奧本海姆信號與系統(tǒng)習題參考答案電子科技大學

1、Chapter 1 Answers1.6 (a.NoBecause when t<0, (1t x =0.(b.NoBecause only if n=0, 2n x has valuable.(c.Yes Because -=-+-+=+k k m n k m n m n x 414444 -=-=k m k n m k n (41(4 -=-=k k n k n 414 N=4.1.9 (a. T=/5Because 0w =10, T=2/10=/5.(b. Not periodic.Because jt t e e t x -=(2, while t e -is not peri

2、odic, (2t x is not periodic.(c. N=2Because 0w =7, N=(2/0w *m, and m=7.(d. N=10Because n j j e e n x 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 -1dtt dx ( is Solution: x(t isBecaus

3、e -=-=k k t t g 2(, dt t dx (=3g(t-3g(t-1 or dtt dx (=3g(t-3g(t+1 1.15. (a. yn=2xn-2+5xn-3+2xn-4Solution: 3212222-+-=n x n x n y321211-+-=n y n y44322134221111-+-+-+-=n x n x n x n x423522111-+-+-=n x n x n xThen, 423522-+-+-=n x n x n x n y(b.No. For it s linearity. the relationship between 1n y an

4、d 2n x is the same in-out relationship with (a. you can have a try.1.16. (a. No.For example, when n=0, y0=x0x-2. So the system is memory. (b. yn=0.When the input is n A ,then, 22-=n n A n y , so yn=0. (c. No.For example, when xn=0, yn=0; when xn= n A , yn=0. So the system is not invertible.1.17. (a.

5、 No.For example, 0(x y =-. So it s not causal.(b. Yes.Because : (sin(11t x t y = ,(sin(22t x t y = (sin(sin(2121t bx t ax t by t ay +=+1.21. Solution:We have known: (a. (b. (c. (d.1.22. Solution: We have known:(a. (b. (e.(g 1.23. Solution:For (21(t x t x t x E v -+= (21(t x t x t x O d -= then,(a. (

6、b. (c .1.24.Solution: For: (21n x n x n x E v -+= (21n x n x n x O d -= then, (a. (b. (c. 1.25. (a. Periodic. T= /2. Solution: T=2 /4= /2. (b. Periodic. T=2. Solution: T=2 / =2. (d. Periodic. T=0.5. Solution: x(t = E v cos( 4t u (t So, T=2 /4 =0.5 1.26. (a. Periodic. N=7 Solution: N= (b. Aperriodic.

7、 Solution: N= 1 = cos(4t u (t + cos(4 (t u (t 2 1 = cos(4t u (t + u (t 2 1 = cos(4t 2 2 * m =7, m=3. 6 / 7 2 * m = 16m , its not rational number. 1/ 8 (e. Periodic. N=16 Solution as follow: xn = 2 cos( n + sin( n 2 cos( n + 4 8 2 6 in this equation, 2 cos( n , its period is N=2 *m/( /4=8, m=1. 4 sin

8、( n , its period is N=2 *m/( /8=16, m=1. 8 2 cos( n + , its period is N=2 *m/( /2=4, m=1. 2 6 So, the fundamental period of xn is N=(8,16,4=16. 1.31. Solution Because x2 (t = x1 (t x1 (t 2, x3 (t = x1 (t + 1 + x1 (t . According to LTI property , y2 (t = y1 (t y1 (t 2, y3 (t = y1 (t + 1 + y1 (t Extra problems: 1. Supp