switch fft, ifft

blochsolver
jan 6 years ago
parent d09eff990f
commit 000cfabd78

@ -173,7 +173,7 @@ def maxwell_operator(k0: numpy.ndarray,
m * hin_n) * k_mag
# divide by epsilon
e_xyz = ifftn(fftn(d_xyz, axes=range(3)) / epsilon, axes=range(3))
e_xyz = fftn(ifftn(d_xyz, axes=range(3)) / epsilon, axes=range(3))
# cross product and transform into mn basis
b_m = numpy.sum(e_xyz * n, axis=3)[:, :, :, None] * -k_mag
@ -187,7 +187,7 @@ def maxwell_operator(k0: numpy.ndarray,
n * b_n[:, :, :, None])
# divide by mu
h_xyz = ifftn(fftn(b_xyz, axes=range(3)) / mu, axes=range(3))
h_xyz = fftn(ifftn(b_xyz, axes=range(3)) / mu, axes=range(3))
# transform back to mn
h_m = numpy.sum(h_xyz * m, axis=3)
@ -227,7 +227,7 @@ def hmn_2_exyz(k0: numpy.ndarray,
m * hin_n) * k_mag
# divide by epsilon
return [ei for ei in numpy.rollaxis(fftn(d_xyz, axes=range(3)) / epsilon, 3)]
return [ei for ei in numpy.rollaxis(ifftn(d_xyz, axes=range(3)) / epsilon, 3)]
return operator
@ -258,7 +258,7 @@ def hmn_2_hxyz(k0: numpy.ndarray,
hin_m, hin_n = [hi.reshape(shape) for hi in numpy.split(h, 2)]
h_xyz = (m * hin_m +
n * hin_n)
return [fftn(hi) for hi in numpy.rollaxis(h_xyz, 3)]
return [ifftn(hi) for hi in numpy.rollaxis(h_xyz, 3)]
return operator
@ -312,7 +312,7 @@ def inverse_maxwell_operator_approx(k0: numpy.ndarray,
n * hin_n[:, :, :, None])
# multiply by mu
b_xyz = ifftn(fftn(h_xyz, axes=range(3)) * mu, axes=range(3))
b_xyz = fftn(ifftn(h_xyz, axes=range(3)) * mu, axes=range(3))
# transform back to mn
b_m = numpy.sum(b_xyz * m, axis=3)
@ -323,7 +323,7 @@ def inverse_maxwell_operator_approx(k0: numpy.ndarray,
m * b_n) / k_mag
# multiply by epsilon
d_xyz = ifftn(fftn(e_xyz, axes=range(3)) * epsilon, axes=range(3))
d_xyz = fftn(ifftn(e_xyz, axes=range(3)) * epsilon, axes=range(3))
# cross product and transform into mn basis crossinv_t2c
h_m = numpy.sum(e_xyz * n, axis=3)[:, :, :, None] / +k_mag

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