奥本海姆 信号与系统 习题参考答案 电子科技大学

Solution:

For:

1(x[n]?x[?n]) 21Od{x[n]}?(x[n]?x[?n])

2Ev{x[n]}? (b).

then, (a).

(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)?Ev{cos(4?t)u(t)}

1?{cos(4?t)u(t)?cos(4?(?t))u(?t)} 21?cos(4?t){u(t)?u(?t)} 21?cos(4?t) 2 So, T=2?/4?=0.5 1.26. (a). Periodic. N=7

Solution: N=

2?*m=7, m=3. 6?/72?*m?16m?, it’s not rational number. 1/8(b). Aperriodic.

Solution: N=

(e). Periodic. N=16 Solution as follow:

x[n]?2cos(?4n)?sin(?8n)?2cos(?2n??6)

in this equation,

2cos(?4n), it’s period is N=2?*m/(?/4)=8, m=1.

sin(?8n), it’s period is N=2?*m/(?/8)=16, m=1.

?2cos(?2n??6), it’s period is N=2?*m/(?/2)=4, m=1.

So, the fundamental period ofx[n] 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. Suppose

Sketch y(t)??t??x(t)dt.

Solution:

2. Suppose

Sketch:

(1). g(t)[?(t?3)??(t?1)?2?(t?1)] (2). g(t)

k?????(t?2k)

?

Solution: (1).

(2).