Generating for loop for toeplitz for Q order analysis.
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b = toeplitz(x,[x(1) zeros(1,Q)])\y;
I have a toeplitz matrix that I want to write a loop for Q=(1:150).
any ideas?
1 Kommentar
Andrei Bobrov
am 20 Feb. 2013
Bearbeitet: Andrei Bobrov
am 20 Feb. 2013
That such x, y.
Antworten (1)
Andrei Bobrov
am 20 Feb. 2013
Bearbeitet: Andrei Bobrov
am 20 Feb. 2013
b{150} = toeplitz(x,[x(1) zeros(1,150)])\y; % THAT SUCH x, y
for jj = 1:150
b{jj} = b{end}(:,1:jj);
end
1 Kommentar
moejobe
am 20 Feb. 2013
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