I Reducing NxN Matrix to 2x2 w/ Physical Constraints

waynewec
Messages
2
Reaction score
0
TL;DR Summary
Reducing an NxN matrix to a 2x2 by application of physical constraints
Gonna preface by saying I never thought linear algebra would be a class I would regret not taking so much... but in short the goal is to reduce an arbitrary symmetric NxN system using a set of auxiliary constraint relationships, e.g. for a 3x3

<br /> \begin{bmatrix}<br /> V_1\\<br /> V_2\\<br /> V_3\\<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> L_{e11}&amp;L_{e12}&amp;L_{e12}\\<br /> L_{e21}&amp;L_{e22}&amp;L_{e23}\\<br /> L_{e31}&amp;L_{e32}&amp;L_{e33}\\<br /> \end{bmatrix}<br /> \cdot<br /> \begin{bmatrix}<br /> i_1\\<br /> i_2\\<br /> i_3\\<br /> \end{bmatrix}\\<br />
using the following constraints
##V_1=V_2=V_p##
##V_3=V_s##
##i_p=i_1+i_2##
##i_s=i_3##
to end up with an equivalent system with L_s, L_p, and M in terms of the starting L_{eij} matrix
<br /> \begin{bmatrix}<br /> V_p\\<br /> V_s\\<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> L_p&amp;M\\<br /> M&amp;L_s\\<br /> \end{bmatrix}<br /> \cdot<br /> \begin{bmatrix}<br /> i_p\\<br /> i_s\\<br /> \end{bmatrix}<br />
For those interested in the context, this is an application specific usage of the method covered in https://onlinelibrary.wiley.com/doi/full/10.1002/eej.23240 but they glossed a bit over some of the key linear math that I don't understand. Eventually I'll be extending this concept to quite large matrices with more complex auxiliary constraints, but for now I'd appreciate some guidance, and some good resources, to get me goin
 
Last edited:
Physics news on Phys.org
Have you tried just eliminating variables? ##V_1 = V_2## relates ##i_n## making it possible to express ##i_1## in terms of ##i_p## and ##i_s##. Clearly, ##i_2 = i_p - i_1## and ##i_3 = i_s## eliminates ##i_2## and ##i_3##.

I get something like,

##V_p = (L_{11}-L_{12})i_1 + L_{12}i_p + L_{13}i_s##
## 0 = (L_{11}-L_{12})i_1 + (L_{12}-L_{22})(i_p-i_1) + (L_{13}-L_{23})i_s##
## V_s = (L_{31}-L_{32})i_1 + L_{32}i_p + L_{33}i_s##

Okay, just use the second equation to eliminate ##i_1##.
 
Paul Colby said:
Have you tried just eliminating variables? ##V_1 = V_2## relates ##i_n## making it possible to express ##i_1## in terms of ##i_p## and ##i_s##. Clearly, ##i_2 = i_p - i_1## and ##i_3 = i_s## eliminates ##i_2## and ##i_3##.

I get something like,

##V_p = (L_{11}-L_{12})i_1 + L_{12}i_p + L_{13}i_s##
## 0 = (L_{11}-L_{12})i_1 + (L_{12}-L_{22})(i_p-i_1) + (L_{13}-L_{23})i_s##
## V_s = (L_{31}-L_{32})i_1 + L_{32}i_p + L_{33}i_s##

Okay, just use the second equation to eliminate ##i_1##.
Absolutely valid, and an approach I've used, but any changes made to constraints or the order of the input matrix requires extremely tedious manual calculations. I was hoping for a direction that relies on matrix mathematics and could be implemented programmatically. 3x3, not so bad - 9x9 will make me want to kill myself
 
Well, okay. The voltage constrains in matrix form,

##\left(\begin{array}{c} V_1 \\ V_2 \\ V_3\end{array}\right) = \left(\begin{array}{cc} 1 & 0 \\ 1 & 0 \\ 0 & 1\end{array}\right)\left(\begin{array}{c} V_p \\ V_s\end{array}\right)##

The current constraints in matrix form,

##\left(\begin{array}{c} i_p \\ i_s \end{array}\right) = \left(\begin{array}{ccc} 1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{c} i_1 \\ i_2 \\ i_3 \end{array}\right)##

More generaly,

##V = C V_c##

and

##I_c = D I##

Clearly,

##V_c = C^{-1} L D^{-1} I_c##

is the solution. All you need to do is figure out what ##C^{-1}## and ##D^{-1}## really mean.
 
Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'
This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...

Similar threads

Replies
4
Views
1K
Replies
9
Views
14K
Replies
1
Views
8K
Replies
3
Views
4K
Back
Top