findbandgap.m

通过计算晶体结构来确定其光谱分布

%function [dth,wh0,dte,we0,dthe,whe0]=findbandgap(OH,OE)
%Find a large bandgap of OH and OE,respectively.
%dth=the bandgap of OH
%dte=the bandgap of OE
%wh0=the midfrequency of the OH's bandgap
%we0=the midfrequency of the OE's bandgap
%----------------------------OH-----------------------------------------
ch=(size(OH));
for n=1:(ch(2)-1)
    dtm=min(OH(:,n+1))-max(OH(:,n));
    if dtm<10e-5
        dtm=0;
    end
    dt1(n)=dtm;
end
dth=max(dt1);            %dth=the bandgap of OH
s=find(dt1==dth);
ohgapmin=max(OH(:,s));
ohgapmax=min(OH(:,s+1));
wh0=ohgapmin+(dth/2);    %midfrequency of the OH's bandgap
%----------------------------OE-----------------------------------------
ce=(size(OE));
for n=1:(ce(2)-1)
    dtme=OE(:,n+1)-OE(:,n);
    dt1e(n)=min(dtme);
end
dte=max(dt1e);             %dte=the bandgap of OE
s1=find(dt1e==dte);
oegapmin=max(OE(:,s1));
oegapmax=min(OE(:,s1+1));
we0=oegapmin+(dte/2);        %midfrequency of the OE's bandgap
%----------------------------OHE-----------------------------------------
OHE=[OH,OE];                 %to find the co-bandgap of OH and OE
che=ch(2)+ce(2);
for m=1:ch(1)
    OHE(m,:)=sort(OHE(m,:));
end

clear dtmhe
dtmhe=0;
dtheee=zeros(1,che-1);
for n=1:(che-1)
    dthee=min(OHE(:,n+1))-max(OHE(:,n));
    if dthee<10e-5
        dthee=0;
    end
    dtheee(n)=dthee;
end
dthe=max(dtheee);            %dthe=the bandgap of OHE
s2=find(dtheee==dthe);
ohegapmin=max(OHE(:,s2));    %the bottom of the gap
ohegapmax=min(OHE(:,s2+1));   %the top of the gap
whe0=ohegapmin+(dthe/2);    %midfrequency of the OHE's bandgap

disp('ohegap end')