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    發(fā)布日期: 2022-09-06
  • 語言: Matlab
  • 標(biāo)簽: PCA??故障診斷??matlab??

資源簡介

matlab 實(shí)現(xiàn)PCA故障診斷功能,帶測試測試數(shù)據(jù),可直接運(yùn)行

資源截圖

代碼片段和文件信息

%建立模型:
Xtrain?=?CC(1:1000:);
Xtrain_Store?=?Xtrain;
%測試數(shù)據(jù)
Xtest?=?CC(1001:1500:);
Xtest_Store?=?Xtest;
size_=?size(Xtest);
test_Count?=?size_(1);
%總輸入變量個(gè)數(shù)
in_count?=?10;
%標(biāo)準(zhǔn)化處理:
X_mean?=?mean(Xtrain);??%按列求Xtrain平均值???????????????????????????
X_std?=?std(Xtrain);????%求標(biāo)準(zhǔn)差??????????????????????
[X_rowX_col]?=?size(Xtrain);?%求Xtrain行、列數(shù)???????????????
%?for?i?=?1:X_col
%Xtrain(:i)=(Xtrain(:i)?-?X_mean(i)./X_std(i));
%Xtest(:i)?=?(Xtest(:i)?-?X_mean(i)./X_std(i));
%?end????????????????????????????????????????????????
Xtrain=(Xtrain-repmat(X_meanX_row1))./repmat(X_stdX_row1);

%求協(xié)方差矩陣
sigmaXtrain?=?cov(Xtrain);
%對協(xié)方差矩陣進(jìn)行特征分解,lamda為特征值構(gòu)成的對角陣,T的列為單位特征向量,且與lamda中的特征值一一對應(yīng):
[Tlamda]?=?eig(sigmaXtrain);????????????????????????????
%?disp(‘特征根(由小到大)‘);
%?disp(lamda);
%?disp(‘特征向量:‘);
%?disp(T);????????????????????????????????????????????

%取對角元素(結(jié)果為一列向量),即lamda值,并上下反轉(zhuǎn)使其從大到小排列,主元個(gè)數(shù)初值為1,若累計(jì)貢獻(xiàn)率小于90%則增加主元個(gè)數(shù)
D?=?flipud(diag(lamda));????????????????????????????
num_pc?=?1;?????????????????????????????????????????
while?sum(D(1:num_pc))/sum(D)?num_pc?=?num_pc?+1;
end?????????????????????????????????????????????????

%取與lamda相對應(yīng)的特征向量
P?=?T(:X_col-num_pc+1:X_col);????????????????????????????

%求置信度為99%、95%時(shí)的T2統(tǒng)計(jì)控制限???????????????????????
T2UCL1=num_pc*(X_row-1)*(X_row+1)*finv(0.999num_pcX_row?-?num_pc)/(X_row*(X_row?-?num_pc));
T2UCL2=num_pc*(X_row-1)*(X_row+1)*finv(0.95num_pcX_row?-?num_pc)/(X_row*(X_row?-?num_pc));

%置信度為99%的Q統(tǒng)計(jì)控制限
for?i?=?1:3
????theta(i)?=?sum((D(num_pc+1:X_col)).^i);
end
h0?=?1?-?2*theta(1)*theta(3)/(3*theta(2)^2);
ca?=?norminv(0.99901);?%正態(tài)累積分布的反函數(shù)
QUCL?=?theta(1)*(h0*ca*sqrt(2*theta(2))/theta(1)?+?1?+?theta(2)*h0*(h0?-?1)/theta(1)^2)^(1/h0);???????????????????????????

%測試:
%標(biāo)準(zhǔn)化處理
n?=?size(Xtest1);
Xtest=(Xtest-repmat(X_meann1))./repmat(X_stdn1);

%求T2統(tǒng)計(jì)量,Q統(tǒng)計(jì)量
[ry]?=?size(P*P‘);
I?=?eye(ry);

T2?=?zeros(n1);
Q?=?zeros(n1);
for?i?=?1:n
????T2(i)=Xtest(i:)*P*inv(lamda(in_count-num_pc+1:in_countin_count-num_pc+1:in_count))*P‘*Xtest(i:)‘;???????????????????????????????????????????
????Q(i)?=?Xtest(i:)*(I?-?P*P‘)*Xtest(i:)‘;????????????????????????????????????????????????????????????????????????????????????
end

%繪圖
????%figure
????subplot(211);
????plot(1:nT2‘k‘);????????????????????????????????????
????title(‘主元分析統(tǒng)計(jì)量變化圖‘);
????xlabel(‘采樣數(shù)‘);
????ylabel(‘T^2‘);
????hold?on;
????line([0n][T2UCL1T2UCL1]‘Linestyle‘‘--‘‘Color‘‘r‘);
????line([0n][T2UCL2T2UCL2]‘Linestyle‘‘--‘‘Color‘‘g‘);
????
????subplot(212);
????plot(1:nQ‘k‘);
????xlabel(‘采樣數(shù)‘);
????ylabel(‘Q‘);
????hold?on;
????line([0n][QUCLQUCL]‘Linestyle‘‘--‘‘Color‘‘r‘);
????
%貢獻(xiàn)圖
%1.確定造成失控狀態(tài)的得分
S?=?Xtest(test_Count:)*P(:1:num_pc);
r?=?[?];
for?i?=?1:num_pc
????if?S(i)^2/lamda(i)?>?T2UCL1/num_pc
????????r?=?cat(2ri);
????end
end

%2.計(jì)算每個(gè)變量相對于上述失控得分的貢獻(xiàn)
cont?=?zeros(length(r)in_count);
for?i?=?length(r)
????for?j?=?1:in_count

?屬性????????????大小?????日期????時(shí)間???名稱
-----------?---------??----------?-----??----

?????文件???????3756??2014-04-25?16:32??Test2.m

?????文件?????355986??2014-04-25?16:34??PCAdata.mat

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

???????????????359742????????????????????2


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