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Asked by Ali Esmaeilpour
on 27 May 2019

- 6 questions asked
- 0 answers
- 0 accepted answers
- reputation: 0

- 6 questions asked
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- 0 accepted answers
- reputation: 0

Accepted Answer by Geoff Hayes

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3 views (last 30 days)

3 views (last 30 days)

Hello people! I want to produce N number of matrices while k=1:N and I want the result in a way like h(k). I mean h(1),h(2) etc which are matrices that depend on k. I mean creating them through a for loop. I tried the following code but it failed to produce N matrices and it just gave one matrix:

clc;

clear;

close all;

hx = [-1/sqrt(5);-2/sqrt(5)];

hu = [0;0];

N = 25;

Ek1 = zeros(N+1,1);

Ek2 = ones(N+1,1);

Ek3 = ones(N,1);

Ek4 = zeros(N,1);

h = zeros(102,1);

for k=1:N

if k==1

h(:,:) = [kron(Ek2,hx);kron(Ek3,hu)];

else

h(:,:) = [kron(Ek1,hx);kron(Ek4,hu)];

end

end

- 0 questions asked
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- 1,456 accepted answers
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**Direct link to this answer:**

Answer by Geoff Hayes
on 27 May 2019

- 0 questions asked
- 2,721 answers
- 1,456 accepted answers
- reputation: 8,058

Accepted Answer

Ali - if the output matrix of each iteration is of dimension 102x1, then you could store each output as a column in a 102xN matrix. For example,

h = zeros(102,N);

for k=1:N

if k==1

h(:,k) = [kron(Ek2,hx);kron(Ek3,hu)];

else

h(:,k) = [kron(Ek1,hx);kron(Ek4,hu)];

end

end

Well instead of h(1) you would use h(:,1).

I want to produce five matrices named h(k) t

It almost sounds as if you want five distinct matrices. That sort of programming (dynamically creating matrices/arrays) is usually discouraged. In this case, a matrix can be used in place of this.

when I use your h(:,k) code separatly the result is h 22*5 and when I copy it to my main program and run h is 22*2 I dont know why and I want it 22*1 I mean for k==1 I need 22*1 matrix I wanna be sure that it replaces a 22*1 matrix when k==1 itself but I think it doesn't do that. I give you my full optimization program maybe you can get better picture of my problem. the following program has dimension problem because of h, but I just wanna show you h 22*2 in my main code and the usage of my question in my main code:

clc;

clear;

close all;

%% Time structure

dt=0.01; %sampling time

%% System structure

A = [1.02 -0.1

0.1 0.98];

B = [0.5 0

0.05 0.5];

G = [0.3 0

0 0.3];

Qf = [50 0

0 50];

[A, B] = c2d(A,B,dt);

[nx,nu] = size(B);

%% MPC parameters

Q = eye(2);

R = eye(2)*50;

N = 5;

%% Building block matrices

I = eye(2,2);

q = 2;

m = 2;

Qbar = blkdiag(Q,Q,Q,Q,Qf,R,R,R,R);

Fq = blkdiag(sqrt(Q),sqrt(Q),sqrt(Q),sqrt(Q),sqrt(Q),sqrt(Qf),sqrt(R),sqrt(R),sqrt(R),sqrt(R),sqrt(R));

Gxx = zeros(2*N , size(A,1));

for i = 1:N

Gxx(q*i-q+1:q*i , :) = A^i;

end

Gxu = zeros(2*N , 2*N);

for i = 1:N

for j = 1:i

Gxu(q*i-q+1:q*i , m*j-m+1:m*j) = A^(i-j)*(B);

end

end

Gxw = zeros(2*N , 2*N);

for i = 1:N

for j = 1:i

Gxw(q*i-q+1:q*i , m*j-m+1:m*j) = A^(i-j)*(I);

end

end

%% Solve using YALMIP

K = sdpvar(repmat(10,1,N),repmat(22,1,N));

alpha = 0.975;

r = 1.96;

Sigmaw = eye(10,10);

Sigma0 = eye(2,2);

xbar0 = ones(2,20);

Cs = [Gxx*xbar0;zeros(12,20)];

hx = [-1/sqrt(5);-2/sqrt(5)];

hu = [0;0];

Ahat = [Gxx*xbar0 eye(10,2);zeros(12,20) zeros(12,2)];

Bhat = [Gxu;eye(12,10)];

Abar = [eye(11,10);zeros(11,10)];

M = Gxx*Sigma0*Gxx' + Gxw*Sigmaw*Gxw';

Mhat = [M zeros(10,12);zeros(12,10) ones(12,12)];

Euhat = ones(22,20);

Ek = [zeros(12,10);eye(10,10)];

g = 3;

Ek1 = zeros(N+1,1);

Ek2 = ones(N+1,1);

Ek3 = ones(N,1);

Ek4 = zeros(N,1);

constraint=[];

objective=0;

for k = 1:N

if k==1

h(:,k) = [kron(Ek2,hx);kron(Ek3,hu)];

else

h(:,k) = [kron(Ek1,hx);kron(Ek4,hu)];

end

b = sqrt(h'*((Abar+Bhat*K{k}*Ek)*M*(Abar+Bhat*K{k}*Ek)')*h);

objective = objective + 0.5*trace(Fq'*(Ahat+Bhat*K{k})*Mhat*(Ahat+Bhat*K{k})'*Fq);

constraint = [constraint, h'*(Cs+Bhat*K{k}*Euhat)+r*(sqrt(h'*(Abar+Bhat*K{k}*Ek)*M*(Abar+Bhat*K{k}*Ek)'*h))<=g, [(h'*(Abar+Bhat*K{k}*Ek)*M*(Abar+Bhat*K{k}*Ek)'*h) (h'*(Abar+Bhat*K{k}*Ek));((Abar+Bhat*K{k}*Ek)'*h) (Gxx*Sigma0*Gxx' + Gxw*Sigmaw*Gxw')]>=0];

end

options = sdpsettings('solver','sedumi');

optimize (constraint,objective,options);

K = value(K);

In these lines of code

b = sqrt(h'*((Abar+Bhat*K{k}*Ek)*M*(Abar+Bhat*K{k}*Ek)')*h);

objective = objective + 0.5*trace(Fq'*(Ahat+Bhat*K{k})*Mhat*(Ahat+Bhat*K{k})'*Fq);

constraint = [constraint, h'*(Cs+Bhat*K{k}*Euhat)+r*(sqrt(h'*(Abar+Bhat*K{k}*Ek)*M*(Abar+Bhat*K{k}*Ek)'*h))<=g, [(h'*(Abar+Bhat*K{k}*Ek)*M*(Abar+Bhat*K{k}*Ek)'*h) (h'*(Abar+Bhat*K{k}*Ek));((Abar+Bhat*K{k}*Ek)'*h) (Gxx*Sigma0*Gxx' + Gxw*Sigmaw*Gxw')]>=0];

you reference h. Should this be h(:,k) instead (since this code is in the for loop)?

I don't know putting this h(:,k) there instead of h could be right because we define an "if else" statement there and h has to be the result of that "if else". when you put h(;,k) there instead of h and run the program you see that h(:,k) is zero and "if else" doesn't work at all.

h(:,k)

ans =

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

if all zeros (for k > 1) is the problem then with this line of code

h(:,k) = [kron(Ek1,hx);kron(Ek4,hu)];

Note that Ek1 and Ek4 are defined to be all zeros...are you sure that this line of code is doing what you expect?

I sent the picture of h(k). I wrote this code to have the matrix h(k) which is in the picture. note that i=1 and hx and hu are known and k=1:N

I don't know if the code is wrong. I've asked you guys for that.

Maybe the problem is with your definition of [ekn] vector (not sure how best to write that). My thinking is that for any k then

e1n = [1; 0; 0; 0; .... 0];

e2n = [0; 1; 0; 0; .... 0];

e3n = [0; 0; 1; 0; .... 0];

...

enn = [0; 0; 0; 0; .... n];]

So on the kth iteration of the for loop, these arrays are all zeros except in the kth position. Your code may look more like

for k=1:N

Ekn1 = zeros(N,1);

Ekn1(k) = 1;

Ekn2 = zeros(N+1,1);

Ekn2(k) = 1;

h(:,k) = [kron(Ekn2,hx);kron(Ekn1,hu)];

% etc.

end

The above may or may not be what you want...it is my interpretation of your above image.

eh... excuse me! how come e1n is not ones(N,1) and it is [1;0;0;0;...;n] ??? and other e2n e3n etc are not zeros(N,1)? because i=1 and when k=i, [ekn]=1 else it's zero.

it says if k=i then [ekn]=1 else it is zero and i=1 is constant. another subject is that the picture says ekn is of dimension N*1 so N rows and one column I suppose!

have you consider these? because I can't see any "if else" in your interpretation...

I think you are right, but if this is the way to construct "h" I should use h(:,k) in my main code or "h"? I mean in those parts h is used like constraint=...

I mean this part:

for k = 1:N

Ekn1 = zeros(N,1);

Ekn1(k) = 1;

Ekn2 = zeros(N+1,1);

Ekn2(k) = 1;

h(:,k) = [kron(Ekn2,hx);kron(Ekn1,hu)];

b = sqrt(h(:,k)'*((Abar+Bhat*K{k}*Ek)*M*(Abar+Bhat*K{k}*Ek)')*h(:,k));

objective = objective + 0.5*trace(Fq'*(Ahat+Bhat*K{k})*Mhat*(Ahat+Bhat*K{k})'*Fq);

constraint = [constraint, ((h(:,k)'*(Cs+Bhat*K{k}*Euhat))+(r*(sqrt(h(:,k)'*(Abar+Bhat*K{k}*Ek)*M*(Abar+Bhat*K{k}*Ek)'*h(:,k)))))<=g, [(h(:,k)'*(Abar+Bhat*K{k}*Ek)*M*(Abar+Bhat*K{k}*Ek)'*h(:,k)) (h(:,k)'*(Abar+Bhat*K{k}*Ek));((Abar+Bhat*K{k}*Ek)'*h(:,k)) (Gxx*Sigma0*Gxx' + Gxw*Sigmaw*Gxw')]>=0];

end

It's Yalmip code not standard matlab code I mean it's the form of a problem you give to Yalmip it solves it. I used h(:,k) and it says no solver applicable but with h it solves it right or wrong it's not obvious yet.

another weird subject is that when I preallocate h and run it takes forever and no result and without preallocating h's dimension becomes 22*2

I don't know I can't figure out what is this progrram do lol...

I pre-allocate with h = zeros((2*N + 1) * size(hx,1), N);

and what is N?

N can be 5 or 10 or 15 or 20 or 25

For the case where N is 5, then

h = zeros((2*N + 1) * size(hx,1), N);

h is a 22x5 array. Is this correct?

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## 1 Comment

## Stephen Cobeldick (view profile)

Direct link to this comment:https://in.mathworks.com/matlabcentral/answers/464248-produce-matrices-through-for-loop#comment_709326

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