Channel capacity - Signal Processing and Optimization for Transceiver Systems - page 223

unless N 1 . Thus, even without any power allocation we obtain improve-

ment in capacity as long as the different parts of the channel are used indepen-
dently. If we allocate different powers to the two subbands then the expression
for capacity becomes

N 0 =

1+ +0 . 5 B log 2 1+ .

p 00
N 0 B

p 01
N 1 B

C alloc =0 . 5 B log 2 (6 . 28)

Since the total power is p 0 we have the constraint
p 00 + p 01
2

= p 0 . (6 . 29)

For fixed bandwidth, noise densities N 1 , and total power p 0 we can

maximize the capacity (6.28) by optimally allocating the numbers p 00 and p 01 .
This is similar to the parallel channel problem discussed in Sec. 22.3 later, and
the solution is again the water-filling allocation:

N 0 and

⎧
⎨

1
λ −N k B

if this is non-negative

p 0 k = (6 . 30)

⎩

0 otherwise.

Here λ is a constant that arises from setting up a Lagrangian for the constrained
optimization problem. It can be found from the power constraint (6.29).
Splitting the channel into a finite number of subbands is a practical way
to approximate the ideal power allocation formula (6.17). A simple way to
mechanize the splitting of a channel into subbands is indicated in Fig. 6.5(a).
Here the transmitter has two ideal filters bandlimited to the two halves of the
channel bandwidth (Fig. 6.5(b)). The same pair of ideal filters is used at the
receiver to split the received signal into subbands. The effect is the creation of
two independent parallel channels with half the original bandwidth, as shown
in Fig. 6.5(c). The idea of power allocation also arises in many other forms in
digital communication, not necessarily in the context of channel capacity.

Example 6.1: Splitting the channel into subbands

Assume H ( f ) is the ideal lowpass channel in Eq. (6.1). In Fig. 6.3, let
B = 1 MHz (total bandwidth 2 MHz), and let

10 − 6 mW / Hz , 10 − 8 mW / Hz .

p 0 =1mW , N 0 =2 × and N 1 =2 ×

So, subband 1 is 100 times less noisier than subband 0. Using the above
numbers we can calculate the capacity of the unsplit channel [Eq. (6.20)],
the capacity of the split channel with equal power in the subbands [Eq.
(6.23)], and the capacity of the split channel with optimum power allocation
[Eq. (6.28)]. The results in kilobits per second are:

C alloc = 3,329 kb / s .
The optimal power allocation obtained from Eq. (6.30) is such that p 00 =
0 . 01 and p 01 =1 . 99, so the powers allocated to the subbands are
p 00 / 2=0 . 005 mW

C unsplit = 993 kb / s , C split = 3,129 kb / s , and

and p 01 / 2=0 . 995 mW .