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Generate HDL code from SVM function in Matlab.

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Mohamed Egila
Mohamed Egila 2017 年 10 月 18 日
コメント済み: Mohamed Egila 2017 年 10 月 24 日
I'm having Matlab 9.0.0.341360 (R2016a), I need to generate HDL code from SVM function in Matlab. Please help. Thanks.

回答 (2 件)

Tim McBrayer
Tim McBrayer 2017 年 10 月 18 日
編集済み: Tim McBrayer 2017 年 10 月 18 日
The R2016a list of built-in MATLAB functions supported for HDL code generation can be found at https://www.mathworks.com/help/releases/R2016a/hdlcoder/ug/functions-supported-for-hdl-code-generation-alphabetical-list.html .
  3 件のコメント
Tim McBrayer
Tim McBrayer 2017 年 10 月 19 日
I don't know what SVM is or does, or how complex it is. There is always the option to create your own MATLAB implementation, using supported functions. Such user-written code should be valid for HDL code generation.
Mohamed Egila
Mohamed Egila 2017 年 10 月 19 日
Hi Tim, SVM stands for "Support Vector Machine", it's a machine learning algorithm.

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Brian Ogilvie
Brian Ogilvie 2017 年 10 月 23 日
Hi Mohamed,
The SVM function or ClassificationSVM object are not directly supported for HDL code generation but you can build your own version fairly easily. We had some internal projects that an intern worked on that involved an SVM for a size 36 feature vector that classified A-Z and 0-9. I can share the code that was created as a starting point for you. You will need to supply your trained Beta and Bias values plus decide on the data types that you need for your algorithm. This can generate a lot of logic and we were not able to fit much more than a detector and classifier for one digit in the smaller FPGA we were using at the time. Obviously, this is not production-ready code and just represents a starting point.
function y = fcn(u)
%#codegen
Beta = zeros(36,1);
Bias = zeros(36,1);
lin_u = zeros(36,1);
% Real Beta and Bias values from training set here
for ii=1:1:36
lin_u(ii) = Beta(ii).*u(ii) + Bias(ii);
end
f_neg = -lin_u;
f = 1-lin_u;
f(f<0) = 0;
f_neg = 1-f_neg;
f_neg(f_neg<0) = 0;
%%Function to perform easy sum
k=0;cnt1=1;
cn1=zeros(36,36-1);
for i36=36-1:-1:1
k=k+1;
for j=1:1:i36
cn1(k,j)=f(1,cnt1);
cnt1=cnt1+1;
end
end
cnt2=1;
for a36=36-1:-1:1
for b36=1:1:a36
cn1(36-a36+b36,a36)=f_neg(1,cnt2);
cnt2=cnt2+1;
end
end
y_tmp=zeros(36,1);
for i=1:1:36
y_tmp(i)=sum(cn1(i,:));
end
y = y_tmp(1); % the value of interest
  1 件のコメント
Mohamed Egila
Mohamed Egila 2017 年 10 月 24 日
Thanks Brian for sharing. I will have a look on it.

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