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).
