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2D circular convolution (part 2)

Posted on 2004-11-07
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Last Modified: 2008-01-09
Two 2D sequences, one is 3 x 4 points & the other is 4 x 3 points in extent, are circularly convolved using (6 x 6)-point 2-D DFTs (Discrete Fourier Transforms). Which samples of the (6 x 6)-output array are identical to the samples of the linear convolution of the two input arrays & which are different??
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Question by:mte01
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    by:GwynforWeb
    I do not entirely follow the question.  Given A=3 by 4  and B=4 by 3   I can see that the convolution is  A**B = 6 by 6  but I do not understand  this statement
     
          "are circularly convolved using (6 x 6)-point 2-D DFTs (Discrete Fourier Transforms)"
     
    do you mean convolving A and B but taking thier DFT multiplying and then doing an inverse DFT ?   If you you do then the outputs are the same as A**B for all inputs A,B  because of the convolution theorem.  
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    by:GwynforWeb
    ......see http://mathworld.wolfram.com/ConvolutionTheorem.html  at bottom of page eqn 7, it applies in discrete domain as well.

    ie         f**g = F^(-1)( F(f)F(g) )  
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    by:mte01
    >> are circularly convolved using (6 x 6)-point 2-D DFTs

    I mean by that zeros are padded to each input (A & B) before making the circular convolution (i.e. before taking the DFTs of A & B and multiplying them, and then taking the IDFT of the product)

    >> then the outputs are the same as A**B for all inputs A,B  because of the convolution theorem

    Yes you are right (I checked it out), for any inputs A & B of the specifications above, the circular convolution & the linear convolution are the same (if the DFT is applied on a 6x6 basis - after padding zeros). However, 2D circular convolution (part 1) wouldn't have the same answer........
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    ai, cs admin
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    by:mte01
    No sorry....this time it's different....please do read the question, these are two completely different questions!!!!!
    Each have a separate answer, and could have been answered by a different expert......
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