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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 hiscomputer. He decides just to add a "T" to the terminal where the cable is connected to his "thin-net" interface, and add on somemore wire. Unfortunately, he is not careful about the coaxial cable he uses, and so he has some 75 Ω TV co-ax instead of the 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 issimilar to what we have done before, just a little more complicated is all.

Cascaded line problem

We will have to do a little more thinking before we can draw out the bounce diagram for this problem. The driverfor ethernet cable coming to Arnold's computer can be modeled as a 10V (open circuit) source with a 50 Ω internal impedance. Since the source does not (initially) know anything about how the line it is driving isterminated, the first signal 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 .

At the junction

V 1 + is incident on the junction. When it hits the junction, there will be a reflected wave V 1 - and also now, a transmitted wave V T 1 + . Since the incident wave can not tell the difference between a 75 Ω resistor and a 75 Ω transmission line, it thinks it is seeing a termination resistor equal to a 50 Ω resistor ( R L 1 ) in parallel with a 75 Ω resistor (the second line). 50 Ω in parallel with 75 Ω is 30 Ω . Let's call this "apparent" load resistor R L ' ), so that we can then calculate 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 30 50 30 50 -0.25
Note that we could have started from scratch and written down KVLs and KCLs for the junction
V 1 + V 1 - V T 1 +
and
I 1 + I 1 - I R L I T 1 +
Then, by re-writing 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
We now have two equations with two unknowns ( V 1 - and V T 1 + ). By solving for V T 1 + and then plugging that into , we could get the ratio of V 1 - to V 1 + , or the voltage reflection coefficient. The interested reader can confirm that indeed, you get the very same resultthis way.

In order to completely solve this problem, we also need to know V T 1 + , the transmitted wave as well. Since says 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 +
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 60 30 50 0.75 T V 1 2
An important thing to note is that
T V 1 Γ V
NOT
T V Γ V 1
We do not "conserve" voltage at a termination, in the sense that the reflected and transmitted voltage have to add up to be theincident 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.

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Source:  OpenStax, Introduction to physical electronics. OpenStax CNX. Sep 17, 2007 Download for free at http://cnx.org/content/col10114/1.4
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