In this video, we're going to talk about
digital communication, the transmition of
bits from one place to another.
It turns out that most channels, or even
digital communication applications are
analog.
So, we have to worry about how we send
those bits.
What is a good choice for the signals, the
analog signals to use, and what are some
bad choices.
Well, since the channel is also analog, we
have to worry about how much bandwidth is
being used to send that bit sequence and
becoming a very important consideration.
So, let's get started and develop what's
called the digital communication model.
So here, I show a situation similar to
what we've already talked about in terms
of analog communication.
In that, I have an analog signal and
message m has to be sent to sink.
Now, what we're going to do as opposed to
the analog situation, where we
automatically send that to the transmitter
to modulate it, let's say, here, we go
through an A to D converter, to create
bits.
Those bits have to somehow be related to
an analog signal that can go through our
analog channel, be it wireline or
wireless.
The receiver then will convert that analog
signal back into a bit sequence, which
then goes into our familiar thing, the D
to A converter and to produce the, the
message that was trying to be transmitted.
So, it's a complicated sequence of events
here, much more so than in an analog case,
but there are some real advantages and
some disadvantages.
And we're going to talk about those as we
go through.
The, there are also situations in which
the message is inherently digital, a
sequence of bits and what I'm thinking
about here is text.
Text, can, is automatically a sequence of
bits, there is no A to D converter at all
so the same model and processing that I
have in both of these are the same.
It doesn't really matter except there's no
A to D or D to A in the digital message
case.
We still have the problem how do you
assign analog signals to bits and how do
you undo that to receive that bit sequence
again.
The channels are all analog in almost
every case.
So, it turns out, assigning bits to analog
signals is a pretty easy task.
What we're basically going to do is assign
a signal, a specific signal for each bit.
We're going to have a signal that
represents a 0, a different signal that
represents a 1.
Those two signals, whatever we choose,
form what we call a signal set.
Let me give you a simple example.
And this is the simplest digital
communication case that we have.
It's called BPSK.
Stands for Binary Phase Shift Keying.
Keying is an old word which comes from the
telegraph days which sending dots and
dashes is the same as sending zeros and
ones in many ways.
So, that's where the terminology is
borrowed from.
In Binary Phase Shift Keying, whatever
signal you use to represent one bit, here
I have a positive [unknown] to represent a
zero.
The other signal is the negative, and that
defines a binary shift keying signal set.
In all cases, one signal is the negative
of the other.
Now, in the case I've shown here, you
could consider this a baseband signal set.
It exists at low frequencies.
If you wanted to modulate it up, you can.
That's also a BPSK signal set because the
two members of the signal set obey this
relationship.
You can see here I have a positive sine
wave and here, I have a negative sine wave
so they are still BPSK.
Now, the, the important thing about
signals in a signal set is that they are
finite duration.
Because what's going to happen is, we're
going to send one bit after another.
And that means each signal represents a
single bit, has to only take a specific
amount of time, and we call that time
capital T and it's know as the bit
interval, time taken to transmit a single
bit.
Well, this has an important consequence,
because now the datarate, the rate at
which we send bits, is given by the
reciprocal of that time interval.
So, whatever the time interval is, the
datarate is the reciprocal, so if you have
a 1 megabit per second transmission rate,
that corresponds to a time of one
microsecond.
One microsecond has been used for each bit
and so there's a direct relationship here.
And we'll see this as important
consequences in just a second.
Alright, now, now here, I'm showing an
example of sending a bit sequence 0110 one
after another.
And you can see how we just attack on each
signal from the signal set one after
another to produce the transmitted wave
form.
It could be modulated or not it doesn't
really matter.
But all, that's why their finite duration,
we simply go one after another until we
finally represent our bit sequence with
analog signals.
And again, the receiver is going to be
worrying about trying to figure out what
the bit sequence was from this analog
signal.
You know, another signal set is the FSK
signal set.
Stands for Frequency- Shift Keying.
And here, we're going to use sine waves to
represent the bits but with two different
frequencies.
Usually, these frequencies are harmonic to
the bit interval and in this example, I
show a frequency of 0 that is 3 over T.
And here, I show an f1 that's 4 over
capital T.
And that is a typical FSK signal set.
They could be different harmonics.
Higher harmonics could be separated by
more than one harmonic number whatever
your choice may be.
The P of t in this formula has to do with
the fact these are finite durations are
multiplying by faults.
So, these are not sine waves for all time.
They again, have duration t.
So, there's a datarate related to the
reciprocal of T just like in the BPSK
case.
And here is what it looks like when you
have that same bit sequence.
It's a bit more subtle, what the different
between a 0 and 1 is.
But hopefully, there are receivers that
can figure this out.
You know, one is the transmission of
bandwidths, how much analog bandwidth in
hertz is taken by sending a sequence of
bits.
Well, the one problem that comes up right
away is that bandwidth depends on what the
bit sequence is itself.
So, let me show you this by a simple
example.
Suppose this is time, and here are my bit
intervals.
And I suppose the bit sequence I'm trying
to represent is 0000.
Well, the signal that represents that for
our signal BPSK baseband signal site is a
constant and that constant has no
bandwidth.
It's a constant basically for all time,
and clearly, that's not going to work very
well.
If I have my alternate bit sequence that I
showed in previous slides, now the
transmitted signal looks like this, which
has a bandwidth, has a wider bandwidth,
certainly than a constant signal.
What signal set, not the signal set, what
bit sequence gives you the worse case
bandwidth?
I've given, it's clearly not the constant.
A bit sequence that has, has no bandwidth,
and the one I showed here is there one
that's worse, that has a higher frequency
content than these.
And the answer, of course, is the
alternating bit sequence is the worst,
because now the signal goes back and
forth, back and forth, back and forth.
We know what the spectrum of this is.
This is a square wave.
So, we right away, using Fourier series,
we know what the spectrum is of this worst
case bit sequence represented by a
baseband BPSK signal set.
We know that its spectrum is given by 2
over j pi k for k odd.
But one thing to note is what the period
is here.
One period here is twice the bit interval.
So, that's why it's a harmonic of 2T
rather than, we use capital T to represent
a period when we talk about square waves.
The period here is 2T, where T is now the
bit interval.
Okay, now, one consequence of this is that
we know that the spectrum is basically
infinitely long.
So, if we have something at the first
harmonic, the third, the fifth, the
seventh, etc., the bandwidth is basically
infinite.
And so, we have a more practical
definition of bandwidth and that is to use
what's called the 90% bandwidth.
So, we want to worry about something
that's similar to what we talked about in
terms of total harmonic distortion.
I want to figure out how many Fourier
series coefficients are needed so that I
grab 90% of the power.
And so, here's the idea.
We have a spectrum of the baseband signal
set for example, which is like this,
right?
First and the third harmonics.
How much, how far do I have to go out so
that the total power in these harmonics is
90% of the total power?
And it turns out when you do the
calculations for the square wave, it turns
out it's almost exactly, it is exactly the
third harmonic.
So, the bandwidth that we care about is
this bandwidth right there.
And so, this harmonic number corresponds
to an absolute frequency of 2 over T so
the bandwidth is 3 over 2T.
So, it's 3 halves of the datarate, the
bandwidth taken to transmit our 1 megabit
per second signal is 1.5 megahertz.
And this is, in general, true.
You need a larger bandwidth in hertz than
you do in datarate measured in bits per
cycle.
When you modulate the baseband signal set,
as you may know, the bandwidth doubles
because now you have to measure things on
either side of this inner frequency.
So now, the bandwidth taken is 3R.
It requires a higher, a larger bandwidth
to transmit, to modulated it than at
baseband.
But as we know, we are forced to use
modulation in most cases anyway because
wireless, wireline channels really work
the best at higher frequencies than
baseband.
Now, the transmission bandwidth for FSK is
a bit more complicated to derive.
So, I'm again going to use my worst case
bit sequence here.
And think about this FSK signal as a
superposition.
So, what I have is a f0 sine wave
multiplied by a square wave and here, I
have an f1 frequency sine wave multiplied
by a square wave, but they're out of phase
with respect to each other.
And all we do is add these up and that
gives us the transmitted signal.
If I can figure out the bandwidth of each,
then I'm in good shape, I can figure it
out what the total bandwidth is by sensor
spectra I have to add.
I think I, I have a way of getting at it.
But it's again, a square wave, okay?
So, I think we can calculate that spectrum
just like we did before.
I'm going to assume that the carrier
frequencies are harmonics of the bit
interval, k0 and k1, to be completely
general.
Now, it turns out, here's a little detail
here and that these square waves, having
nonzero value, their, their average values
is, are not 0, which means c0 is not 0 in
the Fourier series.
So, the spectrum, before you modulate it
up, of that square wave has a value at the
origin, a value at the first harmonic, and
some smaller value given by the formula at
the third harmonic.
So now, what's the bandwidth?
Now, since the there's a, a nonzero
component at the origin, the bandwidth, it
turns out the 90% bandwidth is now this.
It turns out it occurs for just have to go
to the first harmonic.
Now, when you now I want to think about
what the modulated spectrum looks like.
We ought to draw a little picture here.
So, after we modulate, okay, we modulate
up to f0.
We have our center.
You go up to the first harmonic on either
side.
And so, this is a has a bandwidth now of
some amount, but we're worried about the
total bandwidth of the entire FSK signal
set.
So, we draw the other signal, let's say,
it's f1, it, too, has, looks like that.
So, the bandwidth is going to go from the
lowest frequency for f0 up to the highest
frequency for f1.
So, that's what I show in these equations.
We want to, the lowest frequency that's
required for the 90% bandwidth for FSK is
given by this, the highest frequency given
by that.
And putting it in terms of harmonic
numbers, you get these relationships,
subtract those upper and lower frequencies
and you get the following expression when
all is said and done relating it to
datarate, which is 1 over T.
So, it's the difference of the harmonic
numbers plus 1 times the datarate, turns
out to be the bandwidth of FSK, alright?
So, it's interesting you compare BPSK and
FSK.
So, since the FSK is a modulated signal
set, we only need to be fair, we only need
to consider a modulated version of BPSK.
So, we found that the bandwidth for BPSK
is 3 times the datarate, well, it turns
out if you use adjacent harmonic numbers,
the bandwidth for FSK is more efficient.
For the same datarate, you can send a
digital sequence, sequence of bits in a
smaller bandwidth and a significantly
smaller bandwidth.
Of course, if you don't pick adjacent
numbers, harmonic numbers, this bandwidth
goes up.
But it's interesting that you can now
directly compare and worry about the
efficiency of the communication scheme,
the efficiency, at least in terms of
bandwidth required in the channel.
So, in digital communication, what we do
is we assign a signal set to each bit, and
these signals have to be a finite
duration.
That's really important.
So, we can send one bit after another.
The datarate is determined both by the
source and the channel.
Now, what I mean by that is we've
discovered at whatever choice we make for
a signal set, that demands a certain
amount of bandwidth.
Well, that channel has to be able to
support that kind of bandwidth.
In other words, they can not narrow,
narrow band filter the signal so much that
it impinges upon the bandwidth that's
required to send the bits.
So, the datarate determines what the
bandwidth is, in most cases.
There's another issue though, is that this
source is putting out bits at a given
rate.
You probably want to send those bits
through the channel as they emerge from
the source.
This transmission scheme has to keep up,
not getting ahead of what the source is
doing.
If the transmitter works at a lower rate
than the source, there's going to be a
problem.
There's going to be a backlog of bits that
have to be sent.
So, in most cases, you have to keep up.
That's why I say the datarate is
determined by the source ultimately.
How fast is the source putting out those
bits, and then secondly, by the channel in
the terms of its bandwidth, does it have
the bandwidth to send those bits.
So, the signal set choice affects the
bandwidth and we've already seen that FSK
can be a bit more efficient.
But the one other thing that we need to
consider is reception performance.
Don't forget, this channel adds noise and
it attenuates the signal.
This has an affect on how accurately the
bits we get out are the same as the bits
we sent.
And so, it turns out the signal set choice
affects that also.
That's what we're going to talk about in
the next video.