0:00
In this section, we discuss a more general
channel model called the band-limited coloured Gaussian Channel.
[BLANK_AUDIO]
The schematic diagram for this channel is the same
as the schematic diagram for the band-limited white Gaussian Channel.
The only difference is the noise process Z(t).
[BLANK_AUDIO]
Specifically, Z(t) is a zero mean additive coloured Gaussian noise,
meaning that the power spectral density is not necessarily a constant.
X prime of t and Z prime of
t, are filtered versions of X(t) and Z(t), respectively,
both band-limited to the interval [0,W].
[BLANK_AUDIO]
The output process, Y(t) is equal to X prime of t plus Z prime of t.
And Z prime of t is a band-limited coloured Gaussian noise, with the power
spectral density S_{Z prime} of f greater than and or equal to 0
for f from minus W to W and equal to zero otherwise.
As for the case of the white Gaussian channel, we regard X prime of t
as the channel input and Z prime of t as the additive noise process.
We also impose a power constraint P on the input process X prime of t.
[BLANK_AUDIO]
Recall that the capacity of the white Gaussian channel
band-limited to the interval [0,W], is equal to W times
log of 1 plus P over N_0 times W bits per unit time.
For the white Gaussian channel, band-limited to the interval [f_l,f_h],
where f_l is the lowest frequency and f_h is the highest frequency,
and f_l is a multiple of the bandwidth of the channel W equal to f_h minus f_l,
we can apply the bandpass version of the
sampling theorem to obtain the same capacity formula.
This model is called the bandpass white Gaussian Channel.
When f_l, the lowest frequency, is equal to 0, the
bandpass white Gaussian Channel reduces to the band-limited white Gaussian Channel.
[BLANK_AUDIO]
Here is the filter associated with the bandpass white Gaussian Channel.
Note that the lowest frequency f_l is a
multiple of W, the bandwidth of the channel.
And this is the schematic diagram for the bandpass white Gaussian Channel.
[BLANK_AUDIO]
Before we derive the capacity of the band-limited coloured Gaussian
Channel, let us first recall the assumptions of the channel model.
Z(t), the noise process, is a zero mean additive coloured Gaussian noise.
X prime of t and Z prime of t are filtered versions of X(t) and
Y(t) respectively, both band-limited to the interval [0,W].
[BLANK_AUDIO]
And the input power constraints on the input process X prime of t is equal to P.
This is an illustration of the power
spectral density of a band-limited coloured Gaussian Channel.
We now derive the capacity of such a channel.
First, divide the interval [0,W] into k subintervals.
The width of each subinterval is delta_k equal to W over k.
[BLANK_AUDIO]
Let the i-th subinterval be
[f_l^i,f_h^i] for i from one up to k.
As an approximation, assume that the
noise power over the i-th subinterval is a constant S_{Z,i}.
Then the channel consists of k sub-channels with
the i-th sub-channel being a bandpass white Gaussian
Channel occupying the frequency band [f_l^i,f_h^i].
Let P_i be the power allocated to the i-th sub-channel.
Then the capacity of the i-th sub-channel, is delta_k times
log of 1 plus P_i, divided by 2 times S_{Z,i}, times delta_k.
The workout of this expression is as follows.
For a white Gaussian Channel, which is band-limited to the integral [f_l, f_h],
where f_l is a multiple of the bandwidth W prime equals f_h minus f_l,
with noise level N_0 over 2, and power
constraint P, the capacity is W prime log
of one plus P divided by N_0 times W prime.
For the i-th sub-channel,
W prime is equal to delta_k,
P is equal to P_i,
and N_0 over 2 is equal to S_{Z,i}, or N_0 is equal to 2 times S_{Z,i}.
With these, we can obtain the expression in step five.
The noise process Z_i prime of t of the i-th sub-channel is obtained
by passing the noise process Z(t) through the corresponding
ideal bandpass filter band-limited to the interval [f_l^i,f_h^i].
It can be shown that the noise processes Z_i
prime of t, i from 1 up to k, are independent.
We will leave this as an exercise.
By sampling the sub-channels at a Nyquist rate 2 times delta_k,
where delta_k is the bandwidth of each sub-channel,
the k sub-channels can be regarded as a system of parallel Gaussian channels.
Thus the capacity of the channel is equal to the sum of the capacities
of the individual sub-channels when the
power allocation among the k sub-channels is optimal.
Let P_i star be the optimal power allocation for the i-th sub-channel.
[BLANK_AUDIO]
In step five, we obtained the capacity of the i-th sub-channel
when P_i is the power allocated to that sub-channel.
Then the capacity of the system of parallel Gaussian channels
is equal to summation i from 1 up to k, delta_k times log
of 1 plus P_i star, divided by 2 times S_{Z,i} times delta_k,
where the fraction inside the logarithm can be written
as P_i star over 2 times delta_k divided by S_{Z,i}.
And by proposition 11.23, the values of P_i star
over 2 times delta_k is equal to the positive part of nu
minus S_{Z,i}, where the water level nu can be obtained from the
constraint summation P_i star, i from one up to k, equals P.
[BLANK_AUDIO]
As k tends to infinity, where k is the number of subintervals, the capacity
of the system of parallel Gaussian channels given by this summation
tends to the integral from 0 to W log of 1 plus the
positive part of nu minus S_Z(f) divided by S_Z(f) df.
This can be written as one half times the same integral from minus W to W,
because S_Z(-f), is equal to S_Z(f).
That is the power spectral density is symmetrical about f equals 0.
[BLANK_AUDIO]
On the other hand, as k tends to infinity, summation i P_i star is equal
to summation i 2 times delta_k times the positive part of nu minus S_{Z,i}.
And this summation tends
to 2 times the integral
from 0 to W the positive
part of nu minus S_Z(f) df.
Again, this can be written as the integral from minus W
to W, the positive part of nu minus S_Z(f)df because the
power spectral density is symmetrical about f equals 0.
Therefore, the constraint summation i P_i star equals
P, tends to the constraint, the integral from minus
W to W, the positive part of nu minus S_Z(f)df equals P.
So the capacity of the band-limited coloured Gaussian Channel is given by
this formula, where the water level nu can be obtained by this constraint.
The optimal power allocation given in equation one
has a water-filling interpretation given in the next figure.
[BLANK_AUDIO]
Here, the water filling is for the range from minus W to W
with the total water volume equal to P.
And so the water volume on each side is equal to P over 2.