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Hi and welcome to Module 9.3 of Digital 
Signal Processing.

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We are still talking about Digital 
Communication Systems.

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In the previous module we addressed 
bandwidth constraint.

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In this module we will tackle the powered 
constraint so first we will introduce the

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concept of noise and probability of error 
in a communication system.

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We will look at signaling alphabet and 
power and their related power.

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And finally, we'll introduce QAM 
signaling.

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So we have seen that a transmitter sends 
a sequence of symbols a of n.

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Created by the mapper. 
Now we take the receiver into account.

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We don't yet know how, but it's safe to 
assume that the receiver in the end will

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obtain an estimation hat a of n. 
Of the original transmitted symbol

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sequence. 
It's an estimation because even if there

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is no distortion introduced by the 
channel.

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Even if nothing bad happens. 
There will always be a certain amount of

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noise, that will corrupt the original 
sequence.

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When noise is very large, our estimate 
for the transmitted symbol will be off,

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and will incur a decoding error. 
Now, this probability of error will

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depend on the power of the noise, with 
respect to the power of the signal.

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And will also depend on the decoding 
strategies that we've put in place, how

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smart we are in circumventing the effects 
of the noise.

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One we can maximize the probability of 
correctly guessing the transmit symbol,

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is by using suitable alphabets. 
And so we will see in more detail what

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that means. 
Remember the scheme for the transmitter.

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We have a bitstream coming in. 
And then we have the scrambler.

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And then the mapper. 
And here we have a sequence of symbols a

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of n. 
These symbols will have to be sent over

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the channel. 
And to do so, we upsample.

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And we interpolate, and then we transmit. 
Now, how do we go from bitstreams to

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samples in more detail? 
In other words, how does the mapper work?

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The mapper will split the incoming 
bitstreams into chunks and will assign a

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symbol, a of n, from a finite alphabet to 
each chunk.

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The alphabet, we will decide later what 
it is composed of.

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To undo the mapping operation and recover 
the bitstream, the receiver will perform

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a slicing operation. 
So the receiver will a value, hat a of n,

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where hat indicates the fact that noise 
has leaked into the value of the signal.

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And the receiver will decide which symbol 
from the alphabet, which is known to the

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receiver as well, is closest to the 
received symbol.

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And from there, it will be extremely easy 
to piece back the original bitstream.

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As an example, let's look at simple 
two-level signaling.

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This generates signals of the kind we 
have seen in the example so far,

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alternating between two levels. 
The way the mapper works is by splitting

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the incoming bitstream into single bits. 
And the output symbol sequence uses an

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alphabet composed of two symbols, g and 
minus g, and associates g to a bit of

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value 1 and minus g to a bit of value 0. 
And the receiver, the slicer.

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Looks at the sign of the incoming symbol 
sequence which has been corrupted by

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noise. 
And decides that the nth bit will be 1 if

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the sign of the nth symbol is positive, 
and 0 otherwise.

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Lets look at an example, lets assume G 
equal to 1.

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So the two-level signal will alternate 
between plus 1 and minus 1.

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And suppose we have an input bit sequence 
that gives rise to this signal here after

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transmission and after decoding at the 
receiver.

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The resulting symbol sequence will look 
like this, where each symbol has been

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corrupted by a varying amount of noise. 
If we now slice this sequence by

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thresholding, as shown, shown before. 
We recover a simple sequence like this

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where we have indicated in red the errors 
incurred by the slicer because of the

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noise. 
So if you want to analyze in more detail

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what the probability of error is, we have 
to make some hypothesis on the signals

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involved in this toy experiment. 
Assume that each received symbol can be

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modeled as the original symbol plus a 
noise sample.

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Assume also that the bits in the 
bitstream are equiprobable.

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So zero and one appear with probability 
50% each.

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Assume that the noise and the signal are 
independent.

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And assume that the noise is additive 
white Gaussian noise with zero mean and

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known variance sigma 0. 
With this hypothesis, the probability of

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error can be written out as follows. 
First of all, we split the probability of

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errors into 2 conditional probabilities. 
Conditioned by whether the nth bit is

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equal to 1, or the nth bit is equal to 
zero.

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In the first case, when the nth bit is 
equal to 1.

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Remember, the produced symbol will be 
equal to G, so the probability of error

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is equal to the probability for the noise 
sample to be less than minus G.

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Because only in this case the sum of the 
sample plus the noise will be negative.

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Similarly, when the nth bit is equal to 
0, we have a negative sample.

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And the only way for that to change sign 
is if the noise sample is greater than G.

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Since the probability of each occurrence 
is 1 half because of the symmetry of the

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Gaussian distribution function. 
This is equal to the probability for the

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noise sample to be larger than G. 
And we can compute this as the integral

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from G to infinity of the probability 
distribution function for the Gaussian

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distribution with the known variance 
here.

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This function has a standard name. 
It's called the error function.

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And since this integral can not be 
computed in closed form, this function is

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available in most numerical packages 
under this name.

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So the important thing to notice here is 
that the probability of error is some

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function of the ratio between the 
amplitude of the signal and the standard

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deviation of the noise. 
And we can carry this analysis further by

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considering the transmitted power. 
We have a bi-level signal and each level

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occurs with 1 half probability. 
So the variance of the signal, which

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corresponds to the power, is equal to G 
squared time the probability of the nth

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being equal to 1. 
Plus G squared times the probability of

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the nth bit being equal to 0, which is 
equal to G squared.

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And so, if we rewrite the probability 
error function we can write that it is

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equal to the error function of the ratio. 
Between the standard deviation of the

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transmitted signal divided by the 
standard deviation of the noise, which is

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equivalent to saying that it is the error 
function of the square root of the signal

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to noise ratio. 
If we plot this as a function of the

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signal to noise ratio in dBs. 
And I remind here that dBs here mean that

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we compute 10 times the log in base 10 of 
the power of the signal divided by the

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power of the noise. 
And since we are in a log log scale, we

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can see that the probability of error 
decays exponentially with the signal to

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noise ratio. 
This exponential decay is quite the norm

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in communication systems. 
And while the absolute rate of decay

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might change in terms of the linear 
constants involved in the curve.

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The trend will stay the same even for 
more complex signaling schemes.

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So the lesson that we learn from the 
simple example is that in order to reduce

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the probability of error, we should 
increase G, the amplitude of the signal.

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But of course, increasing G also 
increases the power of the transmitted

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signal, and we know that we cannot go 
above the channel's power constraint.

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And so that's how the power constraint 
limits the reliability of transmission.

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The bilevel signalling scheme is very 
instructive, but it's also very limited

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in the sense that we're sending just one 
bit per output symbol.

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So to increase the throughput, to 
increase the number of bits per second

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that we send over a channel, we can use 
multilevel signaling.

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There are very many ways to do so and we 
will just look at a few, but the

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fundamental idea is that we take now. 
Larger chunks of bits and therefore we

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have alphabets that have a higher 
cardinality.

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So more values in the alphabet means more 
bits per symbol and therefore a higher

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data rate. 
But not to give the ending away, we will

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see that the power of the signal will 
also be dependent on the size of the

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alphabet. 
And so, in order not to exceed a certain

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probability of error, given the channel's 
power of constraint, we will not be able

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to grow the alphabet indefinitely. 
But we can be smart in the way we build

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this alphabet and so we will look at some 
examples.

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The first example is PAM, Pulse Amplitude 
Modulation.

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We split the incoming bitstream into 
chunks of M bits so that each chunk

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corresponds to an integer between 0 and 2 
to the M minus 1.

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We can call this sequence of integers k 
of n and this sequence is mapped onto a

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sequence of symbols a of n like so. 
There's a gain factor G, like always.

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And then we use 2 to the n minus 1 odd 
integers around 0.

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So for instance, if M is equal to 2, we 
have 0, 1, 2, and 3 as potential items

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for k of n. 
And a of n will be either.

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Let's assume G is equal to 1. 
Will be either minus 3, or minus 1, or 1,

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or 3. 
We will see why we use the odd integers

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in just a second. 
And the receiver the slicer will work by

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simply associating to the received 
symbol, the closest odd integer, always

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taking the gain into account. 
So graphically, again, PAM for M equal to

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2 and G equal to 1, will look like this. 
Here are the odd integers.

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The distance between two transmitted 
points, or transmitted symbols, is 2G

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right here. 
G Is equal to 1, but it would be in

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general 2 times the gain. 
And using odd integers creates a

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zero-mean sequence. 
If we assume that each symbol is

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equiprobable. 
Which is likely, given that we've used a

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scrambler in the transmitter. 
The the resulting mean is zero.

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The analysis of the probability of error 
for PAM is very similar to what we

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carried out for bilateral signaling. 
As a matter of fact, binary signaling is

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simply PAM with M equal to 1. 
The end result is very similar, and it's

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an exponential decaying function of the 
ratio between the power of the signal and

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the power of the noise. 
The reason why we don't analyze this

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further is because we have an improvement 
in store.

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And the improvement is aimed at 
increasing the throughput, increasing the

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number of bits per symbol that we can 
send without necessarily increasing the

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probability of error. 
So here's a wild idea.

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Let's use complex numbers and build a 
complex valued transmission system.

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This requires certain suspension of 
disbelief for the time being, but believe

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me, it will work in the end. 
The name for this complex valued mapping

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scheme is QAM. 
Which is an acronym for Quadtrature

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Amplitude Modulation, and it works like 
so.

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The mapper takes the income and bit 
stream, and splits it into chunks of M

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bits, with M even. 
Then it uses half of the bits, to define

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a PAM sequence, which we call a of r of 
n, and the reamaining, M over 2 bits, to

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define another independent PAM sequence. 
Ai of n.

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The final symbol sequence is a sequence 
of complex numbers, where the real part

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is the first PAM sequence, and the 
imaginary part is the second PAM

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sequence. 
And of course, in front we have a gain

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factor, G. 
So the transmission alphabet, a, is given

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by points in the complex plane, with 
odd-valued coordinates around the origin.

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At the receiver, the slicer works by 
finding the symbol in the alphabet, which

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is closest in Euclidean distance to the 
received symbol.

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Let's look at this graphically. 
This is a set of points for QAM

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transmission with M equal to 2, which 
corresponds to two bilevel PAM signals on

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the real axis and on the imaginary axis. 
So that results into four points.

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If we increase the number of bits per 
symbol, we set M equal to 4, that

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corresponds to two pam signals with 2 
bits each, which makes for a

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constellation. 
This is how these arrangement of points

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in the complex plain are called. 
A constellation of four by four points at

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the odd-valued coordinates in the complex 
plane.

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If we increase M to 8, then we have a 256 
point constellation, with 16 points per

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side. 
Lets look at what happens when a symbol

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is received, and how we derive an 
expression for the probability of error.

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If this is the nominal constellation, the 
transmitter will choose one of these

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values for transmission, say this one. 
And this value will corrupted by noise in

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the transmission and the receiving 
process.

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And will appear somewhere in the complex 
plane, not necessarily exactly on the

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point it originates from. 
The way the slicer operates, is by

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defining decision regions around each 
point in the constellation.

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So suppose for this point here, the 
transmitted point, the decision region is

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square, of side 2G, centered around which 
is made in point.

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So what happens is that when we receive 
symbols.

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They will now fall on the original point. 
But as long as they fall within the

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decision region, they will be decoded 
correctly.

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So for instance here. 
We will decode this correctly.

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Here we will decode this correctly. 
Same here.

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But this point for instance falls outside 
of the decision region and therefore it

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will be associated to a different 
constellation point, thereby causing an

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error. 
To quantify the probability of error, we

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assume as per usual that each received 
symbol is the sum of the transmitted

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symbol. 
Plus a noise sample theta of n.

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And we further assume that this noise is 
a complex value Gaussian noise of equal

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variance in the complex and real 
components.

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We're working on a completely digital 
system that operates.

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With complex valued quantities. 
So we're making a new model for the

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noise, and we will see later, how to 
translate the physical real noise, into a

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complex variable. 
With these assumptions, the probability

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of error, is equal to the probability 
that the real part of the noise is larger

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than G in magnitude. 
Plus the probability that the imaginary

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part of the noise is larger than G in 
magnitude.

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We assume that real and imaginary 
component of the noise are independent,

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and that's why we can split the 
probability like so.

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Now, if you remember the shape of the 
decision region, this condition is

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equivalent to saying that the noise is 
pushing the real part of the point,

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outside of the decision region, in either 
direction, and same for the imaginary

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part. 
Now if we develop this, this is equal to

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1 minus the probability that the real 
part of the noise is less than G, and the

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00:14:48,715 --> 00:14:52,390
imaginary part of the noise is less than 
G.

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This is the complimentary condition to 
what we just wrote above.

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And so this is equal to 1 minus the 
integral over the decision region d of

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the complex valued probability density 
function for the noise.

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In order to compute this integral, we're 
going to approximate the shape of the

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decision region. 
With the inbound circle.

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So instead of using the square, we're 
going to use a circle centered around the

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transmission point. 
When the constellation is very dense,

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this approximation is quite accurate. 
With this approximation, we can compute

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the integral exactly for a gaussian 
distribution.

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And if we assume that the variance of the 
noise is sigma 0 squared over 2 in each

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component, real or imaginary. 
It turns out that the probability of

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error is equal to each of the minus g 
squared over sigma 0 square.

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Now to obtain a probability of error as a 
function of the signal to noise ratio we

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have to compute the power of the 
transmitted signal.

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So if all symbols are equiprobable and 
independent, it turns out that the

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variance of the signal is G squared times 
1 over 2 to the power of M.

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Which is the probability of each symbol, 
times the sum over all symbols in the

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alphabet of the magnitude of the symbols 
squared.

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Now, it's a little bit tedious but we can 
solve it exactly for M.

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And it turns out that the power to 
transmit the signal is g squared 2 rds3,

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2 to the n to the minus 1. 
Now, if you plug this into the formula

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for the probability of error that we seen 
before.

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00:16:24,165 --> 00:16:28,851
We get that the result is an exponential 
function where the argument is minus 3,

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that multiplies 2 to the minus m plus 1, 
that multiplies the signals to noise

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ratio. 
We can plot this probability of error in

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a log log scale, like we did before. 
And we can paramatrize the curve, as a

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function of the number of points in the 
constellation.

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So here you have the curve for a four 
point constellation, Here's the curve for

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16-points and here's the curve for 
64-points.

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Now you can see that for a given signal 
to noise ratio the probability of error

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increases with the number of points. 
Why is that?

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Well if the signal to noise remains the 
same, and we assume that the noise is

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always at the same level, then it means 
that the power of the signal remains

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00:17:06,580 --> 00:17:10,900
constant as well. 
In that case, if the number of points

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increases, g has to become smaller. 
In order to accomodate a larger number of

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00:17:15,814 --> 00:17:19,354
points for the same power. 
But if g becomes smaller, then the

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00:17:19,354 --> 00:17:23,764
decision regions becomes smaller, the 
separation between points become smaller,

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00:17:23,764 --> 00:17:28,570
and the decision process becomes more 
vulnerable to noise.

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00:17:28,570 --> 00:17:32,490
So in the end here's the final recipe to 
design a QAM transmitter.

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First you pick a probability of error 
that you can live.

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00:17:34,540 --> 00:17:37,600
In general, 10 to the minus 6 is an 
acceptable probability of error at the

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00:17:37,600 --> 00:17:41,116
symbol level. 
Then you find out the signals noise ratio

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00:17:41,116 --> 00:17:44,920
that is imposed by the channel's power 
constraint.

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00:17:44,920 --> 00:17:49,724
Once you have that, you can find the size 
of your constellation, by finding M.

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00:17:49,724 --> 00:17:53,628
Which, based on the previous equations, 
is the log and base 2 of 1 minus 3 over 2

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00:17:53,628 --> 00:17:57,288
times the signal to noise ratio, divided 
by the natural logarithm of the

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00:17:57,288 --> 00:18:02,109
probability of error. 
Of course, you will have to round this to

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00:18:02,109 --> 00:18:05,211
a suitable integer value, and potentially 
to an even power of 2 in order to have a

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00:18:05,211 --> 00:18:09,632
square constellation. 
The final data rate of your system will

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00:18:09,632 --> 00:18:13,658
be M, the number of bits per symbol, 
times W, which, if you remember, is the

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00:18:13,658 --> 00:18:17,816
baud rate of the system, and corresponds 
to the bandwidth allowed for by the

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00:18:17,816 --> 00:18:23,335
channel. 
So we know how to fit the bandwidth

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00:18:23,335 --> 00:18:27,838
constraint via upsampling. 
With QAM, we know how many bits per

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00:18:27,838 --> 00:18:30,860
symbol we can use given the power 
constraint.

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00:18:30,860 --> 00:18:34,600
And so we know the theoretical throughput 
of the transmit for a given reliability

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00:18:34,600 --> 00:18:38,607
figure. 
However, the question remains, how are we

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00:18:38,607 --> 00:18:44,400
going to send complex value symbols over 
a physical channel?

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00:18:44,400 --> 00:18:48,900
It's time, therefore, to Stop the 
suspension of this belief, and look at

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00:18:48,900 --> 00:18:54,902
techniques to do complex signaling over a 
real value channel.