linspace is causing an error

My goal is to plot R, where R=1-T

T is a function of n2 and D, where n2 takes the values

n2 = [1.2 1.75 2.2] and D = linspace(0,1) and the function of T is equal to:

T =

So my code so far is *note(lambda = 1):

n2 = [1.2 1.75 2.2];
D = linspace(0,1);
T= (9.68.*n2.^2)./(11.6964.*n2.^2)+(1-n2.^2).*(5.8564-n2.^2).*(sin((2*pi*D)./1)).^2
>>Arrays have incompatible sizes for this operation.<<

I think what is happening is D which is 1x100 matrix is making this incompatible.

My end goal is to plot this and end up with 3 graphs, where one graph represents a value of n2 going through all of D.

I am very inexperienced at matlab but im trying to figure this out. Any help?

[Steven Lord]

FYI as an aside, if you ever want to compute the sine or cosine of pi times some number consider using sinpi or cospi instead of sin(pi*something) or cos(pi*something).

[Snow]

Thank you that is very helpful.

[Image Analyst]

To clarify, do you want three line plots in a single axes (like @VBBV's answer below)? Or do you want one line plot in three different axes (like my answer below)?

[Snow]

Yes, I was looking for something like @VBBV's answer.

[dpb]

NOTA BENE:

The equation for T used by both @Image Analyst and @VBBV is the same as that originally posted that has a transcription error in not including all the additive terms in the denominator.

See the later Answer I posted that corrects the equation-- the result is much different.

"There be dragons!"

Answer 1

[dpb]

In

%T= (9.68.*n2.^2)./(11.6964.*n2.^2)+(1-n2.^2).*(5.8564-n2.^2).*(sin((2*pi*D)./1)).^2

you're missing a needed set of parentheses around the denominatior; as you've written it only the first term is in the denominator. You can get by by moving that first closing ")" to the end--

%T= (9.68*n2.^2)./(11.6964*n2.^2 + (1-n2.^2).*(5.8564-n2.^2).*(sin((2*pi*D))).^2);

For the other, you need to write for a single value of n2 and evaluate over D in a loop or investigate meshgrid to vectorize over both.

For the beginner, the looping construct first is probably the simpler approach.

ADDENDUM

But, the meshgrid solution is not that bad and for completeness since I did bring it up... :)

all_n = [1.2 1.75 2.2];
[N,D]=meshgrid(all_n(:),linspace(0,1).');
fnT=@(n2,D) 9.68*n2.^2./(11.6964*n2.^2+(1-n2.^2).*(5.8564-n2.^2).*(sin((2*pi*D)./1)).^2);
T=fnT(N,D);
plot(D,T)
grid on
xlabel('D'); ylabel('T')
caption = compose('n2 = %.2f', all_n.');
legend(caption,'location','north');

I would note again my first comment that the equation for T as written in the original posting is in error in that only the first term is incorporated in the denominator whereas the equation as presented in the image clearly has all additive terms in the denominator. This changes the numerical values greatly, although the general shape is similar, it is also flipped upside down from the incorrect version.

[Image Analyst]

Good catch @dpb. We were too trusting that he had got it right. That's also why I suggest breaking things up into simpler terms and combining

term1 = stuff
term2 = whatever
term3 = whatever
numerator = term1 + term2
denominator = term3 + term4 * term5 * term6
T = numerator / denominator

doing it like that can help avoid problems with parentheses missing or in the wrong place.

[dpb]

It was more or less an accident, @Image Analyst; but that first closing parenthesis just happened to catch my eye as I was scanning for the missing dot operator or such that might have been the problem...

I'm somewhat concerned OP never came back and just went on with the wrong solution, though...

[Snow]

@Image Analyst Thank you, that is a good way of going forway with coding.

@dpb I have been meaning to come back. I do believe the answer I accepted is incorect, more so after reading your code and trying to analyze it. I am not familiar with meshgrid so I will need to read up on that.

I am a bit confused though. Because overall I am suppose to be graphing R which R = 1-T. I personally dont see it in your code but maybe I am misunderstanding how your code works?

Edit: I should have been a little more clear, the question states:

"plot R as a function of D∈[0,1] for each of the 3 values of n2"

[dpb]

Oh. Missed that (or forgot about it more accurately) since your code didn't compute R but stopped at T, so did all of us (I think).

That's trivial, of course, once you have T.

Or, you could redefine fnT to return R instead easily enough as well...

...
fnR=@(n2,D) 1-9.68*n2.^2./(11.6964*n2.^2+(1-n2.^2).*(5.8564-n2.^2).*(sin((2*pi*D)./1)).^2);
R=fnR(N,D);
plot(D,R)
grid on
xlabel('D'); ylabel('R')
caption = compose('n2 = %.2f', all_n.');
legend(caption,'location','north');

or, in the spirit of @Image Analyst, keep fnT as it was and add fnR as

fnT=@(n2,D) 9.68*n2.^2./(11.6964*n2.^2+(1-n2.^2).*(5.8564-n2.^2).*(sin((2*pi*D)./1)).^2);
fnR=@(n2,D) 1-fnT(n2,D);
R=fnR(N,D);
...  

for easier debugging (with or without his other suggestion of building the denominator piecewise).

Answer 2

[VBBV]

n2 = [1.2 1.75 2.2];
D = linspace(0,1,100);
for k = 1:length(D)
T(:,k)= (9.68.*n2.^2)./(11.6964.*n2.^2)+(1-n2.^2).*(5.8564-n2.^2).*(sin((2*pi*D(k))./1)).^2;
end
plot(D,T)

Answer 3

[Image Analyst]

It's because n2 and D are different sizes so you can't do it all in one equation. Try a loop

all_n = [1.2 1.75 2.2];
D = linspace(0,1);
whos all_n

  Name       Size            Bytes  Class     Attributes

  all_n      1x3                24  double              

whos D

  Name      Size             Bytes  Class     Attributes

  D         1x100              800  double              

for k = 1 : length(all_n)
    % Get this one value of n
    n2 = all_n(k);
    % Compute T
    T = (9.68.*n2.^2)./(11.6964.*n2.^2)+(1-n2.^2).*(5.8564-n2.^2).*(sin((2*pi*D)./1)).^2;
    % Plot the curve
    subplot(3, 1, k);
    plot(D, T, 'b-', 'LineWidth', 2)
    grid on
    xlabel('D');
    ylabel('T')
    caption = sprintf('n2 = %.2f', n2);
    title(caption);
    fontsize(gca, 12, 'points')
end