# Connexions

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Module by: Bill Wilson. E-mail the author

We can use bounce diagrams to handle somewhat more complicated problems as well.

Arnold Aggie decides to add an additional ethernet interface to the one already connected to his computer. He decides just to add a "T" to the terminal where the cable is connected to his "thin-net" interface, and add on some more wire. Unfortunately, he is not careful about the coaxial cable he uses, and so he has some 75Ω 75 Ω TV co-ax instead of the 50Ω 50 Ω ethernet cable. He ends up with the situation shown here. This kind of problem is called a cascaded line problem because we have two different lines, one hooked up after the other. The analysis is similar to what we have done before, just a little more complicated is all.

We will have to do a little more thinking before we can draw out the bounce diagram for this problem. The driver for ethernet cable coming to Arnold's computer can be modeled as a 10V (open circuit) source with a 50Ω 50 Ω internal impedance. Since the source does not (initially) know anything about how the line it is driving is terminated, the first signal V 1 + V 1 + will be the same as in our initial problem, in this case just a +5V signal headed down the line.

Let's focus on the "T" for a minute Figure 2.

V 1 + V 1 + is incident on the junction. When it hits the junction, there will be a reflected wave V 1 - V 1 - and also now, a transmitted wave V T 1 + V T 1 + . Since the incident wave can not tell the difference between a 75Ω 75 Ω resistor and a 75Ω 75 Ω transmission line, it thinks it is seeing a termination resistor equal to a 50Ω 50 Ω resistor ( R L 1 R L 1 ) in parallel with a 75Ω 75 Ω resistor (the second line). 50Ω 50 Ω in parallel with 75Ω 75 Ω is 30Ω 30 Ω . Let's call this "apparent" load resistor R L ' R L ' ), so that we can then calculate Γ V 1 2 Γ V 1 2 , the first voltage reflection coefficient in going from line 1 to line 2 as:
Γ V 1 2 = R L ' Z 0 1 R L ' + Z 0 1 =305030+50=.25 Γ V 1 2 R L ' Z 0 1 R L ' Z 0 1 30 50 30 50 .25
(1)
Note that we could have started from scratch and written down KVLs and KCLs for the junction
V 1 + + V 1 - = V T 1 + V 1 + V 1 - V T 1 +
(2)
and
I 1 + + I 1 - = I R L + I T 1 + I 1 + I 1 - I R L I T 1 +
(3)
Then, by re-writing Equation 3 in terms of voltage and impedances we have:
V 1 + Z 0 1 V 1 - Z 0 1 = V T 1 + Z 0 2 + V T 1 + R L V 1 + Z 0 1 V 1 - Z 0 1 V T 1 + Z 0 2 V T 1 + R L
(4)
We now have two equations with two unknowns ( V 1 - V 1 - and V T 1 + V T 1 + ). By solving Equation 4 for V T 1 + V T 1 + and then plugging that into Equation 2, we could get the ratio of V 1 - V 1 - to V 1 + V 1 + , or the voltage reflection coefficient. The interested reader can confirm that indeed, you get the very same result this way.

In order to completely solve this problem, we also need to know V T 1 + V T 1 + , the transmitted wave as well. Since Equation 2 says V T 1 + V T 1 + is just the sum of the incident and reflected waves on the first line

V T 1 + = V 1 + + Γ V 1 2 V 1 + V T 1 + V 1 + Γ V 1 2 V 1 +
(5)
We can thus write
V T 1 + V L + =1+ Γ V 1 2 = R L ' + Z 0 1 R L ' + Z 0 1 + R L ' Z 0 1 R L ' + Z 0 1 =2 R L ' R L ' + Z 0 1 =6030+50=0.75 T V 1 2 V T 1 + V L + 1 Γ V 1 2 R L ' Z 0 1 R L ' Z 0 1 R L ' Z 0 1 R L ' Z 0 1 2 R L ' R L ' Z 0 1 60 30 50 0.75 T V 1 2
(6)
An important thing to note is that
T V =1+ Γ V T V 1 Γ V
(7)
NOT
T V + Γ V =1 T V Γ V 1
(8)
We do not "conserve" voltage at a termination, in the sense that the reflected and transmitted voltage have to add up to be the incident voltage. Rather, the transmitted voltage is the sum of the incident voltage and the reflected voltage, so that we can obey Kirchoff's voltage law.

We can now start to make our bounce diagram. We propagate a +5V wave and a -5V wave (separated by 100ns) down towards the junction. Since the line is 40m long, and the waves move at 2×108ms 2 10 8 m s , it takes 200ns for them to get to the junction. There, a -1.25V wave is reflected back towards the source, and a +3.75V wave is transmitted into the second transmission line Figure 3.

Since the load for the second line is 50Ω 50 Ω , and the characteristic impedance, Z 0 2 Z 0 2 for the second line is 75Ω 75 Ω , we will have a reflection coefficient,
Γ V 2 = R L 2 Z 0 2 R L 2 + Z 0 2 =507550+75=-0.2 Γ V 2 R L 2 Z 0 2 R L 2 Z 0 2 50 75 50 75 -0.2
(9)
Thus a -0.75V signal is reflected off of the second load Figure 4.

## Exercise 1

What is the magnitude of the voltage which is developed across the second load?

### Solution

3 Volts!

What happens to the 0.75V pulse when it gets to the "T"? Well there is another mismatch here, with a reflection coefficient Γ V 2 1 Γ V 2 1 given by
Γ V 2 1 =257525+75=-0.5 Γ V 2 1 25 75 25 75 -0.5
(10)
(The 50Ω 50 Ω resistor and the 50Ω 50 Ω transmission line look like a 25Ω 25 Ω termination to the 75Ω 75 Ω line) and a transmission coefficient
T V 2 1 =1+ Γ V 2 1 =0.5 T V 2 1 1 Γ V 2 1 0.5
(11)
and so we add to the bounce diagram Figure 5. We could keep going, but the voltage reflected off of the second load will only be 75mV now, and so let's call it a day.

There are a couple of other interesting applications of bounce diagrams and the transient behavior of transmission lines that we might look at before we move on to other things. The first is called the Charged Line Problem. Here it is:

We have a transmission line with characteristic impedance Z 0 Z 0 and phase velocity v p v p . It is LL long, and for some time has been connected to a battery of potential V g V g Figure 7. At time t=0 t 0 , the switch S, is thrown, which removes the battery from the circuit, and connects the line to a load resistor R L R L . The question is: what does the voltage across the load resistor, V L V L , look like as a function of time? This is almost like what we have done before, but not quite. In the first place, we now have non-zero initial conditions. For t<0 t 0 we will have both voltages and current on the line. In order to match boundary conditions, we must do more than have one voltage and one current, because the voltage on the line must be V g V g , while the current flowing down the line must be 0. So, we will put in both a V + V + and a V - V - and their corresponding currents. Note that x x is going to the left this time. Let's forget about the switch and the load resistor for a minute and just look at the line and battery. We have two equations we must satisfy
V 0 + + V 0 - = V g V 0 + V 0 - V g
(12)
and
I 0 + + I 0 - =0 I 0 + I 0 - 0
(13)
We can use the impedance relationship to change Equation 13 to:
V 0 + Z 0 V 0 - Z 0 =0 V 0 + Z 0 V 0 - Z 0 0
(14)
I hope most of you can then see by inspection that we must have
V 0 + = V 0 - = V g 2 V 0 + V 0 - V g 2
(15)
OK, the switch S is thrown at t=0 t 0 . Now the end of the line looks like this. We have anticipated the fact that we are going to need another voltage and current wave if we are going to be able to match boundary conditions when the load resistor is connected, and have added a V 1 + V 1 + and a V 1 - V 1 - to the line. These are new voltage and current waves which originate at the load resistor position in order to satisfy the new boundary conditions there. Now we do KVL and KCL again.
V 0 + + V 0 - + V 1 + = V L V 0 + V 0 - V 1 + V L
(16)
and
V 0 + Z 0 V 0 - Z 0 + V 1 + Z 0 = V L R L V 0 + Z 0 V 0 - Z 0 V 1 + Z 0 V L R L
(17)
We have already made the impedance substitution for the current equation in Equation 17. We know what the sum and difference of V 0 + V 0 + and V 0 - V 0 - are, so let's substitute in.
V g + V 1 + = V L V g V 1 + V L
(18)
and
V 1 + Z 0 = V L R L V 1 + Z 0 V L R L
(19)
From this we get
V L =( R L Z 0 V 1 + ) V L R L Z 0 V 1 +
(20)
which we substitute back into Equation 18
V g + V 1 + =( R L Z 0 V 1 + ) V g V 1 + R L Z 0 V 1 +
(21)
which we can solve for V 1 + V 1 +
V 1 + = V g 1+ R L Z 0 =( Z 0 R L + Z 0 V g ) V 1 + V g 1 R L Z 0 Z 0 R L Z 0 V g
(22)
The voltage on the load is given by Equation 18 and is clearly just:
V L = V g Z 0 R L + Z 0 V g V L V g Z 0 R L Z 0 V g
(23)
and in particular, when R L R L is chosen to be Z 0 Z 0 (which is usually done when this circuit is used), we have
V L = V g 2 V L V g 2
(24)
Now what do we do? We build a bounce diagram! Let us stay with the assumption that R L = Z 0 R L Z 0 , in which case the reflection coefficient at the resistor end is 0. At the open circuit end of the transmission line ΓΓ is +1. So we have this. Note that for this bounce diagram, we have added an additional voltage, V g V g , on the baseline, to indicate that there is an initial voltage on the line, before the switch is thrown, and tt starts on the bounce diagram.

If we concentrate on the voltage across the load, we add V g V g and V g 2 V g 2 and find that the voltage across the load resistor rises to V g 2 V g 2 at time t=0 t 0 Figure 10. The V g 2 V g 2 voltage wave travels down the line, hits the open circuit, reflects back, and when it gets to the load resistor, brings the voltage across the load resistor back down to zero. We have made a pulse generator!

In today's digital age, this might seem like a strange way to go about creating a pulse. Imagine however, if you needed a pulse with a very large potential (100s of thousands or even millions of volts) for say, a particle accelerator. It is unlikely that a MOSFET will ever be built which is up to the task! In fact, in a field of study called pulsed power electronics just such circuits are used all the time. Sometimes they are built with real transmission lines, sometimes they are built from discrete inductors and capacitors, hooked together just as in the distributed parameter model. Such circuits are called pulse forming networks or PFNs for short.

Finally, just because it affords us a good opportunity to review how we got to where we are right now, let's consider the problem of a non-resistive load on the end of a line. Suppose the line is terminated with a capacitor! For simplicity, let's let R s = Z 0 R s Z 0 , so when S is closed a wave V 1 + = V g 2 V 1 + V g 2 heads down the line Figure 11. Let's think about what happens when it hits the capacitor. We know we need to generate a reflected signal V 1 - V 1 - , so let's go ahead and put this in the figure, along with its companion current wave.

The capacitor is initially uncharged, and we know we can not instantaneously change the voltage across a capacitor (at least without an infinite current!) and so the initial voltage across the capacitor should be zero, making V 1 - 0= V 1 + V 1 - 0 V 1 + , if we make time t=0 t 0 be when the initial wave just gets to the capacitor. So, at t=0 t 0 , Γ V 0=-1 Γ V 0 -1 . Note that we are making ΓΓ a function of time now, as it will change depending upon the charge state of the capacitor.

The current into the capacitor, I C I C is just I 1 + + I 1 - t I 1 + I 1 - t .

I C 0= I 1 + + I 1 - 0= V g Z 0 I C 0 I 1 + I 1 - 0 V g Z 0
(25)
since
I 1 + = V 1 + Z 0 = V g 2 Z 0 I 1 + V 1 + Z 0 V g 2 Z 0
(26)
and
I 1 - 0= V 1 Z 0 = V g 2 Z 0 I 1 - 0 V 1 Z 0 V g 2 Z 0
(27)
How will the current into the capacitor I C t I C t behave? We have to remember the capacitor equation:
I C t=Cd V C td t =C( V 1 + + V 1 - t) t =Cd V 1 - td t I C t C t V C t C t V 1 + V 1 - t C t V 1 - t
(28)
since V 1 + V 1 + is a constant and hence has a zero time derivative. Well, we also know that
I C t= I 1 + + I 1 - t= V 1 + Z 0 V 1 - t Z 0 I C t I 1 + I 1 - t V 1 + Z 0 V 1 - t Z 0
(29)
So we equate Equation 28 and Equation 29 and we get
Cd V 1 - td t = V 1 + Z 0 V 1 - t Z 0 C t V 1 - t V 1 + Z 0 V 1 - t Z 0
(30)
or
d V 1 - td t +1 Z 0 C V 1 - t=1C V 1 + Z 0 t V 1 - t 1 Z 0 C V 1 - t 1 C V 1 + Z 0
(31)
which gets us back to another Diff-E-Q!

The homogeneous solution is easy. We have

d V 1 - td t +1 Z 0 C V 1 - t=0 t V 1 - t 1 Z 0 C V 1 - t 0
(32)
for which the solution is obviously
V 1 homo - t= V 0 et Z 0 C V 1 homo - t V 0 t Z 0 C
(33)
After a long time, the derivative of the homogeneous solution is zero, and so the particular solution (the constant part) is the solution to
1 Z 0 C V 1 part - =1C V 1 + Z 0 1 Z 0 C V 1 part - 1 C V 1 + Z 0
(34)
or
V 1 part - = V 1 + V 1 part - V 1 +
(35)
The complete solution is the sum of the two:
V 1 - t= V 1 homo - t+ V 1 part - = V 0 et Z 0 C+ V 1 + V 1 - t V 1 homo - t V 1 part - V 0 t Z 0 C V 1 +
(36)
Now all we need to do is find V 0 V 0 , the initial condition. We know, however, that V 1 - 0= V 1 + V 1 - 0 V 1 + , so that makes V 0 =-2 V 1 + V 0 -2 V 1 + ! So we have:
V 1 - t=2 V 1 + et Z 0 C+ V 1 + = V 1 + (12et Z 0 C) V 1 - t -2 V 1 + t Z 0 C V 1 + V 1 + 1 2 t Z 0 C
(37)
Since V 1 + = V g 2 V 1 + V g 2 we can plot V 1 - t V 1 - t as a function of time from which we can make a plot of Γ V t Γ V t The capacitor starts off looking like a short circuit, and charges up to look like an open circuit, which makes perfect sense. Can you figure out what the shape would be of a pulse reflected off of the capacitor, given that the time constant Z 0 C Z 0 C was short compared to the width of the pulse?

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