Digital Signal Processing Reference
In-Depth Information
The variance is the second central moment:
2
4
3
5
¼
0
:
5
1
ð
1
0
2
s
X
2
x
2
e
jxj
d
x ¼
0
:
5
x
2
e
x
d
x þ
x
2
e
x
d
x
¼ m
2
¼ ðX m
X
Þ
¼ X
2
1
1
0
2
4
3
5
¼
¼
0
:
5
1
1
1
x
2
e
x
d
x þ
x
2
e
x
d
x
x
2
e
x
d
x:
0
0
0
(2.473)
Using integral 1 from Appendix
A
and the value of the Gamma function from
Appendix
D
, we arrive at:
s
X
2
¼ Gð
3
Þ¼
2
!
¼
2
:
(2.474)
The standard deviation is:
p
2
s
X
¼
:
(2.475)
From (
2.437
), given
k ¼
2 and using (
2.471
) and (
2.475
), we have:
n
o
jXj
2
2
p
PfjX m
X
jks
X
g¼P
1
=
4
:
(2.476)
This probability is the shaded area below the PDF in Fig.
2.61
.
Next, we find the exact probability using the symmetry property of the PDF:
o
¼
2
1
2
2
n
5e
x
d
x ¼
e
2
jXj
2
2
p
p
P
0
:
¼
0
:
0591
:
(2.477)
p
This probability is less than the probability (0.25) of (
2.476
), thus confirming the
Chebyshev inequality.
Exercise E.2.9
The input signal
X
is uniform in the interval [0,
a
], where
a
is
constant. The PDF is shown in Fig.
2.62a
. The signal is quantized where the
characteristic of the quantizer is given as (Fig.
2.62b
):
Y ¼ nD
for
nD<X ðn þ
1
Þ D;
n ¼
1
;
...
;
4
(2.478)
Search WWH ::
Custom Search