Combine convolution filters (bandpass) into a single kernel

조회 수: 3 (최근 30일)
Doctor G
Doctor G 2015년 1월 13일
댓글: Doctor G 2015년 1월 13일
Is there a way to take the low pass and high pass filters in the following code and combine them into a single kernel and apply one conv2() function?
*note*: length(lfilter) = 21, length(hfilter) = 81.
what we are basically doing in the last step is saying to remove the large objects from the image (after already removing the very small objects with a Gaussian blur).
properties (Constant)
minStar = 2; % min star radius
maxStar = 8; % max star radius
threshold = 12;
end
----------
function filter2(this)
normalize = @(x) x/sum(x);
lfilter = normalize(exp(-((-ceil(5*this.minStar):ceil(5*this.minStar))/(2*this.minStar)).^2));
hfilter = normalize(exp(-((-ceil(5*this.maxStar):ceil(5*this.maxStar))/(2*this.maxStar)).^2));
this.low = conv2(lfilter',lfilter,this.raw,'same');
this.high = conv2(hfilter',hfilter,this.raw,'same');
this.filtered = this.low - this.high;
this.foreground = this.filtered > this.threshold;
end

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Alfonso Nieto-Castanon
Alfonso Nieto-Castanon 2015년 1월 13일
Convolution is a linear operation so yes, you can combine the two filtering operations into one. Just make the filters the same size and add/subtract them. For example:
minmax=max(this.minStar,this.maxStar);
lfilter = normalize(exp(-((-ceil(5*minmax):ceil(5*minmax))/(2*this.minStar)).^2));
hfilter = normalize(exp(-((-ceil(5*minmax):ceil(5*minmax))/(2*this.maxStar)).^2));
lhfilter = lfilter - hfilter;
this.filtered = conv2(lhfilter',lhfilter,this.raw,'same');
  댓글 수: 2
Doctor G
Doctor G 2015년 1월 13일
편집: Doctor G 2015년 1월 13일
Thanks for the info about them being linear. I am new to matlab, and as your probably saw there is a narrow peak in the high-pass filter, and and a broad exp for the low. So by default they are different lengths. How do you shift an array sideways? (zero filled of course).
Doctor G
Doctor G 2015년 1월 13일
I found my answer padarray(A, [0 paddingAmount]) which does padding from both sides. Thank you for your help.

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