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)?ee?t?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]?A?[n]?[n?2], so y[n]=0.
2(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))
have known:
1.21. Solution: We (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.
Solution:
For:
1(x[n]?x[?n]) 21Od{x[n]}?(x[n]?x[?n])
2Ev{x[n]}? (b).
then, (a).