A problem in writing a code

4 vues (au cours des 30 derniers jours)
Hamed P. Azad
Hamed P. Azad le 9 Déc 2011
I have a problem in a code.
y_d = zeros (N,1)
for cont = 1:N
y_d (cont,1) = u (cont);
end
u_minus = zeros (N,1);
Output = zeros (N,1);
for count = 1:N
Output (count,1) = y_sin_real (count);
end
for c = 0:N
mat2 = y_d - ((h_minus * u_minus) - (Output - y_d));
u_plus = mat1\(h_plus_t * Q)* mat2;
Output = (h_plus * u_plus) + (h_minus * u_minus);
u_minus = [u_minus ; u_plus(1)];
u_minus = u_minus (2 : N+1);
error = abs(y_d) - abs(Output);
end
as you can see, i first give an initial state to Output to use in calculation, but in the loop, i'm changing it's value N times. but it's not workin and in the end of the loop i still have the initial numbers for Output.
what is the problem???
thanks.
  1 commentaire
Dr. Seis
Dr. Seis le 9 Déc 2011
You should also show how the other variables (i.e., the one in your third "for" loop) are created above. Or at least tell us their dimensions.

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Réponses (3)

Hamed P. Azad
Hamed P. Azad le 9 Déc 2011
yes, the problem is in the 3rd for loop, but the variables are set correctly, and the code runs fine; except for the "Output" that stays the way it was initiated.
the code is long and the variables are set in different loops.
  2 commentaires
Sean de Wolski
Sean de Wolski le 9 Déc 2011
Can you provide us with enough of the input data that we can replicate this?
Dr. Seis
Dr. Seis le 9 Déc 2011
Okay... then what are their dimensions? Are they also Nx1 vectors?

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Daniel Shub
Daniel Shub le 9 Déc 2011
One issue is that in the 3rd loop you overwrite Output. All you work in the second loop is lost on the first iteration of the 3rd loop. The line:
Output = (h_plus * u_plus) + (h_minus * u_minus);
is potentially problematic.

Dr. Seis
Dr. Seis le 9 Déc 2011
The first line in your third "for" loop is:
mat2 = y_d - ((h_minus * u_minus) - (Output - y_d));
which is basically:
mat2 = h_minus * u_minus - Output;

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