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Unfortunately, since we don't know what value the phasor V + has, these equations do not do us a whole lot of good! One way to deal with this is to simply divide this equation into this equation . That gets rid of V + and the s and so we now come up with a new variable, which we shall call line impedance , Z s .

Z s V s I s Z 0 1 -2 s 1 -2 s
Z s represents the ratio of the total voltage to the total current anywhere on the line. Thus, if we have a line L long, terminated with a load impedance Z L , which gives rise to a terminal reflection coefficient , then if we substitute and L into , the Z L which we calculate will be the "apparent" impedance which we would see looking into the input terminals to the line!

There are several ways in which we can look at . One is to try to put it into a more tractable form, that we might be able to use to find Z s , given some line impedance Z 0 , a load impedance Z L and a distance, s away from the load. We can start out by multiplying top and bottom by s , substituting in for , and then multiplying top and bottom by Z L Z 0 .

Z s Z 0 Z L Z 0 s Z L Z 0 s Z L Z 0 s Z L Z 0 s
Next, we use Euler's relation, and substitute s s for the exponential. Lots of things will cancel out, and if we do the math carefully, we end up with
Z s Z 0 Z L Z 0 s Z 0 Z L s
For some people, this equation is more satisfying than , but for me, both are about equally opaque in terms if seeing how Z s is going to behave with various loads, as we move down the line towards the generator. does have the nice property that it is easy to calculate, and hence could be put into MATLAB or aprogrammable calculator. (In fact you could program just as well for that matter.) You could specify a certain set of conditions and easily find Z s , but you would not get much insight into how a transmission line actually behaves.

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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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