概率论第十一章习题解答

仍用上述推证方法，可知为使 E[W (t )W (t + τ )] 只依赖于 τ ，有 R X (τ ) = RY (τ ) ,且
R XY (τ ) = − RYX (τ ).
在以上条件下， W (t ) 平稳且 µ W = 0, RW (τ ) = R X (τ ) cos ω 0 τ + R XY (τ ) sin ω 0τ . (2) 利用傅氏变换的位移性质可得
∫
c
⎛ τ ⎜1 − −T ⎜ ⎝ T
+T
⎞ 2 ⎟ cos ωτdτ = (1 − cos ωT ). ⎟ Tω 2 ⎠
7.
R X (τ ) =
1 2π
∫ ω C cosτωdω =
−
ωc
C sin ω cτ
πτ
, R X (0) =
Cω c
π
.
于是有：当 τ = 0 时，结论显然成立，当 τ ≠ 0 时， lim
a2
=
另法：
a2
α
2
+
b . α (α + 1)
2b 1 , H (ω ) = , 2 α + iω ω +1
S X (ω ) = 2πa 2δ (ω ) +
2b ⎞ 1 ⎛ , S Y (ω ) = ⎜ 2πa 2δ (ω ) + 2 ⎟ 2 ω + 1⎠ ω + α 2 ⎝
RY (0 ) =
18. 先计算 RY (τ ).
3
第 3 页 共 7 页
由此可知
S X (ω ) = ∑ ai2 S X i (ω ).
i =1
N
10. RY (τ ) = E{[ X (t ) + X (t − τ 0 )][ X (t + τ ) + X (t + τ − τ 0 )]}
= 2 R X (τ ) + R X (τ − τ 0 ) + R X (τ + τ 0 ),
(3) 由题设， 显然有 R XZ (τ ) = R X (τ ) ,因而 S XZ (ω ) = S X (ω ) = 16. 由于 R XX τ ) = R′ τ ), 于是由傅氏变换的微分性质可得 & ( X (
16 . ω + 16
2
S XX τ )] = iωF [R X (τ )] = iωS X (ω ). & (ω ) = F [R ′ X (
(
)
E[W (t )W (t + τ )]
= E{[X (t ) cos ω 0 t + Y (t )sin ω 0 t ][X (t + τ ) cos ω 0 (t + τ ) + Y (t + τ )sin ω 0 (t + τ )]} = R X (τ ) cos ω 0 t cos ω 0 (t + τ ) + RY (τ )sin ω 0 t sin ω 0 (t + τ ) + R XY (τ )sin ω 0 (t + τ ) cos ω 0 t + RYX (τ )sin ω 0 t cos ω 0 (t + τ )
（3） S X (ω ) =
∫
+∞
−∞
be
−
−
τ2 2α 2
e
2
−iωτ
dτ = be
−
α 2ω 2
2
∫
+∞
−∞
e
−
(τ +iα ω )
2
2
2α
2
dτ = 2πabe
−
α 2ω 2
2
, (a > 0)
其中
∫
+∞
(τ +iα ω )
2
−∞
e
2a
2
dτ = 2πa
由复变函数的围道积分得到。
（4） S X (ω ) =
17.
RY (0) = ∫
+∞ +∞
−∞ −∞ +∞
∫ h(τ )h(τ )R (τ
1 2 X
1
− τ 2 )dτ 1dτ 2
− τ 1 −τ 2
=∫
=
0
∫
+∞
0
e −ατ1 e −ατ 2 a 2 + be
(
)dτ dτ
1
2
+∞ τ1 +∞ + b ∫ e −ατ1 dτ 1 ⎡ ∫ e −ατ 2 e −(τ 1 −τ 2 ) dτ 2 + ∫ e −ατ 2 e −(τ 2 −τ 1 ) dτ 2 ⎤ ⎢ ⎥ 0 0 τ 1 ⎣ ⎦ α 2
10
=
4
π
+
20
π
∫
10
0
4 ⎛ sin 2 5τ ⎞ ⎛ ω⎞ ⎟ ⎜1 + τω ω d − = 1 cos ⎜ ⎟ 2 ⎟ π⎜ τ ⎝ 10 ⎠ ⎠ ⎝
.
9. X (t ) 的相关函数为
⎫ N ⎧ ⎤⎪ ⎤⎡ N ⎪⎡ N R X (τ ) = E ⎨⎢∑ a i X i (t )⎥ ⎢∑ a j X j (t + τ )⎥ ⎬ = ∑ ai2 R X I (τ ), ⎪ ⎦ ⎣ j =1 ⎦⎪ ⎭ i =1 ⎩⎣ i =1
1 2π
2b ⎞ 1 a2 b ⎛ 2 ( ) + π δ ω ω = + 2 . a d ⎟ 2 2 2 2 ∫−∞ ⎜ α (α + 1) ω + 1⎠ ω + α α ⎝
+∞
S X (ω ) =
2
16 3 , H (ω ) = , 3 + iω ω + 16
2
因而
RY (ω ) = H (ω ) S X (ω ) =
S X (ω ) = lim
E FT (ω , ς ) 2T
2
T → +∞
=
a2 2
⎡ sin 2 (ω − ω 0 )T sin 2 (ω + ω 0 )T ⎤ + lim lim ⎢T →+∞ ⎥ (ω − ω 0 )2 T T →+∞ (ω + ω 0 )2 T ⎥ ⎢ ⎣ ⎦
= 1 2π
a 2π [δ (ω − ω 0 ) + δ (ω + ω 0 )] 2
∫
+∞
−∞
e
−α τ
cos ωτdτ =
（2）仍由例 7.2 有 S X (ω ) =
1 + ∞ −α τ e cos ω − ω 0 τ + cos(ω + ω 0 )τ dτ 2 ∫−∞
[ (
2α ; ω +α 2
2
)
]
⎤ ⎡ 1 1 ; =α⎢ + 2 2 2 2 ⎥ (ω + ω 0 ) + α ⎦ ⎥ ⎢ (ω − ω 0 ) + α ⎣
(
144 144 ⎛ 1 1 ⎞ = − 2 ⎜ 2 ⎟. 2 7 ⎝ ω + 9 ω + 16 ⎠ ω + 16 ω + 9
2
)(
)
利用例 7.2 的结果可得 RY (τ ) =
1 −3 τ −4 τ 24e − 18e , 7
(
)
因而 RY (τ ) 在 τ = 0, τ = 0.5, τ = 1 的值分别为
= 1 [R X (τ ) + RY (τ )]cos ω 0τ + 1 [R X (τ ) − RY (τ )]cos ω 0 (2t + τ ) 2 2 1 1 + [R XY (τ ) + RYX (τ )]sin ω 0 (2t + τ ) + [R XY (τ ) − RYX (τ )]sin ω 0τ 2 2
a2 cos ωτ 2
[
2
(t )] = E[a 2 cos 2 (ω 0 t + Θ)] = a
1 ⎛ ⎜1 + 2 ⎝ π
2
∫
π
0
2 ⎞ a cos(2ω 0 t + 2θ )dθ ⎟ = , ⎠ 2
1 T → ∞ 2T lim
∫ [
T −T
E X 2 (t ) dt =
T
]
a2 2
T
0 0
(3) FT (ω , ς ) = a ∫−T cos(ω 0 t + Θ )e −iωt dt = a ∫−T [e i (ω t +Θ ) + e −i (ω t +Θ ) ]e −iωt dt
1
第 1 页 共 7 页
第七章
1
习题解答
(1) E ( X (t )) = 1 ∫0 a cos(ω 0 t + θ )dθ π
π
=−
2a
π
sin ω 0 t ,
E [X (t )X (t + τ )] =
可见 X (t ) 非平稳. (2) E X
1
π
∫
π
0
a 2 cos(ω 0 t + θ ) cos[(ω 0 (t + τ ) + θ )]dθ =
[(
)]
+ E FT (ω , ς )
2 sin [(ω − ω 0 )T ]sin [(ω + ω 0 )T ]cos 2Θ ω − ω 02
2
}
[
2
1 ⎪ ]= a ⎧ ⎨ ⎪ (ω − ω )