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Higher-order generalized cosine windows define a useful parametric family of windows functions, embedding Hamming, Hann, Blackman, Nutall, etc.: $$w(n) = \sum_{k = 0}^{K} a_k\; \cos\left( \frac{2 \pi k n}{N} \right)$$ Are there non-separable 2D similar forms, i.e. not a tensor product of the 1D form, usable on the standard rectangular lattice (excluding designs such as in Multidimensional windows over arbitrary lattices and their application to FIR filter design)?

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    $\begingroup$ Why do you need them to be "non-separable"? What are the properties of the 1D generalized cos windows that you're after? $\endgroup$ – Marcus Müller Sep 18 '15 at 13:23
  • $\begingroup$ I first wanted to exclude the mere tensor product of 1D windows. Plus, non-separable windows, used for tappering or modulation in filter banks, can be more appropriate in dealing with directions in images. I would like to have both clean sample tappering (zeroing) on the border of the image, and an easy discrete Fourier implementation. $\endgroup$ – Laurent Duval Sep 19 '15 at 14:09
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The form:

$$w(x, y) = \sum_{j = -K}^{K}\sum_{k = -K}^{K} a_{|j|,|k|}\; \cos(\pi jx + \pi ky),$$

where:

$$a_{j,k} = a_{k,j},\\ \sum_{j = -K}^{K}\sum_{k = -K}^{K} a_{|j|,|k|} = 1,\\ x \in -1..1,\\ y \in -1..1$$

yields window functions that are 2D similar. Window functions of that form are not generally separable, but some of them are, for example:

$$w(x, y) = \frac{\cos(\pi x - \pi y)}{8} + \frac{\cos(\pi x + \pi y)}{8} + \frac{\cos(\pi x)}{4} + \frac{\cos(\pi y)}{4} + \frac{1}{4}\\ = \left(\frac{\cos\left(\pi x\right)}{2} + \frac{1}{2}\right)\left(\frac{\cos\left(\pi y\right)}{2} + \frac{1}{2}\right)$$

Separable window function

For this window function, if the balance between the zero-frequency term and the higher terms is altered (as in going from Hann to Hamming), separability breaks.

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    $\begingroup$ In my earlier answer I argued that it is difficult to make all four boundaries go to zero value, but when I tried a numerical optimization of the window function it turned out that the boundaries can go to exactly zero with only a few cosine terms. So I deleted that false answer. $\endgroup$ – Olli Niemitalo Sep 19 '15 at 19:50

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