69 lines
2.3 KiB
Python
69 lines
2.3 KiB
Python
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import numpy, scipy, gridlock, fdfd_tools
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from fdfd_tools import bloch
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from numpy.linalg import norm
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import logging
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logging.basicConfig(level=logging.DEBUG)
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logger = logging.getLogger(__name__)
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dx = 40
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x_period = 400
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y_period = z_period = 2000
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g = gridlock.Grid([numpy.arange(-x_period/2, x_period/2, dx),
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numpy.arange(-1000, 1000, dx),
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numpy.arange(-1000, 1000, dx)],
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shifts=numpy.array([[0,0,0]]),
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initial=1.445**2,
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periodic=True)
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g.draw_cuboid([0,0,0], [200e8, 220, 220], eps=3.47**2)
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#x_period = y_period = z_period = 13000
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#g = gridlock.Grid([numpy.arange(3), ]*3,
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# shifts=numpy.array([[0, 0, 0]]),
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# initial=2.0**2,
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# periodic=True)
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g2 = g.copy()
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g2.shifts = numpy.zeros((6,3))
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g2.grids = [numpy.zeros(g.shape) for _ in range(6)]
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epsilon = [g.grids[0],] * 3
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reciprocal_lattice = numpy.diag(1000/numpy.array([x_period, y_period, z_period])) #cols are vectors
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#print('Finding k at 1550nm')
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#k, f = bloch.find_k(frequency=1/1550,
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# tolerance=(1/1550 - 1/1551),
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# direction=[1, 0, 0],
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# G_matrix=reciprocal_lattice,
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# epsilon=epsilon,
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# band=0)
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#
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#print("k={}, f={}, 1/f={}, k/f={}".format(k, f, 1/f, norm(reciprocal_lattice @ k) / f ))
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print('Finding f at [0.25, 0, 0]')
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for k0x in [.25]:
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k0 = numpy.array([k0x, 0, 0])
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kmag = norm(reciprocal_lattice @ k0)
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tolerance = 1e-4 * (1/1550) # df = f * dn_eff / n
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logger.info('tolerance {}'.format(tolerance))
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n, v = bloch.eigsolve(4, k0, G_matrix=reciprocal_lattice, epsilon=epsilon, tolerance=tolerance)
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v2e = bloch.hmn_2_exyz(k0, G_matrix=reciprocal_lattice, epsilon=epsilon)
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v2h = bloch.hmn_2_hxyz(k0, G_matrix=reciprocal_lattice, epsilon=epsilon)
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ki = bloch.generate_kmn(k0, reciprocal_lattice, g.shape)
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z = 0
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e = v2e(v[0])
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for i in range(3):
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g2.grids[i] += numpy.real(e[i])
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g2.grids[i+3] += numpy.imag(e[i])
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f = numpy.sqrt(numpy.real(numpy.abs(n))) # TODO
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print('k0x = {:3g}\n eigval = {}\n f = {}\n'.format(k0x, n, f))
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n_eff = norm(reciprocal_lattice @ k0) / f
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print('kmag/f = n_eff = {} \n wl = {}\n'.format(n_eff, 1/f ))
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