答案习题解答第一章

33222111-3-2- 4 图2n 4n x(n) 0n0 042 42 n84 42 42 n （c）ej16

1j(n)1 25解：（a）1.2,（c）1.6,

20，这是有理数，故是周期序列，且周期为T205，这是有理数，故是周期序列，且周期为T54

3, 3,

14，这是有理数，故是周期序列，且周期为T14314，这是有理数，故是周期序列，且周期为T143121681111解：(a)xn的波形如下图所示：))x1nxe(nxo(nx1(nxe(n)xo(nxn。另外，由波 xenxonxnxenxon。

n

（e）0.3nu(n)（f）0.3nu(n1 解：(a)由于n0h(0)0， N1，h(n)

n0

Nn0n而正项级数2n

h(n)，由于n0h(0)0， 1limN1，

N

n0 收敛，根据正项级数收敛的充要条件，可知h(n)，所以该系统 由于n0h(0)0 Nh(n)3nlim3n N由于n0h(0)0 1 h(n)3n n03 1 3由于n0h(0)0

h(n)

0.3n 1

10

h(n)0.3n n13由于n0h(0)0

h(n)(n4)1（a）y(n)3x(n) （b）y(n)x(n1)（c）y(n)x(n)0.5x(n1 y(nn0)x(nn0)0.5x(nn0,,设h1n40.5nunun3图解：（a）hnh1nh2nh3nh4nh5nh2nh3nh4nn1unnununh1nh2nh3nh4n4n2n1n2uhn7un24n6n1n4n或者hn7un2nn14nynxnhn7xnun2xnxn14xn3图h2nunun2nn1，hnh1nn2n1n2h1n2h1n1h1n2h1nhn2h1n1h1n2已知hn，求h1n。上式是一个递推 系统，最后可得，当n0,1,2,3,4,5,6,,"时，h1n1,3,3,2,1,0,0,"。(b)ynxnh1nh2nh2nh1nxnh2nh2h1nnn1n2n1nh1nnn1n2nh1nh1n1h1n2h1n图y2nxnhn"0,0,6,11,15,18,14,10,6,3,1,0,0,"y3nxnhn"0,0,1,2,22,1,0,0,"y4nxnhn"0,0,1,2,2,2,1,0,0,"（a）x(n)h(n)h(n)x(n（b）x(n)[h1(n)h2(n)][x(n)h1(n)]h2(n（c）x(n)[h1(n)h2(n)]x(n)h1(n)x(n)h2k令knkxnhnk

xnkhkhnxxn[h1nh2n]xn[h2nh1n]

xkh2nmh1n

k xn[h1nh2n]h2kxmh1nmk k

kh2n[xnh1n][xnh1n]h2xn[h1nh2n]xk[h1nkh2nkk xkh1nkxkh2nk k kxnh1nxnh2 h(n)2R4(n（d）x(n)R5(n

x(n)1

0n h(n) 2n

3n 0n（f）x(n)

h(n) ①n0，y(n)0n②0n3，y(n)1n3③4n7，y(n) 18n④7ny(n0

n0,n0n4n（b）ynxnhn2R4nnn22R4n2R4n（c）ynxnhnn20.5nRn0.5n2Rn240.5nRn （d）ynxnhnRm0.5nmunm0.5nRm0.5mun ①n0，y(n)0nn②0n4，y(n)0.5n4

m10.5n10.5n20.5n10.5 1 ③5n，y(n)0.5n0.5m10.510.5n31

n0n5 3

3由题意，xnanR9n hnR5ynxnhnamRm3Rn mn①3n0，y(n) am

1nn②1n5，y(n)am55③6n9，y(n)am

11 a3an1 3n 1 1 1nan41 6n n n y1(n) n n 2解：（a）y1nx1nh1nx1mh1nm3h1n2 , ,n1, 3（b）y2nx2nh2nx2mh2nmh2n

n1,2,

n4,3,2

y(n)0.5y(n1)x(n)0.5x(n1求系统的单位脉冲响应hnhn0.5hn1n0.5nhn0.5n1un1ny(n)ay(n1)x(n 0a nn

y20ay2110

n1,y21ay200（3）xnnn1y3ny3nay3n1nnn

n1,y31ay3010a1而习题1-14给出了初始条件？8"3 3 ynxnhnn0,yn38

8

xm uum xm ,n88n

yn31x0.51m30.5xxm8m

mn2,yn3x0.52m30.52x0.5xxm8

3 8 8

xnmhnanuna1，试求系统的单位阶跃响应。解：ynanununamumun。m。n非零值区间为0m,mnynam

11h(n)anu(n ynxnhnanu(nu(nanu(nu(n 1 1 利用习题1.17的结果，可得y a 1 1 1y(n)0.7y(n1)2x(n)x(n2yn0.7yn12nnn

y00.7y1202n1,y10.7y021120.7n

n

y40.7y30.02hnyn0.020.7n2un22n1.4nhunhnu1 0.1360.09520.7n1

T

2

10.05s (b)xatsin40nTtnT xn的数字频率为0.82，xnN5xnsin0.8n3x1tcos(2t)，x2tcos(6t)，x3tcos(10t采样频率为s8rads3个采样输出序列，并比较这些结果，画出波形及采样点位置， 解：因为s8s2s6s10，x1 xˆ xtitnTcosnitn 4 nitn 4xˆ(t)xtinitn 42 2 xˆ(t)