Implement following matrix using for loop

I want implement the following matrix and expression:
%Big_lambda
where theta,r,t are 1*3 matrix and a,b,p,q are 1*1 matrix.
My approach:
theta = [ 45 46 48] ;
t= [ 1 2 3 ];
r= [200 210 220];
for i=1:3
dif_x = [(-cos(theta(i))) (r(i).*sin(theta(i))) (2*a.*t(i)+b);...
(-sin(theta(i))) (-r(i).*cos(theta(i))) (2*p.*t(i)+q)]
W = dif_x(i)*Big_lambda(i)*dif_x(i)'
end
Is the approach correct or if it is wrong what should be done!
Thanks in advance.

댓글 수: 7

Bob Thompson
Bob Thompson 2019년 5월 15일
Your approach generally seems fine, although I'm not really sure why you have W indexed. Do you want to multiply individual elements, or the entire matrix?
Is there something wrong with what you have written? Are you getting any error messages? Please explain more, or post the entire error message.
Tipu Sultan
Tipu Sultan 2019년 5월 15일
@madhan ravi This is not the same expression that you thought here I am implementing the follwing:
21.17.png
for which I require the value of 'dif_x'.
Tipu Sultan
Tipu Sultan 2019년 5월 15일
@Bob W is indexed as I want to use the value of W in a for loop and subscript denotes the value for each iteration of i.
Bob Thompson
Bob Thompson 2019년 5월 15일
You have still not indicated what exactly your problem is.
Tipu Sultan
Tipu Sultan 2019년 5월 15일
편집: Tipu Sultan 2019년 5월 15일
I want to implement the following expression:
the expression contains partial derivative and big lambda which are basically the following expression:
where big lamda in the above can be taken as a identity matrix of 3*3.
W having index means it it will be inside a loop,I have implemented as follows:
theta = [ 45 46 48] ;
t= [ 1 2 3 ];
r= [200 210 220];
x = [ r; theta ; t];
Big_lambda = eye(2);
dif_x = zeros(2, 3, 3); % Pre-allocation
for i=1:3
dif_x(:, :, i) = [-cos(theta(i)), r(i).*sin(theta(i)), 2*a.*t(i)+b;...
-sin(theta(i)), -r(i).*cos(theta(i)), 2*p.*t(i)+q];
W = dif_x(:,:,i)*Big_lambda(i)*dif_x(:,:,i)'
end
I want to know I am going in the correct direction or I need to change some thing.
Bob Thompson
Bob Thompson 2019년 5월 15일
편집: Bob Thompson 2019년 5월 15일
As far as I can tell you are going in the correct direction. I am not a SME on your specific problem, but for the code aspect it seems to be functional.
I noticed you adjusted your indexing on dif_x. Don't forget to index W as well if you want to retain the results from each loop, instead of just the final loop.
W(i) = dif_x(:,:,i)*Big_lambda(i)*dif_x(:,:,i)'

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2019년 5월 15일

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2019년 5월 15일

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