# Andrews plot function for multivariate control chart limits UCL and LCL

조회 수: 4 (최근 30일)
Nainsi Gupta 2023년 7월 1일
댓글: Paul 2023년 7월 2일
I am working on Andrews plot for quality control chart. So I need to define UCL and LCL for that. I have attached the images of the paper for the reference and better clarity.
Please tell me how can I develop a vector for A(theta) in the attached image. I also need to use this function for the monitoring the out-of-control points at the later stage.

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### 채택된 답변

Paul 2023년 7월 1일
Brute force approach ...
Use symbolic for illustration
syms theta
p = 5;
A = reshape([sin((1:p)*theta) ; cos((1:p)*theta)],1,[])
A =
A = [1/sqrt(sym(2)) A];
A = A(1:p)
A =
Also, you may be interested in andrewsplot
##### 댓글 수: 2없음 표시없음 숨기기
Nainsi Gupta 2023년 7월 2일
How can I define theta for a limit -pi to pi? And then how to plot it.
Paul 2023년 7월 2일
x = 1:6; % an observation
% non-vectorized approach, but easy(?) to follow
syms theta
p = size(x,2);
A = reshape([sin((1:p)*theta) ; cos((1:p)*theta)],1,[]);
A = [1/sqrt(sym(2))+0*theta A];
A = A(1:p)
A =
A = matlabFunction(A)
A = function_handle with value:
@(theta)[sqrt(2.0)./2.0,sin(theta),cos(theta),sin(theta.*2.0),cos(theta.*2.0),sin(theta.*3.0)]
theta = linspace(-pi,pi,100);
aplot1 = 0*theta;
for ii = 1:numel(aplot1)
aplot1(ii) = sum(x.*A(theta(ii)),2);
end
figure
plot(theta,aplot1)
% non-symbolic, vectorized approach
neven = ceil((p-1)/2);
nodd = floor((p-1)/2);
aplot2 = x(1)/sqrt(2) + sum(x(2:2:end).*sin(theta(:).*(1:neven)),2) + sum(x(3:2:end).*cos(theta(:).*(1:nodd)),2);
figure
plot(theta,aplot2)
norm(aplot1(:)-aplot2(:))
ans = 6.3429e-15

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### 추가 답변 (1개)

KALYAN ACHARJYA 2023년 7월 1일
syms theta
% P variate vector
%Cross check here, just random data, it can be variable
p=10;
data_sin=sin((1:round(p/2)+1)*theta);
data_cos=cos((1:round(p/2))*theta);
data(1:2:2*numel(data_sin))=data_sin;
data(2:2:end)=data_cos;
data=[1/sqrt(2),data]
data =
##### 댓글 수: 2없음 표시없음 숨기기
Nainsi Gupta 2023년 7월 1일
How can I define a limit for theta (-pi<theta<pi)
Nainsi Gupta 2023년 7월 1일
Also there is some issue with the code. if p=3, then data should contain 3 elements. similarly for other values of p

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