信号与系统课后答案西南交大.pdf

习题 1 参考答案 1.1非周期信号 周期信号，周期为 2 非周期信号 非周期信号 周期信号，周期为 0.5 周期信号，周期为 3 2 周期信号，周期为 1 非周期信号 1.2功率信号 2 1 P =功率信号1P =既不是功率信号也不是能量信号 能量信号1E =功率信号1P =功率信号1P = 1.5(a)4()3()3()2()24() 1()1()()(+=tututtuttuttutf (b)4()4()3()3()1()1()()(+=tuttuttutttutf (c)2()1()()( 1 =tututftf,其中) 2 cos(10)( 1 +=ttf (d)2()2 ()1(2)()(+=tuttuttutf (e)5()1(3)(+=kukukf (f)5()1()1()(+=kukukf k 1.7)2(t)10( 100 te a 2 )(t -1 0 4 1.10线性系统 非线性系统 线性系统 线性系统 非线性系统 非线性系统 线性系统 非线性系统 1.11时变系统 非时变系统 时变系统 时变系统 非时变系统 1.12非线性，非时变，因果系统 线性，时变，因果系统系统 线性，时变，因果系统系统 线性，非时变，因果系统 1.130 sin32)( 2 += ttety t 1.140 56)( 2 += teety tt f 0 1011)( 2 += teety tt 习题 2 参考答案 2.1 254 12 )( 23 2 + + = ppp pp pH 65 1 )( 2 + + = pp p pH 2.2)(4)()(2)(3)(tftftytyty+=+ + )()(3)(8)(4)(tftftytyty+=+ + 2.3 232 1 )( 23 2 + + = ppp pp pH 2.4 232 23 )( 2 2 + + = pp pp pH 2.5 44 3 )( 2 + = pp p pH 2.6 3)22( 3)( )( 23 + + = pppp pp pH 2.70 222)( 22 = tteeety ttt 2.80 )( 3 = teety tt x 0 )1()(+= tetty t x 0 ) 4 2(cos2)( 2 = ttety t x 0 ) 4 3(cos2)(=tttyx 0 )1017(20)( 2 += tetety tt x 2.90 57)( 2 = teety tt x 0 )54(6)( 3 += tetety tt x 0 5 . 025 . 1)( 2 += teety tt x 2.10)( )5 . 05 . 0()( 3 tueeth tt +=)( )(tuteth t = )()2(sin 2 3 )( 2 tuteth t = )()3(sin 3 1 )(tutth=)( )()( 22 tueteeth ttt = 2.11)( )( )22()( 2 ttueeth tt += )( )( 3 tuteth t = )( )32()( 2 tueeth tt += 2.120 22)( 32 = teety tt x 0 6 1 2 1 3 2 )( 2-3 += teety tt f 0 6 1 2 5 3 8 )( 2-3 += teety tt 0 22)( -32 = teety tt x 0 22)( -322 += teetety ttt f 0 44)( 3-22 += teetety ttt 2.14 + 32 4 32 21 4 12 10 4 1 2 2 t tt t t tt + + + 32 6 9 20 6 12 01- 6 1 36 1 2 2 t t t t t t t 2.150 33)( -2 = teety tt x )( )( )2()( 2 ttueeth tt += 0 2 5 2 2 1 )( -2 += teety tt f 0 2 5 2 5 )( 2- += teety tt 2.160 31)(+= ttety t 0 )72(1)(+= tetty t 0 )41()(+= tetty t 0 )52()(+= tetty t 2.170 2 5 2 1 32)( 2 += teetty tt 0 6 1 2 3 3 2 )( 32 += teeety ttt 2.180 12)(+= ttetety tt 2.19 )( 3 1 )( 3 1 )( 2 tuetuety tt f += 2.20 )() 3 4 ()( 3 1 )( 22 tueetuety ttt f += 2.21 )()(tuety t f = 2.22 )( 3 tut )() 4 1 4 1 2 1 ( 2 tuet t + )() 2 1 2 cossin (tue tt t + )3( )3( 2 tueteu t + )45(cos+t 2.23)2( 1cos )( cos1 + tu t tu t )1()1( 2 1 )( 2 1 22 tuttut )1( 2 34 2 + + tu tt )2( 2 1026 + tu ee t )2()2()1()1(2)(+tuttutttu )()( 2 tuee tt 2.240 cossin)( 32 += teettty tt 0 23)( 32 = teety tt x 0 34cossin)( 32 += teettty tt f 自由响应0 32 + tee tt 强迫响应ttcossin+ 暂态响应0 32 + tee tt 稳态响应ttcossin+ 0 2)( 322 += teetety ttt 0 23)( 32 = teety tt x 0 )( 322 += teetety ttt f 自由响应 2 32tt ee 强迫响应 t te 2 暂态响应0 2 322 + teete ttt 稳态响应0 习题 3 参考答案 3.10 0 =a 2 sin 4 n n an=0= n b )7cos 7 1 5cos 5 1 3cos 3 1 (cos 4 )(+=tttttf 0= n a n b n n 2 )1( 1+ = )sin4 4 1 sin3 3 1 sin2 2 1 sin( 2 )(+=tttttf 3.2 A F= 0 2 cos )1( 2 n n A Fn = tjn n e n n A tf 0 2 1 2 cos )( = = A F 2 0 = 2 1 sin )1(2 1 sin )1( + + + = n n An n A Fn 3.3)(tf为偶函数、奇谐波函数，0= n b，只有 0 a、 2 a、 4 a、 )(tf为奇函数，只有 n b 3.5 tjtj eetf 00 22 4 1 4 1 2 1 )( += )( 2 1 )( 2 1 )( 2 1 )( ) 3 6() 3 6() 4 4() 4 4( 22 + += tjtjtjtj tjtj ee j eeee j tf 3.6 tjtjtjtjtjtj eeeeeetf 000000 3223 2 1 2 3 1 2 3 2 1 )(+= 3.72)10cos(3)302cos(2)204cos()( 000 +=ttttf 3.8)2( 2 22 2 jj ee A j A e a A a A ja + 22 A 2 cos 2 222 3.9 3 3 + j e j 2)(+j e )()( 00 + 3 )(33 9 j e j +)( )1( 23 2 2 jj eej ) 31 ( 12 22 j e j j + + 3.10)8Sa(2)(2 2 2 0 2 2 0 2 )( )( + + j j e 1)(Sa(-1)(Sa(+j 3.11) 2 ( 2 1 jF -j ejF 3 )( ) 2 ( 2 1 jF d d j)(3-)(jFjF d d j -j ejF)( -j e jF 2 3 ) 2 ( 2 1 +)()(jF d d jF) 3 2 ( 3 1 jF j e jF F -j 2 ) 2 ( )()0( )(jF d d je-j 3.12)( 2 sin 4 0 n n n = )( 2 )1( 0 1 n n n n = + 3.13) )(Sa( cc c t-tk ) 2 (sin 2 2 t t k c 3.14)(2)(3)(2 5 tuetuet tt +t 2 1 )2()2( 2 1 +tt)3()3( 2 1 +tutu )Sa( 3 t t 1 )( )(5sgn. 0)( )(5sgn. 0 )( + tuettuet tt )sin( j 1 0t ) 1t ( j e )1t( 1)sin( t )()( 2 tuetue tt 3.15 )( sin 22 2 3.16 65j)j( 1j 2 + + )(2)( 32 tuetue tt + 3.17 )()( 42 tuetue tt )( 9 1 )( )( 9 8 )( 3 2 )( 42- tuetuetuetutety tttt f += 3.18 13 5cos312sin3tt 3.19 )cos(t )sin(3t 3.20 )()( 2 tuetue tt )(tute t )()(sgn 2 1 2 1 tuet t +)( 2 1 )( 2 1 tuetue tt + 3.22 )(sat )(sat)(sat 3.23 160 Ts = 160 fs= 100 Ts = 100 fs= 160 Ts = 160 fs= 1000 Ts = 1000 fs= 3.24 0 2 0 2 0 0 0 0 2 3.25 16 Ts = 16 fs= 4 Ts = 4 fs= 习题 4 参考答案 4.1 s e1 -2s + 2 3)(s 1 + s 3e 2e -3s 2s- + 1s 3s 2 + )sin1scos1(e-1 1s 1 s- 2 + + 84ss 2 2 + 2526ss 13ss 4 3 + + 4s 2-s 2 2 2 + 4.2 22 4)(s 4s + 3 e s 3 1 - 4)-(s- 2 e 1(s 2s ）+ + 5s2s 2 2 + 2s2s )1s(s 2 2 + + 22 2 5)s4(s 3s4s + + 1)(s 2 3 - 3 2 e 14(s 29s309s + + + ） s50s10s 5 23 + 4.3 (a)Tsee-(1 Ts 1 Ts-Ts- 2 (b)e2e-(1 Ts 2 Ts- s 2 T 2 + (c)e(1 4s 2 2s- 22 + + (d)see(-1 s 1 s-s 2 + 4.4 (a) s - e1 1 + (b)e-(1 )es(1 1 s 2 T s 2 T + (c) )e-1)(1(s 1 s-2 + (d)e2e-(1 )e-(1Ts 2 Ts- s 2 T Ts-2 + 4.5 )()( 3 tuetue tt +)(5)(2)( 2 tuetuet tt + )()3t(sin)t(u)3(cos3 22 tuete tt )( 2 5 )(2)( 2 1 2 tuetuetu tt + )()3t(sin)t(u)3(cos)(tuttu+)(4)(t)t(2u)(tuetuet tt + )(2)(3)t(2u)( 2 tuetuet tt +)()()t(tutuetu t + )()2t(sin 2 1 )t(u)2(cos2)(2tuetetu tt )( 2 1 )(2 2 tuettute tt 4.6 )()cos(2tutt)()33tt 2 1 ()( 2 tueeet ttt + )1()(tutu)1()1()()( )13()12(32 + tuetuetuetue tttt )1()()sin(tutut)2( 3 tue t 4.7 910ss 32s 2 + + 56s5ss 12s3s 23 2 + + 4.8 )()2t(sine 2 7 )t(u)2(cos3e)(3 t -t tuttu+ )3()3)-2(t(sine 10 7 )3-t(u)32(cose 5 3 )3( 5 3 3)-(t-3)- t ( + tuttu 4.9 )()e23e-(3e -3t-3t2t tut )2()e 2 1 2ee 2 3 (e 2)- t (42)-3(t-2)-2(t-8 + tu 4.10)()3e3e-(-2te -t-2t2t tu+ 4.11 0 t)(1)(+= tety t x 0 4 1 4 1 2 t )( 3- += teeety ttt f 0 4 1 )t 2 3 4 3 ()( 3- += teety tt 0 3t)(1)(+= tety t x 0 )sin(3 25 2 -)cos(3 50 3 50 3 t 10 3 )( - += ttteety tt f 0 )sin(3 25 2 -)cos(3 50 3 )3t 50 53 (t 10 3 )( - += ttteety tt 4.12 0 t 2 3 3 14 4 29 12 43 )( 4-54-2 += teeeety tttt 0 9 4 9 5 t 3 7 )( 4- += teeety ttt 4.13)()sin( - tutete tt 4.140 (A)0.02)( 40 = teti t 0 (V)2.4)(24. 0)( 40 = tettu t L 4.150 2 3 2 3 1)( 32 = teetu tt 4.16)()2(3 - tue t 4.17 )2()2(2)1()1(4)(2)(+=tuttutttutyf 4.18 321232121 321 1 )( HHHGHHGHH HHH sH + = 4.20 (a) 214 2 32243213 54321 1 )( GGHHHGHHHGH HHHHH sH + = (b) 32ss 12s3s 2 2 + + 1 (c) 2431 1687168711654716547168321654321 1GHHG GHHHHHHHHGHHHHHHHHHHHHHHHHHHHHH + (d) 121 GHL=, 242 GHL=, 363 GHL=, 47654324 GHHHHHHL= )()()(1 43232143423231214321 LLLLLLLLLLLLLLLLLLLL+= 876543211 HHHHHHHPH=1 1 = = 87654321 HHHHHHH H(s) H 4.21)(-GHHL 1321 =,)(-GHHL 2432 =,)(-GHL 353 =, 315323524313231321 GGHHHGHGHHGHH1)(1+=+=LLLLL 1HHHH1P 543211 =H1 1 = 1111P2= 243132212 GHHGHH1)(1+=+=LL + = 24313254321 GHHGHH1HHHH H(s) H 4.28 系统稳定 4.29 系统不稳定系统稳定 系统不稳定系统临界稳定 系统不稳定系统不稳定 4.30 系统不稳定系统不稳定 系统不稳定系统临界稳定 4.314k 时系统稳定 4.32 系统稳定系统不稳定 系统临界稳定 第五章的答案 5.3(a)1(5 . 0)1(5 . 0)( 1 +=kkkf (b)3(5 . 0)2(5 . 0)1(5 . 0)( 2 +=kkkkf 或)4(5 . 0)1(5 . 0)( 2 +=kukukf (c)8()5(2)5()(2)( 3 +=kukukukukkf 5.4)4(3)3(2)2()1()(2)(+=kkkkkkf 5.5(1)是，周期14=N(2)不是(3)是，周期30=N(4)是，周期3=N 5.6(1)、(2)、(3)、(4)、(7)、(8)、(9)是线性系统，(5)、(6)不是。. 5.7(2)、(5)、(6)因果系统，(1)、(2)、(5)、(6)是时不变系统。 5.11 (a)()(3 . 0)(kfkyky=+(b)()(7 . 0)1(kfkyky=+ (c)()2(2) 1(3)(kfkykyky=+ (d)1()()2(6)1(5)(+=+kfkfkykyky 5.12 (1) k x ky)2(5 . 0)(=0k(2) k x ky)5 . 0(2)(=0k (3) k x ky)7 . 0(5 . 0)(=0k(4) kk x ky)4 . 0(3 . 0)3 . 0(4 . 0)(=0k (5) 11 32)( + = kk x ky0k(6) k x ky)3(3 . 01 . 0)(+=0k (7) k x ky22)(=0k(8) k x kky)3 . 0)(2 . 01 . 0()(=0k 5.13(1) kk x ky)5 . 0()2 . 0(2)(=0k (2) kk x kky)4 . 0(3)1 . 0(2)(=0k改：94 . 0 )2(= x y 5.14 (1)()1 . 0()(kukh k =(2)1()3 . 0()( 1 = kukh k (3)()2( 3 2 )1( 3 1 )(kukh kk +=(4)()3()(kukkh k = (5)()2(3)3(2)3()( 1 kukkh kkk += (6)() 4 3 sin()2()()1()1( 2 )(kukkujj j kh kkk =+= 5.15(1)(2ku k (2)1( 6 )1( 3 cos+kuk (3)( 11 1 ku a a a a k + (4)1( 11 1 )1 . 0( 11 10 1 + ku k (5) 2 sin( 2 1 ) 2 cos( 2 1 2 1 kk +(6) kk )3(3)2(2+ 5.16 k ,.0 , 0 , 2 , 6 ,12,12,12,10 , 6 5.17 (1)()2()1()(kukh kk =(2)()2( 3 2 )1( 2 1 6 1 ku kk + 5.18 (1)(*)()( 21 khkhkh= (2)()5 . 0()( 1 kukh k =)()1()( 2 kukh k =)()1(2)5 . 0()(kukh kk += 5.19系统的零状态响应)()8 . 0(2)4 . 0()(kuky kk f += 5.20 系统的零状态响应 kkk f ky)1()2( 9 2 )7 . 0( 9 7 )(+= 改:右边的是 H2(E),其输出进入加法器输入端加-号. 5.21 (1)系统的单位阶跃响应)(10)5(32)(kukg kk += (2)系统的零状态响应 )3(10)5(32)(10)5(322 )3()(2)( 33 += = kuku kgkgky kkkk f 5.22 系统的完全响应 )()3 . 0(5 . 1)1 . 0(25 . 0 )5 . 0(25. 1)1 . 0()5 . 0()(kuky kkkkk +=0k 5.23 系统的完全响应 )( 2 cos)2(41 3 1 )2()2()(kukjjky kkk += 0k 5.244.3)0 (=y82.3) 1 (=y036.4)2 (=y 5.25)(2)1()3()4(2)2(kfkfkfkfky+=+ 5.26 系统的零状态响应)()1()(kukTkyf+= 5.27 系统的单位响应)()3 . 0(2)(kujjkh n = 5.28 系统的单位响应)(*)()(*)()( 4321 khkhkhkhkh+= 5.29 系统的单位响应)()2 . 0()(kukh k = 5.30(1)-3(2) k 9 , 13,14, 5 , 1 (3)()2(2ku k 5.31)(1)2 . 0(5 . 0)3 . 0()( 2 kuky kkk += 5.32)(2)2 . 0(5 . 0)3 . 0()(kuky kk +=0k 5.33 k NN N k NN N ku)2( 5 . 02 )5 . 0(10 )5 . 0( 5 . 02 )2(10 )( =k=0,1,2,3,N 第六章习题答案 6.1 (1) 3 2 z(2) 1 7 . 0 4 z z (3) 321 32 +zzz(4) 21 1 . 02 . 05 . 0 +zz (5) 09 . 0 2 2 z z (6) 1 2 2 z z (7) 1 5 . 0 4 4 z z (8) 22 az az 6.2(1) 1 1 +z(2) 1 1 + z zz (3) 5 . 0+z z (4) 2 2 + z z (5) 2 )1(z z (6) 2 1 )1( 2 z zz (7) 2 )3 . 0( 3 . 0 + z z (8) 1 )12(5 . 0 2 + zz zz (9) 1 2 2 +z z (10) 12 )226( 4 1 2 + + zz zz (11) 12 2 +zz z 注:改为:)2cos(kk (12) 22 2 )1( )1( + z zz (13) 04 . 0 2 . 0 2 +z z (14) 5 . 0+z z 6.3 (1)()2( 5 2 )3( 5 3 ku kk +(2)()2 . 0()2 . 0(kuk kk + (3)2(2)2()2(2+kukuk (4)2()4()1()4( 21 + kuku kk 6.5(1)1)0 (=f0)(=f(2)0)0 (=f0)(=f 6.6(1) k )2 . 0(2) kk )2()1(3)1()5 . 0( 1 ku k (4)1()4 . 0(3)()3 . 0(2 1 kuku kk (5) kk e3)( 2 + (6)2(25 . 0 )1()(+kkk(7) kk ku)3( 4 5 )( 4 1 )2(+ (8)43()43(5 . 1 kk jj+注:23 改成 9 (9)()5 . 0( 1 kuk k (10)()5 . 0()5 . 0(2)(kukkf kk += 6.7(1)1)0 (=f、2.0) 1 (=f、25.0)2 (=f (2)2)0(=f、8)1(=f、29)2(=f (3)1)0 (=f、2) 1 ( =f、1)2 (=f 6.8(1)1( )2( ku k k (2)1( )4 . 0( ku k k (3)1( )3( )1( 1 + ku k k k (4)1( )5 . 0( )1( 1 + ku k k k 6.9(1) k ky)3()(=(2) k ky)2()(=(3) k ky)2(5 . 05 . 0)(= (4) kk ky)3 . 0()1 . 0(2)(= (5)()3 . 0( 13 6 )2 . 0( 2 1 26 25 )3 . 0(5)2 . 0(6)(kuky kkkk += (6)()3( 10 1 )3( 2 5 )2( 5 8 )3(12)2(10)(kuky kkkkk += 6.10(1) 6 . 0 )( = z z zH)(6 . 0)(kukh k = (2) 1 . 0 1 )( + = z zH)1()1 . 0()( 1 = kukh k (3) 04 . 0 4 . 0 )( 2 + = zz z zH)()2 . 0()(kukkh k = (4) 03 . 0 2 . 0 2 . 0 )( 2 + + = zz z zH)()1 . 0( 2 15 )()3 . 0( 6 5 )( 3 20 )(kukukkh kk += 6.11(1)()1(6 . 0)(kfkyky=+ (2)答案不唯一，)(1 . 0)1()2(06 . 0 )1(7 . 0)(kfkfkykyky+=+ 6.13(a) 2 )( + = z z zH(b) 3 . 01 . 0 15 . 0 )( 2 + + = zz z zH (c) 45 )( 2 2 + + = zz zz zH(d) 08. 09 . 0 8 . 09 . 0 )( 2 2 + + = zz zz zH 6.14 )3)(2( )1( )( + + = zz zz zH)()3(2)2()(kukh kk += 6.15(1)()2 . 0(2)1 . 0(ku kk + (2)()5 . 0()( 1 kukky k f = (3)()8 . 0( 9 5 9 5 kuk k + (4)()4 . 0( 27 50 )5 . 0( 27 2 9 16 ku kk +注:题中 0.6f(k)改为 0.6f(k-2) (5)()6 . 0( 9 40 )3 . 0( 36 25 )5 . 0( 4 15 ku kkk + (6)()3( 2 7 )2(6)3( 4 1 )3( 4 23 kuk kkkk + 6.16 系统的零状态响应)()3 . 0(2 3 5 )2 . 0( 3 1 ku kk +注:题中 0.3k改为 (-0.3)k 6.17 系统的单位响应)()4 . 0(5 . 2)()2 . 0(15)( 5 . 12)(kukukkh kk += 6.18 系统的单位响应)()2( 10 3 )()3( 15 2 )( 6 1 )(kukukkh kk = 6.19 系统的输入序列)()2 . 0()(kukf k = 6.20 该系统的差分方程)2(2)()2(2) 1()(+=kfkfkykyky 6.21 此时输入序列)(24)(kukf k = 6.22 (1) 系统的单位响应)(2)2(3)(kukh k = (2) 系统的零输入响应)(3)2(6)(kuky k x =0k (3) 系统的零状态响应)(1)2(6)3(6)(kuky kk f += 6.23 零输入响应分量 k )3(，零状态响应分量)( 3 )3(3kue e e e kk + + 6.24 系统的输出： )()249 . 1 cos(3162 . 0 )2( 15 1 )1( 6 1 )1()2()(kukky kkkk += 6.25 (1)、(2)、(4)、(5)、(6)不稳定(3)稳定。 6.26 (1) 5 . 0 13 )( 2 + + = kzz z zH(2) 2 3 k 6.27 系统的稳态响应)781. 0 3 sin(4 . 0)1475. 0 6 cos(469 . 1 )(+=kkkyss 6.28(1) 系统函数 18 . 0 3 . 0 )( 2 = zz z zH 单位响应)()6 . 0( 9 10 )3 . 0( 9 10 )(kukh k += (2)稳定