clc
clear
close?all

n1=13;??????%取樣本?

sample=high_guiyi（1:n1:）;
...................................
%計(jì)算各種氣體的比值
%sanratio1=C2H2/C2H4;?sanratio2=CH4/H2;?sanratio3=C2H4/C2H6

sanratio1=sample（:5）./sample（:4）;
sanratio2=sample（:2）./sample（:1）;
sanratio3=sample（:4）./sample（:3）;
....................................
%計(jì)算比值的范圍
%可以調(diào)整比值的范圍?如：0.1與3的值
for?kk1=1:n1
if??sanratio1（kk1）<0.1
????sanratio1（kk1）=0;
end
if??sanratio1（kk1）<3&sanratio1（kk1）>0.1
?????sanratio1（kk1）=1;
end
if??sanratio1（kk1）>3
????sanratio1（kk1）=2;
end
??kk1=kk1+1;
end
..................................................
for?kk1=1:n1
if??sanratio2（kk1）<0.1
????sanratio2（kk1）=1;
end
if??sanratio2（kk1）<1&sanratio2（kk1）>0.1
?????sanratio2（kk1）=0;
end
if??sanratio2（kk1）>1
????sanratio2（kk1）=2;
end
?kk1=kk1+1;
end
............................................
for?kk1=1:n1
if??sanratio3（kk1）<1
????sanratio3（kk1）=0;
end
if??sanratio3（kk1）<3&sanratio3（kk1）>1
?????sanratio3（kk1）=1;
???
end
if??sanratio3（kk1）>3
????sanratio3（kk1）=2;
????
end
kk1=kk1+1;
end
.............................
sanratio=[sanratio1?sanratio2?sanratio3];?
%得出編碼結(jié)果

result=zeros（n11）;
?for?kk2=1:n1
if??sanratio（kk21）==0&sanratio（kk22）==0&sanratio（kk23）==0
????result（kk21）=10;
end
if??sanratio（kk21）==0&sanratio（kk22）==1&sanratio（kk23）==0
????result（kk21）=1;
end
if??sanratio（kk21）==1&sanratio（kk22）==1&sanratio（kk23）==0
????result（kk21）=2;
end
if??sanratio（kk21）==1&sanratio（kk22）==0&sanratio（kk23）==1
????result（kk21）=3;
end
if??sanratio（kk21）==1&sanratio（kk22）==0&sanratio（kk23）==2
????result（kk21）=3.1;
end
if??sanratio（kk21）==2&sanratio（kk22）==0&sanratio（kk23）==1
????result（kk21）=3.2;
end
if??sanratio（kk21）==2&sanratio（kk22）==0&sanratio（kk23）==2
????result（kk21）=3.3;
end
if??sanratio（kk21）==1&sanratio（kk22）==0&sanratio（kk23）==2
????result（kk21）=4;
end
if??sanratio（kk21）==0&sanratio（kk22）==0&sanratio（kk23）==1
????result（kk21）=5;
end
if??sanratio（kk21）==0&sanratio（kk22）==2&sanratio（kk23）==0
????result（kk21）=6;
end
if??sanratio（kk21）==0&sanratio（kk22）==2&sanratio（kk23）==1
????result（kk21）=7;
end
if??sanratio（kk21）==0&sanratio（kk22）==2&sanratio（kk23）==2
????result（kk21）=8;
end
end
..............................
cout1=0;
cout21=0;?cout22=0;?cout23=0;cout24=0;cout25=0;cout26=0;
cout3=0;
cout41=0;?cout42=0;
cout51=0;cout52=0;
%通過編碼設(shè)定故障類型

for?kk3=1:n1?
%?if?result（kk3）==0
%??disp（‘缺碼?‘）??
%?end????

if?result（kk3）==10
?disp（‘無故障?‘）;??
cout1=cout1+1;
end

if?result（kk3）==1
%disp（‘低放‘）;
cout21=cout21+1;
end
if?result（kk3）==3
%disp（‘低放‘）;
cout22=cout22+1;
end
if?result（kk3）==3.1
%disp（‘低放‘）;
cout23=cout23+1;
end
if?result（kk3）==3.2
%disp（‘低放‘）;
cout24=cout24+1;
end
if?result（kk3）==3.3
%disp（‘低放‘）;
cout25=cout25+1;
end

if?result（kk3）==4
%disp（‘低放‘）;
cout26=cout26+1;
end

if?result（kk3）==2
%disp（‘高放‘）;
cout3=cout3+1;
end

if?result（kk3）==5

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----

?????文件???????3748??2010-03-06?14:52??threerateIEC1.m

-----------?---------??----------?-----??----

?????????????????3748????????????????????1