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Are you seeking a solution (you seem to have found one) or understanding as well? Do you know complex numbers? Your technique is based on the properties of the complex unit circle. Suppose you have a point on the complex unit circle: $$u = a + bi,\; a^2+b^2=1$$ If you square that point: $$u^2 = ( a^2 - b^2 ) + 2abi$$ The result will also lay on the ...

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With the help of a colleague we come up to this answer: import math import matplotlib.pyplot as plt import numpy window_size = 31 hann = numpy.zeros(window_size) expected = numpy.hanning(window_size) window_wT = math.pi / (window_size - 1) window_k1 = math.cos(window_wT) window_k2 = math.sin(window_wT) window_k3 = 0 hann[1] = window_k2 * window_k2 for i ...

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i think this generates a sinusoid: $$y[n] = 2 \cos(\omega_0) y[n-1] - y[n-2]$$ if you initialize $$y[-2] = \cos(2 \omega_0)$$ $$y[-1] = \cos(\omega_0)$$ then the result is $$y[n] = \cos(\omega_0 n) \qquad n \ge 0$$ the period of this cosine function is $N=\frac{2\pi}{\omega_0}$ and you want $N-1$ to be the non-zero width of the window. finally you ...

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I have just found this paper that calculates the $\lambda_i$ parameters on an L-surface: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.135.5969&rep=rep1&type=pdf

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