A discrete magnetostatic solver in Matlab/Octave

Transcription

1 discrete magnetostatic solver in Matlab/Octave Tobias Schäfer 1st March 01 Contents 1 Introduction 1 Theory 1 Boundary conditions 5 3 Solution 5 31 Iterative solution Hint 5 3 FEM Solution 6 4 Matlab/Octave code 6 1 Introduction In the open-rtms project a complex magnetic field is applied to the human brain The whole setup consists of several coils placed around the skull I assume, that the properties of the brain are LTI (linear, time invariant), so that a simple superposition of the fields from the different coils can be calculated the get the field distribution inside the brain The coils I assume to be small compared to the distance between them and the field strength is small as well, so that the influence of the coils towards each other can also be neglected These assumptions should work out well, if following types of coils are used: a simple core made from a screw with some turns of copper ferrite core is not needed, because of the low frequency flat coils without core In this case also overlapping coils are not a problem 1

2 THEORY Theory To calculate the magnetic fieldlines we first calculate the vector potential fter is computed we can calculate the magnetic field B = rot() = To calculate we start with the Poisson equation in 3D (: permeability of material, j: current density in coil): ( 1 x ) ( 1 x y ) ( 1 y z ) = j (1) z To simplify the solution of equation (1) the problem is simplified to a D rotational system (Figure ) r Mesh cells ir z Core Coil Figure 1: Rotational symmetric setup in this rotational case the Poisson equation (1) becomes: ( ) 1 (r ) ( 1 r r r z ) z = j () This equation is solved for a grid of cells The terms in () are rewritten as finite differences: The inner partial differentials according to figure will be and (r ) r (r r/) (r r/) (r r/) (r r/) r z (z z/) (z z/) z (3) (4)

3 THEORY (r r,z) (r r,z) (r,z z) (r, z) (r,z z) (r,z z) (r,z) (r,z z) (r r,z) (r r,z) Figure : Cells in the grid For the whole expression () we get ( ) 1 (r ) r r r (r r) (r r) r r (r r) (r r) (r r/) (r r/) r (r r/) (r r/) r (5) and ( 1 z ) z (z z) (z z/) z (z z) (z z/) z z (z z) (z z/) z (z z) (z z/) z (6) s it can be seen from figure is used in equations (5) and (6) on a 45 rotated grid between the cells For this grid is hard to define in Matlab/Octave without making errors, we define at the very same points as (as shown in figure ) and use terms like = (z z/) = (z) (z z) (7) 3

4 THEORY Figure 3: inbetween Equations (5) and (6) become and (r r) (r r) r (r r) (r r) r (r r/) ( (r r)) r (r r/) ( (r r)) r (8) (z z) (z z) ( (z z)) z ( (z z)) z (9) Everything put together and all the different taken apart: (r r) (r r,z) (r r/) ( (r r)) r (r r) (r r,z) (r r/) ( (r r)) r (r,z z) ( (z z)) z (r,z z) ( (z z)) z ( r (r,z) (r r/) ( (r r)) r r (r r/) ( (r r)) r ) ( (z z)) z ( (z z)) z = j (10) 4

5 3 SOLUTION 1 Boundary conditions This lengthy equation (10) is solved for (r,z) 1 Boundary conditions In theory the magnetic field lines (or actually field planes) extent into infinity The simulated volume is finite 3 Solution 31 Iterative solution - and a j-matrix are given The -matrix is initialized to zero The matrix is calculated iteratively For every step k = 1,,3, Calculate k (r,z) from equation (10) using the values of the surrounding cells from the previous step ( k 1 (r r,z), k 1 (r r,z), k 1 (r,z z), k 1 (r,z z)) Calculate the error in cell (r,z): err = k (r,z) k 1 (r,z) Correct the error in cell (r,z): k (r,z) = k 1 (r,z) err V Repeat V is a twiddel-factor: V < 0 instable V 0 exact solution V 1 fast solution 1 < V < oscillates but might become stable eventually V > instable 311 Hint If one looks closely at the error calculation and correction one can see a discrete 1st order lowpass filter (PT-1) here Transfer behavior from x k to y k : Transfer-function: y k = y k 1 V (x k y k 1 ) Y (z) = Y (z) z 1 V (X(z) Y (z) z 1 Y (z) X(z) = V 1 (V 1) z 1 G(z) = V z z (V 1) The filter has a zero at 0 ı0 and a pole at (1 V ) ı0 For V < 0 or V > the pole leaves the unit circle and the filter becomes instable 5

6 3 FEM Solution 4 MTLB/OCTVE CODE 3 FEM Solution The equation 1 can be rewritten in the form C = j with a vector of the vectorpotentials in very cell, j a vector of the current density in every cell and C a matrix of constants With a simple inversion of C we get the solution for : = C 1 j 4 Matlab/Octave code 6