奥本海姆 信号与系统 习题参考答案 电子科技大学

Chapter 1 Answers

1.6 (a).No

Because when t<0, x1(t)=0.

(b).No

Because only if n=0, x2[n] has valuable.

(c).Yes

Because x[n?4m]?k?????{?[n?4m?4k]??[n?4m?1?4k]}

?

?k?????{?[n?4(k?m)]??[n?1?4(k?m)]} ?{?[n?4k]??[n?1?4k]}

?k??? N=4. 1.9 (a). T=?/5

Because w0=10, T=2?/10=?/5.

(b). Not periodic.

Because x2(t)?e?te?jt, while eis not periodic, x2(t) is not periodic.

?t(c). N=2

Because w0=7?, N=(2?/w0)*m, and m=7.

(d). N=10

Because x4(n)?3ej3?/10ej(3?/5)n, that is w0=3?/5, N=(2?/w0)*m, and m=3.

(e). Not periodic.

Because w0=3/5, N=(2?/w0)*m=10?m/3 , it’s not a rational number.

1.14 A1=3, t1=0, A2=-3, t2=1 or -1

Because g(t)?Solution: x(t) is

dx(t) is dtk?????(t?2k),

?dx(t)dx(t)=3g(t)-3g(t-1) or =3g(t)-3g(t+1) dtdt1.15. (a). y[n]=2x[n-2]+5x[n-3]+2x[n-4]

Solution:

y2[n]?x2[n?2]?1x2[n?3] 21 ?y1[n?2]?y1[n?3]

2 ?{2x1[n?2]?4x1[n?3]}?1{2x1[n?3]?4x1[n?4]} 2 ?2x1[n?2]?5x1[n?3]?2x1[n?4]

Then, y[n]?2x[n?2]?5x[n?3]?2x[n?4]

(b).No. For it’s linearity.

the relationship between y1[n] and x2[n] is the same in-out relationship with (a).

you can have a try. 1.16. (a). No. For example, when n=0, y[0]=x[0]x[-2]. So the system is memory. (b). y[n]=0.

When the input is A?[n],

then, y[n]?A2?[n]?[n?2], so y[n]=0.

(c). No.

For example, when x[n]=0, y[n]=0; when x[n]= A?[n], y[n]=0.

So the system is not invertible. 1.17. (a). No.

For example, y(??)?x(0). So it’s not causal. (b). Yes.

Because : y1(t)?x1(sin(t)) ,

y2(t)?x2(sin(t))

ay1(t)?by2(t)?ax1(sin(t))?bx2(sin(t))

1.21. Solution: We have known: (a).

(b).

(c).

(d).

1.22. Solution: We have known:

(a).

(b).

(e).

(g)

1.23. Solution:

For

1Ev{x(t)}?[x(t)?x(?t)]

21Od{x(t)}?[x(t)?x(?t)]

2

then, (a).

(b).

(c). 1.24.