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Characteristic impedance
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== Derivation == === Using the telegrapher's equation === {{main|telegrapher's equation}} [[File:Transmission line element.svg|thumb|Consider one section of the transmission line for the derivation of the characteristic impedance. The voltage on the left would be <math>\ V\ </math> and on the right side would be <math>\ V + \operatorname{d} V ~.</math> This figure is to be used for both the derivation methods.]] The differential equations describing the dependence of the [[voltage]] and [[Electric current|current]] on time and space are linear, so that a linear combination of solutions is again a solution. This means that we can consider solutions with a time dependence <math>\ e^{j \omega t}</math>. Doing so allows to factor out the time dependence, leaving an ordinary differential equation for the coefficients, which will be [[phasor]]s, dependent on position (space) only. Moreover, the parameters can be generalized to be frequency-dependent.<ref name="Miano">{{cite book | last=Miano | first=Giovanni | last2=Maffucci | first2=Antonio | title=Transmission Lines and Lumped Circuits | publisher=Academic Press | publication-place=San Diego | date=2001 | isbn=0-12-189710-9 | pages=130-135}}</ref><ref>{{cite book | last=Mooijweer | first=H. | title=Microwave Techniques | publisher=Macmillan Education UK | publication-place=London | date=1971 | isbn=978-1-349-01067-7 | doi=10.1007/978-1-349-01065-3 | pages=74-79}}</ref> Consider a [[steady-state]] problem such that the voltage and current can be written as: <math display="block">\begin{align} v(x,t) &= V(x) e^{j \omega t}\\[.5ex] i(x,t) &= I(x) e^{j \omega t} \end{align} </math> Take the positive direction for <math> V </math> and <math> I </math> in the loop to be clockwise. Substitution in the telegraph equations and factoring out the time dependence <math>\ e^{j \omega t}</math> now gives: <math display="block">\begin{align} \frac{\mathrm{d}V}{\mathrm{d}x} &= -\left( R + j \omega L \right) I = -Z I,\\[.5ex] \frac{\mathrm{d}I}{\mathrm{d}x} &= -\left( G + j\ \omega C \right) V = -Y V, \end{align}</math> with impedance <math>Z</math> and [[admittance]] <math>Y</math>. Derivation and substitution of these two [[first-order differential equation]]s results in two uncoupled second-order differential equations: <math display="block">\begin{align} \frac{\mathrm{d}^2 V}{\mathrm{d}x^2} &= k^2 V,\\[.5ex] \frac{\mathrm{d}^2 I}{\mathrm{d}x^2} &= k^2 I, \end{align}</math> with <math> k^2 = Z Y = ( R + j \omega L)(G + j \omega C )= (\alpha + j \beta)^2 </math> and <math> k = \alpha + j \beta </math> called the [[propagation constant]]. The solution to these type of equations can be written as: <math display="block">\begin{align} V(x) &= A e^{-k x} + B e^{k x}\\[.5ex] I(x) &= A_{1} e^{-k x} + B_{1} e^{k x} \end{align}</math> with <math>A</math>, <math>A_1</math>, <math>B</math> and <math>B_1</math> the [[constant of integration|constants of integration]]. Substituting these constants in the first-order system gives: <math display="block">\begin{align} A_1 &= A \frac{k}{R+ j \omega L}\\[.5ex] B_1 &= -B \frac{k}{R + j \omega L} \end{align}</math> where <math display="block"> \frac{A}{A_1} = -\frac{B}{B_1} = \frac{R + j \omega L}{k} = \sqrt{\frac{R+ j\omega L}{G + j \omega C}} = \sqrt{\frac{Z}{Y}} = Z_{0}. </math> It can be seen that the constant <math>Z_0,</math> defined in the above equations has the dimensions of impedance (ratio of voltage to current) and is a function of primary constants of the line and operating frequency. It is called the ''characteristic impedance'' of the transmission line.<ref name=":1"/> The general solution of the telegrapher's equations can now be written as: <math display="block">\begin{align} v(x,t) &= V(x) e^{j \omega t} = A e^{-\alpha x}e^{j(\omega t - \beta x)} + B e^{\alpha x}e^{j(\omega t + \beta x)} \\[.5ex] i(x,t) &= I(x) e^{j \omega t} = \frac{A}{Z_0} e^{-\alpha x}e^{j(\omega t - \beta x)} - \frac{B}{Z_0} e^{\alpha x}e^{j(\omega t + \beta x)} \end{align}</math> Both the solution for the voltage and the current can be regarded as a superposition of two travelling waves in the <math>x_{+}</math> and <math>x_{-}</math> directions. For typical transmission lines, that are carefully built from wire with low loss resistance <math>\ R\ </math> and small insulation leakage conductance <math>\ G\ ;</math> further, used for high frequencies, the inductive reactance <math>\ \omega L\ </math> and the capacitive admittance <math>\ \omega C\ </math> will both be large. In those cases, the [[phase constant]] and characteristic impedance are typically very close to being real numbers: <math display="block">\begin{align} \beta & \approx \omega \sqrt{L C} \\[.5ex] Z_{0} & \approx \sqrt{\frac{L}{C}} \end{align}</math> Manufacturers make commercial cables to approximate this condition very closely over a wide range of frequencies. === As a limiting case of infinite ladder networks === ==== Intuition ==== {{see also|Iterative impedance|Constant k filters}} {{Multiple image| |align=right |direction=vertical <!--image 1--> | 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 | width=300px <!--image 2--> | image2 = L-section iterative impedance.svg | alt2 = Iterative impedance of the equivalent finite L-circuit | caption2 = Iterative impedance of the equivalent finite L-circuit }} 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.
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