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frameLPCC.m
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frameLPCC.m
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% Copyright (c) 2012, Jianxia Xue, jxue@cs.olemiss.edu
% All rights reserved.
%
% Redistribution and use in source, with or without
% modification, are permitted provided that the following conditions are
% met:
%
% * Redistributions of source code must retain the above copyright
% notice, this list of conditions and the following disclaimer.
% * Redistributions in binary form must reproduce the above copyright
% notice, this list of conditions and the following disclaimer in
% the documentation and/or other materials provided with the distribution
%
% THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
% AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
% IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
% ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT OWNER OR CONTRIBUTORS BE
% LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR
% CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF
% SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS
% INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN
% CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
% ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE
% POSSIBILITY OF SUCH DAMAGE.
% function [a, h] = frameLPCC(frame, p, N)
% compute the LPC coefficients and the corresponding
% frequency spectrum with a fixed
% @param[in] frame - the given frame in time
% @param[in] N - the number of LPCC coefficents
% @param[in] p - the number of LPC coeffecients,
% default value is 14
% @param[out] result - the resulting LPCC coeffecients
% @param[out] a - the resulting LPC coeffecients
% @param[out] h - the resulting LPC frequency response in dB
%
% @author Jianxia Xue
% @version 0.20120229
function [result, a, h] = frameLPCC(frame, N, p)
if (nargin<3)
p = 14;
end
[a, h] = frameLPC(frame, p);
a = a(:);
G = sum(frame.*frame);
result = zeros(N,1);
result(1) = G;
for i=2:N
idx = max(1,i-p) : (i-1);
f = frame(idx) .* idx(:)/i;
aidx = idx - idx(1) + 1;
result(i) = sum(f.*a(flipud(aidx)));
if (i<p)
result(i) = result(i)+a(i+1);
end
end
end