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'
%
%
% for 3 speech corpi, this function converts their SNRenv to dprime values
% and to percent correct
%
% 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.org/amt-1.5.0/doc/data/data_green1964.php
% #Author: Søren Jørgensen (2010)
% This file is licensed unter the GNU General Public License (GPL) either
% version 3 of the license, or any later version as published by the Free Software
% Foundation. Details of the GPLv3 can be found in the AMT directory "licences" and
% at <https://www.gnu.org/licenses/gpl-3.0.html>.
% You can redistribute this file and/or modify it under the terms of the GPLv3.
% This file is distributed without any warranty; without even the implied warranty
% of merchantability or fitness for a particular purpose.
if nargin < 2
amt_disp('Specify the spech material: CLUE, DANTALE or DANTALEII');
end
% ---------- Determine the model parameters based on the speech material used
switch material
case 'CLUE'XXX Description is missing
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;