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alternate src formulation

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This commit is contained in:
Jan Petykiewicz 2019-08-07 02:27:04 -07:00
parent 07c94617fe
commit aade81c21e
1 changed files with 18 additions and 5 deletions
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23
meanas/fdfd/waveguide_mode.py
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@ -169,9 +169,9 @@ def compute_source(E: field_t,
M = numpy.zeros_like(E, dtype=complex)
src_order = numpy.roll(range(3), -axis)
exp_iphi = numpy.exp(1j * polarity * wavenumber * dxes[1][axis][slices[axis]])
rollby = -1 if polarity > 0 else 0
# exp_iphi = numpy.exp(1j * polarity * wavenumber * dxes[1][axis][slices[axis]])
# rollby = -1 if polarity > 0 else 0
# J[src_order[1]] = +exp_iphi * H[src_order[2]] * polarity / dxes[1][axis][slices[axis]]
# J[src_order[2]] = -exp_iphi * H[src_order[1]] * polarity / dxes[1][axis][slices[axis]]
# M[src_order[1]] = +numpy.roll(E[src_order[2]], rollby, axis=axis) / dxes[0][axis][slices[axis]]
@ -181,14 +181,16 @@ def compute_source(E: field_t,
s2[axis] = slice(slices[axis].start, slices[axis].stop)
s2 = (src_order, *s2)
J[s2] = numpy.roll(functional.curl_h(dxes=dxes)(H), rollby, axis=axis+1)[s2] * polarity * exp_iphi
M[s2] = numpy.roll(functional.curl_e(dxes=dxes)(E), -rollby, axis=axis+1)[s2]
rollby = 1 if polarity > 0 else 0
exp_iphi = numpy.exp(-1j * polarity * wavenumber * dxes[1][axis][slices[axis]])
J[s2] = numpy.roll(functional.curl_h(dxes=dxes)(H.conj()), -rollby, axis=axis+1)[s2] * polarity * exp_iphi
M[s2] = -numpy.roll(functional.curl_e(dxes=dxes)(E.conj()), rollby, axis=axis+1)[s2]
m2j = functional.m2j(omega, dxes, mu)
Jm = m2j(M)
Jtot = J + Jm
return Jtot.conj()
return Jtot
def compute_overlap_e(E: field_t,
@ -371,6 +373,17 @@ def compute_source_e(QE: field_t,
else:
start = slices[aa].start
stop = slices[aa].stop
# if start is not None or stop is not None:
# if start is None:
# start = 1
# stop -= 1
# elif stop is None:
# stop = E.shape[aa + 1] - 1
# start += 1
# else:
# start += 1
# stop -= 1
# slices_reduced[aa] = slice(start, stop)
slices_reduced = (slice(None), *slices_reduced)
# Don't actually need to mask out E here since it needs to be pre-masked (QE)