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. 2018 Aug 16;32(4):e3945. doi: 10.1002/nbm.3945

Figure 6.

Figure 6

Violation of the fODF unit integral constraint. One the left, a two‐fibre crossing configuration (volume fractions f 1 = 0.5, f 2 = 0.5). On the right, the same two‐fibre crossing configuration with additional CSF partial volume (volume fractions f 1 = 0.5, f 2 = 0.3, f csf = 0.2). By comparing the two fODFs it is easy to verify that the two fODF profiles cannot be constrained to have the same (unit) integral while also preserving the same amplitude for the lobes with f = 0.5