2022年信号与系统奥本海姆习题答案 .pdf

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1、Chapter 1 Answers 1.6 (a).No Because when t0, )(1tx=0. (b).No Because only if n=0, 2nxhas valuable. (c).Yes Because kkmnkmnmnx414444kmknmkn)(41)(4kknkn414N=4. 1.9 (a). T=/5 Because 0w=10, T=2/10=/5. (b). Not periodic. Because jtteetx)(2, while teis not periodic, )(2txis not periodic. (c). N=2 Becaus

2、e 0w=7, N=(2/0w)*m, and m=7. (d). N=10 Because njjeenx)5/3(10/343)(, that is 0w=3/5, N=(2/0w)*m, and m=3. (e). Not periodic. Because 0w=3/5, N=(2/0w)*m=10m/3 , it s not a rational number. 1.14 A1=3, t1=0, A2=-3, t2=1 or -1 dttdx )(is Solution: x(t) is Because kkttg)2()(, dttdx )(=3g(t)-3g(t-1) or dt

3、tdx)(=3g(t)-3g(t+1) 1.15. (a). yn=2xn-2+5xn-3+2xn-4 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 1 页，共 24 页Solution: 3212222nxnxny 321211nyny44322134221111nxnxnxnx42 35 22111nxnxnxThen, 423522nxnxnxny(b).No. For it s linearity. the relationship between 1nyand 2nxis the same in-out relationship wit

4、h (a). you can have a try. 1.16. (a). No. For example, when n=0, y0=x0 x-2. So the system is memory. (b). yn=0. When the input is nA, then, 22nnAny, so yn=0. (c). No. For example, when xn=0, yn=0; when xn=nA, yn=0. So the system is not invertible. 1.17. (a). No. For example, )0()(xy. So it s not cau

5、sal. (b). Yes. Because : )(sin()(11txty, )(sin()(22txty)(sin()(sin()()(2121tbxtaxtbytay1.21. Solution: We have known: (a). 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 2 页，共 24 页(b). (c). (d). 1.22. Solution: We have known: (a). (b). (e). 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 3 页，共 24 页(g

6、) 1.23. Solution: For )()(21)(txtxtxEv)()(21)(txtxtxOdthen, (a). (b). (c). 1.24. 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 4 页，共 24 页Solution: For: )(21nxnxnxEv)(21nxnxnxOdthen, (a). (b). 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 5 页，共 24 页(c). 1.25. (a). Periodic. T=/2. Solution: T=2/4=/2

7、. (b). Periodic. T=2. Solution: T=2/=2. (d). Periodic. T=0.5. Solution: )()4cos()(tutEtxv)()(4cos()()4cos(21tuttut)()()4cos(21tutut)4cos(21tSo, T=2/4=0.5 1.26. (a). Periodic. N=7 Solution: N=m*7/62=7, m=3. (b). Aperriodic. Solution: N=mm16*8/12, it s not rational number. (e). Periodic. N=16 Solution

8、 as follow: 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 6 页，共 24 页)62cos(2)8sin()4cos(2nnnnxin this equation, )4cos(2n, it s period is N=2*m/(/4)=8, m=1. )8sin(n, it s period is N=2*m/(/8)=16, m=1. )62cos(2n, it s period is N=2*m/(/2)=4, m=1. So, the fundamental period ofnxis N=(8,16,4)=16. 1.31.

9、 Solution Because )()1()(),2()()(113112txtxtxtxtxtx. According to LTI property , )() 1()(),2()()(113112tytytytytytyExtra problems: Sketch tdttxty)()(. 1.Suppose Solution: 2. Suppose Sketch: (1). )1(2)1()3()(ttttg(2). kkttg)2()(精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 7 页，共 24 页Solution: (1). (

10、2). Chapter 2 2.1 Solution: Because xn=(1 2 0 1)0, hn=(2 0 2)1, then (a). So, 4222 124 121nnnnnny(b). according to the property of convolutioin: 212nyny(c). 213nyny精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 8 页，共 24 页2.3 Solution: *nhnxnyknhkxkkkknuku22)21(2211)21()21()21(12)2(0222nununnkk)21(1

11、21nunthe figure of the yn is: 2.5 Solution: We have known: elsewherennx.090.1，, elsewhereNnnh.00.1，,(9N) Then, 10nununx, 1Nnununhkknukhnhnxny*kknuknuNkuku)10)(1(So, y4 kkukuNkuku)64)(1(4,.14,.1400NNkNk=5, then 4NAnd y14 kkukuNkuku)414)(1(精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 9 页，共 24 页14,.1

12、14,.11455NNkNk=0, then 5N4N2.7 Solution: 2 ky nx k g nk（a） 1x nn, 2 1 2 2kky nx k g nkkg nkg n(b) 2x nn, 2 2 2 4kky nx k g nkkg nkg n(c)S is not LTI system. (d) x nu n,0 2 2 2 kkky nx k g nku k g nkg nk2.8 Solution: )1(2)2(*)()(*)()(tttxthtxty)1(2)2(txtxThen, 精选学习资料 - - - - - - - - - 名师归纳总结 - - - -

13、- - -第 10 页，共 24 页That is, otherstttttttty,.010,.2201,.41.,.412,.3)(2.10 Solution: (a). We know: Then, )()()(ttth)()(*)()(*)()(tttxthtxty)()(txtxthat is, So, otherstttttty,.011,.11,.0,.)(b). From the figure of )(ty, only if 1, )(tywould contain merely there discontinuities. 2.11 Solution: (a). )(*)5

14、()3()(*)()(3tuetututhtxtytdtueudtueutt)()5()()3()(3)(3ttttdetudetu5)(33)(3)5()3(精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 11 页，共 24 页5,.353,.313.,.0315395)(33)(3393)(3teededetedetttttttttt(b). )(*)5() 3()(*)/ )()(3tuettthdttdxtgt)5() 3()5(3)3(3tuetuett(c). It s obvious that dttdytg/)()(. 2.12 S

15、olutionktktkttuekttuety)3(*)()3(*)()(kktktue)3()3(Considering for 30t,we can obtain 303311 )3()(eeeektueetytkktkkt. (Because kmust be negetive ,1)3(ktufor 30t). 2.19 Solution: (a). We have known: 121nxnwnw(1) 1nwnyny(2) 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 12 页，共 24 页from (1), 21)(1EEEHfro

16、m (2), EEEH)(2then, 212212)21(1)21)()()()(EEEEEEHEHEH22 1)21(nxnynynybut, 143281nxnynyny143)21(:.812or141(b). from (a), we know )21)(41()()()(221EEEEHEHEH21241EEEE)41()21(2nunhnn2.20 (a). 1 1)0cos()cos()()cos()(0dtttdtttu(b). 0 dttt) 3()2sin(50has value only on 3t, but 5,03dttt) 3()2sin(50=0 (c). 0

17、641551)2cos()()2cos()1(dtttudu64)2cos()(dttt0|)2(scott0|)2sin(20tt精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 13 页，共 24 页2.23 Solution: kthkTtthtxty)(*)()(*)()(kkTth)(2.27Solution ( )yAy t dt,( )xAx t dt,( )hAh t dt. ( )( )*( )( ) ()y tx th txx td( )( ) ()( ) ()( )()( )( )( )( )yxhAy t dtxx td dt

18、xx tdtdxx tdtdxxd dxdxdA A精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 14 页，共 24 页2.40Solution (a) ()( )(2)tty texd,Let ( )( )x tt,then ( )( )y th t. So , 2()(2)(2)( )(2)( )(2)ttttth tededeu t(b) (2)( )( )*( ) (1)(2)*(2)ty tx th tu tu teu t(2)(2)(1)(2)(2)(2)ttueu tdueu td22(2)(2)12(1)(4)ttttu tedu

19、 ted(2)2(2)212(1)|(4)|ttttu teeu tee(1)(4)1 (1) 1 (4)tteu teu t2.46 SolutionBecause )1(2)1(2)(33tudtdetuedtdtxdtdtt) 1(2)(3) 1(2)(333tetxtetxt. From LTI property ,we know )1(2)(3)(3thetytxdtdwhere )(this the impulse response of the system. So ,following equation can be derived. )()1(223tuethetFinall

20、y, ) 1(21)()1(23tueetht2.47 Soliution According to the property of the linear time-invariant system: (a). )(2)(*)(2)(*)()(000tythtxthtxty(b). )(*)2()()(*)()(00thtxtxthtxty)(*)2()(*)(0000thtxthtx012y(t)t4精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 15 页，共 24 页)2()(00tyty(c). ) 1() 1(*)(*)2() 1(*)2(

21、)(*)()(00000tytthtxthtxthtxty(d). The condition is not enough. (e). )(*)()(*)()(00thtxthtxtydthx)()(00)()()(000tydmmthmx(f). )()()(*)()(*)()(*)()(000000tytythtxthtxthtxtyExtra problems: 1. Solute h(t), hn (1). )()(6)(5)(22txtytydtdtydtd(2). 12 122nxnynynySolution: (1). Because 3121)3)(2(1651)(2PPPPP

22、PPHso )()()()3121()(32tueetPPthtt(2). Because )1)(1(1) 1(22)(22iEiEEEEEEEEHiEEiiEEi1212so )1()1(21212nuiiikiEEiiEEinhnn精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 16 页，共 24 页Chapter 3 3.1 Solution: Fundamental period 8T.02/8/400000000033113333( )224434cos()8sin()44jktjtjtjtjtkkjtjtjtjtx ta ea ea

23、ea ea eeejejett3.2 Solution: for, 10a, 4/2jea, 4/2jea, 3/42jea, 3/42jeanNjkkNkeanx)/2(njnjnjnjeaeaeaeaa)5/8(4)5/8(4)5/4(2)5/4(20njjnjjnjjnjjeeeeeeee)5/8(3/)5/8(3/)5/4(4/)5/4(4/221)358cos(4)454cos(21nn)6558sin(4)4354sin(21nn3.3 Solution: for the period of)32cos(tis 3T, the period of)35sin(tis 6Tso th

24、e period of)(txis 6 , i.e. 3/6/20w)35sin(4)32cos(2)(tttx)5sin(4)2cos(21200twtw)(2)(21200005522twjtwjtwjtwjeejeethen, 20a, 2122aa, ja25, ja253.5 Solution: (1). Because ) 1()1 ()(112txtxtx, then )(2txhas the same period as )(1tx, that is 21TTT, 12ww(2). 212111( )(1)(1)jkw tjkw tkTTbxt edtxtx tedtT精选学习

25、资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 17 页，共 24 页111111(1)(1)jkw tjkw tTTxt edtx tedtTT111)(jkwkkjkwkjkwkeaaeaea3.8 Solution: ktjwkkeatx0)(while: )(txis real and odd, then 00a, kkaa2T, then 2/20wand 0kafor 1kso ktjwkkeatx0)(tjwtjweaeaa00110)sin(2)(11taeeatjtjfor 12)(2121212120220aaaadttx2/21a)s

26、in(2)(ttx3.13 Solution: Fundamental period 8T.02/8/4ktjwkkeatx0)(tjkwkkejkwHaty0)()(00004,.0sin(4)()0,.0kkH jkkk000( )()4jkw tkky ta HjkweaBecause 48004111( )1( 1)088Tax t dtdtdtTSo ( )0y t. 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 18 页，共 24 页3.15 Solution: ktjwkkeatx0)(tjkwkkejkwHaty0)()(0dte

27、jkwHtyTatjkwTk0)()(10for 100,.0100,.1)(wwjwHif 0ka, it needs 1000kwthat is 12100,.1006/2kkand kis integer, so 8K3.22 Solution: 021)(1110tdtdttxTaTdttedttedtetxTatjktjktjkwTk1122112121)(10tjktdejk1121111121jketejktjktjkjkeeeejkjkjkjkjk)()(21jkkkjk)sin(2)cos(221kjkkjkjkk) 1()cos()cos(2210. k精选学习资料 - -

28、 - - - - - - - 名师归纳总结 - - - - - - -第 19 页，共 24 页3.34 Solution: 404402()( )1184416tjtj ttjttjtHjh t edteedte edteedtjjA periodic continous-signal has Fourier Series:. 0( )jktkkx ta eT is the fundamental period of ( )x t.02/TThe output of LTI system with inputed ( )x tis 00( )()jktkky ta HjkeIts coeff

29、icients of Fourier Series: 0()kkba H jk(a)( )()nx ttn.T=1, 0211kaT. 01/221/ 21( )( )1jkw tjktkTax t edtt edtT（Note：If ( )()nx ttnT,1kaT）So 2282(2 )16(2)4()kkba Hjkkk(b)( )( 1)()nnx ttn.T=2, 0,11kaT01/ 23/ 21/ 21/ 2111( )( )( 1) (1)2211( 1) 2jkw tjk tjk tkTkax t edtt edttedtTSo 241 ( 1) ()16()kkkba H

30、 jkk, (c) T=1, 02精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 20 页，共 24 页01/ 421/ 4sin()12( )jktjktkTkax t edtedtTk28sin()2()16(2) kkkba Hjkkk3.35 Solution: T= /7, 02/14T. ktjwkkeatx0)(tjkwkkejkwHaty0)()(00()kkba H jkwfor otherwisewjwH,.0250,.1)(,01,.17()0,.kHjkwotherwisethat is 0250250,.14kk, and

31、 kis integer, so 18.17kork. Let ( )( )y tx t,kkba, it needs 0ka,for 18.17kork. 3.37 Solution: 101() ( )212( )21312411511cos224njj nj nnnnjnnjnnnjjjH eh n eeeeeeeA periodic sequence has Fourier Series:2() jknNkkNx na e. N is the fundamental period of x n. The output of LTI system with inputed x nis 2

32、2() ()jkjknNNkkNy na H ee. 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 21 页，共 24 页Its coefficients of Fourier Series: 2()jkNkkba H e(a) 4 kx nnk.N=4, 14ka.So 2314()524cos()44jkNkkba H ek3165cos()42kbk3.40 Solution: According to the property of fourier series: (a). )2cos(2)cos(20000000tTkatkwaeaea

33、akktjkwktjkwkk(b). Because 2)()()(txtxtxEv2kvkkkaEaaa(c). Because 2)(*)()(txtxtxRe2*kkkaaa(d). kkkaTjkajkwa220)2()(e). first, the period of )13( txis 3TTthen 3)(1)13(131213120dmemxTdtetxTamTjkTtTjkTkTjkkmTjkTTjkTjkmTjkTeadmemxTedmeemxT221122211)(1)(13.43 (a) Proof:（ i） Because ( )x tis odd harmonic,

34、(2/)( )jkT tkkx ta e,where 0kafor every non-zero even k.精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 22 页，共 24 页(2/)()2(2/)(2/)()2TjkTtkkjkjkT tkkjkT tkkTx ta ea eea eIt is noticed that k is odd integers or k=0.That means ( )()2Tx tx t（ii）Because of ( )()2Tx tx t,we get the coefficients of Fourier

35、 Series222/200/ 222(/ 2)/2/ 20022/2/ 200111( )( )( )11( )(/ 2)11( )( )( 1)jktjktjktTTTTTTkTjktjktTTTTTjktjktTTkTTax t edtx t edtx t edtTTTx t edtx tTedtTTx t edtx tedtTT2/ 2011( 1) ( )jktTkTx t edtTIt is obvious that 0kafor every non-zero even k. So ( )x tis odd harmonic,(b) Extra problems: kkTttx)(

36、)(, T(1). Consider )(ty, when )( jwHis ( )x t112211t0.精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 23 页，共 24 页(2). Consider )(ty, when )( jwHis Solution: kkTttx)()(11T, 220Tw(1). ktjkktjkwkkekjHaejkwHaty20)2(1)()(02(for k can only has value 0) (2). ktjkktjkwkkekjHaejkwHaty20)2(1)()(0teetjtj2cos2)(122(for k can only has value 1 and 1) 精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 24 页，共 24 页