11由于当K0和K2时,有A0K所以当K2时,A0且A022JTJT33因此,XTAEAE11又由XT为实信号可知,AAA1113121222由条件5,6可知XTDTAA2A1116231所以AA112则,A1,B,C03348解2222JKNJKMJKNJKN110NNNNAAXNNEXMEEAEK0KNNNNMN22JKNJK1NNBAXNXN1EAAEKKKNNN2JKN0KEVEN1NJKNCAXNXNEAAEKKKN22AKODDNNKD4JKN2NNAXNXNEKN2NN/2244N1N/21N1J2KNJKNJKN2NNNNXNEXNEXNEN2N0N0NN/24N4NN/21/21JKMJKNN2NNN2N令MN2AXNEXME2K2N22NM0044NN/21/21JKNJKN2NNJK2NN2AXNEXNEE2KN22NN002A2KAA2KK222JKNJKN11NNEAXNEXNEAKKNNNNNN2N2JKNJKN11NN2NFA1XNEXNEAKNKNNNNNN2K0,1,2,N1JKNJKN11NJNNNGA1XNEXNEEK2N2NN2NN2N2KN2KNN12N1JNJN1N2N2XNEXNE2NN0NN2KN2KNN1N1JNJN1JKNN2N2XNEXNNEE2NN0N01JKNA1EKN22A,K为奇数KN20,K为偶数1NHYNXN1XN2对信号XN周期为偶数时,YN的周期大小不变,仍为N,直接利用变换性质即可,1AAAKKNK22对信号XN周期为奇数时,此时YN的周期性发生了变换,周期为2N,傅立叶级数系数为2JKN1N2NAXN1XNEK2NNN2再分成前后两部分,22NN121JKNJKN1NN22NNAXN1XNEXN1XNEK2NN0NN经整理后得22NN11JKNJKN1NNK22NNAXN1XNEXN1XNE1K2NNN001AAKK2352解AXN是实信号,XNXN222N1N1N1JKNJKNJKNNNN而XNAEAEXNXNAEKKKK0K0K0AA或写为AAKKKK令ABJC,则有ABJC,从而有KKKKKKBB,CCKKKKNB当N为偶数时,为一整数。22NN1N1N1JN111JNNN2AXNEXNE1XNN2NNN00N0显然,A是一个实数。N222N1N1JKNJKNNNCXNAEBJCEKKKK0K0当N为奇数时,上式可写为N1222JKNJKNNNXNAAEAE0KKK1NN1122222JKNJKNJKNNNNXNAAEAEA2REAE00KKKKK11N1222A2BCOSKNCSINKN0KKNNK1N1222JKNJKNNNN当N为偶数时,有XNAA1AEAE0NKK2K1N1222JKNJKNNNNXNAA1AEAE0NKK2K1N1222NAA12BCOSKNCSINKN0NKK2NNK1JKD由C知,当AAE时,有KKN为奇数时N1N1222JKN2NXNA2REAEA2ACOSKN0K0KKNK1K1N为偶数时N122JKNNNXNAA12REAE0NK2K1N122NAA12ACOSKN0NKK2NK1322KNKNEYNAD2DCOSFCSIN00KKK77K1分别求出信号XN,ZN的偶部与奇部3322,EXNA2BCOSKNOXN2CSINKN0KDK77K1K13322EZND2DCOSKN,OZN2FSINKN0KDK77K1K1YNA2DEZNOXNOZN00VDD第四章1SINT410A解XTSINTTSINT令XTSINT,XT12TXJ111JJ则0,1XJ211,J,202J所以XJ,0220,其他SINT12124BATDTXJD3T2211J1411证明GJXJHYJ33391J因为Y3TY3311J所以Y3TYGJ39311即GTY3TA,B3331JTJ5T413AXT1EEXT非周期2BFXTHTXJHJ11J225EJJ1110JJ5T则XTHTFXJHJ1EE10J2因此XTHT是周期的,周期为。5C由B可知,XT和HT都不是周期的,但卷积周期。这说明两个非周期信号的卷积有可能是周期的。A2T414解由条件2得FAEUT2JA所以1JXJ2JA11即XJA2J1J1J2JTT2XTAEEUT22TT22由条件3知XTDT1AEEDT102A12,由XT0A23TT2从而有XT23EEUT。AT421解AECOSWTUT,A001ATEUT,A0AJW111XJW2AJWWAJWW00C1COSTT,1XT0,T111JTJT1COST,T11,T1E,T1E,T1XT220,TT10,10,TT10,12SINWSINWSINWXWWWW2TETESIN4TUT,1112TESIN4TUTAW2J2JW42JW4DAWJ11XWJ22DW22JW42JW4G如图所示1,T1DXTTT22DT0,T1T2T22COS2W1,T12SINWW0,T1DXT2SINWAW2COS2W而且A00DTWAW2SINWXWA0WCOS2WJWJWW2SIN3W2AXJWW21,T32SIN3W422解W0,T3JT2ET,3XT0,T3J,1W01,1W00,1W0CJJJ,0W10,0W11,0W12SIN05TSIN05T2SIN05TJ05TJ05TATJEJETTTWW,10DAT1SINT1COST2WW,01DTTTJW3WE,1W01SINTT31COS32JW3TT33WE,0W11SINTT31COS3XT2TT331,2WW11,12DXWEWW33DW0,其他0,其他11SIN05TTSIN05J3TJ3TJ15TJ15TJTXTEEEE22TTCOS3T2SIN05TCOS15TCOS3TSINTSIN2TXT22JTJTJTJT424解A1说明XT是实奇的,满足的有A,D2说明XT是实偶的,满足的有E,F3存在A使得XTA为实偶的4说明X00,满足的有A,B,C,D,FDXT5说明0,满足的有B,C,E,FT0DT6说明XT是离散的,满足的有BB要求只满足145,其余不满足。例如JW425AXT1是实偶的,则EXW为实偶的XWWBXJ0XTDT7C2X0XWDW71,3T12SINWJW2DATE0,OTHERWISEW2SINWJW2XWEDW2XTAT7T0W22EXWDW2XTDT26FXTREXWEXT如下图所示E427解A设XTYT2,012JCOS1JTJT则FYTEDTEDT102JCOS1J2所以XJE2KJCOS14KJ1JKTTTBAXTEDTEKTTK12KAXJKTT1431AHWW,1JW5HW2,22JW2HW321JW经计算得YWYWYWXWHWXWHWXWHW123112233WW11JYTSINTB例如HT05HT05HT可以证明当组合系412数总和为1时,HT,HT,HT的线性组合均符合题意123SIN4T1432HT,T11,W4SIN4TT0,W4JWEW,4HW0,W4AXTCOS6T,12JW2COSTEW1W1,2JW12XWEW6W616YWXWHW011YT011KBXTSIN3KT22K01KXWW3KW3K2J2K011133JJJWYWEW3EW3EW3W32JJ222YT05SIN3T12SIN4T1CXT,3T11,W4SIN4TT0,W4JWEW,4HW0,W41,W4YW30,W4SIN4TYT3TSIN2T2DXT,4T1,W2SIN2TT0,W2W4,0W1,WW21,218XW42W40,WW20,2,0W8SIN2T12YTXT144T1YWJW421434HW2XW6W5JW2JW3JW2DXTDYTDYTA4XT6YT5,2DTDTDT23TTBHT2EEUT44TTXTETEUT3JWXW24JW0505CYW42JWJW42TTYT05E05EUT435A22AJWAJWAWHW,HW1,22AJWAJWAW2WW11HWTG2TG22AWAT2BYTCOSCOSTCOS3T3233436解A由于YJXJHJ22YJ14JJ93J因此HJ212XJ86J13JJ311BHJ22JJ4324TT所以HTEEUT2YJ93JC由于HJ2XJ86JJ所以关联该系统的输入和输出的微分方程为2DYTDYTDXT68YT39XT2DTDTDT1,1SINT443解设YJFT0,1SINT2对GTXTCOST两边取傅立叶变换,其中1。T12GJFXTCOSTYJFXTCOS2T1YJ211XJFXTCOS