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Read Back Signals in Magnetic Recording - Research Group Fidler

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Numerical Methods<br />

Here A ij are 3× 3-submatrices<br />

describ<strong>in</strong>g the <strong>in</strong>teraction of the j-th tetrahedron of the<br />

conductor model with the i-th node of the magnetic model. Thus the matrix A= ( A ) is a<br />

( 3n) ( 3m)<br />

× -Matrix, where n denotes the number of nodes of the magnetic model and m the<br />

number of tetrahedrons <strong>in</strong> the conductor model. Usually the matrix A is fully populated and<br />

requires a lot of memory. The multiplication with such a matrix is very time consum<strong>in</strong>g. To<br />

speed up the calculation and save memory an adaptive cross-approximation technique can be<br />

used (see chapter 4.3). Unfortunately this technique needs a lot of memory when calculat<strong>in</strong>g<br />

the approximation of the <strong>in</strong>teraction matrix A for large magnetic and conductor models.<br />

4.4.5 Hybrid FEM/BEM for Current Field<br />

To overcome the storage problem we proposed an alternative way to calculate the magnetic<br />

field <strong>in</strong> the magnetic model. Here we use a common mesh for the magnet and the conductor<br />

model. Similar to the stray field calculation (see Section 2.5.2) we can reduce the <strong>in</strong>teraction<br />

matrix to a boundary <strong>in</strong>teraction matrix.<br />

Start<strong>in</strong>g from Equation (2.29)<br />

curl H = j, (4.49)<br />

apply<strong>in</strong>g the curl-operator on both sides, and consider<strong>in</strong>g that div H = 0 lead to a Poisson<br />

equation for each component of H.<br />

Δ H =−curl<br />

j (4.50)<br />

with boundary condition (see Appendix B)<br />

out <strong>in</strong><br />

∂H ∂H<br />

− = n× j<br />

∂n ∂n<br />

∂V<br />

. (4.51)<br />

Now the magnetic field can be calculated quite similar to the stray field, described <strong>in</strong> Section<br />

4.2. We split the magnetic field <strong>in</strong>to two parts<br />

H = H1+ H 2,<br />

(4.52)<br />

where H 1 is the solution of<br />

Δ H =− j (4.53)<br />

<strong>in</strong><br />

1 curl<br />

ij<br />

52

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