Digital Signal Processing Reference
In-Depth Information
From here,
p
2ln
ð
0
x
s
¼
:
5
Þ
:
(5.206)
From (
5.206
), the median is found to be:
x ¼
1
:
774
s:
(5.207)
Exercise 5.3
The random variable
X
is a Rayleigh variable with a parameter of
s
.
Find the variance of the random variable
Y ¼
2
þ X
2
:
(5.208)
Answer
The variance of the random variable
Y
is
Y
2
s
2
Y
¼ Y
2
:
(5.209)
From (
5.208
), the mean and the mean squared values are:
Y ¼
2
þ X
2
;
(5.210)
2
Y
2
¼ð
2
þ X
2
¼
4
þ X
4
þ
4
X
2
Þ
:
(5.211)
From (
5.35
), the mean squared value of Rayleigh variable
X
is
¼
2
s
2
X
2
:
(5.212)
x
2
2
s
2
d
x ¼
1
1
s
2
1
s
2
8
s
6
x
5
e
¼
8
s
4
X
4
¼
:
(5.213)
0
Placing (
5.212
) into (
5.210
), we obtain:
Y ¼
2
þ
2
s
2
:
(5.214)
Similarly,
¼
4
þ
8
s
4
þ
8
s
2
Y
2
¼
4
þ X
4
þ
4
X
2
:
(5.215)
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