So in this video, I want to go through an
extended example that compares analog and
digital communication.
So, what I'm going to assume is that I
have an analog signal that I want to send
from one place to another.
I have two choices, I can use amplitude
modulation, completely analog system.
Or I can run the analog signal, sample it
with an A to D converter, converts it to
bits.
Use digital communication schemes and then
put it through a D to A to get back your
original signal.
Which is better?
Particularly, when you are trying to use
the same channel for both.
So, I'm trying to make a very fair
comparison.
So, we're going to send, compare an analog
and digital scheme, who are on the same
channel.
Our criterion for quality, our criterion
for which one wins, is going to be the one
that produces the largest signal-to-noise
ratio in the final result.
That's going to be our test to see which
one is better.
Now as we've seen in previous videos,
analog communication because of the
channel noise is going to result in a
noisy signal.
And we already know what that
signal-to-noise ratio is.
And I'm just going to, recall that result
and use it.
We're going to have to do a bit of
calculation though for the digital
channel.
And I want to point out some things before
we go into the details.
Recall that because of the amplitude
quantization in the A to D converter.
That introduces an unrecoverable error.
Once you quantize to bits, you can not get
back the original analog signal perfectly.
So that results in a, a unrecoverable
error that we won't be able to do anything
about.
Also the channel introduces error because
the probability of error in receiving a
bit correctly is not zero.
So that also introduces noise into the
result, which makes the SNR go down.
So guess what?
No matter if it's analog or digital the
result is going to be noisy.
So we have to learn how to live with that
and that's just a fact of life.
More than that, which one results in the
smallest amount of noise?
Which one is better?
And, so here's the conditions for my
little test here.
I'm assuming I have a baseband signal.
That has a bandwidth of 8 kHz I'm sorry 4
kHz.
I'm just using this as an example a 4 kHz
signal.
And for the digital side it's going to be
sampled with a B-bit converter.
I'm going to let the number of bits be
your variable here to see what the effect
of that is.
This happen can be little bit surprised in
how long how that effects the results.
Well, so lets go through the analog
system.
I have to use in modulated scheme and so
we know that if you have a bandwidth of w
the transmission band width.
That you use in doing analog is twice that
bandwidth so we have an 8 kHz transmission
bandwidth.
And we also know from previous results
that the signal-to-noise ratio that
received a message signal has that
formula.
I do want to point out that I have not
written the alpha from the channel in
here, the channel insinuation.
I'm just merging that with A.
As you know, the correct expression isn't
alpha squared A squared.
And so, I'm just going to merge the two
into one symbol so.
So we simplify the expressions a little
bit.
Okay, now for digital communication we
have to sample, and we're going to have to
sample at a sampling frequency of 8 kHz,
twice the highest frequency.
Okay, and we also know that the
transmission bandwidth.
Is 3 times r for modulated bpsk which is a
very good signal set.
So the data rate is going to be the
sampling rate, 8 kilohertz times the
number of bits in that a to d converter.
Because we have to send the same signal
every capital T seconds, T being the
sampling interval.
You use more bits, you're going to have to
have a higher transmission bandwidth
because the data rate goes up.
So we're going to use this result a little
bit later.
So again to repeat what I said on the
first line, the digital side of things,
the final signal you get out of the
communication system is the original
signal plus the quantization error.
Which is due to the A to D converter's
amplitude quantization plus the
communication error, the effect of Pe not
being zero.
So we're going to f-figure out each of
these and just add them up and then
compute a signal to noise ratio.
So the quantization error we've gone
through.
So some more little details are we're
going to assume that the signal.
As an amplitude that's less than, 1.
Just, so I can scale things appropriately.
And if you express the final result out of
the A to D converter.
And then the D to A in terms of bits.
You get this, for the formula.
So to understand this let's suppose we
draw our, our diagram that we've been
doing.
So signal goes between minus 1 and 1.
And we're use, we're dividing this up into
quantization intervals somehow.
So if you set all the bits in this
expression to 0, what you get for a result
is minus 1.
So 0, 0, 0, 0, 0, our remaining bits we
use, is going to correspond to an
amplitude of minus 1.
And when you set it equal to all 1s,
you'll get something very close to 1, if
you evaluate this expression, so that's
the, this forum formula does not really
come out of the air.
It's just trying to relate bits to an
amplitude that's in this range, okay.
Well, when it's received we're going to
have a quantiza-, we're going to have an
error because the received bits aren't
necessarily the same.
And that's called a Channel Error.
But because the fact, back in the analog
and digital converter, we did not get back
exactly the answer that we started with.
We have already figured out in a previous
video.
That that's the power in the error.
2 to the minus 2b, B being the number of
bits in the converter, divided by 3.
And this result we derive previously.
But now we're going to look at the channel
errors.
Now, the channel errors, The resulting
error because of errors in the
communication just revolve around the fact
that the transmitted bit and the receive
bit are not the same.
So we want the power in that error and I'm
going to write that in kind of a new way.
These angle brackets mean average.
So that's a typical notation for main
average.
So we're going to square the error in
average, to get an idea what the average
power in the error is.
If you go through the calculations, what
you discover is just the average Of the
square of the difference between the 2
bits.
So we have to figure out now what that
average is and it's a pretty easy
calculation.
So, it's pretty clear, when the bits agree
that, that difference is 0 and there's
nothing that contributes to the.
Average, you know term that we'll get is
when the bits disagree.
So we could have sent a zero, and received
a one.
And we could have sent a one and received
a zero.
Well these errors occur with a probability
of pe.
But the probability we send to zero or
send to one is a half.
So that makes the total probability of
that term being a half Pe.
Total probability for this term being a
half and they all add up to just be Pe so
it turns out this averaging term here is
really quite easy to see.
I do wnat to piont out something else
before we go on.
Some bits matter more than others.
Notice this exponent out here.
This has to do with the fact that the
higher bit, the one that has a bigger
value of k here, you make an error in that
bit, it really effects the result for the
error.
If the smaller bit were k0, doesn't has,
have as much effect.
So some errors and some bits matter more
than others and we want to see the overall
effect of channel errors.
Okay so, so I'm using my result that Pe
was the average value here, it's a
constant, and to sum this term up it turns
out that's a finite geometric series,
which we've already talked about.
And once you do all those calculations,
this is what you wind up with.
Now, 2 to the 2b, I'm going to assume, is
a whole lot bigger than 1.
Which is normally the case.
And that's where the approximation,
approximate sign comes in.
And once you drop that one.
And to simplify, you get 4 3rds Pe.
So you get a rather simple result for the
channel error.
So that's the, power and the channel error
we have the power and the quantization
error.
And so the final result for the SNR, our
digital communications scheme.
Is, there's the signal power, of course,
and I just add up the two powers from
quantization and channel errors.
Okay.
So, there's a little detail here, and that
is we have to be fair about the
transmitter power I want to make the
transmitter power the same.
So, I'm going to recall something here.
So, recall for digital that the energy per
bit which si the critical part of
probability of error expression is A
squared T.
Well T is the dur-, is the bit interval
duration.
And so, what's T?
And T, as I wrote in the previous slide,
is 1 over R.
So, R, the data rate, is the sampling
frequency times the number of bits in the
A to D converter.
So relating it back to bandwidth and bits
this is expression we get and so you
figure it all our amplitude squared which
is from the transmitter in the effect of
the channel Its related to the energy per
bit through this expression.
So I'm going to take my analog expression,
and convert it into something that's
related to what I call Es.
Which is the energy per bit times the
number of bits.
So Es is the energy per sample that's used
in the communications scheme.
I want to keep that the same between the
analog and digital.
Systems so I can compare them fairly.
So our signal to noise ratio here, is
given by this.
Notice the power in s disappeared because
its going to be the same in both
expressions.
I'm assuming basically it's one, just to
keep it simple.
And here's Pe.
And here's Pe.
So it's for a QPSK thing.
The result was 2Eb over N not.
And i've simply substituted in Es for that
so I can explicitly take into account how
many bits are being So what I'm going to
do.
I'm going to plot this as a, the analog
system as a function in Es or N not.
I'm going to plot the result of this
communication schemes a function of Es
over N not.
And Es over N not, remember is the
basically summarizes how much and
attenuation was used in the, the
transmitter and the channel compared to
the channel noise.
So this is the axis we want to compare on.
And I've plotted the analog and digital
cases here the digital for two different
bits in the A to D converter.
And as we showed, the analog is the
simplest because the overall
signal-noise-ratio for the at the output
is the same as Es over N not, so you get a
straight line.
And that's the equality line.
So.
For 4 bits is the dash curve.
For 8 bits you get the solid red curve.
It's kind of interesting.
Let's see if we can understand this.
First of all, as the Channel
Signal-to-noise ratio goes up.
Pe basically is going to 0.
So what happens is you're left with the,
with the quantization errors.
So these are the limits of the
quantization error, and as we know, the
more bits you use, the smaller that error
gets, which, since this is the signal to
noise ratio That means the SNR goes up.
So that part I understand.
You go down to the other extreme, Pe is
basically a half.
No matter how many bits you use, because
the channel is really, really, really
noisy.
So that's why these two curves converege.
In between it's interesting, that the 4
bit result gives you a higher SNR than the
8 bit result.
And the reason for that is, because Pe for
the four bit case is actually smaller than
it is for the eight bit case.
Because you're dividing up time less from
dividing it by 4 rather than 8.
That has an effect of producing a slightly
bigger SNR in the result using 4 bits.
But only for a while.
Once the SNR gets big enough the 8 bit
becomes far superior.
However, can't lose sight of the fact,
that the digital curves, lie above the
analog curves.
So, in this comparison, the analog system
is inferior, it's worse than digital
systems.
This is despite the fact we introduce
error in the A to D converters, that we
can never get around.
It's clear that at least in this
comparison that analog is not as good as
digital.
So, digital wins.
And that's the kind of analysis that you
want to do as an engineer to figure out
which scheme is going to work.
You use some criterion, like the signal to
noise ratio for the, final receive
message.
And then you pick your design criterion
here is analog or digital according to
that choice.
There is a little caveat here, I do want
to point out, that one thing that isn't
fair with this comparison is that channel
bandwidth needed for analog is much
smaller that viewed for the digital
schemes we just talked about.
Digital transmission bandwidth, like I
said, is twice the highest frequency of
the message so it's just 8 kHz.
However, for 4 bits you need 96 kHz and
for 8 bits you need twice that much so in
some sense this isn't a very fair
comparison.
So what that means.
Is that, what happens if you make them the
same?
What happens if you constrain the
transmission bandwidth to be the same for
both cases.
Well, it turns out that if you constrain,
digital schemes to work within a 8 kHz
regime.
You can show that.
Amplitude modulation, the analog scheme
now wins.
It always results in a higher SNR, it's
kind of interesting.
However, if you use more bandwidth than
that, something like this, it turns out
you basically cannot effectively use that
extra bandwidth for analog communication.
It doesn't have that flexibility.
And then it turns out digital ones.
So, in the grand scheme of things, overall
this is one of the reasons why, in the
modern uses, everything is being converted
to bits and sent in a digital way.
Now, I'm going to talk about a situation
In a upcoming videos, we'll find out, that
you can actually send bits through a noisy
channel with Pe winding up being zero.
And we're going to figure out how that
works.
And it involves some very clever
engineering.
And we'll see that in, in the coming,
upcoming videos.