0:00:00.880,0:00:04.220
Hi and welcome to Module 9.3 of Digital [br]Signal Processing.

0:00:04.220,0:00:07.520
We are still talking about Digital [br]Communication Systems.

0:00:07.520,0:00:10.790
In the previous module we addressed [br]bandwidth constraint.

0:00:10.790,0:00:13.964
In this module we will tackle the powered [br]constraint so first we will introduce the

0:00:13.964,0:00:17.830
concept of noise and probability of error [br]in a communication system.

0:00:17.830,0:00:22.264
We will look at signaling alphabet and [br]power and their related power.

0:00:22.264,0:00:25.400
And finally, we'll introduce QAM [br]signaling.

0:00:25.400,0:00:28.600
So we have seen that a transmitter sends [br]a sequence of symbols a of n.

0:00:28.600,0:00:32.500
Created by the mapper. [br]Now we take the receiver into account.

0:00:32.500,0:00:36.340
We don't yet know how, but it's safe to [br]assume that the receiver in the end will

0:00:36.340,0:00:41.200
obtain an estimation hat a of n. [br]Of the original transmitted symbol

0:00:41.200,0:00:43.364
sequence. [br]It's an estimation because even if there

0:00:43.364,0:00:45.700
is no distortion introduced by the [br]channel.

0:00:45.700,0:00:49.591
Even if nothing bad happens. [br]There will always be a certain amount of

0:00:49.591,0:00:52.780
noise, that will corrupt the original [br]sequence.

0:00:52.780,0:00:56.225
When noise is very large, our estimate [br]for the transmitted symbol will be off,

0:00:56.225,0:01:00.140
and will incur a decoding error. [br]Now, this probability of error will

0:01:00.140,0:01:04.470
depend on the power of the noise, with [br]respect to the power of the signal.

0:01:04.470,0:01:07.620
And will also depend on the decoding [br]strategies that we've put in place, how

0:01:07.620,0:01:11.900
smart we are in circumventing the effects [br]of the noise.

0:01:11.900,0:01:15.512
One we can maximize the probability of [br]correctly guessing the transmit symbol,

0:01:15.512,0:01:20.412
is by using suitable alphabets. [br]And so we will see in more detail what

0:01:20.412,0:01:24.400
that means. [br]Remember the scheme for the transmitter.

0:01:24.400,0:01:26.570
We have a bitstream coming in. [br]And then we have the scrambler.

0:01:26.570,0:01:34.756
And then the mapper. [br]And here we have a sequence of symbols a

0:01:34.756,0:01:37.204
of n. [br]These symbols will have to be sent over

0:01:37.204,0:01:42.305
the channel. [br]And to do so, we upsample.

0:01:42.305,0:01:47.955
And we interpolate, and then we transmit. [br]Now, how do we go from bitstreams to

0:01:47.955,0:01:52.586
samples in more detail? [br]In other words, how does the mapper work?

0:01:52.586,0:01:56.870
The mapper will split the incoming [br]bitstreams into chunks and will assign a

0:01:56.870,0:02:02.260
symbol, a of n, from a finite alphabet to [br]each chunk.

0:02:02.260,0:02:05.900
The alphabet, we will decide later what [br]it is composed of.

0:02:05.900,0:02:09.890
To undo the mapping operation and recover [br]the bitstream, the receiver will perform

0:02:09.890,0:02:14.252
a slicing operation. [br]So the receiver will a value, hat a of n,

0:02:14.252,0:02:21.660
where hat indicates the fact that noise [br]has leaked into the value of the signal.

0:02:21.660,0:02:25.536
And the receiver will decide which symbol [br]from the alphabet, which is known to the

0:02:25.536,0:02:29.355
receiver as well, is closest to the [br]received symbol.

0:02:29.355,0:02:33.880
And from there, it will be extremely easy [br]to piece back the original bitstream.

0:02:33.880,0:02:36.730
As an example, let's look at simple [br]two-level signaling.

0:02:36.730,0:02:40.150
This generates signals of the kind we [br]have seen in the example so far,

0:02:40.150,0:02:45.157
alternating between two levels. [br]The way the mapper works is by splitting

0:02:45.157,0:02:51.657
the incoming bitstream into single bits. [br]And the output symbol sequence uses an

0:02:51.657,0:02:57.175
alphabet composed of two symbols, g and [br]minus g, and associates g to a bit of

0:02:57.175,0:03:05.270
value 1 and minus g to a bit of value 0. [br]And the receiver, the slicer.

0:03:05.270,0:03:09.622
Looks at the sign of the incoming symbol [br]sequence which has been corrupted by

0:03:09.622,0:03:13.410
noise. [br]And decides that the nth bit will be 1 if

0:03:13.410,0:03:18.750
the sign of the nth symbol is positive, [br]and 0 otherwise.

0:03:18.750,0:03:22.110
Lets look at an example, lets assume G [br]equal to 1.

0:03:22.110,0:03:25.980
So the two-level signal will alternate [br]between plus 1 and minus 1.

0:03:25.980,0:03:30.944
And suppose we have an input bit sequence [br]that gives rise to this signal here after

0:03:30.944,0:03:35.715
transmission and after decoding at the [br]receiver.

0:03:35.715,0:03:39.940
The resulting symbol sequence will look [br]like this, where each symbol has been

0:03:39.940,0:03:44.790
corrupted by a varying amount of noise. [br]If we now slice this sequence by

0:03:44.790,0:03:49.724
thresholding, as shown, shown before. [br]We recover a simple sequence like this

0:03:49.724,0:03:54.790
where we have indicated in red the errors [br]incurred by the slicer because of the

0:03:54.790,0:03:57.812
noise. [br]So if you want to analyze in more detail

0:03:57.812,0:04:01.508
what the probability of error is, we have [br]to make some hypothesis on the signals

0:04:01.508,0:04:07.321
involved in this toy experiment. [br]Assume that each received symbol can be

0:04:07.321,0:04:11.230
modeled as the original symbol plus a [br]noise sample.

0:04:11.230,0:04:14.520
Assume also that the bits in the [br]bitstream are equiprobable.

0:04:14.520,0:04:19.270
So zero and one appear with probability [br]50% each.

0:04:19.270,0:04:21.810
Assume that the noise and the signal are [br]independent.

0:04:21.810,0:04:25.842
And assume that the noise is additive [br]white Gaussian noise with zero mean and

0:04:25.842,0:04:30.370
known variance sigma 0. [br]With this hypothesis, the probability of

0:04:30.370,0:04:34.206
error can be written out as follows. [br]First of all, we split the probability of

0:04:34.206,0:04:38.860
errors into 2 conditional probabilities. [br]Conditioned by whether the nth bit is

0:04:38.860,0:04:41.700
equal to 1, or the nth bit is equal to [br]zero.

0:04:41.700,0:04:44.600
In the first case, when the nth bit is [br]equal to 1.

0:04:44.600,0:04:48.376
Remember, the produced symbol will be [br]equal to G, so the probability of error

0:04:48.376,0:04:53.616
is equal to the probability for the noise [br]sample to be less than minus G.

0:04:53.616,0:04:58.398
Because only in this case the sum of the [br]sample plus the noise will be negative.

0:04:58.398,0:05:02.991
Similarly, when the nth bit is equal to [br]0, we have a negative sample.

0:05:02.991,0:05:08.450
And the only way for that to change sign [br]is if the noise sample is greater than G.

0:05:08.450,0:05:13.730
Since the probability of each occurrence [br]is 1 half because of the symmetry of the

0:05:13.730,0:05:18.508
Gaussian distribution function. [br]This is equal to the probability for the

0:05:18.508,0:05:23.146
noise sample to be larger than G. [br]And we can compute this as the integral

0:05:23.146,0:05:26.916
from G to infinity of the probability [br]distribution function for the Gaussian

0:05:26.916,0:05:30.582
distribution with the known variance [br]here.

0:05:30.582,0:05:34.960
This function has a standard name. [br]It's called the error function.

0:05:34.960,0:05:38.110
And since this integral can not be [br]computed in closed form, this function is

0:05:38.110,0:05:41.550
available in most numerical packages [br]under this name.

0:05:41.550,0:05:45.326
So the important thing to notice here is [br]that the probability of error is some

0:05:45.326,0:05:48.984
function of the ratio between the [br]amplitude of the signal and the standard

0:05:48.984,0:05:54.212
deviation of the noise. [br]And we can carry this analysis further by

0:05:54.212,0:05:59.187
considering the transmitted power. [br]We have a bi-level signal and each level

0:05:59.187,0:06:03.200
occurs with 1 half probability. [br]So the variance of the signal, which

0:06:03.200,0:06:06.905
corresponds to the power, is equal to G [br]squared time the probability of the nth

0:06:06.905,0:06:11.700
being equal to 1. [br]Plus G squared times the probability of

0:06:11.700,0:06:14.710
the nth bit being equal to 0, which is [br]equal to G squared.

0:06:14.710,0:06:18.120
And so, if we rewrite the probability [br]error function we can write that it is

0:06:18.120,0:06:23.136
equal to the error function of the ratio. [br]Between the standard deviation of the

0:06:23.136,0:06:26.528
transmitted signal divided by the [br]standard deviation of the noise, which is

0:06:26.528,0:06:30.132
equivalent to saying that it is the error [br]function of the square root of the signal

0:06:30.132,0:06:34.810
to noise ratio. [br]If we plot this as a function of the

0:06:34.810,0:06:39.875
signal to noise ratio in dBs. [br]And I remind here that dBs here mean that

0:06:39.875,0:06:44.348
we compute 10 times the log in base 10 of [br]the power of the signal divided by the

0:06:44.348,0:06:49.773
power of the noise. [br]And since we are in a log log scale, we

0:06:49.773,0:06:54.388
can see that the probability of error [br]decays exponentially with the signal to

0:06:54.388,0:06:59.800
noise ratio. [br]This exponential decay is quite the norm

0:06:59.800,0:07:02.882
in communication systems. [br]And while the absolute rate of decay

0:07:02.882,0:07:07.580
might change in terms of the linear [br]constants involved in the curve.

0:07:07.580,0:07:12.110
The trend will stay the same even for [br]more complex signaling schemes.

0:07:12.110,0:07:15.197
So the lesson that we learn from the [br]simple example is that in order to reduce

0:07:15.197,0:07:19.905
the probability of error, we should [br]increase G, the amplitude of the signal.

0:07:19.905,0:07:23.265
But of course, increasing G also [br]increases the power of the transmitted

0:07:23.265,0:07:28.140
signal, and we know that we cannot go [br]above the channel's power constraint.

0:07:28.140,0:07:33.380
And so that's how the power constraint [br]limits the reliability of transmission.

0:07:33.380,0:07:37.760
The bilevel signalling scheme is very [br]instructive, but it's also very limited

0:07:37.760,0:07:41.350
in the sense that we're sending just one [br]bit per output symbol.

0:07:41.350,0:07:44.290
So to increase the throughput, to [br]increase the number of bits per second

0:07:44.290,0:07:47.990
that we send over a channel, we can use [br]multilevel signaling.

0:07:47.990,0:07:50.790
There are very many ways to do so and we [br]will just look at a few, but the

0:07:50.790,0:07:56.382
fundamental idea is that we take now. [br]Larger chunks of bits and therefore we

0:07:56.382,0:08:00.210
have alphabets that have a higher [br]cardinality.

0:08:00.210,0:08:04.434
So more values in the alphabet means more [br]bits per symbol and therefore a higher

0:08:04.434,0:08:07.576
data rate. [br]But not to give the ending away, we will

0:08:07.576,0:08:10.408
see that the power of the signal will [br]also be dependent on the size of the

0:08:10.408,0:08:13.614
alphabet. [br]And so, in order not to exceed a certain

0:08:13.614,0:08:16.857
probability of error, given the channel's [br]power of constraint, we will not be able

0:08:16.857,0:08:21.200
to grow the alphabet indefinitely. [br]But we can be smart in the way we build

0:08:21.200,0:08:24.480
this alphabet and so we will look at some [br]examples.

0:08:24.480,0:08:27.580
The first example is PAM, Pulse Amplitude [br]Modulation.

0:08:27.580,0:08:31.651
We split the incoming bitstream into [br]chunks of M bits so that each chunk

0:08:31.651,0:08:36.720
corresponds to an integer between 0 and 2 [br]to the M minus 1.

0:08:36.720,0:08:40.248
We can call this sequence of integers k [br]of n and this sequence is mapped onto a

0:08:40.248,0:08:46.169
sequence of symbols a of n like so. [br]There's a gain factor G, like always.

0:08:46.169,0:08:50.770
And then we use 2 to the n minus 1 odd [br]integers around 0.

0:08:50.770,0:08:58.300
So for instance, if M is equal to 2, we [br]have 0, 1, 2, and 3 as potential items

0:08:58.300,0:09:02.685
for k of n. [br]And a of n will be either.

0:09:02.685,0:09:10.970
Let's assume G is equal to 1. [br]Will be either minus 3, or minus 1, or 1,

0:09:10.970,0:09:15.870
or 3. [br]We will see why we use the odd integers

0:09:15.870,0:09:19.180
in just a second. [br]And the receiver the slicer will work by

0:09:19.180,0:09:23.200
simply associating to the received [br]symbol, the closest odd integer, always

0:09:23.200,0:09:28.600
taking the gain into account. [br]So graphically, again, PAM for M equal to

0:09:28.600,0:09:32.350
2 and G equal to 1, will look like this. [br]Here are the odd integers.

0:09:32.350,0:09:37.790
The distance between two transmitted [br]points, or transmitted symbols, is 2G

0:09:37.790,0:09:42.288
right here. [br]G Is equal to 1, but it would be in

0:09:42.288,0:09:47.730
general 2 times the gain. [br]And using odd integers creates a

0:09:47.730,0:09:49.224
zero-mean sequence. [br]If we assume that each symbol is

0:09:49.224,0:09:51.472
equiprobable. [br]Which is likely, given that we've used a

0:09:51.472,0:09:55.500
scrambler in the transmitter. [br]The the resulting mean is zero.

0:09:55.500,0:09:58.855
The analysis of the probability of error [br]for PAM is very similar to what we

0:09:58.855,0:10:03.829
carried out for bilateral signaling. [br]As a matter of fact, binary signaling is

0:10:03.829,0:10:08.231
simply PAM with M equal to 1. [br]The end result is very similar, and it's

0:10:08.231,0:10:11.427
an exponential decaying function of the [br]ratio between the power of the signal and

0:10:11.427,0:10:15.300
the power of the noise. [br]The reason why we don't analyze this

0:10:15.300,0:10:18.530
further is because we have an improvement [br]in store.

0:10:18.530,0:10:21.743
And the improvement is aimed at [br]increasing the throughput, increasing the

0:10:21.743,0:10:25.580
number of bits per symbol that we can [br]send without necessarily increasing the

0:10:25.580,0:10:29.990
probability of error. [br]So here's a wild idea.

0:10:29.990,0:10:34.130
Let's use complex numbers and build a [br]complex valued transmission system.

0:10:34.130,0:10:37.298
This requires certain suspension of [br]disbelief for the time being, but believe

0:10:37.298,0:10:41.394
me, it will work in the end. [br]The name for this complex valued mapping

0:10:41.394,0:10:44.790
scheme is QAM. [br]Which is an acronym for Quadtrature

0:10:44.790,0:10:47.736
Amplitude Modulation, and it works like [br]so.

0:10:47.736,0:10:52.176
The mapper takes the income and bit [br]stream, and splits it into chunks of M

0:10:52.176,0:10:56.842
bits, with M even. [br]Then it uses half of the bits, to define

0:10:56.842,0:11:00.934
a PAM sequence, which we call a of r of [br]n, and the reamaining, M over 2 bits, to

0:11:00.934,0:11:06.850
define another independent PAM sequence. [br]Ai of n.

0:11:06.850,0:11:11.800
The final symbol sequence is a sequence [br]of complex numbers, where the real part

0:11:11.800,0:11:14.473
is the first PAM sequence, and the [br]imaginary part is the second PAM

0:11:14.473,0:11:18.202
sequence. [br]And of course, in front we have a gain

0:11:18.202,0:11:21.985
factor, G. [br]So the transmission alphabet, a, is given

0:11:21.985,0:11:29.490
by points in the complex plane, with [br]odd-valued coordinates around the origin.

0:11:29.490,0:11:33.195
At the receiver, the slicer works by [br]finding the symbol in the alphabet, which

0:11:33.195,0:11:37.293
is closest in Euclidean distance to the [br]received symbol.

0:11:37.293,0:11:42.170
Let's look at this graphically. [br]This is a set of points for QAM

0:11:42.170,0:11:47.120
transmission with M equal to 2, which [br]corresponds to two bilevel PAM signals on

0:11:47.120,0:11:55.000
the real axis and on the imaginary axis. [br]So that results into four points.

0:11:55.000,0:11:58.864
If we increase the number of bits per [br]symbol, we set M equal to 4, that

0:11:58.864,0:12:02.590
corresponds to two pam signals with 2 [br]bits each, which makes for a

0:12:02.590,0:12:07.547
constellation. [br]This is how these arrangement of points

0:12:07.547,0:12:12.136
in the complex plain are called. [br]A constellation of four by four points at

0:12:12.136,0:12:15.900
the odd-valued coordinates in the complex [br]plane.

0:12:15.900,0:12:22.725
If we increase M to 8, then we have a 256 [br]point constellation, with 16 points per

0:12:22.725,0:12:26.840
side. [br]Lets look at what happens when a symbol

0:12:26.840,0:12:31.420
is received, and how we derive an [br]expression for the probability of error.

0:12:31.420,0:12:35.182
If this is the nominal constellation, the [br]transmitter will choose one of these

0:12:35.182,0:12:40.310
values for transmission, say this one. [br]And this value will corrupted by noise in

0:12:40.310,0:12:43.629
the transmission and the receiving [br]process.

0:12:43.629,0:12:47.919
And will appear somewhere in the complex [br]plane, not necessarily exactly on the

0:12:47.919,0:12:52.100
point it originates from. [br]The way the slicer operates, is by

0:12:52.100,0:12:56.500
defining decision regions around each [br]point in the constellation.

0:12:56.500,0:13:01.505
So suppose for this point here, the [br]transmitted point, the decision region is

0:13:01.505,0:13:07.150
square, of side 2G, centered around which [br]is made in point.

0:13:07.150,0:13:10.890
So what happens is that when we receive [br]symbols.

0:13:10.890,0:13:14.302
They will now fall on the original point. [br]But as long as they fall within the

0:13:14.302,0:13:17.110
decision region, they will be decoded [br]correctly.

0:13:17.110,0:13:19.710
So for instance here. [br]We will decode this correctly.

0:13:19.710,0:13:22.520
Here we will decode this correctly. [br]Same here.

0:13:22.520,0:13:26.396
But this point for instance falls outside [br]of the decision region and therefore it

0:13:26.396,0:13:29.987
will be associated to a different [br]constellation point, thereby causing an

0:13:29.987,0:13:33.916
error. [br]To quantify the probability of error, we

0:13:33.916,0:13:37.574
assume as per usual that each received [br]symbol is the sum of the transmitted

0:13:37.574,0:13:42.156
symbol. [br]Plus a noise sample theta of n.

0:13:42.156,0:13:47.634
And we further assume that this noise is [br]a complex value Gaussian noise of equal

0:13:47.634,0:13:52.640
variance in the complex and real [br]components.

0:13:52.640,0:13:57.860
We're working on a completely digital [br]system that operates.

0:13:57.860,0:14:01.614
With complex valued quantities. [br]So we're making a new model for the

0:14:01.614,0:14:05.326
noise, and we will see later, how to [br]translate the physical real noise, into a

0:14:05.326,0:14:09.764
complex variable. [br]With these assumptions, the probability

0:14:09.764,0:14:13.604
of error, is equal to the probability [br]that the real part of the noise is larger

0:14:13.604,0:14:18.606
than G in magnitude. [br]Plus the probability that the imaginary

0:14:18.606,0:14:21.850
part of the noise is larger than G in [br]magnitude.

0:14:21.850,0:14:24.640
We assume that real and imaginary [br]component of the noise are independent,

0:14:24.640,0:14:27.598
and that's why we can split the [br]probability like so.

0:14:27.598,0:14:31.738
Now, if you remember the shape of the [br]decision region, this condition is

0:14:31.738,0:14:35.947
equivalent to saying that the noise is [br]pushing the real part of the point,

0:14:35.947,0:14:40.570
outside of the decision region, in either [br]direction, and same for the imaginary

0:14:40.570,0:14:45.376
part. [br]Now if we develop this, this is equal to

0:14:45.376,0:14:48.715
1 minus the probability that the real [br]part of the noise is less than G, and the

0:14:48.715,0:14:52.390
imaginary part of the noise is less than [br]G.

0:14:52.390,0:14:57.200
This is the complimentary condition to [br]what we just wrote above.

0:14:57.200,0:15:00.737
And so this is equal to 1 minus the [br]integral over the decision region d of

0:15:00.737,0:15:05.680
the complex valued probability density [br]function for the noise.

0:15:05.680,0:15:09.600
In order to compute this integral, we're [br]going to approximate the shape of the

0:15:09.600,0:15:13.430
decision region. [br]With the inbound circle.

0:15:13.430,0:15:16.454
So instead of using the square, we're [br]going to use a circle centered around the

0:15:16.454,0:15:19.838
transmission point. [br]When the constellation is very dense,

0:15:19.838,0:15:24.118
this approximation is quite accurate. [br]With this approximation, we can compute

0:15:24.118,0:15:27.570
the integral exactly for a gaussian [br]distribution.

0:15:27.570,0:15:32.498
And if we assume that the variance of the [br]noise is sigma 0 squared over 2 in each

0:15:32.498,0:15:37.820
component, real or imaginary. [br]It turns out that the probability of

0:15:37.820,0:15:41.800
error is equal to each of the minus g [br]squared over sigma 0 square.

0:15:41.800,0:15:45.496
Now to obtain a probability of error as a [br]function of the signal to noise ratio we

0:15:45.496,0:15:49.450
have to compute the power of the [br]transmitted signal.

0:15:49.450,0:15:53.415
So if all symbols are equiprobable and [br]independent, it turns out that the

0:15:53.415,0:15:58.819
variance of the signal is G squared times [br]1 over 2 to the power of M.

0:15:58.819,0:16:02.979
Which is the probability of each symbol, [br]times the sum over all symbols in the

0:16:02.979,0:16:07.220
alphabet of the magnitude of the symbols [br]squared.

0:16:07.220,0:16:11.432
Now, it's a little bit tedious but we can [br]solve it exactly for M.

0:16:11.432,0:16:15.152
And it turns out that the power to [br]transmit the signal is g squared 2 rds3,

0:16:15.152,0:16:20.610
2 to the n to the minus 1. [br]Now, if you plug this into the formula

0:16:20.610,0:16:24.165
for the probability of error that we seen [br]before.

0:16:24.165,0:16:28.851
We get that the result is an exponential [br]function where the argument is minus 3,

0:16:28.851,0:16:33.324
that multiplies 2 to the minus m plus 1, [br]that multiplies the signals to noise

0:16:33.324,0:16:37.590
ratio. [br]We can plot this probability of error in

0:16:37.590,0:16:41.741
a log log scale, like we did before. [br]And we can paramatrize the curve, as a

0:16:41.741,0:16:45.550
function of the number of points in the [br]constellation.

0:16:45.550,0:16:49.645
So here you have the curve for a four [br]point constellation, Here's the curve for

0:16:49.645,0:16:53.520
16-points and here's the curve for [br]64-points.

0:16:53.520,0:16:56.912
Now you can see that for a given signal [br]to noise ratio the probability of error

0:16:56.912,0:17:01.000
increases with the number of points. [br]Why is that?

0:17:01.000,0:17:03.790
Well if the signal to noise remains the [br]same, and we assume that the noise is

0:17:03.790,0:17:06.580
always at the same level, then it means [br]that the power of the signal remains

0:17:06.580,0:17:10.900
constant as well. [br]In that case, if the number of points

0:17:10.900,0:17:15.814
increases, g has to become smaller. [br]In order to accomodate a larger number of

0:17:15.814,0:17:19.354
points for the same power. [br]But if g becomes smaller, then the

0:17:19.354,0:17:23.764
decision regions becomes smaller, the [br]separation between points become smaller,

0:17:23.764,0:17:28.570
and the decision process becomes more [br]vulnerable to noise.

0:17:28.570,0:17:32.490
So in the end here's the final recipe to [br]design a QAM transmitter.

0:17:32.490,0:17:34.540
First you pick a probability of error [br]that you can live.

0:17:34.540,0:17:37.600
In general, 10 to the minus 6 is an [br]acceptable probability of error at the

0:17:37.600,0:17:41.116
symbol level. [br]Then you find out the signals noise ratio

0:17:41.116,0:17:44.920
that is imposed by the channel's power [br]constraint.

0:17:44.920,0:17:49.724
Once you have that, you can find the size [br]of your constellation, by finding M.

0:17:49.724,0:17:53.628
Which, based on the previous equations, [br]is the log and base 2 of 1 minus 3 over 2

0:17:53.628,0:17:57.288
times the signal to noise ratio, divided [br]by the natural logarithm of the

0:17:57.288,0:18:02.109
probability of error. [br]Of course, you will have to round this to

0:18:02.109,0:18:05.211
a suitable integer value, and potentially [br]to an even power of 2 in order to have a

0:18:05.211,0:18:09.632
square constellation. [br]The final data rate of your system will

0:18:09.632,0:18:13.658
be M, the number of bits per symbol, [br]times W, which, if you remember, is the

0:18:13.658,0:18:17.816
baud rate of the system, and corresponds [br]to the bandwidth allowed for by the

0:18:17.816,0:18:23.335
channel. [br]So we know how to fit the bandwidth

0:18:23.335,0:18:27.838
constraint via upsampling. [br]With QAM, we know how many bits per

0:18:27.838,0:18:30.860
symbol we can use given the power [br]constraint.

0:18:30.860,0:18:34.600
And so we know the theoretical throughput [br]of the transmit for a given reliability

0:18:34.600,0:18:38.607
figure. [br]However, the question remains, how are we

0:18:38.607,0:18:44.400
going to send complex value symbols over [br]a physical channel?

0:18:44.400,0:18:48.900
It's time, therefore, to Stop the [br]suspension of this belief, and look at

0:18:48.900,0:18:54.902
techniques to do complex signaling over a [br]real value channel.