Abstract
We present a stable, recursive algorithm for the Gabor (1946) filter that achieves-to within a multiplicative constant-the fastest possible implementation. For a signal consisting of N samples, our implementation requires O(N) multiply-and-add (MADD) operations, that is, the number of computations per input sample is constant. Further, the complexity is independent of the values of /spl sigma/, and /spl omega/ in the Gabor kernel, and the coefficients of the recursive equation have a simple, closed-form solution given /spl sigma/ and /spl omega/. Our implementation admits not only a "forward" Gabor filter but an inverse filter that is also O(N) complexity.
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