
Simulating a simple equation
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I am new fairly new to Matlab - a light user for a few years and know only the basics.
I am trying to use it to simulate a trend in two parameters with a third and to then plot the result.
All i want to do, at least to start with, is to simulate the trend in Co and Cr with E according to:
c=0.057;% a constant
E0=0;
E=-0.5:0.05:0.5;
Co=linspace(0,1,0.01); %dependent variable
Cr=linspace (0,1,0.01); %second dependent variable
E=E0+(c*log(Co/Cr));
plot(E,Co,'b*-')
i think lines 3&4 are equivalent ways of specifying a range of numbers.
This fails with an error related to the vectors being of unequal length. f, by "plotting vectors" the software means E and Co then i don't understand why these would need to be "the same length" (indeed this error remains if i confine them to the same numerical range anyway).
My questions are:
(i) am i going about this supposedly simple task in the right way and (ii) what specifically am i doing wrong
any help hugely appreciated
1 件のコメント
John BG
2018 年 5 月 6 日
Hi Jason
1.
you are plugging the function output E into what plot expects to be the reference vector of the plotting, therefore, the plot shows up a vertical line:
c=0.057; % obvious constant
E0=0;
E=-0.5:0.05:0.5;
Co=[.01:.01:1]
Cr=[.01:.01:1]
E=E0+(c*log(Co./Cr));
plot(E,Co,'b*-')
so instead of
plot(E,Co,'b*-')
E, the function, has to be the 2nd input field of plot, or the only input field of plot, setting parameters aside.
plot(Co,E,'b*-')
2.
This is still pretty flat, because Co==Cr
isequal(Co,Cr)
ans =
logical
1
don't you really mean E to be something like this?
c=0.057;% a constant
E0=0;
E=-0.5:0.05:0.5;
Co=[.01:.01:1]
Cr=[.01:.01:1]
E=E0+(Co.*log(c./Cr));
plot(Co,E,'b*-');grid on

採用された回答
Walter Roberson
2012 年 11 月 18 日
In your line
E=E0+(c*log(Co/Cr));
the / operation is matrix division, not element-by-element division. Change the line to
E = E0 + (c * log(Co ./ Cr));
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