Hi, welcome to module 9.3 of digital
signal processing.
We're still talking about digital
communication systems.
In the previous module we addressed the
bandwidth constraint,
and in this module we will tackle the
power constraint.
So first we will introduce the concept of
noise and probability of error in a
communication system.
We will look at signalling alphabet and
the related power.
And finally we'll introduce QAM
signalling.
So, we have seen the transmitter
sends a sequence of symbols, a of n,
created by the mapper.
Now we take the receiver into account.
We don't yet know how, but it's safe to
assume
that the receiver in the end, will obtain
an estimation.
Hat a of n, of the original transmitted
symbol sequence.
It's an estimation.because even if there
is no distortion
introduced by the channel, even if nothing
bad happens.
There will always be a certain amount
of noise that will corrupt the original
sequence.
When noise is very large, our estimate for
the transmitted
symbol will be off and will incur a
decoding error.
Now this probability of error will depend
on the power of
the noise with respect to the power of the
signal and
will also depend on the decoding
strategies that we put in
place, how smart we are in circumvent and
defects of the noise.
One way we can maximize the probabilty of
correctly guessing the transmitted symbol
is by using
suitable alphabets.
And so we'll see in more detail what that
means.
Remember the scheme for the transmitter.
We have a bitstream coming in and then we
have the scrambler,
and then the mapper.
And here we have a sequence of symbols
a of n.
These symbols will have to be sent over
the channel and to do so we up sample and
we interpolate and
then we transmit.
Now, how do we go from bitstreams to
samples in more detail.
In other words, how does the mapper work?
The mapper will split the incoming
bitstreams into chunks.
And will assign a symbol, a of n, from a
finite alphabet to each chunk.
The alphabet we will decide later what it
is composed of.
To undo the mapping operation and recover
the bitstream.
The receiver will perform a slicing
operation.
So, the receiver will receive a value of
hat a of n where
hat indicates the fact that noise has
leaked into the value of the signal.
And the receiver will decide which symbol
from the alphabet
which is known to the receiver as well is
closest
to the received symbol, and from there, it
will be
extremely easy to piece back the original
bitstream as an example,
let's look at simple two level signalling.
This generates signals of the kind we has
seen
in the examples so far alternating between
two levels.
The way the mapper works is by splitting
the incoming bitstream into single bits.
And the output symbol sequence uses an
alphabet composed of two symbols
G and minus G, and associates G to bit
value
one, and minus G to a bit of value zero.
At the receiver the slicer, looks at the
sign of
the incoming symbol sequence which has
been corrupted by noise
and decides that the nth bit will be one,
if
the sign of the nth symbol is positive and
zero otherwise.
Let's look at an example, let's assume g
equal to 1,
so the two level signal will alternate
between plus 1 and
minus 1.
And suppose we have an input bit sequence
that gives rise to this signal here.
After transmission and after decoding at
the receiver.
The resulting symbol sequence will look
like this, where each
symbol has been corrupted by a varying
amount of noise.
If we now slice this sequence by
thresholding, as
shown before, we recover a symbol sequence
like this.
Where we
have indicated in red the errors incurred
by the slicer because of the noise.
So if you want to analyze in more detail
what the probability of error is.
We have to make some hypothesis on the
signals involved in this toy experiment.
Assume that each received symbol can be
modelled
as the original symbol plus a noise
sample.
Assume also that the bits in the bits
stream are equiprobable.
So zero and one
appear with probability 50% each.
Assume that the noise and the signal are
independent and assume that the
noise is additive white Gaussian noise
with
zero mean and known variance, sigma zero.
With this hypothesis the probability of
error can be written out as follows.
First of all, we split the probability
of errors into two conditional
probabilities conditioned by
whether the n-th bit is equal to one or
the n-th bit is equal to zero.
In the first case when the n-th bit is
equal
to one, remember the produce symbol will
be equal to G.
So the probability of error is equal to
the probability
for the noise sample to be less than minus
G.
Because only in this case, the sum of the
sample plus the noise, will be negative.
Similarly, when the amplitude is equal to
zero, we have a negative sample.
And the only way for that to change sine
is if the noise
sample is greater than G.
Since the probability of each occurrence
is one
half, because of the symmetry of the
Gaussian
distribution function, this is equal to
the probability
for the noise sample to be larger than G.
And we can compute this as the integral
from G to infinity of
the probability distribution function for
the
Gaussian distribution with the known
variance here.
So what we have here is the
tail probability of a Gaussian with
standard deviation sigma zero.
So, this is the integral that we're trying
to compute.
Usually the tail probability of a unit
variance
Gaussian is indicated by the notation Q of
G.
The Q function of G.
Now because our variance is actually sigma
zero squared, we
have to normalize the Q function argument
by the standard deviation.
And we find that the probability
of error is equal to the Q function of G
over sigma 0.
Since this integral is not computable
exactly, if we want to calculate the Q
function, we have to resort to numerical
packages, or tabulated versions of this
function.
And usually, what you would find in
numerical packages in, is a derived
function
called the error function, which is
related
to the Q function by this formula here.
But the point that is really important
at the end of this derivation is that the
probability of error is equal to some
function of the ratio between the
amplitude of
the signal, and the standard deviation of
the noise.
Now we can carry this analysis further by
considering the transmitted power.
We have a bi-level signal, and each level
occurs with one half probability.
So the variance of the signal, which
corresponds
to the power is equal to G squared times
the probability of the n-th bit being
equal to 1, plus G squared times the
probability of the n-th bit being equal to
0, which is equal to G squared.
And so if we rewrite the probability of
error, we can see that it is equal to the
Q function of the ratio between the
standard deviation
of the signal and the standard deviation
of the noise.
But this is really equivalent to saying
that the probability of error is equal
to the Q function of the square root of
the signal to noise ratio of
the transmitted signal.
If we plot this as a function of the
signal noise to ratio in dBs and I remind
here that dBs here mean that we compute 10
times
the log in base 10 of the power of the
signal divided by the power of the noise.
And since we are in a log, log scale.
We can see that the probability of error
decays exponentially with the signal to
noise ratio.
This exponentially decay is quite a norm
in communication systems
and while the absolutely rate of decay
might change in terms
of the linear constants involved in the
curve, the trend
will stay the same even for more complex
signal and schemes.
So the lesson that we learned from the
simple example is that in order
to reduce the probability of error, we
should increase G, the amplitude of the
signal.
But of course,
increasing G also increases the power of
the transmitted signal, and
we know that we cannot go above the
channel's power constraint.
And so that's how the power constraint
limits the reliability of transmission.
The bi-level signal is keen, is very
instructive, but it's also very
limited in the sense that we're sending
just one bit per output symbol.
So to increase the throughput, to increase
the number
of bits per second that we send over a
channel,
we can use multilevel signaling.
There are very many ways to do so, we'll
just look at a few.
But the fundamental idea is that we take
now, larger chunks of
bits, and therefore we have alpha bits
that have a higher cardinality.
So more values in the alpha bit means more
bits per symbol and therefore a higher
data rate.
But, not to give the ending away.
We will see that the power of the signal
will also be dependent on the size of the
alphabet and so in order not exceed in the
probability of error, given the
channels power of constraint, we will not
be able to grow the alphabet indefinitely.
But we can be smart in a way we build this
alphabet.
And so we will look in some examples.
The first example is PAM, Pulse-amplitude
Modulation.
We split the incoming bitstream into
chunks of M bits.
So that the each chunk corresponds to an
integer between 0 and 2 to the m minus 1.
We can
call this sequence of integers, k of n and
this sequence is mapped onto a
sequence of symbols a of n like so There's
a gain factor G, like always.
And then we use 2 to the M minus 1, odd
integers around 0.
So for instance, if M is equal to 2, we
have 0, 1, 2,
and 3 as potential items for k of n.
And a of n will be either lets assume
G is equal to 1 will be either minus 3 or
minus or 1 or 3.
We will see why we used the odd integers
in just a second.
At the receiver, the slicer will work by
simply associating to the
received symbol, the closest odd integer,
always taking the gain into account.
So graphically again,
PAM for M equal to 2 and G equal to 1 will
look like this.
Here are the odd integers.
The distance between two transmitted
points, or
transmitted symbols, is 2G, right here G
is equal to 1, but it would be in general,
2 times the gain.
And using odd integers creates a zero-mean
sequence.
If we assume that each symbol is equally
probable, which
is likely, given that we've used a
scrambler in the transmitter,
then the resulting mean is zero.
The analysis of the probability of error
for PAM is
very similar to what we carried out for
bi-level signaling.
As a matter of fact, bi-level signaling is
simply PAM with m equal to 1.
The end result is very similar and its an
exponential decaying of the
ratio between the power of the signal and
the power of the noise.
The reason why we don't analyze this
further
is because we have an improvement in
store.
And the improvement
is aimed at increasing the throughput,
increasing the numbers of bit per
symbol that we can send without
necessarily increasing the probability of
error.
So here's a wild idea, let's use complex
numbers and build a complex valued
transmission system.
This requires certain suspension of
disbelief for the time
being, but believe me, it will work in the
end.
The name for this complex valued mapping
scheme
is QAM, which is an acronym for quadrature
amplitude modulation.
And it works like so.
The mapper takes the incoming bitstream
and splits
it into chunks of M bits, with M even.
And then it uses half of the bits to
define
a PAM sequence which we call a of r of n.
And the remaining m over two bits to
define
an another independent PAN sequence a i of
n.
The final symbol sequence is a sequence of
complex numbers
where the real part is the first PAM
sequence
and the imaginary part is a second PAM
sequence.
And of course in front we have again
factor G.
So the transmission of alphabet A is given
by points
in the complex plane with odd valued
coordinates around the origins.
And the receiver just lies through works
by finding the
symbol in the alphabet which is closest in
Euclidean distance
to the received symbol.
Let's look at this graphically.
This is the set of points for QAM
transmission with M equal to two, which
corresponds to two bi-level PAM signals on
the real axis and on the imaginary axis.
So that results into four points.
If we increase the number of bits per
symbol, we set M equal to four.
That corresponds to two PAM
signals with two bits each.
Which makes for a constellation.
This is how these arrangement of points in
the complex plane are called.
A constellation of four by four points at
the odd valued coordinates in the complex
plane.
If we increase M to 8, then we have
a 256 point constellation with 16 points
per side.
Lets look at what happens
when a symbol is received and how we
derive an expression from the probability
of the error.
If this is the nominal constellation the
transmitter will
choose one of these values for
transmission, say this one.
And this value will be corrupted by
noise in the transmission and the
receiving process.
And will appear somewhere in the complex
plane,
not necessarily exactly on the point it
originates from.
The way the slicer operates
is by defining decision regions around
each point in the constellation.
So suppose for this point here, the
transmitted point, the decision
region is a square of side, 2G, centered
around the transmitted point.
So what happens, is that when we receive
symbols, they will not fall on the
original
point, but as long as they fall within
the decision region, they will be decoded
correctly.
So for instance, here.
We will decode this correctly.
Here we will decode this correctly, same
here.
But this point for instance falls outside
of the decision region
and therefore it will be associated to a
different constellation point.
Thereby, causing an error.
To quantify the probability of error, we
assume as per usual that each
received symbol is the sum of the
transmitted symbol plus a noise sample.
Inter of n.
And we further assume that this noise is a
complex value
Gaussian noise of equal variance in the
complex and real components.
We're working on a completely digital
system that operates with complex valued
quantities.
So we're making a new model for the noise.
And we will see later how to translate
the physical real noise into a complex
variable.
With this assumptions.
The probability of error is equal to the
probability
that the real part of the noise is larger
than
G in magnitude, plus the probability that
the imaginary
part of the noise is larger than G in
magnitude.
We assume that real and imaginary
components of the noise are
independent, and that's why we can split
the probability like so.
Now, if you remember the shape of
the decision region, this condition is
equivalent to
saying that the noise is pushing the real
part of the point
outside of the decision region in either
direction and same for the majority part.
Now if we develop this, this is to equal 1
minus the probability the real part of
the noise is less than G and the imaginary
part of the noise is less than G.
This is the complimentary condition to
what we just wrote above.
And so this is equal to 1
minus the integral over the decision
region, d, of
the complex valued probability density
function for the noise.
In order to compute this integral, we're
going to approximate
the shape of the decision region with the
inbound circle.
So instead of using the square.
We're going to a circle centered around
the transmission point.
When the conciliation is very dense, this
approximation is quite accurate.
With this approximation,
we can compute the integral exactly for a
Gaussian distribution.
And if we assume that the variance of the
noise is
sigma 0 squared over 2 in each component,
real and imaginary.
It turns out that the probability of error
is equal
to e to the minus g square over sigma 0
square.
Now to obtain a probability of error as a
function of the signal
to noise ratio, we have to compute the
power of the transmitted signal.
So if all symbols are equiprobable and
independent.
It turns out that the variance of the
signal is G
square times 1 over 2 to the power of M,
which
is a probability of each symbol times the
sum of over
all symbols in the alphabet of the
magnitude of the symbol squared.
Now it's a little bit tedious but we can
solve it exactly for M
and it turns out that the power of the of
the transmitted signal is G
square, 2 3rds, 2 the M minus 1.
Now if we plug this in to the formula for
the probability of error, that we've
seen before, we get that the result is
an exponential function where the argument
is minus 3.
That multiplies 2 to the minus m plus 1,
that multiplies the signal's noise ratio.
We can plot this probability of error in a
log log scale, like we did before.
And we can parameterize the curve as a
function of the number of points in the
constellation.
So here you have the curve for a four
point constellation.
Here's the curve for 16 points.
And here's the curve for 64 points.
Now, you can see that for a given signal
to noise
ratio, the probability of error increases
with the number of points.
Why is that?
Well, if the signal to noise remains the
same,
and we assume that the noise is always at
the
same level.
Then it means that the power of the signal
remains constant as well.
In that case, if the number of points
increases G has to become
smaller in order to accommodate a larger
number of points for the same power.
But if G becomes smaller then the decision
reaches become smaller.
The separation between points becomes
smaller and
the decision process becomes more
vulnerable to noise.
So in the end, here is the final recipe
to design a QAM transmitter.
First you pick a probability of error that
you can live with.
In general 10 to the minus 6 is
an acceptable probability of error at the
symbol level.
Then you find out the signal to noise
ratio that is imposed by the channel's
power constraint.
Once you have that, you can find the size
of your constellation by finding m which
based on the previous equations is the log
in base 2 of 1 minus 3 over 2
times the signal to noise ratio divided by
the natural logarithm of the probability
of error.
Of course you will have to round this to a
suitable integer value and
potentially to an even power of 2 in order
to have the square constellation.
The final data rate of your system will be
m the number of bits
per symbol, times w which if you remember
is the board rate of system.
And corresponds to the bandwidth allowed
for by the channel.
So, we know how to fit the bandwidth
constraint by upsampling.
With QAM, we know how many bits per symbol
we can use given the power constraint.
And so we know the theoretical throughput
of the transmitter, for a given
reliability figure.
However, the question remains, how are we
going
to send complex value symbols over a
physical channel.
It's time therefore to stop the suspension
of disbelief and
look at techniques to do complex signaling
over a real value channel.