In this video, we're going to talk about the Wireline Communication Channel.
What that means is that the transmitter and the receiver, are connected by
literally a wire. It also could turn out it means it could
be optic fiber. It just depends on which situation.
This has, these kind of channels have very different characteristics than the
wireless channels, discussed in the previous video.
And lets see how that word. The wireline channel I have a dedicated
piece of wire connecting them, that's the idea, could be more than one transmitter
and one receiver on a given piece of wire. It turns out to be the way ethernet works.
We're not going to talk much about that but basically, that is also a wireline
channel. In designing a wireline channel Maxwell's
equation come into the act again because you have to minimize what's called
crosstalk. So, let me take a simple example which is
known as the two, two-wire pair. You need two wires because lets assume a
sending station is a voltage source and you can't apply a voltage except if you
have two places to apply that voltage. So, you need a, that's why you have a two
wire pair. And let's assume the receiving system
looks like a simple load resistor, for example.
Well, suppose there is another two-wire pair sitting over here.
Well it turns out both of these pairs look like antennas.
What is sent by one will be picked up by the other.
That's called crosstalk, and in general this is a very bad situation.
It's like you picked up your telephone and all of the telephone calls, currently
active in your neighborhood, you could hear them all.
Particularly, that's not probably what you want.
What it turns out a simple change in that two-wire system can mitigate the
crosstalk, and that is by twisting the pairs together.
So you take the two wires and nearly twist them, it's called a twisted pair and
greatly reduces the crosstalk. It turns out if you solve masses equations
for this much more complicated geometry, you can see that the crosstalk is greatly
minimized. An even better way of minimizing crosstalk
is to use what's known as a coaxial cable. And here I show a photograph of a piece of
coaxial cable. You might recognize this as an TV cable
that you use to connect to especially cable TV, but it's also used for
electronic cable suite, cabling televisions together.
Here I show a cross section of what it looks like, and here's a real live cross
section of my piece of coaxial cable. So, the black part here is insulation to
keep other wires from touching it. And inside you see this copper, the bright
orange just gives away copper. So, that's one of the conductors.
In between this central conductor, here's the other one.
In between those two is the, what's here white, that's the dielectric.
This case that I'm showing you is Teflon. It's a dielectric, especially designed for
the piece of coax. And so a voltage is applied between the
outer conductor and the inner conductor. And that's how the signals are sent.
And it turns out if you solve Maxwell's equations for this geometry you'll see
that it basically does not radiate at all and leads to a very good approximation so,
multiple. Coaxial cables can carry separate
communication signals without interfering with each other and that's good for
wireline channels. Now, we need to explore the properties of
the wireline channel, it's very interesting.
And again going back to Maxwell's equations, you can show that a, this
circuit model I show here this well describes the Maxwell's equations for the
coaxial cable and for the twisted pair, for that matter.
So here's the idea, this is the model for a very small section of coaxial cable of
length delta x. Each section of coaxial cable is described
by a resistor, in series with an inductor, and then there's a parallel chain of
another resistor and a capacitor. And we notice the tildas over the RLGC are
tilda is resistance per unit length, so that's why the resistance here is r tilda
times delta x same for the inductor. This resistor that's in parallel it turns
out its conductance is going to be a lot more useful than our former formula so I
showed as conductance per unit length. This conductance per unit length here is
different from this resistance. They're not the same at all.
So what we need to do is figure out how the complex amplitude of the current at
position x is related to the complex amplitude of the voltage at position x.
So I'm assuming that the voltage, let's say, at position x is equal to the real
part of some complex amplitude. Times a complex, complex exponential.
So this should be, voltage is a function of x int.
So, right away assuming sinusoidal source and everything we know about impedance
will help us figure out the relationship between the voltage and the current.
So, well what I'm going to do first is go back to fundamentals remember, Kirchoff's
Current Law, KCL? Well, we need it right now.
So I want to write KCL at that node, so the current coming into that node is I of
x minus delta x. The current going south is given by the
sum of those two currents, and that's given by the voltage across them divided
times their conductance and emittances with the delta x, and that's equal to the
current that's leaving in the I of x, so that's KCL at that node.
We're going to write KVL along the loop and we'll pick that loop.
So, this voltage is equal to this voltage, which is given by this plus the voltage it
crosses parallel combination, which is just the voltage at there, so I think you
see where this is headed. And now, I'm going to rearrange the
equation let delta x go to 0 and arrive at they said are different two equations, one
for the KCL equation, one for the KVL equation.
Well I want to get one relationship here, so what I'm going to do is take a
derivative of this equation which gives me a second derivative here, for the left
side and on this side that gives me a derivative with respect to x.
And now I can substitute in that equation down below in a while arrived at the
simpler equation for the complex amplitude of the voltage at a given location x.
And it turns out to be a very simple differential equation as we're going to
see in a second where the second derivative of the voltage is related to
the voltage times a frequency dependent constant.
It does not end on x. That's why I considered a constant, but it
is frequency dependent. We'll see how that works out in just a
second. So, turns out the solution of this
equation says, that voltage propagates just like in as you might expect because
it's governed by Maxwell's equations. So, what's going on here is I'm going to
call this gamma squared. And if you substitute this assumed
equation into this equation. I'm pretty sure, I'm sure you can see,
that when I take a second derivative of the voltage, that brings down gamma twice,
giving me a gamma squared, and I get two solutions.
The reason for the V plus and V minus we'll get to in a second, but you can have
either a plus or a minus in the signs on those exponents and that solves this
equation. So that's the differential equation.
We see it's a solution looks like this and gamma is this rather complicated function.
It turns out simplifying this is really not worth it.
It doesn't really simplify very well at all.
You can certainly multiply these two complex numbers together, but taking the
square root means converting to polar form and trying to simplify and it just doesn't
simplify. I suggest you try it for fun if you have a
very strange sense of fun. It's not very nice at all.
That's suffice to say we know it's a complex number so it has a real part and
it has an imaginary part and those are the critical things we're going to need.
Now we're going to assume that we always pick and you can prove that this is true
that the a variable is positive. So, if we go back up to our formulas and,
and stick in the solution here for what gamma is.
If you had a positive, this says for x greater than zero that this will decay,
all right, even minus ax. If you assume that for x positive you had
the other term here, that would mean it would blow up.
It would get infinitely big, that's a nonphysical answer so we throw it away.
It doesn't, it mathematically fits, but it doesn't fit the physics of the situation.
Power has to be conserved[INAUDIBLE]. So, that's why the plus and minus.
This is the signal you get when the positive x axis from where a source might
be let's say, the supplying a voltage and for the negative side you get that for
approach. All right, so to put it all together, a
voltage for a given point on the coaxial cable and in turn, notice this is a lower
case v so I'm losing my real part. This is the critical part.
I'm about to show you that when you have a solution that depends on 2 pi ft minus bx
that corresponds to a propagating signal. And that's really important so let's
proceed to show you that if this formula does means that voltage propagates.
What I am going to look at is this exponent up here and I am going to think
about it, evaluate it at t time t2 and let's say location x2.
I am going to add and subtract t1 here and what you find is that this expression
equals the same quantity at a different time and a different location x1.
The separation between x1 and x2 is given by this quantity.
So, lets plot what's happening here. Lets say that's x and that's t.
We write x1 and here is t1 of course, if we go about to x2, we're at time t2 and
signal here at x1, t1 is the same as the signal at a different location and a
different time. That means the signal propagates.
It's moving from x1, t1 location to the x2, t2 location.
So any time you see a signal that depends only on time and space like this it means
its a propagating signal. So, let's examine this more carefully.
How fast is it propagating? You look at the units inside here, you see
that's distance and that's time. So, that means this quantity has to have
units of speed of velocity. So, the speed of propagation just by unit
analysis is related to b. And once we have, when you know that, we
now can calculate the wavelength and the wavelength is given from these classic
formulas which applies to all propagating waves.
That, the wavelength times the frequency is just the speed of propagation so we
find that the wavelength is 2 pi over b. Well what's b?
And if you recall, b is this equals this imaginary part of the, of the square root
of gamma squared is equal to the imaginary part of gamma and it's a complicated
expression. It turns out if you analytically try to
find that square root, it's really quite difficult.
And what I want to do is point out that it really simplifies though, if we consider a
special case and that is when frequency gets big.
When frequency is big, you can drop G tilde with a respect to that term.
You can drop R tilde with respect to that term.
And so what you wind up with is j2 pi f c tilde times j 2pi f L tilde.
And that equals minus 4 pi squared f squared L tilde c tilde and we now have to
take the square root. And the square root of this because of the
minus sign is pure imaginary and as j 2 pi f square root L tilde C tilde.
So the imaginary part is just j is 2 pi f square root of L tilde C tilde and the 2
pi f is cancel. And lo behold, we to find out the speed of
propagation is equal, is depend only on the characteristics of the cable.
And is equal to that quantity depends only on the inductance per unit length and on
the capacitance per unit length of the cable.
Most well designed coaxial cables the speed of propagation is a that of third to
have of the speed of lighting through space so its not as certainly not through
space but its not far away from it just a small factor difference.
Well, that's the propagating part. How about the attenuating part?
What is a? Well, we know that a is given by the real
part of, of gamma and It's a little bit trickier to show that the real part is not
0. We've already shown that as frequency gets
big, it looks like the imaginary part dominates.
And that's true, because it turns out that imaginary part is proportional to
frequency. But there is a small real part which is
does not depend on frequency. It's a constant as frequency gets big and
is easily seen. Once you do the analysis which is in the
notes that a becomes the this quantity depends on the resistance per unit length
and the conductance per unit length of the cable.
And Z nought not is a quantity we're going to talk about in a later slide later part
of this lecture. And this is square root of L tilde over C
tilde. And this quantity shows once you figure
out what it is for standard pieces of coaxial cable, this can be large.
This can be 100 db per kilometer if you're not too careful at high frequencies.
It turns out, this attenuation factor is smaller at low frequencies but if you
operate the cable at low frequencies, the speed of propagation will now depend on
frequency. And you'll get different delays for
different frequencies. And we've already seen that, that, that
won't be a very good situation. So wireline channels have the problem that
the signal will attenuate, can attenuate very rapidly.
Now it turns out these equations also apply to optical fiber and you'll find the
attenuation factor is much smaller and you can get 1 db per kilometer or even less
for very well designed optic cable, but for coaxial cable, it's a different story.
Now, I just want to point out that of course the current propagates also, so
we've been talking about the voltage here. And remember one of the differential
equations was that for relationship between voltage and current.
So I can just take the derivative of my expression and I'll solve for the current.
And what you get for an answer, after all is said and done, plugging in for gamma,
is that the current is given by a very similar expression.
The important parts are the, the exponents here.
So we get the same kind of thing in that 1, the plus side goes for positive x and
the negative root goes for the negative x. Similar kind of things happen.
The current also does not diminish, so this goes to one, and a high frequency
limit and this corresponds to a propagation, so both current and voltage
propagate, which you'd not expect. What was really interesting is this little
result. If we're going to talk about a little bit
later. Just like the complex amplitude of the
current voltage. Looks like the, the impedance which is
that ratio depends on frequency but we have a little surprise for you.
So, let's summarize. The voltage on a piece of coax at any
given point in space. When you apply a sinusoidal source, is
given by this expression. And if things are designed right, this
place goes to 1. So and that always occurs at high
frequencies. So at sufficiently high frequencies
determined by the characteristics of the coaxial cable signals propagate without
loss at a given speed which is a pretty it's a fraction of the speed of light in,
through space. Now let's worry about that ratio energy of
the impedance. So this ratio which we just saw, is called
the characteristic impedance of the cable. In it earliest frequency dependent and
also depends on the component values that well describe each of coaxial cable of
interest. Well what happens is frequency gets big?
Well this frequency gets big, we can drop this term and drop that term.
Ratio does not depend on frequency at all, becomes a constant.
And I think this is one of the most amazing results about coaxial cable and
twist of pair. Was that, at high frequencies, the
characteristic impedance. The, a piece of coax or twisted pair of
high frequencies looks like a resistor. Notice there's no j's in there.
It looks like a resistor. I find it really, kind of, cool that
regarding that resistor depends on the inductance per unit length and the
capacitance per unit length. That's the way it all works out for this
model is that at high frequencies the piece of coax, if you had to use a
theravadon equivalent, you look down that piece of coax, at high frequencies it
looks like a resistor. Well, this has very important consequences
in actually[UNKNOWN] wireline channels. Now, there's a detail that I should have
emphasized but didn't, is that the circuit model and the KCL and KVL equations were
used to derive them assume the cable is very, very long.
In other words that KCL equations apply at every point in space.
That was very important. So, suppose you have a physical situation
where you have some length of coax and you just cut the end off and you leave it
open. Well at this point that current has to be
0 because it can't flow between the 2 conductors.
That's what the cutting it means. What happens now, suppose the voltage, the
source from the transmitter sends this little blue pulse.
And we know it's going to propagate, but look what happens because of the improper,
and just cut it, it bounces back. And it bounces back theoretically with no
attenuation. This greatly effects the transmission
properties of the coax. You get the same thing back again, and
guess what? If the impedance looking back that way is
not right, it's going to bounce again and again and again.
It's going to go on forever. Now let's consider the situation where I
attach a resistor between the inner and outer conductors.
Where it has a resistance that's equal to the characteristic impedance of the coax.
As I said, that is the evident equivalent for an infinite piece of coax.
So it looks like the cable goes on forever, just because you attach this
resistor. Now, that means that when you send that
pulse down the cable, it disappears. It gets absorbed, if you will, by the
impedance. So it is very important to terminate the
cable correctly. You'll find this in all well designed
systems that is at make sure that both the load and that our termination resistor,
the value equal to Z nought is used with nought, you're going to get reflections
and it's not going to work at all. It turns out, for most coaxial cables, at
least that I know of, of course this is in impedance in the range of 50 to 75 ohms.
It's not totally big and just need to look up the specifications for the coax you use
. I've seen some coaxial cables where the
coaxial impedance is actually labeled on the surface of the coax.
So it's pretty easy to find out what's going on.
So let's summarize the characteristics of the wireline channels.
They are especially designed to minimize interference using twisted pair of coaxial
cable, no cross talk. So you have privacy in a wired
communication system. Moreover, the bad news is, they have to
be. Receiver and transmitter must be connected
to the PC cable. Like I said, you could have multiple
receivers in the, the transmitters. But they have to be connected together.
There's no flexibility in moving around. That it's sort of tethered to the cable.
If you design it well and pick your frequencies right, you have very little
attenuation as long as you stay at high frequencies.
If you don't, you can live that way, you'll discover though, that the amplitude
will decrease exponentially with distance. And it depends on that factor a.
How far you can go. Early, cables for the telegraph system
back in the 19th century. Worked at low frequencies.
And, they discovered that they couldn't transmit very far at all.
And your only solution. Since this is before the electronics era.
As they just applied a very big voltage to the telegraph wires in order to get the
signals across, interesting. Let's also now compare wireless and
wireline. Wireless has the great advantage let's you
connect receiver and transmitter At will almost as long as they're within a proper
distance of each other. There's no real limitations, no
connections have to be made, just have to know, understand each others
characteristics. The marked downside of a wireline is that
it does require this direct connection. You're tethered.
You can't move around. However, you have no noise, no
interference. Wireless is definitely prone to noise and
interference because of the linearity. Maxwell's equations, you, you have an
antenna, you receive everybody else's transmissions.
We haven't talked about noise yet. What noise also comes in and in a you get
that too whereas in the wireline case transmitter and receiver is basically the
reason of interference if there is no noise.
We're going to talk about how we design now a communication system using these
results.