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2020-11-22T07:46:28Z

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== Motivation ==

[[Image:Generic_bergeron.png|right|thumb|500px|Figure 1. A generic single-phase distributed parameter line model]]

In this section, we try to answer the question - why bother with a model for frequency dependent lines?

Consider the generic single-phase [[Distributed Parameter Line Model|distributed parameter line model]] of length <math>l \,</math> metres shown in Figure 1. Note that the model is represented as functions of both distance and time.

We can represent the series elements as an impedance and the shunt elements as an admittance as follows:

: <math> \boldsymbol{z} = R + j \omega L = R + jX \, </math>
: <math> \boldsymbol{y} = \frac{1}{R_{sh}} + j \omega C = G + jB \, </math>

Where <math> R \, </math> is the series [http://en.wikipedia.org/wiki/Electrical_resistance resistance] (<math>\Omega /m </math>)
:: <math> X \, </math> is the series [http://en.wikipedia.org/wiki/Electrical_reactance reactance] (<math>\Omega /m </math>)
:: <math> G \, </math> is the shunt [http://en.wikipedia.org/wiki/Electrical_conductance conductance] (<math>S /m </math>)
:: <math> B \, </math> is the shunt [http://en.wikipedia.org/wiki/Susceptance susceptance] (<math>S /m </math>)

If we only consider this model in the steady-state (i.e. the frequency domain), then time can be neglected and the model is exactly like the conventional single-phase [[Distributed Parameter Line Model|distributed parameter line model]].

Recall that the solution to the voltage and current of the steady-state model is as follows:

: <math> \boldsymbol{V}_{k} = \left( \frac{\boldsymbol{V}_{m} - \boldsymbol{I}_{m,k} \boldsymbol{Z}_{c}}{2} \right) e^{\boldsymbol{\gamma} l} + \left( \frac{\boldsymbol{V}_{m} + \boldsymbol{I}_{m,k} \boldsymbol{Z}_{c}}{2} \right) e^{-\boldsymbol{\gamma} l} \, </math> ... Equ. (1)

: <math> \boldsymbol{I}_{k,m} \boldsymbol{Z}_{c} = \left( \frac{\boldsymbol{V}_{m} - \boldsymbol{I}_{m,k} \boldsymbol{Z}_{c}}{2} \right) e^{\gamma l} - \left( \frac{\boldsymbol{V}_{m} + \boldsymbol{I}_{m,k} \boldsymbol{Z}_{c}}{2} \right) e^{-\gamma l} \, </math> ... Equ. (2)

Where <math> \boldsymbol{\gamma} = \sqrt{\boldsymbol{zy}} = \sqrt{(R + j \omega L)(G + j \omega C)} </math>

:: <math> \boldsymbol{Z}_{c} = \sqrt{\boldsymbol{\frac{z}{y}}} = \sqrt{\frac{(R + j \omega L)}{(G + j \omega C)}} </math>

'''Things to note:'''
:1) Voltage and current are complex phasors hence they are in bold.
:2) The equations have slightly different signs to those derived in the [[Distributed Parameter Line Model|distributed parameter line model]]. This is because the direction of current <math>I_{m,k}(t) \, </math> in Figure 3 is set in the opposite direction in this model.

We can add Equ. (1) to Equ. (2) to get:

: <math> \boldsymbol{V}_{k} + \boldsymbol{I}_{k,m} \boldsymbol{Z}_{c} = \left( \boldsymbol{V}_{m} - \boldsymbol{I}_{m,k} \boldsymbol{Z}_{c} \right) e^{\boldsymbol{\gamma} l} \, </math> ... Equ. (3)

Similarly, we can subtract Equ. (2) from Equ. (1) to get:

: <math> \boldsymbol{V}_{k} - \boldsymbol{I}_{k,m} \boldsymbol{Z}_{c} = \left( \boldsymbol{V}_{m} + \boldsymbol{I}_{m,k} \boldsymbol{Z}_{c} \right) e^{-\boldsymbol{\gamma} l} \, </math> ... Equ. (4)

Re-arranging Equ. (3) and Equ. (4) to solve for the currents <math>\boldsymbol{I}_{m,k} \, </math> and <math>\boldsymbol{I}_{k,m} \, </math>:

: <math> \boldsymbol{I}_{m,k} = \frac{\boldsymbol{V}_{m}}{\boldsymbol{Z}_{c}} - \frac{\boldsymbol{V}_{k}}{\boldsymbol{Z}_{c}} e^{-\boldsymbol{\gamma} l} - \boldsymbol{I}_{k,m} e^{-\boldsymbol{\gamma} l} \, </math> ... Equ. (5)

: <math> \boldsymbol{I}_{k,m} = \frac{\boldsymbol{V}_{k}}{\boldsymbol{Z}_{c}} - \frac{\boldsymbol{V}_{m}}{\boldsymbol{Z}_{c}} e^{-\boldsymbol{\gamma} l} - \boldsymbol{I}_{m,k} e^{-\boldsymbol{\gamma} l} \, </math> ... Equ. (6)

From [https://en.wikipedia.org/wiki/Fourier_transform#Functional_relationships Fourier Transform theory], we know that multiplying by a complex exponential in the frequency domain is equivalent to a time shift in the time domain. Therefore, in the lossless case (i.e. R = G = 0.), the exponential part of <math> e^{-\boldsymbol{\gamma} l} \, </math> is:

:: <math> \boldsymbol{\gamma} l = \sqrt{\boldsymbol{zy}} l = \sqrt{(j \omega L)(j \omega C)} l = j \omega \sqrt{LC} l = j \omega \tau </math>

Where <math> \tau \, </math> is the travel time that we saw in the [[Bergeron_Line_Model#Lossless_Lines|Bergeron model]].

The characteristic impedance is also a scalar frequency-independent constant:

:: <math> \boldsymbol{Z}_{c} = \sqrt{\boldsymbol{\frac{z}{y}}} = \sqrt{\frac{(j \omega L)}{(j \omega C)}} = \sqrt{\frac{L}{C}}</math>

However if the line is not lossless, then we can see that the propagation constant and characteristic impedance are frequency dependent, i.e.

:: <math> \boldsymbol{\gamma} = \sqrt{\boldsymbol{zy}} = \sqrt{(R + j \omega L)(G + j \omega C)} </math>

:: <math> \boldsymbol{Z}_{c} = \sqrt{\boldsymbol{\frac{z}{y}}} = \sqrt{\frac{(R + j \omega L)}{(G + j \omega C)}} </math>

Since these quantities are complex and frequency dependent, the model does not simplify to a simple time shift in the time domain as we saw in the lossless case.

Moreover, the line parameters themselves are frequency dependent rather than constant across the full frequency range, i.e. <math>L(\omega) \, </math>, <math>C(\omega) \, </math>, <math>R(\omega) \, </math> and <math>G(\omega) \, </math>. Therefore, a constant parameter line model such as the [[Bergeron Line Model|Bergeron model]] is not suitable for cases where the frequency dependencies are significant (e.g. zero sequence impedances). This is the motivation for developing models that are able to capture the frequency dependencies of lines.

== Single-Phase Lines ==

[[Image:Under_construction.png|300px]]

== Multi-Conductor Lines ==

[[Image:Under_construction.png|300px]]

== Related Topics ==

:* [[Overhead Line Models]]
:* [[Multi-conductor Line Models]]
:* [[Travelling Wave Line Model]]
:* [[Bergeron Line Model]]

[[Category:Modelling / Analysis]]

Frequency Dependent Line Model - Revision history

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