for some cutoff value
0
, where
j
=
|H
j
|
2
S/(N
0
B) is the SNR associated with the jth channel assuming
it is allocated the entire power budget. This optimal power allocation is illustrated in Figure 4.11. The
cutoff value is obtained by substituting the power adaptation formula into the power constraint, so
0
must satisfy
j
1
0
-
1
j
= 1.
(4.25)
The capacity then becomes
C =
j:
j
0
B log
2
(
j
/
0
).
(4.26)
This capacity is achieved with by sending at different rates and powers over each subchannel. Multicarrier
modulation uses the same technique in adaptive loading, as discussed in more detail in Chapter 12.
f
2
1
=
0
1 N B
0
j
j
j
S
S
S|H |
Figure 4.11: Water-Filling in Block Frequency-Selective Fading
When H(f ) is continuous the capacity under power constraint S is similar to the case of the block-
fading channel, with some mathematical intricacies needed to show that the channel capacity is given by
[5]
C =
max
S(f ): S(f )df S
log
2
1 +
|H(f)|
2
S(f )
N
0
df.
(4.27)
The equation inside the integral can be thought of as the incremental capacity associated with a given
frequency f over the bandwidth df with power allocation S(f ). The optimal power allocation over
frequency, S(f ), found via the Lagrangian technique applied to (4.27), is again water-filling:
S(f )
S
=
1
0
-
1
(f )
(f )
0
0
(f ) <
0
(4.28)
This results in channel capacity
C =
f :(f )
0
log
2
((f )/
0
)df.
(4.29)
Example 4.7: Consider a time-invariant block fading frequency-selective fading channel consisting of
three subchannels of bandwidth B = 1 MHz.The frequency response associated with each channel is
H
1
= 1, H
2
= 2 and H
3
= 3. The transmit power constraint is S = 10 mW and the noise power spectral
density is N
0
= 10
-9
W/Hz. Find the Shannon capacity of this channel and the optimal power allocation
that achieves this capacity.
100
Summary :
The cutoff value is obtained by substituting the power adaptation formula into the power constraint, so 0 must satisfy j 1 0 - 1 j = 1. The optimal power allocation over frequency, S(f ), found via the Lagrangian technique applied to (4.27), is again water-filling: S(f ) S = 1 0 - 1 (f ) (f ) 0 0 (f ) <
Tags :
capacity,channel,allocation,fading,associated,constraint,optimal,block,log,oer,frequency,technique,cutoff