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How to build a matrix with "sum" function in its elements?

2 vues (au cours des 30 derniers jours)
Mohamed Abdullah
Mohamed Abdullah le 9 Avr 2020
Hello everyone, I am building an mpc controller in matlab rather than using the built in 'mpc command'. I am facing some difficulties in building the mpc's matrices I thought about using for loop.my question is how to use for loop to generate these matrices? ...thanks in advance
  2 commentaires
Walter Roberson
Walter Roberson le 9 Avr 2020
Is it imposed on you that you must use a for loop, rather than a vectorized solution like I show?
Mohamed Abdullah
Mohamed Abdullah le 10 Avr 2020
yes

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Walter Roberson
Walter Roberson le 10 Avr 2020
Modifié(e) : Walter Roberson le 10 Avr 2020
Outline, not checked in detail against Su,b matrix requirements
CAn = Cb;
TCAn = 0;
TCAB = 0;
for K = 1 : p
%Sub
CAB = CAn * Bu;
TCAB = TCAB + CAB;
%now store TCAB into appropriate locations in Sub
insert storage code here
%Sxb
CAn = CAn * A;
TCAn = TCan + CAn;
%now store TCAN into appropriate locations in Sxb
end

Plus de réponses (1)

Walter Roberson
Walter Roberson le 9 Avr 2020
sxb = C(b) .* cumsum(A.^(1:p)) .';
You might then want to
Sub = flipud(hankel(flipud(sxb)))
  3 commentaires
Walter Roberson
Walter Roberson le 9 Avr 2020
If Cb is a matrix, then Sx,b would not be p x 1 as required by the diagram -- not even if A is also a matrix.
If A is a matrix, then for A^2 to exist then A would have to be a square matrix, same number of rows and columns. Call that m x m . Then if Cb is q x r, in order for Cb * A^2 to be scalar, either Cb and A are scalar, or else r = m and q x m * m x m -> q x m would have to be scalar, which could only happen if q = m = 1 so both scalar.
The logic is a bit different if A^n is intended ot indicate A.^n (each element individually raised to the power.) In such a case, if A is m x n then A.^2 would be m x n, and Cb * A^2 could be scalar if Cb is 1 x m and A is m x 1. Is that what is happening?
Mohamed Abdullah
Mohamed Abdullah le 10 Avr 2020
Modifié(e) : Mohamed Abdullah le 10 Avr 2020
I think the only way to make the dimension of 'Sxb' as (PX1) is to deal with the whole block 'Sxb' as a cellarray block containing matrices.for example, let's say p=3, A matrix has size of (mXm) with m=3 and Cb matrix has size of (qXm) with q= 2. if we dealt with 'Sxb' as a Matrix, the final size of it will be (qXp,m):
Sxb=[C*A;
C*A^2+C*A;
C*A^3+C*A^2+C*A];
I have succeeded to build 'Sxb' with for loop as follows:
Sxb=cell(3,1);
>> Sxb{1,1}=zeros(2,3);
>> for p=1:1:3
Sxb{p,1}= C*A^p+Sxb{1,1};
Sxb{1,1}=Sxb{p,1};
end
but I still need some help to figure out how to build the other two blocks 'Sub' and 'Sdb'.

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