Element multiplication and subtraction

I'd like to run the following though I keep getting element multiplication/division errors:

Ron = 0.012;

RL = 0.003;

R = 6.667;

D = 0:0.001:.999;

M = (D/(1-D))*(1/( (2*Ron*D/(R*((1-D)^2)))+1+(RL/R)*(1/((1-D)^2)) ))

plot (D,M);

I've several permutations of placing dots in front of multiplication, division and raise to the power signs though I just cannot seem to get this to work.

How should "M" be written so that Matlab will accept it?

Thank you

Answer 1

[Adam Danz]

This line below avoids error but that doesn't mean it does what it is intended to do.

   %(D/(1-D))*(1 /( (2*Ron*D/(R*((1-D) ^2)))+1+(RL/R)*(1 /((1-D) ^2)) ))  %old version 
M = (D/(1-D))*(1./( (2*Ron*D/(R*((1-D).^2)))+1+(RL/R)*(1./((1-D).^2)) ))  %new version
%               _                     _                 _       _         %underline differences

James Tursa's solution is also error-free (copied below) but produces totally different results.

M2 = (D./(1-D)).*(1./( (2.*Ron.*D./(R.*((1-D).^2)))+1+(RL./R).*(1./((1-D).^2)) ))
isequal(M,M2)
%ans =
%  logical
%   0

This is why it's critical to understand what the equation is supposed to do. I haven't put much time into making sense of it so maybe James' version is the correct version or maybe neither answers are correct.

Answer 2

[James Tursa]

If you are unsure, then just put dots next to every * and / and ^ operation. If a scalar happens to be involved it will still work as expected. E.g.,

M = (D./(1-D)).*(1./( (2.*Ron.*D./(R.*((1-D).^2)))+1+(RL./R).*(1./((1-D).^2)) ))

[Adam Danz]

The approach of putting dots in every multiplication, division, and exponent just to eliminate the error is a bad approach. Our two solutions are a great example why it's a bad approach. Your solution

M2 = (D./(1-D)).*(1./( (2.*Ron.*D./(R.*((1-D).^2)))+1+(RL./R).*(1./((1-D).^2)) ))

and my solution

M = (D/(1-D))*(1./( (2*Ron*D/(R*((1-D).^2)))+1+(RL/R)*(1./((1-D).^2)) ))

both produce a vector of the same size and shape but the values aren't equal. Both of our solutions "work" in that they avoid error but at least one of them must be incorrect and there's a good chance both are incorrect.

[James Tursa]

"... is a bad approach ..."

I will respectfully disagree on this one. Just looking at the variables there was obviously only one vector, namely D. So IMO it was quite likely he was just looking for element-wise operations in calculating M ... he was just unsure where to put the dots. Again, I am talking about this particular post only!

[Adam Danz]

You took the time to realize that and it was a good call. My caution is against the preface "If you are unsure...". I don't think it is a good idea to throw a dot in, test it, and if there's no error, accept it without being certain that it's the correct correction. My answer is a good example of that.

Answer 3

[CD]

The second solution (M2) matches my results via excel.

Thank you!

[Adam Danz]

Ok, good! Don't forget to accept his answer. Having a set of expected results helps to decide if you're equation is correct. Otherwise, avoid throwing in dots just to avoid error and assuming the calculation is correct.