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Let say if we wish to filter an input sequence x[n1, n2, n3] of NxNxN points using an Linear Shift Invariance system with impulse response h[n1, n2, n3], where the filter is a separable sequence, h[n1, n2, n3] = a[n1]b[n2]c[n3].

How can we develop a computationally efficient way to obtain the output y[n1, n2, n3]?

My thought: since we can represent the Fourier Transform of the Filter as: $$ H(w_1, w_2, w_3) = \sum \sum \sum h[n_1,n_2,n_3]*e^{-jw_1n_1} *e^{-jw_2n_2}*e^{-jw_3n_3} $$ $$ H(w_1, w_2, w_3) = A(w_1)B(w_2)C(w_3)$$, where

$$ A(w_1) = \sum h[n_1,n_2,n_3]*e^{-jw_1n_1} $$ $$ B(w_2) = \sum h[n_1,n_2,n_3]*e^{-jw_2n_2} $$ $$ C(w_3) = \sum h[n_1,n_2,n_3]*e^{-jw_3n_3}$$

then,

$$ y(n_1, n_2, n_3) = x(n_1, n_2, n_3)*h(n_1, n_2, n_3)$$ $$ Y(w_1, w_2, w_3) = X(w_1, w_2, w_3)H(w_1, w_2, w_3)$$ $$ Y(w_1, w_2, w_3) = X(w_1, w_2, w_3)A(w_1)B(w_2)C(w_3)$$

I got stuck in this point and I am not too sure how to proceed through the process. I don't quite see how to separate the sequence and how to multiply through the vectors to get the convolution sum of $Y(w_1, w_2, w_3)$

can anyone please guide me to the right direction? Thanks!

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x * h = IFT[FT[x * h] and FT[x * h] is Ft[x] FT[h]. Since h is separable, it might be computationally efficient to expand its transform as a convolution Ft[x]*FT[b]*FT[c].

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