Digital Signal Processing Reference
In-Depth Information
per sample in the sequence
s
(
n
)
.
Using the fact (see below) that
1)
2
=
M
(
M
2
−
1)
1
2
+3
2
+5
2
+
...
+(
M
−
,
(2
.
2)
6
we therefore have
E
ave,P AM
=
(
M
2
−
1)
A
2
=
(2
2
b
1)
A
2
−
,
(2
.
3)
3
3
so that
A
=
3
E
ave,P AM
2
2
b
.
−
1
Since
b
=log
2
M,
the quantity
=
(
M
2
−
1)
A
2
3log
2
M
E
b,P AM
=
E
ave,P AM
log
2
M
,
(2
.
4)
is called the
energy per bit
.
Derivation of Eq. (2.2).
It is well known that
k
=1
k
2
=
n
(
n
+1)(2
n
+1)
/
6
.
Now, for even
M
we can write
1+3
2
+
...
+(
M
1)
2
−
M
k
2
−
(2
2
+4
2
+
...
+
M
2
)
=
k
=1
4
1+2
2
+
...
+(
M
2
)
2
M
k
2
−
=
k
=1
M
2
+1
(
M
+1)
M
(
M
+ 1)(2
M
+1)
6
1
6
×
4
M
2
=
−
2
M
+1
(
M
+2)
=
M
(
M
2
−
M
(
M
+1)
6
1)
=
−
,
6
which proves Eq. (2.2).
2.2.2 Average energy in a QAM constellation
Now consider a
b
-bit QAM constellation. There are
M
=2
b
codewords of the
form
z
=
x
+
jy,
where
x
and
y
belong to (
b/
2)-bit constellations with
√
M
words each
. T
hus the
average energy in the real part is as in Eq. (2.3) with
M
replaced by
√
M
=2
b/
2
:
1)
A
2
E
ave,x
=
(
M
−
.
3
Search WWH ::
Custom Search