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This document looks at terminated transmission lines.

If all we did was launch signals down semi-infinite transmission lines, we would not get very much useful work done. We reallyneed to have a finite length line, and put something at the end...like a termination . So let's take a look at a terminated transmission line .

At the load end of the line

The line has characteristic impedance Z 0 and we assume that it is terminated with a load resistor R L . If we have connected a source to the other end of the line, then we will have launched a voltage wave V 1 + and a current wave I 1 + down the line. If the line is L long, then it will take a time L v p , where v p 1 L C , for the signal to get to the end of the line.

What happens when the signal gets to the load? We can assume some voltage V L will appear across the load resistor, and hence a current I L will flow through it. The most logical thing to assume would be that V L V 1 + . But , we quickly run into a contradiction. If V L V 1 + , then

I L V L R L V 1 + R L
But Kirchoff says the sum of the currents into the output terminal must equal zero, thus I 1 + must equal I L . This then says
I 1 + V 1 + Z 0 I L V 1 + R L
which can only be true if Z L Z 0 , which while possible, will not be the case in general.

What are we to do? We have an obvious contradiction. The telegrapher's equations permit two solutions to the transmission line problem: a signal going in the x direction, V 1 + , and a signal going in the x direction, V 1 - . The only way out of our problem is to assume that when the V 1 + signal gets to the load, a new signal , going in the x direction, is created, which then heads back towards the load. So, let's put in a V 1 - and I 1 - .

Reflected waves are generated

Now there is a little problem with signs here that we have todeal with. We can either draw I 1 - so that it points in the x direction, and say
I 1 - V 1 - Z 0
or we can draw I 1 - going in the x direction, and say
I 1 - V 1 - Z 0
Since most of the time, it is better if we define all currents going in the same direction, and since this equation says that V - and I - are related by Z 0 the latter choice seems to be the better one.

Going by that convention then, we equate voltage on either side of the termination

V 1 + V 1 - V L
and we sum currents into the output terminal
I 1 + I 1 - I L
or V 1 + Z 0 V 1 - Z 0 V L R L This makes
V L R L Z 0 V 1 + V 1 -
which we can substitute into to get:
V 1 + V 1 - R L Z 0 V 1 + V 1 -
or V 1 + 1 R L Z 0 V 1 - R L Z 0 1 and this can be solved for V 1 - as
V 1 - R L Z 0 R L Z 0 V 1 + vL V 1 +
where vL is called the load voltage reflection coefficient .

We could also have solved for I 1 + in terms of I 1 - and we would have found:

I 1 - Z 0 R L Z 0 R L V 1 + IL V 1 +
where IL is the load current reflection coefficient . Note that IL vL

Let's take a break from equation manipulating, and think about what we have here. First of all, although theresult we have obtained is very important, the method we used to get there was even more so. What did we do? We postulated avoltage and current on the line, and then took a look to see if that solution resulted in a reasonable result. In this case itdid not. We were in gross violation of Kirchoff Laws! We rescued ourselves by taking the only possible escape route: we added anadditional voltage and current to the solution. Since V - and I - are related to each other in a different manner than V + and I + (by Z 0 rather than Z 0 ), this gave us an additional degree of freedom so that we could simultaneously satisfy both the transmission line I-V relationships as well as the load I-Vrelationship.

Let's take a look at vL for a minute. Over what range can it vary? A glance at shows that it depends on the range of R L . If we exclude the possibility of negative resistance (a reasonable exclusion) then 0 R L . As R L varies over this range, the voltage reflection coefficient goes from -1 vL 1 . When R L is less than Z 0 , the reflection coefficient is negative. When R L is greater than Z 0 , the reflection coefficient is positive. When R L Z 0 the reflection coefficient is zero, and the line is said to be matched . In a matched transmission line, a signal traveling down it is completely absorbed by the load,and nothing is reflected from it. In this case we can have the incident voltage and current signals just equal load voltage andcurrents without the need to add reflected waves. For an unmatched transmission line, a signal incident on the load is(partially) reflected, and a new signal starts moving back down the line towards the source. An example of a transmission lineis a buss in a computer. What would be the implications of unmatched terminations on various connections between the bussand the computer circuitry?

Two special cases of terminated transmission lines that are of interest are R L 0 (a shorted line) and R L (an open line). For R L 0 , vL -1 and so a signal with the same amplitude but opposite sign as the incident wave, is reflected back down the line. For R L , a signal with the same amplitude and sign gets reflected back down the line. It is easy toremember which is which, and also to make sure you have the right order in the equation for the reflection coefficient, whenyou keep in mind that the voltage across the load is the sum of V 1 + and V 1 - . If the line is terminated with a short circuit, V L must equal 0. Since

V 1 - vL V 1 +
and vL R L Z 0 R L Z 0 From these equations we can see that with R L 0 (a short), vL does indeed -1 , and so V 1 - V 1 + and hence V L V 1 + V 1 - 0 as it should.

Now, let's go back to our terminated transmission line. What is going to happen to V 1 - after it leaves the load? It obviously travels back down the line towards the generator. What happens when it getsback there? Time to look at .

Back at the generator

At the generator end of the line we can not leave out V 1 + , as it is still here, so long as the source remains connected. Once it is launched, it stays on the line forever. Wejust have to add in the new V 1 - . What happens now, if we test Kirchoff's voltage law here at the generator end?
V S I S R S V 1 + V 1 - 0
Substituting I S I 1 + I 1 - and then using the impedance relationship between the current and the voltages we get:
V S R S Z 0 V 1 + V 1 - V 1 + V 1 - 0
Note that again, we used V 1 + Z 0 for I 1 + and V 1 - Z 0 for I 1 - .

We know

V 1 + Z 0 Z 0 R S V S
which if we substitute into and re-arrange a little bit we get
V S 1 R S Z 0 R S Z 0 Z 0 R S R S Z 0 V 1 - V 1 - 0
The stuff inside the big parentheses sums to zero (as in fact itshould, this is just the solution to the initial V 1 + generation problem) and we are left with the uncomfortable conclusion that V 1 - must be zero! What are we going to do? We will just have to add another wave V 2 + to the solution! ( )

Yet another wave is formed

Now we have two ways to proceed from here. The first would be the dumb way, and try to solve this whole problem. But then wecould also be smart and note that all of the equations we have relating voltages and currents in this problem are linear equations and hence we can use the principle of superposition .

Using superposition

Thus, the top sketch in can be broken into the two circuits (a) and (b) at the bottom of the figure. Then, everything is trivial. We have alreadysolved (a) - that was just our initial V 1 + launching problem. Part (b) is just a reflection problem again, with a wave of V 1 - incident on a load of value R S through a transmission line with characteristic impedance Z 0 . The reflected wave in this instance is V 2 + , which is why we had to use numbered subscripts in the first place.

If you do not believe using superposition is valid, you can try doing the problem over again, but it shouldbe pretty obvious that we can write

V 2 + R S Z 0 R S Z 0 V 1 - vS V 1 -
with
vS R S Z 0 R S Z 0

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Source:  OpenStax, Communications b : filters and transmission lines. OpenStax CNX. Nov 30, 2012 Download for free at http://cnx.org/content/col11169/1.2
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