Element multiplication and subtraction

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CD
CD on 26 Mar 2019
Commented: Adam Danz on 26 Mar 2019
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

Accepted Answer

Adam Danz
Adam Danz on 26 Mar 2019
Edited: Adam Danz on 26 Mar 2019
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.

More Answers (2)

James Tursa
James Tursa on 26 Mar 2019
Edited: James Tursa on 26 Mar 2019
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)) ))
  3 Comments
James Tursa
James Tursa on 26 Mar 2019
Edited: James Tursa on 26 Mar 2019
"... 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
Adam Danz on 26 Mar 2019
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.

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CD
CD on 26 Mar 2019
The second solution (M2) matches my results via excel.
Thank you!
  1 Comment
Adam Danz
Adam Danz on 26 Mar 2019
Edited: Adam Danz on 26 Mar 2019
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.

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