its not taking the values of y when y value is ranging from 0.25 to 0.45 with an interval of 0.05,but it is able to take the value when y value is constant. and its coming in some matrix but with constant values.
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Y1 = 7.7054
Y2 = 9.6584
I(3)= -0.0043
I(4)=0.4141
n=4 ;
m=2;
y=0.25:0.05:0.45;
for j=1:m
for i=1:n
if (j==1)
ud(i+1,1)=y;
taub(i+1,1)=y;
ud(i+1,2)=(I(3)*exp(Y2*y)+I(4)*exp(-Y2*y)+k2)^0.5;
taub(i+1,2)=(f*ro*ud(i+1,2)^2)/8;
end
end
end
ud
taub
plot(ud(i,1),ud(i,2))
sir here y will be my first column and ud will be my second column,bt ud is dependent on y,y is varying from 0.25 to 0.45 by having an interval of 0.05. here i just want to change the value of y in each iteration.
7 件のコメント
ES
2014 年 1 月 23 日
Shouldn't you be using y(i) instead of y alone in the lines
ud(i+1,1)=y;
taub(i+1,1)=y;
ud(i+1,2)=(I(3)*exp(Y2*y)+I(4)*exp(-Y2*y)+k2)^0.5;
Also what is k2?
ellora
2014 年 1 月 24 日
Image Analyst
2014 年 1 月 24 日
But we don't know if y should really be y(j) or y(i). y has 5 elements, not 2 or 4 like m and n.
ellora
2014 年 1 月 24 日
Image Analyst
2014 年 1 月 24 日
Is "the table" what you call "taub"? And by "find" you really mean "assign"? Or do you really mean "find" as in search for, or identify the location of?
ellora
2014 年 1 月 25 日
ellora
2014 年 1 月 26 日
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