Distributed Parameter Line Model

This article describes the steady-state or frequency domain distributed parameter line model. By steady-state, we mean that the voltage and current along the line are steady and do not change with time.

Single-Phase Distributed Parameter Model

Figure 1. Distributed parameter line model (single-phase): Overall line representation (top); Model for a small line segment (bottom)

In a real transmission line, the R, L and C circuit elements are not lumped together, but are uniformly distributed along the length of the line. In order to capture the distributed nature of the circuit parameters, consider the single-phase line model in Figure 1. In particular, consider a small line segment of length $\Delta x$ metres, which is located at a distance $x$ metres from the receiving end bus.

The top diagram of Figure 1 shows the complete transmission line and the small line segment is represented by the dotted box. The bottom diagram is an expanded view of the line segment showing a typical line segment model with series and shunt circuit elements. The series elements can be represented as an impedance and the shunt elements can be represented as an admittance as follows:

${\boldsymbol {z}}=R+j\omega L=R+jX\,$

${\boldsymbol {y}}={\frac {1}{R_{sh}}}+j\omega C=G+jB\,$

Where Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle R\,} is the series resistance ($\Omega /m$)

$X\,$ is the series reactance ($\Omega /m$)

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle G\,} is the shunt conductance ($S/m$)

$B\,$ is the shunt susceptance ($S/m$)

Note that the impedance and admittance are denoted in per-length values (i.e. per metre). Therefore, the series impedance of a line segment of length $\Delta x$ metres is Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\boldsymbol {z}}\Delta x\,} . The same logic applies for the shunt admittance.

Derivation of Voltage and Current Equations

Analysing this circuit using Kirchhoff's voltage law:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\boldsymbol {V}}(x+\Delta x)={\boldsymbol {V}}(x)+{\boldsymbol {z}}\Delta x{\boldsymbol {I}}(x)\,}

Re-arranging this equation, we get:

${\frac {{\boldsymbol {V}}(x+\Delta x)-{\boldsymbol {V}}(x)}{\Delta x}}={\boldsymbol {z}}{\boldsymbol {I}}(x)\,$

The left-hand side is Newton's difference quotient and the limit as $\Delta x\to 0$ is by definition the derivative of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\boldsymbol {V}}(x)} , i.e.

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{\Delta x\to 0}{\frac {{\boldsymbol {V}}(x+\Delta x)-{\boldsymbol {V}}(x)}{\Delta x}}={\frac {d{\boldsymbol {V}}(x)}{dx}}={\boldsymbol {z}}{\boldsymbol {I}}(x)\,} ... Equ. (1)

Similarly, analysing the circuit using Kirchhoff's current law:

${\boldsymbol {I}}(x+\Delta x)={\boldsymbol {I}}(x)+{\boldsymbol {y}}\Delta x{\boldsymbol {V}}(x+\Delta x)\,$

Re-arranging this equation, we get:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {{\boldsymbol {I}}(x+\Delta x)-{\boldsymbol {I}}(x)}{\Delta x}}={\boldsymbol {y}}{\boldsymbol {V}}(x+\Delta x)\,}

Taking the limit of both sides as $\Delta x\to 0$:

${\frac {d{\boldsymbol {I}}(x)}{dx}}={\boldsymbol {y}}{\boldsymbol {V}}(x)\,$ ... Equ. (2)

Differentiating Equations (1) and (2) again with respect to $x$, we get:

${\frac {d^{2}{\boldsymbol {V}}(x)}{dx^{2}}}={\boldsymbol {z}}{\frac {d{\boldsymbol {I}}(x)}{dx}}\,$

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d^{2}{\boldsymbol {I}}(x)}{dx^{2}}}={\boldsymbol {y}}{\frac {d{\boldsymbol {V}}(x)}{dx}}\,}

We can re-substitute Equations (1) and (2) into the above to get:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d^{2}{\boldsymbol {V}}(x)}{dx^{2}}}={\boldsymbol {zy}}{\boldsymbol {V}}(x)\,} ... Equ. (3)

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d^{2}{\boldsymbol {I}}(x)}{dx^{2}}}={\boldsymbol {zy}}{\boldsymbol {I}}(x)\,} ... Equ. (4)

The pair of equations above can be re-arranged to form homogenous second-order linear ordinary differential equations, i.e.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^{2} \boldsymbol{V}(x)}{dx^{2}} - \boldsymbol{zy} \boldsymbol{V}(x) = 0 \, }

The general solution to this differential equation is:

${\boldsymbol {V}}(x)=A_{1}e^{{\boldsymbol {\gamma }}x}+A_{2}e^{-{\boldsymbol {\gamma }}x}\,$

Where ${\boldsymbol {\gamma }}={\sqrt {\boldsymbol {zy}}}$ is known as the propagation constant (with units Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle m^{-1}} )

Plugging this solution back into Equ. (1), we can also solve for ${\boldsymbol {I}}(x)$, i.e.

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d}{dx}}\left[A_{1}e^{{\boldsymbol {\gamma }}x}+A_{2}e^{-{\boldsymbol {\gamma }}x}\right]={\boldsymbol {z}}{\boldsymbol {I}}(x)\,}

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \Rightarrow {\boldsymbol {I}}(x)={\frac {A_{1}e^{\gamma x}-A_{2}e^{-\gamma x}}{{\boldsymbol {Z}}_{c}}}\,}

Where ${\boldsymbol {Z}}_{c}={\sqrt {\boldsymbol {\frac {z}{y}}}}$ is known as the characteristic impedance or surge impedance (with units $\Omega$)

We can solve for the constants Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle A_{1}\,} and $A_{2}\,$ by using the boundary conditions at the receiving end of the line, i.e. at the receiving end, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=0\,} , Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\boldsymbol {V}}(0)={\boldsymbol {V_{r}}}\,} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\boldsymbol {I}}(0)={\boldsymbol {I_{r}}}\,} . The solution is as follows (see the full derivation here):

$A_{1}={\frac {{\boldsymbol {V_{r}}}+{\boldsymbol {I_{r}}}{\boldsymbol {Z}}_{c}}{2}}\,$

$A_{2}={\frac {{\boldsymbol {V_{r}}}-{\boldsymbol {I_{r}}}{\boldsymbol {Z}}_{c}}{2}}\,$

Substituting these constants, we get the final distributed parameter transmission line equations:

${\boldsymbol {V}}(x)=\left({\frac {{\boldsymbol {V_{r}}}+{\boldsymbol {I_{r}}}{\boldsymbol {Z}}_{c}}{2}}\right)e^{{\boldsymbol {\gamma }}x}+\left({\frac {{\boldsymbol {V_{r}}}-{\boldsymbol {I_{r}}}{\boldsymbol {Z}}_{c}}{2}}\right)e^{-{\boldsymbol {\gamma }}x}\,$ ... Equ. (5)

${\boldsymbol {I}}(x)=\left({\frac {{\boldsymbol {V_{r}}}+{\boldsymbol {I_{r}}}{\boldsymbol {Z}}_{c}}{2{\boldsymbol {Z}}_{c}}}\right)e^{\gamma x}-\left({\frac {{\boldsymbol {V_{r}}}-{\boldsymbol {I_{r}}}{\boldsymbol {Z}}_{c}}{2{\boldsymbol {Z}}_{c}}}\right)e^{-\gamma x}\,$ ... Equ. (6)

Hyperbolic Form of Transmission Line Equations

Equations (5) and (6) can be re-arranged as follows:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\boldsymbol {V}}(x)=\left({\frac {e^{{\boldsymbol {\gamma }}x}+e^{-{\boldsymbol {\gamma }}x}}{2}}\right){\boldsymbol {V_{r}}}+{\boldsymbol {Z}}_{c}\left({\frac {e^{{\boldsymbol {\gamma }}x}-e^{-{\boldsymbol {\gamma }}x}}{2}}\right){\boldsymbol {I_{r}}}}

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\boldsymbol {I}}(x)={\frac {1}{{\boldsymbol {Z}}_{c}}}\left({\frac {e^{{\boldsymbol {\gamma }}x}-e^{-{\boldsymbol {\gamma }}x}}{2}}\right){\boldsymbol {V_{r}}}+\left({\frac {e^{{\boldsymbol {\gamma }}x}+e^{-{\boldsymbol {\gamma }}x}}{2}}\right){\boldsymbol {I_{r}}}}

Given the exponential forms of the hyperbolic functions $\sinh x$ and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \cosh x} :

$\sinh x={\frac {e^{x}-e^{-x}}{2}}$

$\cosh x={\frac {e^{x}+e^{-x}}{2}}$

We can substitute these into the transmission line equations to get the well-known hyperbolic form of the equations:

${\boldsymbol {V}}(x)=\cosh({\boldsymbol {\gamma }}x){\boldsymbol {V_{r}}}+{\boldsymbol {Z}}_{c}\sinh({\boldsymbol {\gamma }}x){\boldsymbol {I_{r}}}$

${\boldsymbol {I}}(x)={\frac {1}{{\boldsymbol {Z}}_{c}}}\sinh({\boldsymbol {\gamma }}x){\boldsymbol {V_{r}}}+\cosh({\boldsymbol {\gamma }}x){\boldsymbol {I_{r}}}$

For a line of length $l\,$ metres, the ABCD parameters of the above equations can be represented in matrix form as follows:

$\left[{\begin{matrix}{\boldsymbol {V_{s}}}\\\\{\boldsymbol {I_{s}}}\end{matrix}}\right]=\left[{\begin{matrix}\cosh({\boldsymbol {\gamma }}l)&{\boldsymbol {Z}}_{c}\sinh({\boldsymbol {\gamma }}l)\\\\{\frac {1}{{\boldsymbol {Z}}_{c}}}sinh({\boldsymbol {\gamma }}l)&\cosh({\boldsymbol {\gamma }}l)\end{matrix}}\right]\left[{\begin{matrix}{\boldsymbol {V_{r}}}\\\\{\boldsymbol {I_{r}}}\end{matrix}}\right]\,$

where the sending end quantities are Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\boldsymbol {V_{s}}}={\boldsymbol {V}}(l)\,} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\boldsymbol {I_{s}}}={\boldsymbol {I}}(l)\,}

Multi-conductor Distributed Parameter Model

The distributed parameter model can be extended to a multi-conductor line of n conductors by replacing the voltage and current phasors with $n\times 1$ vectors, e.g. for a three-phase, three-conductor line:

${\boldsymbol {V}}=\left[{\begin{matrix}{\boldsymbol {V_{a}}}(x)\\{\boldsymbol {V_{b}}}(x)\\{\boldsymbol {V_{c}}}(x)\end{matrix}}\right]\,$, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\boldsymbol {I}}=\left[{\begin{matrix}{\boldsymbol {I_{a}}}(x)\\{\boldsymbol {I_{b}}}(x)\\{\boldsymbol {I_{c}}}(x)\end{matrix}}\right]\,}

And the impedance and admittance with $n\times n$ complex matrices, e.g. for a three-phase, three-conductor line:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle [Z]=\left[{\begin{matrix}Z_{aa}&Z_{ab}&Z_{ac}\\Z_{ba}&Z_{bb}&Z_{bc}\\Z_{ca}&Z_{cb}&Z_{cc}\end{matrix}}\right]\,}

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle [Y]=\left[{\begin{matrix}Y_{aa}&Y_{ab}&Y_{ac}\\Y_{ba}&Y_{bb}&Y_{bc}\\Y_{ca}&Y_{cb}&Y_{cc}\end{matrix}}\right]\,}

Equations (1) and (2) above in the single-phase model can now be re-written in the multi-conductor model as:

${\frac {d{\boldsymbol {V}}}{dx}}=[Z]{\boldsymbol {I}}\,$ ... Equ. (7)

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d{\boldsymbol {I}}}{dx}}=[Y]{\boldsymbol {V}}\,} ... Equ. (8)

Differentiating these equations with respect to $x$, we get:

${\frac {d^{2}{\boldsymbol {V}}}{dx^{2}}}=[Z]{\frac {d{\boldsymbol {I}}}{dx}}\,$ ... Equ. (9)

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d^{2}{\boldsymbol {I}}}{dx^{2}}}=[Y]{\frac {d{\boldsymbol {V}}}{dx}}\,} ... Equ. (10)

Substituting Equ. (8) into Equ. (9) and Equ. (7) into Equ. (10), we end up with the multi-conductor equivalent of equations (3) and (4):

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d^{2}{\boldsymbol {V}}}{dx^{2}}}=[Z][Y]{\boldsymbol {V}}\,} ... Equ. (11)

${\frac {d^{2}{\boldsymbol {I}}}{dx^{2}}}=[Y][Z]{\boldsymbol {I}}\,$ ... Equ. (12)

The issue with the multi-conductor case is that because the matrices [Z] and [Y] are full, then their product [Z][Y] or [Y][Z] is also full. For example, if we were to expand out Equ. (11):

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \left[{\begin{matrix}{\frac {d^{2}{\boldsymbol {V_{a}}}(x)}{dx^{2}}}\\\\{\frac {d^{2}{\boldsymbol {V_{b}}}(x)}{dx^{2}}}\\\\{\frac {d^{2}{\boldsymbol {V_{c}}}(x)}{dx^{2}}}\end{matrix}}\right]=\left[{\begin{matrix}Z_{aa}&Z_{ab}&Z_{ac}\\Z_{ba}&Z_{bb}&Z_{bc}\\Z_{ca}&Z_{cb}&Z_{cc}\end{matrix}}\right]\left[{\begin{matrix}Y_{aa}&Y_{ab}&Y_{ac}\\Y_{ba}&Y_{bb}&Y_{bc}\\Y_{ca}&Y_{cb}&Y_{cc}\end{matrix}}\right]\left[{\begin{matrix}{\boldsymbol {V_{a}}}(x)\\{\boldsymbol {V_{b}}}(x)\\{\boldsymbol {V_{c}}}(x)\end{matrix}}\right]\,}

The first line of the system of equations above can be expanded as follows:

${\frac {d^{2}{\boldsymbol {V_{a}}}(x)}{dx^{2}}}=\left(Z_{aa}Y_{aa}+Z_{ab}Y_{ba}+Z_{ac}Y_{ca}\right){\boldsymbol {V_{a}}}(x)+\left(Z_{aa}Y_{ab}+Z_{ab}Y_{bb}+Z_{ac}Y_{cb}\right){\boldsymbol {V_{b}}}(x)+\left(Z_{aa}Y_{ac}+Z_{ab}Y_{bc}+Z_{ac}Y_{cc}\right){\boldsymbol {V_{c}}}(x)\,$

Unlike in the single-phase case above, there is no closed-form general solution to this second-order differential equation (because of the cross-coupling between phases). The so-called modal transformation is required to decouple the phases from equations (11) and (12).

Modal Transformation

The modal transformation is a technique for decoupling the phases from equations (11) and (12) based on an Eigenvalue decomposition. In this section, we will develop the modal transformation from first principles. Alternative derivations can be found in [1], [2] and [3].

Consider linear transforms of the voltage and current vectors denoted ${\boldsymbol {V'}}$ and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\boldsymbol {I'}}} , where by convention, the subscripts change from phases (abc) to modes (012), i.e. for a three-phase, three-conductor line:

${\boldsymbol {V'}}=\left[{\begin{matrix}{\boldsymbol {V_{0}}}(x)\\{\boldsymbol {V_{1}}}(x)\\{\boldsymbol {V_{2}}}(x)\end{matrix}}\right]\,$, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\boldsymbol {I'}}=\left[{\begin{matrix}{\boldsymbol {I_{0}}}(x)\\{\boldsymbol {I_{1}}}(x)\\{\boldsymbol {I_{2}}}(x)\end{matrix}}\right]\,}

Suppose that the original quantities are related to the transformed quantities as follows:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\boldsymbol {V}}=[T_{v}]{\boldsymbol {V'}}} ... Equ. (13)

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\boldsymbol {I}}=[T_{i}]{\boldsymbol {I'}}} ... Equ. (14)

Where $[T_{v}]\,$ and $[T_{i}]\,$ are $n\times n$ transformation matrices. We won't define the transformation matrices just yet, but it will become obvious later on that they are actually the eigenvectors of $[Z][Y]\,$ and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle [Y][Z]\,} respectively.

Substituting these transformed quantities into equations (7) and (8):

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d[T_{v}]{\boldsymbol {V'}}}{dx}}=[Z][T_{i}]{\boldsymbol {I'}}\,}

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d[T_{i}]{\boldsymbol {I'}}}{dx}}=[Y][T_{v}]{\boldsymbol {V'}}\,}

Supposing that the transformation matrices $[T_{v}]\,$ and $[T_{i}]\,$ are independent of x, we can re-arrange the equations above as follows:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d{\boldsymbol {V'}}}{dx}}=[T_{v}]^{-1}[Z][T_{i}]{\boldsymbol {I'}}\,}

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d{\boldsymbol {I'}}}{dx}}=[T_{i}]^{-1}[Y][T_{v}]{\boldsymbol {V'}}\,}

Following the same procedure as before, i.e. differentiating the equations by x and substituting, we end up with:

${\frac {d^{2}{\boldsymbol {V'}}}{dx^{2}}}=[T_{v}]^{-1}[Z][Y][T_{v}]{\boldsymbol {V'}}\,$ ... Equ. (15)

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d^{2}{\boldsymbol {I'}}}{dx^{2}}}=[T_{i}]^{-1}[Y][Z][T_{i}]{\boldsymbol {I'}}\,} ... Equ. (16)

In order for us to decouple the phases in equations (15) and (16), we need $[T_{v}]^{-1}[Z][Y][T_{v}]\,$ and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle [T_{i}]^{-1}[Y][Z][T_{i}]\,} to be diagonal matrices, i.e.

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle diag({\boldsymbol {\lambda _{v}}})=[T_{v}]^{-1}[Z][Y][T_{v}]\,}

$diag({\boldsymbol {\lambda _{i}}})=[T_{i}]^{-1}[Y][Z][T_{i}]\,$

It is apparent from inspection that the transformation matrix $[T_{v}]\,$ and diagonal matrix $diag({\boldsymbol {\lambda _{v}}})$ are the eigenvectors and eigenvalues of $[Z][Y]\,$ respectively.

Similarly, $[T_{i}]\,$ and $diag({\boldsymbol {\lambda _{v}}})$ are the eigenvectors and eigenvalues of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [Y][Z] \,} .

Furthermore, since the matrices Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [Z] \,} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [Y] \,} are both symmetric, then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [Z][Y] = \left( [Y][Z] \right)^{T} \,} . We know that the eigenvalues of a matrix and its transpose are the same, so therefore Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [Z][Y] \,} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [Y][Z] \,} also share the same eigenvalues, i.e.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle diag(\boldsymbol{\lambda_{v}}) = diag(\boldsymbol{\lambda_{i}}) = diag(\boldsymbol{\lambda})\,}

We can now re-write equations (15) and (16) as follows:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} \frac{d^{2} \boldsymbol{V_{0}}(x)}{dx^{2}} \\ \\ \frac{d^{2} \boldsymbol{V_{1}}(x)}{dx^{2}} \\ \\ \frac{d^{2}\boldsymbol{V_{2}}(x)}{dx^{2}} \end{matrix} \right] = \left[ \begin{matrix} \lambda_{0} & & \\ & \lambda_{1} & \\ & & \lambda_{2} \end{matrix} \right] \left[ \begin{matrix} \boldsymbol{V_{0}}(x) \\ \boldsymbol{V_{1}}(x) \\ \boldsymbol{V_{2}}(x) \end{matrix} \right] \, } ... Equ. (17)

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} \frac{d^{2} \boldsymbol{I_{0}}(x)}{dx^{2}} \\ \\ \frac{d^{2} \boldsymbol{I_{1}}(x)}{dx^{2}} \\ \\ \frac{d^{2}\boldsymbol{I_{2}}(x)}{dx^{2}} \end{matrix} \right] = \left[ \begin{matrix} \lambda_{0} & & \\ & \lambda_{1} & \\ & & \lambda_{2} \end{matrix} \right] \left[ \begin{matrix} \boldsymbol{I_{0}}(x) \\ \boldsymbol{I_{1}}(x) \\ \boldsymbol{I_{2}}(x) \end{matrix} \right] \, } ... Equ. (18)

Recall from the single-phase case that the propagation constant is defined as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{\gamma} = \sqrt{\boldsymbol{zy}} } . In multi-conductor matrix form, this can be written as a propagation matrix Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [\Gamma] \, } as follows:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [\Gamma] = ([Z][Y])^{\frac{1}{2}} \, }

Let the eigenvalues of the propagation matrix Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [\Gamma] \, } be Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \gamma_0, \gamma_1, \gamma_2 \, } .

A property of eigenvalues is that if matrix Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A \, } has the eigenvalues Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_1, \lambda_2, ... , \lambda_n \,} , then matrix Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A^{k} \, } has the eigenvalues Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_1^{k}, \lambda_2^{k}, ... , \lambda_n^{k} \,} .

Therefore, since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_0, \lambda_1, \lambda_2 \, } are the eigenvalues of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [Z][Y] \, } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \gamma_0, \gamma_1, \gamma_2 \, } are the eigenvalues of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ([Z][Y])^{\frac{1}{2}} \, } , we can say that:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} \lambda_{0} & & \\ & \lambda_{1} & \\ & & \lambda_{2} \end{matrix} \right] = \left[ \begin{matrix} \gamma_{0}^{2} & & \\ & \gamma_{1}^{2} & \\ & & \gamma_{2}^{2} \end{matrix} \right] \, }

We can now re-write equations (17) and (18) as follows:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} \frac{d^{2} \boldsymbol{V_{0}}(x)}{dx^{2}} \\ \\ \frac{d^{2} \boldsymbol{V_{1}}(x)}{dx^{2}} \\ \\ \frac{d^{2}\boldsymbol{V_{2}}(x)}{dx^{2}} \end{matrix} \right] = \left[ \begin{matrix} \gamma_{0}^{2} & & \\ & \gamma_{1}^{2} & \\ & & \gamma_{2}^{2} \end{matrix} \right] \left[ \begin{matrix} \boldsymbol{V_{0}}(x) \\ \boldsymbol{V_{1}}(x) \\ \boldsymbol{V_{2}}(x) \end{matrix} \right] \, } ... Equ. (19)

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} \frac{d^{2} \boldsymbol{I_{0}}(x)}{dx^{2}} \\ \\ \frac{d^{2} \boldsymbol{I_{1}}(x)}{dx^{2}} \\ \\ \frac{d^{2}\boldsymbol{I_{2}}(x)}{dx^{2}} \end{matrix} \right] = \left[ \begin{matrix} \gamma_{0}^{2} & & \\ & \gamma_{1}^{2} & \\ & & \gamma_{2}^{2} \end{matrix} \right] \left[ \begin{matrix} \boldsymbol{I_{0}}(x) \\ \boldsymbol{I_{1}}(x) \\ \boldsymbol{I_{2}}(x) \end{matrix} \right] \, } ... Equ. (20)

Each of these decoupled modal differential equations above can now be solved using the general solution:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{V_{0}}(x) = A_{0} e^{\gamma_0 x} + B_{0} e^{-\gamma_0 x} \, }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{V_{1}}(x) = A_{1} e^{\gamma_1 x} + B_{1} e^{-\gamma_1 x} \, }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{V_{2}}(x) = A_{2} e^{\gamma_2 x} + B_{2} e^{-\gamma_2 x} \, }

Taking the first derivative of the equations above with respect to x, we get:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} \frac{d \boldsymbol{V_{0}}(x)}{dx} \\ \\ \frac{d \boldsymbol{V_{1}}(x)}{dx} \\ \\ \frac{d\boldsymbol{V_{2}}(x)}{dx} \end{matrix} \right] = \left[ \begin{matrix} \gamma_{0} & & \\ & \gamma_{1} & \\ & & \gamma_{2} \end{matrix} \right] \left[ \begin{matrix} A_{0} e^{\gamma_0 x} - B_{0} e^{-\gamma_0 x}\\ A_{1} e^{\gamma_1 x} - B_{1} e^{-\gamma_1 x} \\ A_{2} e^{\gamma_2 x} - B_{2} e^{-\gamma_2 x} \end{matrix} \right] \, } ... Equ. (21)

Let's denote equation (21) above in the following shorthand notation:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d\boldsymbol{V'}}{dx} = [\gamma] \boldsymbol{V_x} \, } ... Equ. (22)

Recall from earlier that:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d \boldsymbol{V'}}{dx} = [T_{v}]^{-1} [Z] [T_{i}] \boldsymbol{I'} \, }

Equating this with equation (22) and solving for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{V_x} } , we get:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{V_x} = [\gamma]^{-1} [T_{v}]^{-1} [Z] [T_{i}] \boldsymbol{I'} \, }

We define the modal characteristic impedance matrix (or modal surge impedance matrix) as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [Z_c] = [\gamma]^{-1} [T_{v}]^{-1} [Z] [T_{i}] \, } ... Equ. (23)

Using this definition, the modal current vector can be expressed as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{I'} = [Z_c]^{-1} \boldsymbol{V_x} \, } ... Equ. (24)

It can be shown that if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [\gamma] \, } is diagonal, then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [Z_c] \, } is also diagonal (see the derivation here). Therefore, when expanded, equation (24) looks like this:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} \boldsymbol{I_{0}}(x) \\ \boldsymbol{I_{1}}(x) \\ \boldsymbol{I_{2}}(x) \end{matrix} \right] = \left[ \begin{matrix} \frac{1}{\boldsymbol{Z_0}} & & \\ & \frac{1}{\boldsymbol{Z_1}} & \\ & & \frac{1}{\boldsymbol{Z_2}} \end{matrix} \right] \left[ \begin{matrix} A_{0} e^{\gamma_0 x} - B_{0} e^{-\gamma_0 x}\\ A_{1} e^{\gamma_1 x} - B_{1} e^{-\gamma_1 x} \\ A_{2} e^{\gamma_2 x} - B_{2} e^{-\gamma_2 x} \end{matrix} \right] \, } ... Equ. (25)

Boundary Conditions

We now have the following six modal equations that we can solve using boundary conditions:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} \boldsymbol{V_{0}}(x) \\ \boldsymbol{V_{1}}(x) \\ \boldsymbol{V_{2}}(x) \end{matrix} \right] = \left[ \begin{matrix} A_{0} e^{\gamma_0 x} + B_{0} e^{-\gamma_0 x} \\ A_{1} e^{\gamma_1 x} + B_{1} e^{-\gamma_1 x} \\ A_{2} e^{\gamma_2 x} + B_{2} e^{-\gamma_2 x} \end{matrix} \right] \, }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} \boldsymbol{I_{0}}(x) \\ \boldsymbol{I_{1}}(x) \\ \boldsymbol{I_{2}}(x) \end{matrix} \right] = \left[ \begin{matrix} \frac{1}{\boldsymbol{Z_0}} & & \\ & \frac{1}{\boldsymbol{Z_1}} & \\ & & \frac{1}{\boldsymbol{Z_2}} \end{matrix} \right] \left[ \begin{matrix} A_{0} e^{\gamma_0 x} - B_{0} e^{-\gamma_0 x}\\ A_{1} e^{\gamma_1 x} - B_{1} e^{-\gamma_1 x} \\ A_{2} e^{\gamma_2 x} - B_{2} e^{-\gamma_2 x} \end{matrix} \right] \, }

The boundary conditions at the receiving end Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{V_r} } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{I_r} } must be transformed to modal quantities as follows:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{V_r}' = [T_v]^{-1} \boldsymbol{V_r} }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \boldsymbol{I_r}' = [T_i]^{-1} \boldsymbol{I_r} }

At the receiving end (i.e. x = 0), the modal equations reduce as follows:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} \boldsymbol{V_{0r}} \\ \boldsymbol{V_{1r}} \\ \boldsymbol{V_{2r}} \end{matrix} \right] = \left[ \begin{matrix} A_{0} + B_{0} \\ A_{1} + B_{1} \\ A_{2} + B_{2} \end{matrix} \right] \, }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} \boldsymbol{Z_0} \boldsymbol{I_{0r}} \\ \boldsymbol{Z_1} \boldsymbol{I_{1r}} \\ \boldsymbol{Z_2} \boldsymbol{I_{2r}} \end{matrix} \right] = \left[ \begin{matrix} A_{0} - B_{0} \\ A_{1} - B_{1} \\ A_{2} - B_{2} \end{matrix} \right] \, }

Solving for the constants of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [A] \, } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [B] \, } :

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} A_{0} \\ A_{1} \\ A_{2} \end{matrix} \right] = \frac{1}{2} \left[ \begin{matrix} \boldsymbol{V_{0r}} - \boldsymbol{Z_0} \boldsymbol{I_{0r}} \\ \boldsymbol{V_{1r}} - \boldsymbol{Z_1} \boldsymbol{I_{1r}} \\ \boldsymbol{V_{2r}} - \boldsymbol{Z_2} \boldsymbol{I_{2r}} \end{matrix} \right] \, } ... Equ. (26)

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} B_{0} \\ B_{1} \\ B_{2} \end{matrix} \right] = \frac{1}{2} \left[ \begin{matrix} \boldsymbol{V_{0r}} + \boldsymbol{Z_0} \boldsymbol{I_{0r}} \\ \boldsymbol{V_{1r}} + \boldsymbol{Z_1} \boldsymbol{I_{1r}} \\ \boldsymbol{V_{2r}} + \boldsymbol{Z_2} \boldsymbol{I_{2r}} \end{matrix} \right] \, } ... Equ. (27)

These constants can be plugged back into the modal voltage and current equations to get the final expressions:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} \boldsymbol{V_{0}}(x) \\ \boldsymbol{V_{1}}(x) \\ \boldsymbol{V_{2}}(x) \end{matrix} \right] = \frac{1}{2} \left[ \begin{matrix} (\boldsymbol{V_{0r}} - \boldsymbol{Z_0} \boldsymbol{I_{0r}}) e^{\gamma_0 x} + (\boldsymbol{V_{0r}} + \boldsymbol{Z_0} \boldsymbol{I_{0r}}) e^{-\gamma_0 x} \\ (\boldsymbol{V_{1r}} - \boldsymbol{Z_1} \boldsymbol{I_{1r}}) e^{\gamma_1 x} + (\boldsymbol{V_{1r}} + \boldsymbol{Z_1} \boldsymbol{I_{1r}}) e^{-\gamma_1 x} \\ (\boldsymbol{V_{2r}} - \boldsymbol{Z_2} \boldsymbol{I_{2r}}) e^{\gamma_2 x} + (\boldsymbol{V_{2r}} + \boldsymbol{Z_2} \boldsymbol{I_{2r}}) e^{-\gamma_2 x} \end{matrix} \right] \, } ... Equ. (28)

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} \boldsymbol{I_{0}}(x) \\ \boldsymbol{I_{1}}(x) \\ \boldsymbol{I_{2}}(x) \end{matrix} \right] = \left[ \begin{matrix} \frac{1}{2 \boldsymbol{Z_0}} & & \\ & \frac{1}{2 \boldsymbol{Z_1}} & \\ & & \frac{1}{2 \boldsymbol{Z_2}} \end{matrix} \right] \left[ \begin{matrix} (\boldsymbol{V_{0r}} - \boldsymbol{Z_0} \boldsymbol{I_{0r}}) e^{\gamma_0 x} - (\boldsymbol{V_{0r}} + \boldsymbol{Z_0} \boldsymbol{I_{0r}}) e^{-\gamma_0 x} \\ (\boldsymbol{V_{1r}} - \boldsymbol{Z_1} \boldsymbol{I_{1r}}) e^{\gamma_1 x} - (\boldsymbol{V_{1r}} + \boldsymbol{Z_1} \boldsymbol{I_{1r}}) e^{-\gamma_1 x} \\ (\boldsymbol{V_{2r}} - \boldsymbol{Z_2} \boldsymbol{I_{2r}}) e^{\gamma_2 x} - (\boldsymbol{V_{2r}} + \boldsymbol{Z_2} \boldsymbol{I_{2r}}) e^{-\gamma_2 x} \end{matrix} \right] \, } ... Equ. (29)

Hyperbolic Form of Multi-conductor Line Equations

As in the single-phase distributed parameter model, the final modal voltages and currents shown in equations (28) and (29) can be converted to the following hyperbolic forms:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} \boldsymbol{V_{0}}(x) \\ \boldsymbol{V_{1}}(x) \\ \boldsymbol{V_{2}}(x) \end{matrix} \right] = \left[ \begin{matrix} \cosh{(\gamma_0 x)} & & \\ & \cosh{(\gamma_1 x)} & \\ & & \cosh{(\gamma_2 x)}\end{matrix} \right] \left[ \begin{matrix} \boldsymbol{V_{0r}} \\ \boldsymbol{V_{1r}} \\ \boldsymbol{V_{2r}} \end{matrix} \right] + \left[ \begin{matrix} \boldsymbol{Z_0} \sinh{(\gamma_0 x)} & & \\ & \boldsymbol{Z_1} \sinh{(\gamma_1 x)} & \\ & & \boldsymbol{Z_2} \sinh{(\gamma_2 x)} \end{matrix} \right] \left[ \begin{matrix} \boldsymbol{I_{0r}} \\ \boldsymbol{I_{1r}} \\ \boldsymbol{I_{2r}} \end{matrix} \right] \, } ... Equ. (30)

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \begin{matrix} \boldsymbol{I_{0}}(x) \\ \boldsymbol{I_{1}}(x) \\ \boldsymbol{I_{2}}(x) \end{matrix} \right] = \left[ \begin{matrix} \frac{1}{\boldsymbol{Z_0}} \sinh{(\gamma_0 x)} & & \\ & \frac{1}{\boldsymbol{Z_1}} \sinh{(\gamma_1 x)} & \\ & & \frac{1}{\boldsymbol{Z_2}} \sinh{(\gamma_2 x)}\end{matrix} \right] \left[ \begin{matrix} \boldsymbol{V_{0r}} \\ \boldsymbol{V_{1r}} \\ \boldsymbol{V_{2r}} \end{matrix} \right] + \left[ \begin{matrix} \cosh{(\gamma_0 x)} & & \\ & \cosh{(\gamma_1 x)} & \\ & & \cosh{(\gamma_2 x)} \end{matrix} \right] \left[ \begin{matrix} \boldsymbol{I_{0r}} \\ \boldsymbol{I_{1r}} \\ \boldsymbol{I_{2r}} \end{matrix} \right] \, } ... Equ. (31)

Derivations of these forms are straightforward and left to the interested reader to prove.

References

• [1] Wedepohl, L. M., "Application of matrix methods to the solution of travelling-wave phenomena in polyphase systems", Proceedings of the IEE, Vol. 110(12), 1963

• [2] Hedman, D. E., "Propagation on Overhead Transmission Lines I - Theory of Modal Analysis ", IEEE Transactions on Power Apparatus and Systems, Vol 84(3), 1965

• [3] Bowman, W. I., and McNamee, J. M., "Development of Equivalent Pi and T Matrix Circuits for Long Untransposed Transmission Lines", IEEE Transactions on Power Apparatus and Systems, Vol 83(6), 1964

Related Topics

• Overhead Line Models
• Multi-conductor Line Models
• Travelling Wave Line Model
• Bergeron Line Model