Editing Characteristic impedance (section)

=== As a limiting case of infinite ladder networks ===

==== Intuition ====
{{see also|Iterative impedance|Constant k filters}}

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| image1            = Ladder iterative impedance.svg
| alt1              = Iterative impedance of an infinite ladder of L-circuit sections
| caption1          = Iterative impedance of an infinite ladder of L-circuit sections
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| image2            = L-section iterative impedance.svg
| alt2              = Iterative impedance of the equivalent finite L-circuit 
| caption2          = Iterative impedance of the equivalent finite L-circuit 
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Consider an infinite [[ladder network]] consisting of a series impedance <math>\ Z\ </math> and a shunt admittance <math>\ Y ~.</math> Let its input impedance be <math>\ Z_\mathrm{IT} ~.</math> If a new pair of impedance <math>\ Z\ </math> and admittance <math>\ Y\ </math> is added in front of the network, its input impedance <math>\ Z_\mathrm{IT}\ </math> remains unchanged since the network is infinite. Thus, it can be reduced to a finite network with one series impedance <math>\ Z\ </math> and two parallel impedances <math>\ 1 / Y\ </math> and <math>\ Z_\text{IT} ~.</math> Its input impedance is given by the expression<ref>{{cite book |title=The Feynman Lectures on Physics|title-link=The Feynman Lectures on Physics |volume=2 |first1=Richard |last1=Feynman |author1-link=Richard Feynman |first2=Robert B. |last2=Leighton |author2-link=Robert B. Leighton |first3=Matthew |last3=Sands |author3-link=Matthew Sands |section=Section 22-6. A ladder network |section-url=https://www.feynmanlectures.caltech.edu/II_22.html#Ch22-S6}} </ref><ref name=lee2004/>

:<math>\ Z_\mathrm{IT} = Z + \left( \frac{\ 1\ }{ Y } \parallel Z_\mathrm{IT} \right)\ </math>

which is also known as its [[iterative impedance]]. Its solution is:

:<math>\ Z_\mathrm{ IT } = {Z \over 2} \pm \sqrt { {Z^2 \over 4} + {Z \over Y} }\ </math>

For a transmission line, it can be seen as a [[Mathematical limit|limiting case]] of an infinite ladder network with [[infinitesimal]] impedance and admittance at a constant ratio.<ref name="feynman">{{cite book|title=The Feynman Lectures on Physics|title-link=The Feynman Lectures on Physics|volume=2|first1=Richard|last1=Feynman|author1-link=Richard Feynman|first2=Robert B.|last2=Leighton|author2-link=Robert B. Leighton|first3=Matthew|last3=Sands|author3-link=Matthew Sands|section=Section 22-7. Filter |section-url=https://www.feynmanlectures.caltech.edu/II_22.html#Ch22-S7 |quote=If we imagine the line as broken up into small lengths Δℓ, each length will look like one section of the L-C ladder with a series inductance ΔL and a shunt capacitance ΔC. We can then use our results for the ladder filter. If we take the limit as Δℓ goes to zero, we have a good description of the transmission line. Notice that as Δℓ is made smaller and smaller, both ΔL and ΔC decrease, but in the same proportion, so that the ratio ΔL/ΔC remains constant. So if we take the limit of Eq. (22.28) as ΔL and ΔC go to zero, we find that the characteristic impedance z0 is a pure resistance whose magnitude is √(ΔL/ΔC). We can also write the ratio ΔL/ΔC as L0/C0, where L0 and C0 are the inductance and capacitance of a unit length of the line; then we have <math>\sqrt{\frac{L_0}{C_0}}</math>}}.</ref><ref name=lee2004>{{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=2.5 Driving-point impedance of iterated structure |page=44 }}</ref> Taking the positive root, this equation simplifies to:

:<math>\ Z_\mathrm{IT} = \sqrt{ \frac{\ Z\ }{ Y }\ }\ </math>

==== 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.