function [Pc_est SNRenv] = data_green1964(SNRenv_lin,material)
%DATA_GREEN1964 Converts the overall SNRenv to percent correct
% Usage: [Pc_est SNRenv ] = data_green1964(SNRenv_lin,material);
%
% Input parameters:
% SNRenv_lin : matrix with the SNRenv values (not in dB),
% one for each SNR and processing combination.
%
% material : Specify the speech material used.
% Current: 'CLUE', 'DANTALE' or 'DANTALE II'
%
% XXX Description is missing
%
% References:
% D. M. Green and T. G. Birdsall. The effect of vocabulary size. In J. A.
% Swets, editor, Signal Detection and Recognition by Human Observers,
% pages 609--619. John Wiley & Sons, 1964.
%
%
% Url: http://amtoolbox.sourceforge.net/amt-0.10.0/doc/data/data_green1964.php
% Copyright (C) 2009-2020 Piotr Majdak and the AMT team.
% This file is part of Auditory Modeling Toolbox (AMT) version 0.10.0
%
% This program is free software: you can redistribute it and/or modify
% it under the terms of the GNU General Public License as published by
% the Free Software Foundation, either version 3 of the License, or
% (at your option) any later version.
%
% This program is distributed in the hope that it will be useful,
% but WITHOUT ANY WARRANTY; without even the implied warranty of
% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
% GNU General Public License for more details.
%
% You should have received a copy of the GNU General Public License
% along with this program. If not, see <http://www.gnu.org/licenses/>.
% AUTHOR: Søren Jørgensen august 2010
if nargin < 2
amt_disp('you have to specify the spech material used: CLUE, DANTALE or DANTALEII')
end
% ---------- Determine the model parameters based on the speech material used
switch material
case 'CLUE'
m = 8000;
sigma_s = .6;
case 'DANTALE'
m = 8000 ;
sigma_s = .9;
case 'DANTALEII'
m = 50;
sigma_s = .9;
end
% -------- The general conversion constants. They are the same for alle materials
k = .82;
q = .5;
%-------- Only used for output:
SNRenv = 10*log10(SNRenv_lin);
% ---------- Converting from SNRenv to d_prime --------------
d_prime = k*(SNRenv_lin).^q; %SNRm_lin; %
%----------- Converting from d_prime to Percent correct, Green and Birdsall (1964)----------
Un = 1*norminv(1-(1/m));
mn = Un + (.577 /Un);% F^(-1)[1/n] Basically gives the value that would be drawn from a normal destribution with probability p = 1/n.
sig_n= 1.28255/Un;
Pc_est = normcdf(d_prime,mn,sqrt(sigma_s.^2+sig_n.^2))*100;