Editing Characteristic impedance (section)

==== Derivation ====

Using this insight, many similar derivations exist in several books<ref name="feynman"/><ref name=lee2004/> and are applicable to both lossless and lossy lines.<ref>{{cite book |first=Thomas H. |last=Lee |author-link=Thomas H. Lee (electronic engineer) |year=2004 |title=Planar Microwave Engineering: A practical guide to theory, measurement, and circuits |publisher=Cambridge University Press |section=2.6.2. Characteristic impedance of a lossy transmission line |page=47}}</ref>

Here, we follow an approach posted by Tim Healy.<ref name=":2">{{cite web |url=http://www.ee.scu.edu/eefac/healy/char.html |title=Characteristic impedance |website= ee.scu.edu |access-date=2018-09-09 |archive-date=2017-05-19 |archive-url=https://web.archive.org/web/20170519040949/http://www.ee.scu.edu/eefac/healy/char.html |url-status=dead }}</ref> The line is modeled by a series of differential segments with differential series elements <math>\ \left( R\ \operatorname{d}x,\ L\ \operatorname{d}x \right)\ </math> and shunt elements <math>\ \left(C\ \operatorname{d}x,\ G\ \operatorname{d}x\ \right)\ </math> (as shown in the figure at the beginning of the article). The characteristic impedance is defined as the ratio of the input voltage to the input current of a semi-infinite length of line. We call this impedance <math>\ Z_0 ~.</math> That is, the impedance looking into the line on the left is <math>\ Z_0 ~.</math> But, of course, if we go down the line one differential length <math>\ \operatorname{d}x\ ,</math> the impedance into the line is still <math>\ Z_0 ~.</math> Hence we can say that the impedance looking into the line on the far left is equal to <math>\ Z_0\ </math> in parallel with <math>\ C\ \operatorname{d}x\ </math> and <math>\ G\ \operatorname{d}x\ ,</math> all of which is in series with <math>\ R\ \operatorname{d}x\ </math> and <math>\ L\ \operatorname{d}x ~.</math> Hence:
<math display="block">\begin{align}
 Z_0 &= (R + j\ \omega L)\ \operatorname{d}x + \frac{ 1 }{\ (G + j \omega C)\ \operatorname{d}x + \frac{1}{\ Z_0\ }\ } \\[1ex]
 Z_0 &= (R + j\ \omega L)\ \operatorname{d}x + \frac{\ Z_0\ }{Z_0\ (G + j \omega C)\ \operatorname{d}x + 1\, } \\[1ex]
 Z_0 + Z_0^2\ (G + j\ \omega C)\ \operatorname{d}x &= (R + j\ \omega L)\ \operatorname{d}x + Z_0\ (G + j\ \omega C)\ \operatorname{d}x\ (R + j\ \omega L)\ \operatorname{d}x + Z_0
\end{align} </math>

The added <math>\ Z_0\ </math> terms cancel, leaving
<math display="block">\ Z_0^2\ (G + j\ \omega C)\ \operatorname{d}x = \left( R + j\ \omega L \right)\ \operatorname{d}x + Z_0\ \left( G + j\ \omega C \right)\ \left( R + j\ \omega L \right)\ \left( \operatorname{d}x \right)^2 </math>

The first-power <math>\ \operatorname{d}x\ </math> terms are the highest remaining order. Dividing out the common factor of <math>\ \operatorname{d}x\ ,</math> and dividing through by the factor <math>\ \left( G + j\ \omega C \right)\ ,</math> we get
<math display="block">\ Z_0^2 = \frac{ \left( R + j\ \omega L \right) }{\ \left( G + j\ \omega C \right)\ } + Z_0\ \left( R + j\ \omega L \right)\ \operatorname{d}x ~.</math>

In comparison to the factors whose <math>\ \operatorname{d}x\ </math> divided out, the last term, which still carries a remaining factor <math>\ \operatorname{d}x\ ,</math> is infinitesimal relative to the other, now finite terms, so we can drop it. That leads to
<math display="block">\ Z_0 = \pm \sqrt{\frac{\ R + j\ \omega L\ }{G + j\ \omega C}\ } ~.</math>

Reversing the sign {{math|±}} applied to the square root has the effect of reversing the direction of the flow of current.