(完整版)信号与系统奥本海姆_习题答案
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{ [n 4m 4k ] [n 4m 1 4k ]}
{ [n 4(k m)] [n 1 4(k m)]}
{ [n 4k ] [n 1 4k ]}
s
Because g (t ) (t 2k ) ,
Chapter 1 Answers 1.6 (a).No
Because when t<0, x (t ) =0. 1 (b).No
Because only if n=0, x [n] has valuable. 2 (c).Yes
Because x[n 4m]
k
k
k
N=4.
1.9 (a). T= /5
Because w =10, T=2 /10= /5. 0 (b). Not periodic.
Because x (t ) e t e jt , while e t is not periodic, x (t ) is not periodic. 2 2 (c). N=2
Because w =7 , N=(2 / w )*m, and m=7. 0 0 (d). N=10
Because x (n) 3e j 3 / 10 e j (3 / 5)n , that is w =3 /5, N=(2 / w )*m, and m=3. 4 0 0 (e). Not periodic.
Because w =3/5, N=(2 / w )*m=10 m/3 , it’not a rational number. 0 0 1.14 A1=3, t1=0, A2=-3, t2=1 or -1
Solution: x(t) is dx(t ) dt is
k 1.15. (a). y[n]=2x[n-2]+5x[n-3]+2x[n-4] dx(t ) dx(t ) =3g(t)-3g(t-1) or =3g(t)-3g(t+1) dt dt 2 2
2 1
2
Solution:
y [n] x [n 2] 1 x [n 3]
2 2
1 y [n 2] y [n 3] 1
1 {2 x [n 2] 4 x [n 3]} {2 x [n 3] 4 x [n 4]} 1 1 1 1
2 x [n 2] 5x [n 3] 2 x [n 4] 1 1 1
Then, y[n] 2 x[n 2] 5x[n 3] 2 x[n 4]
(b).No. For it’s linearity.
the relationship between y [n] and x [n] is the same in-out relationship with (a). 1 2 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 2 [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 : y (t ) x (sin(t )) , y (t ) x (sin(t )) 1 1 2 2
ay (t ) by (t ) ax (sin(t )) bx (sin(t )) 1 2 1 2 1.21. Solution: We
(a). have known:
(b).
(c).
(d).
1.22. Solution:
We have known:
(a).
(b).
(e). 2
2 E {x(t )}
(g)
1.23. Solution:
For 1 [ x(t ) x(t )]
v
1 O {x(t )} [ x(t ) x(t )] d then,
(a).
(b).
(c) .
1.24. 2
Solution: For: E {x[n]} v 1
2 ( x[n] x[n])
1 O {x[n]} ( x[n] x[n]) d then, (a).
(b). Solution: x(t ) E {cos(4 t )u(t )}
s
(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.
v
1 {cos(4t )u(t ) cos(4 (t ))u(t )} 2 1 cos(4 t ){u(t ) u(t )} 2 1 cos(4 t ) 2
So, T=2 /4 =0.5
1.26. (a). Periodic. N=7
Solution: N= 2 * m =7, m=3. 6 / 7
(b). Aperriodic. Solution: N= 2
1/ 8 * m 16m , it’not rational number.
(e). Periodic. N=16
Solution as follow:
