How to estimate an ODF from single orientation measurements.

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ODF Estimation |

Automatic halfwidth selection |

Effect of halfwidth selection |

These EBSD datasets consist of two phases, Iron, and Magnesium. The ODF of the Iron phase is computed by the command

`odf = calcODF(ebsd('fo').orientations)`

odf = ODF crystal symmetry : Forsterite (mmm) specimen symmetry: 1 Harmonic portion: degree: 28 weight: 1

The function calcODF implements the ODF estimation from EBSD data in MTEX. The underlying statistical method is called kernel density estimation, which can be seen as a generalized histogram. To be more precise, let's be a radially symmetric, unimodal model ODF. Then the kernel density estimator for the individual orientation data is defined as

The choice of the model ODF and in particular its halfwidth has a great impact in the resulting ODF. If no halfwidth is specified the default halfwidth of 10 degrees is selected.

MTEX includes an automatic halfwidth selection algorithm which is called by the command calcKernel. To work properly, this algorithm needs spatially independent EBSD data as in the case of this dataset of very rough EBSD measurements (only one measurement per grain).

% try to compute an optimal kernel psi = calcKernel(ebsd('fo').orientations)

estimate optimal kernel halfwidth: 100% psi = deLaValeePoussinKernel bandwidth: 130 halfwidth: 2.3°

In the above example, the EBSD measurements are spatial dependent and the resulting halfwidth is too small. To avoid this problem we have to perform grain reconstruction first and then estimate the halfwidth from the grains.

% grains reconstruction grains = calcGrains(ebsd); % correct for to small grains grains = grains(grains.grainSize>5); % compute optimal halfwidth from the meanorientations of grains psi = calcKernel(grains('fo').meanOrientation) % compute the ODF with the kernel psi odf = calcODF(ebsd('fo').orientations,'kernel',psi)

estimate optimal kernel halfwidth: 100% psi = deLaValeePoussinKernel bandwidth: 48 halfwidth: 6° odf = ODF crystal symmetry : Forsterite (mmm) specimen symmetry: 1 Radially symmetric portion: kernel: de la Vallee Poussin, halfwidth 6° center: 2922 orientations, resolution: 3° weight: 1

Once an ODF is estimated all the functionality MTEX offers for ODF analysis and ODF visualization is available.

h = [Miller(1,0,0,odf.CS),Miller(1,1,0,odf.CS),Miller(1,1,1,odf.CS)]; plotPDF(odf,h,'antipodal','silent')

As mentioned above a proper halfwidth selection is crucial for ODF estimation. The following simple numerical experiment illustrates the dependency between the kernel halfwidth and the estimated error.

Let's start with a model ODF and simulate some individual orientation data.

```
modelODF = fibreODF(Miller(1,1,1,crystalSymmetry('cubic')),xvector);
ori = calcOrientations(modelODF,10000)
```

ori = orientation size: 10000 x 1 crystal symmetry : m-3m specimen symmetry: 1

Next we define a list of kernel halfwidth ,

hw = [1*degree, 2*degree, 4*degree, 8*degree, 16*degree, 32*degree];

estimate for each halfwidth an ODF and compare it to the original ODF.

e = zeros(size(hw)); for i = 1:length(hw) odf = calcODF(ori,'halfwidth',hw(i),'silent'); e(i) = calcError(modelODF, odf); end

progress: 100%

After visualizing the estimation error we observe that its value is large either if we choose a very small or a very large halfwidth. In this specific example, the optimal halfwidth seems to be about 4 degrees.

close all plot(hw/degree,e) xlabel('halfwidth in degree') ylabel('esimation error')

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