随机信号分析(赵淑清 郑薇)第四章习题解答

(0)
Leabharlann Baidu
RAS
(0)
1
2
SAC ()d
1
2
2 2a cos( )d
2a
2
cos(
)d
0
2
RAC
(0)
RAS
(0)
2a
2
8
a cos[ ( 0 ) ]
SX () a cos[ ( 0 ) ]
0
2 0 2 2 0 2
其它
试求: (2) AC (t)和AS (t)是否正交?
其它
式中, a, ,0皆为正常数,且0 . 试求 :
(1) AC (t), AS (t)的功率谱密度和平均功率.
(2) AC (t)和AS (t)是否正交?
(1)
S
AC
()
S
As
()
S
X
(
0
) 0
S
X
(
0
)
2
其它
6
a cos[ ( 0 ) ]
SX () a cos[ ( 0 ) ]
4.2 设复随机过程X (t)是广义平稳的,试证明:
RX ( ) RX* ( ) 并证明功率谱密度SX ()是实函数.
解: 设复随机过程X (t) A(t) jB(t),其中A(t)和B(t) 都是实平稳随机过程,且是联合平稳的.
RX ( ) E[ X *(t) X (t )] E[{A(t) jB(t)}{A(t ) jB(t )}]
0
2 0 2 2 0 2
其它
试求: (1) AC (t), AS (t)的功率谱密度和平均功率.
S
AC
(
)
S
As
(
)
a
c
os
(
)
a
c
os
(
)
0
2 其它
S
AC
()
S
As
(
)
2a
cos(
)
2
0
其它
7
S
AC
(
)
S
As
(
)
2a
cos(
)
0
2 其它
RAC
3
4.3 设有一窄带信号x(t) xc (t) cos(0t) xs (t) sin(0t), 其中的xc (t)与xs (t)的带宽远小于0.设X c ()和X s () 分别为xc (t)与xs (t)的傅里叶变换, Z ()为x(t)的解析
函数z(t) x(t) jxˆ(t)的傅里叶变换, 试证 :
X
c
()
1 2
[Z
(
0
)
Z
* (
0
)]
X
s ()
1 2j
[Z (
0 )
Z *(
0 )]
x(t) xc (t) cos(0t) xs (t) sin(0t) xˆ(t) xc (t) sin(0t) xs (t) cos(0t) z(t) x(t) jxˆ(t) [xc (t) jxs (t)]e j0t
数学期望为零的窄带平稳随机过程
X (t) AC (t) cos(0t) AS (t)sin(0t) AC (t)与AS (t)正交的条件是: SX ()的单边谱关于0对称
本题中的 SX ()的单边谱是关于 0对称的
AC (t)与AS (t)正交
9
RX ( ) RA ( ) RB ( ) j[RBA( ) RAB ( )]
RX* ( ) RA ( ) RB ( ) j[RBA( ) RAB ( )]
RX
( )
RX*
( )
RA( )
RB ( )
j[RAB( )
RBA
(
)]
2
4.2 设复随机过程X (t)是广义平稳的,试证明:
Z *( 0 ) X c () jX s ()
5
4.4 数学期望为零的窄带平稳随机过程
X (t) AC (t) cos(0t) AS (t) sin(0t),其功率谱密度为:
a cos[ ( 0 ) ]
SX () a cos[ ( 0 ) ]
0
2 0 2 2 0 2
RA ( ) RB ( ) j[RAB ( ) RBA( )] 1
4.2 设复随机过程X (t)是广义平稳的,试证明:
RX ( ) RX* ( ) 并证明功率谱密度SX ()是实函数.
RX ( ) RA ( ) RB ( ) j[RAB ( ) RBA( )]
RX ( ) RA( ) RB ( ) j[RAB ( ) RBA( )]
4
4.3
X
c
()
1 2
[Z
(
0
)
Z
* (
0
)]
Xs
()
1 2j
[Z (
0 )
Z *(
0 )]
z(t) [xc (t) jxs (t)]e j0t z(t)e j0t xc (t) jxs (t)
Z ( 0 ) X c () jX s ()
Z ( 0 ) X c () jX s ()
RX ( ) RX* ( ) 并证明功率谱密度SX ()是实函数.
RX ( ) RA( ) RB ( ) j[RAB( ) RBA( )] RX ( ) RA( ) RB ( ) j[RAB( ) RAB( )]
SX () FT[RX ( )] SA() SB () 2Im[ SAB()] SA() SB () j[SAB() SAB()] SA() SB () j 2 j Im[ SAB()]