Evaluating a matrix at different time steps

2017 3 29
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更新时间：2017 3 29
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Dear all, I have a matrix which is a transition matrix. I need to calculate the time evolution of the associated probability distribution. The matrix is raised to 't', the time variable and I want to evaluate the matrix at different time steps. Is it possible in Matlab? The code to generate the square matrix is given below which is working well for a given value if t.
if true
N=3;
w=0.2;
t=1;
G=zeros(N);
for j=1:N %rows
for n=1:N %columns
for k=1:N %summation
G(j,n)=G(j,n)+1/N*(((1-w)+w*cos((2*pi*k)/N)).^t*cos((2*pi*(j-n)*k)/N));
end
end
end
disp(G)
end

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Star Strider
Star Strider 2017-3-29
编辑：Andrei Bobrov 2017-3-29

0 个投票

Add a separate loop for ‘t’, and a third dimension to your ‘G’ matrix:
N=3;
w=0.2;
tv = linspace(0, 10, 10); % Define Time Vector
G=zeros(N,N,length(tv)); % Preallocate Here
for t = 1:length(tv) % ‘Time’ Loop
for j=1:N %rows
for n=1:N %columns
for k=1:N %summation
G(j,n,t)=G(j,n,t)+1/N*(((1-w)+w*cos((2*pi*k)/N)).^tv(t)*cos((2*pi*(j-n)*k)/N));
end
end
end
% disp(G)
end
Define the vector of times you want to evaluate your matrix in ‘tv’. The rest of the code will then work. Each ‘page’ (third dimension elements) of ‘G’ will be ‘G’ at the corresponding times.
Also, move the preallocation step to be before the loop! Otherwise, it resets all previous values of ‘G’ to zero.

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Vipin Padinjarath
Vipin Padinjarath 2017-3-29
That's great! It's working. Thank you very much.:-)
Vipin Padinjarath
Vipin Padinjarath 2017-3-29
编辑：Vipin Padinjarath 2017-3-29
But I have a doubt. When I set the value of t and evaluate the code (my old code) the values of the elements of the matrix are different from the corresponding matrix this new code generates. Why is that so? I am sorry, I am quite new to matlab.
Star Strider
Star Strider 2017-3-29
My pleasure!
I see the problem. I forgot to include the ‘t’ dimension on the right-hand side reference to ‘G’. This should work correctly:
G(j,n,t)=G(j,n,t)+1/N*(((1-w)+w*cos((2*pi*k)/N)).^tv(t)*cos((2*pi*(j-n)*k)/N));
↑ ← ADDED ‘t’ HERE
(My apologies for the error.)
Vipin Padinjarath
Vipin Padinjarath 2017-3-29
It works perfectly now!! Cheers!Thanks again!!

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更多回答（1 个）

Andrei Bobrov
Andrei Bobrov 2017-3-29
编辑：Andrei Bobrov 2017-3-29

0 个投票

R2016b and later
N = 3;
w = 0.2;
k = 1:N;
k = reshape(k,1,1,1,[]);
jj = (1:N)';
n = 1:N;
t = (0:10);
t = reshape(t,1,1,[]);
g = ((1-w+w*cos(2*pi*k/N)).^t.*cos(2*pi*(jj-n).*k/N))/N;
G = sum(g,4);
R2016a and earlier
N = 3;
w = 0.2;
k = 1:N;
k = reshape(k,1,1,1,[]);
jj = (1:N)';
n = 1:N;
t = (0:10);
t = reshape(t,1,1,[]);
g0 = cos(bsxfun(@times,bsxfun(@minus,jj,n),2*pi*k)/N);
g1 = bsxfun(@power,1-w+w*cos(2*pi*k/N),t);
g = bsxfun(@times,g0,g1)/N;
G = sum(g,4);

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Vipin Padinjarath
Vipin Padinjarath 2017-3-29
Thank you Andrei Bobrov. But for my standards, this is too much. I don't understand much of it. But will learn for sure. Thanks again.

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