help me with this for loop please

x = -10:0.1:10;
y = 0:0.005:1;
minNmax1 = 0:.005:1;
upper_1 = 0:.005:1;
y2 = 0:.005:1;
f1 = trapmf(x,[-2 0 0 2].');
minNmax1 = zeros(length(x), 2);
for k = 1 :length(y)
minNmax1(k,:) = [min(x),max(x)];
end
I'm trying to get [min(x),max(x)] for 201 values of y but this will only give me 1 answer [-10 10] which i assume is the first iteration?
help!

6 commentaires

Katalin
Katalin le 23 Juin 2015
Sorry it is not clear what you would like in your variable. min(x) and max(x) are both constants, it will fill your 201x2 matrix with -10 in the first column and 10 in the second. Could you please explain more what you would like to achieve?
soloby
soloby le 23 Juin 2015
Modifié(e) : soloby le 23 Juin 2015
hi, thanks for the reply.
I'm trying to get minimum and maximum values of my function f1 (triangular tooth) for 200 different values of y.
i.e.,
@ y = 0 min(x), max(x)
@ y = 0.005 min(x), max(x)
@ y = 0.01 min(x), max(x)
and so forth. all the way up to y = 1
Jan
Jan le 23 Juin 2015
Modifié(e) : Jan le 23 Juin 2015
I cannot understand your notation. What does "@ y = 0.01 min(x), max(x)" mean? There is no "y" in this expression on the right hand side. What does the "y=0.01" part mean? Which y-values do you want to check and do you think that any y-value is exactly 0.01?
Perhaps you want "f1" to appear anywhere?
Katalin
Katalin le 23 Juin 2015
How does f1 depend on y?
soloby
soloby le 23 Juin 2015
Modifié(e) : soloby le 23 Juin 2015
my function f1 ranges from 0 to 1. if you plot(x,f1) it can be seen.
I need to split my f1 values (0 to 1) into 201 pieces and find the upper and lower x values for each iterations of y. Think of fuzzy logic, it is similar to how it is solved.
Easy concept, but I'm afraid a lot harder to execute.
so my f1 is 0 to 1
I want to divide it up into equal numbers (201 in my case)
at each iteration, I want to find the x values Please someone help!
soloby
soloby le 23 Juin 2015
Check my picture please, I need to find those starred points for all the red lines (201 of them)

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Réponses (1)

Muhannad
Muhannad le 24 Juin 2015

0 votes

will min(x) always be the same number as x is not changing? you can use repmat(min(x), 201,1)) if that is the case

1 commentaire

soloby
soloby le 25 Juin 2015
min(x) will be the same for each iterations of y but I need it for all 201 y's.

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le 23 Juin 2015

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