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I How do we derive an Algebraic Companion Form equation set?

  1. Mar 9, 2016 #1
    Hello,

    I'm trying to study a settingless relaying scheme for Power System protection. The method is fairly well known and requires that one model a piece of equipment (like a transformer) with differential equation sets and compare that computational model with the actual operating behavior. If the two differ beyond assumptions of error, then tripping is performed.

    The Basic form of the equation relies on "through" variables (current) and "across" variables (voltages). So if you have a piece of equipment that has three terminals through which current flows and across which a potential difference is measured (the voltages and current are time-varying), the system of equations can be written as...
    ##\begin{bmatrix}\textbf{i}\\ 0\\\end{bmatrix}=\begin{bmatrix}\textbf{f}_{1}(\dot{\textbf{v}}, \dot{\textbf{y}}, ... , \int\textbf{v}, \int\textbf{y}, ...,\textbf{u}, \textbf{v}, \textbf{y}, t )\\ \textbf{f}_{2}(\dot{\textbf{v}}, \dot{\textbf{y}}, ... , \int\textbf{v}, \int\textbf{y}, ...,\textbf{u}, \textbf{v}, \textbf{y}, t )\end{bmatrix}##

    f1 and f2 are arbitrary vector functions, i is the vector for the "through" variables (in this case I would assume i = (i1, i2, i3) for a three terminal device), v is the vector for the "across variables," y is a vector for internal state variables, and u is a vector for independent controls.

    The literature I found says that to attain the Algebraic Companion Form you have to integrate over the simulation period 'h'.

    ##\int_{0}^{h}\begin{bmatrix}\textbf{i}\\ 0\\\end{bmatrix} = \int_{0}^{h} \begin{bmatrix}\textbf{f}_{1}(\dot{\textbf{v}}, \dot{\textbf{y}}, ... , \int\textbf{v}, \int\textbf{y}, ...,\textbf{u}, \textbf{v}, \textbf{y}, t )\\ \textbf{f}_{2}(\dot{\textbf{v}}, \dot{\textbf{y}}, ... , \int\textbf{v}, \int\textbf{y}, ...,\textbf{u}, \textbf{v}, \textbf{y}, t )\end{bmatrix}##


    If you do then you attain the below expression...

    ##\begin{bmatrix}\textbf{i}(t)\\ 0\\\end{bmatrix} =\textbf{G}(\textbf{v}(t), \textbf{v}(t - h), \textbf{i}(t), \textbf{i}(t - h), \textbf{y}(t), \textbf{y}(t - h), t)\begin{bmatrix}\textbf{v}(t)\\ \textbf{y}(t)\\\end{bmatrix} + \frac{1}{2}\begin{bmatrix}\textbf{K}(\textbf{v}(t), \textbf{v}(t-h), \textbf{i}(t), \textbf{i}(t-h), \textbf{y}(t), \textbf{y}(t-h), t)\\ \textbf{Q}(\textbf{v}(t), \textbf{v}(t-h), \textbf{i}(t), \textbf{i}(t-h), \textbf{y}(t), \textbf{y}(t-h), t)\\\end{bmatrix} - \begin{bmatrix}\textbf{O}(t-h)\\\textbf{P}(t-h)\\\end{bmatrix} + \begin{bmatrix}\beta\\\alpha \\\end{bmatrix}##

    Where G is the Jacobian matrix, O and P are vectors depending only on past history values of through, across or internal states. ##\beta## and ##\alpha## denote cubic and higher order terms, and vectors K and Q represent the quadratic term.


    My concern is that I'm not entirely sure how to come to the ACF (Algebraic Companion Form). I suppose this would be more along the lines of graduate level stuff, but I would like to be able to understand how one goes from the general form to the ACF. If anyone could provide an explanation, or perhaps literature on the subject, I would be much obliged.
     
    Last edited by a moderator: Mar 9, 2016
  2. jcsd
  3. Mar 14, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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