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