Matlab Graph Solved

Question

Hello,

I need to plot a function using Matlab that represents the path of a light ray based on several formulas involving temperature, height, the angle at which the ray arrives, etc.
I have managed to create this function and I get a graph that is (in my opinion) correct.
Now I need to display several functions on the same graph by varying the initial angle (between -1° and 0.1°).

I tried using the command
For i0=-1:0.1:0.1;
function(...);
plot(x,z);
end

But this only gives me the function for the last angle, that is, when i0 is equal to 0.1.

If someone could advise/guide me, it would be really kind of them.

Configuration: Windows XP Firefox 3.0.7

Fee Fay · Accepted Answer

Good evening my friend!

You just need to add the hold on command before your loop.
If you plan to relaunch your program or plot other curves on another figure, use the hold off command beforehand; otherwise, you will have the new curves on top of those from the figure produced by the loop.
Finally, to summarize, do something like this:

hold on; for i0=-1:0.1:0.1 function(...); plot(x,z); end hold off;

I wish you a very good evening!
--
All the animals shouted very loudly
That he was the most beautiful toad when he played the banjo.

Pallas · Answer

Thank you very much for your relevant and very quick response!

Now, I just need to explain to the software that the rays are absorbed by the ground, meaning that the portions of the function that go below 0 must be erased.

Have a good evening to you too.

Fee Fay · Answer

There’s no need to mention it, my friend! ;-)
I’m not quite sure how your data is organized, but I think that afterward you can do what you mentioned through logical indexing.

Since you're just starting with Matlab, I’ll try to detail my point as clearly as possible.

Let’s imagine that you have, for example, a vector v=[2;-3;4;6;-1;0;3;5];.
You can access the elements of this vector using the usual indexing, I mean v(1), v(2), etc...
But you can also access the elements of this vector through logical indexing, using a boolean vector (let’s call it z) of the same size as v.
z is therefore made up of 0 and 1. A 0 in z means that you will not take the element of v at the same position as the 0 in z. Conversely, a 1 means that you will take it.
Let’s illustrate what I just said and what you might be thinking: "What is this weirdo talking about!" :-D

Let’s define, for example, z=[0;0;1;0;0;0;0;0];.
By doing v(z), it will return the third element of v, which is v(3)=4.
Now, if we define z=[1;0;1;0;1;0;1;0];, by doing v(z), you will get the vector containing the elements v(1), v(3), v(5), v(7). We have taken one element out of two.

The question you are probably asking yourself now is: But how on earth do I use the nonsense you just told me to select only the strictly positive elements of v, for goodness' sake? :-D

It’s very simple! ;-)
It’s very easy to define a boolean vector from v that contains 1 at the positions where the corresponding elements of v are negative and 0 at the positions where the corresponding elements of v are strictly positive.
You just need to do z=(v<=0);, the boolean vector is (v<=0);.
This famous vector z (you can test it) is then equal to [0;1;0;0;1;1;0;0];

Be careful, this is not the subject here, but to construct a boolean vector by equality, the syntax is ==. For example, to get the boolean vector with 1 at the positions where the elements of v are zero, the syntax is z=(v==0);, the boolean vector here is (v==0);.

Finally, you just need to use the logical indexing I mentioned at the beginning to remove all negative values by defining v(z) as an empty matrix. The syntax is v(z)=[];, the vector v is then shortened, we have removed three values from it, those that were negative.

Here, I have somewhat detailed the steps, but to keep it short, you just need to do v(v<=0)=[]; to trim v.

I wish you a great weekend!
--
All the animals were crying out loud
That he was the most beautiful toad, when he played the banjo.

Pallas · Answer

After 3 readings, I think I have understood your reasoning well.
So I simply introduce v(v<=0)=[] after defining my vector v (which, for me, is the vector that designates the value of the ordinates of each point). This should replace the values <=0 with an empty value.

I don't understand why it doesn't work: I'm no longer getting any graph at all and my x-axis is only showing values from 0 to 1, whereas it used to go from 0 to 1000 (the y-axis has also shrunk).

Fee Fay · Answer

Sorry, I have to rush Pallas, can we discuss this tomorrow? I'll quickly explain what comes to mind while reading your message... If you shorten the y vector, you also need to shorten the x vector so that both vectors have the same length (I'm actually surprised you didn't get an error message when you asked Matlab to plot the curves). For example, if you want to plot v as a function of x and you shorten v like this: v(v

Pallas · Answer

Arf sorry, you are right, Matlab was sending me an error message saying that the vectors were not of the same length.
I suspected it was a mistake of this kind.

Now it works.
That said, I still don’t have exactly what I was looking for because the curves of the rays that touch the ground end up going back up, so Matlab connects with a straight line the moment the ray hits the ground and comes out again. Whereas the ray is supposed to be absorbed by the ground and never come out.

If you have any idea, don't hesitate because I currently see no other solution than to use something like Find followed by a loop (like I was taught to do in Pascal).

In any case, thank you very much for all the time you have already given me.

Have a good night as well.

Fee Fay · Answer

Good evening Pallas!

Since I don't have any visual support, it's not impossible that I might tell you nonsense, but let's try anyway. :-D
If I understood correctly what you want to do, there is a trick that is worth what it's worth, which is to set all the values you don't want to be plotted to NaN (Not-a-Number). Since these are values that are not numbers by definition, Matlab doesn't plot them.
If you want some information about this famous NaN, you can type help NaN or doc NaN. The summarized translation of the beginning of the help is approximately as follows...
NaN is for Matlab an arithmetic representation of Not-a-Number.
NaN is obtained for results of undefined mathematical operations such as 0/0 and inf-inf.
inf is, for Matlab, the arithmetic representation of +∞, obtained for example by doing exp(1000), 1/0, etc...

Having finished this digression, here’s how you can proceed (if I'm not completely off regarding your request), it's best to show you an example. By doing this:

x=(-2:0.001:2); y=x.^2-1; plot(x,y);

You get a parabola that goes below the x-axis. Now, let's replace all the negative values of y with NaN and plot the curve again:

y(y<=0)=NaN; plot(x,y);

Everything that was going below the x-axis has been removed, and nothing has been plotted. Is this something like what you wanted to do?

I wish you a very good night!
--
All the animals cried out loud
That he was the most beautiful toad when he played the banjo.

Pallas · Answer

It's not exactly what I was looking for because there's still the part of the curve that "sticks out" from the ground when it should be completely erased.

I based my work on what you told me before to erase my rays. I just used the "find" command to locate the index of the first negative value in the z vector, then in the case where r has at least one value, I delete everything from the first negative value to the end of the z vector.
Thus, I find the correct solution.

Thanks again for your help. Your explanations are really clear, enjoyable to read, and appropriate.

I wish you a good night as well and I'll see you next time in case I have major difficulties with the rest of my work and you choose to answer (again) my questions.

Fee Fay · Answer

Hello Pallas!

I'm sorry if I was somewhat off the mark then. That said, I'm really glad that you appreciated the few explanations from the previous messages and that they helped you solve your problem. It was a real pleasure for me to exchange with you as well.

I'll bother you for a few more moments, however, hoping you'll read this last message anyway. :-D
I remember that when starting with Matlab, it can be quite difficult to know all the options related to a function; you need to get into the habit of checking the help thoroughly, etc... you don't always have the time or desire to do so when a program works correctly as it is. However, sometimes, you end up coding something that is slightly less compact than it could be by making the best use of the options of the different functions, which isn't necessarily a bad thing because over-compacting can sometimes lead to losing readability in the program.
Anyway... all this to say that if I understood correctly what you wanted to do this time (we can always hope), I think three lines suffice to achieve what you were aiming for. I don't know if that's the case with what you wrote; it depends on how you used find. ;-)
These lines should therefore be enough to remove all the values following the first negative value of the vector y and plot its representative curve as a function of x:

x(find(y<=0,1,'first'):end)=[]; y(find(y<=0,1,'first'):end)=[]; plot(x,y);

find(y<=0,1,'first') provides the index of the first negative value of y, and from that index to the end (end), we remove everything by replacing it with an empty matrix. Maybe I'm not completely off this time... :-D

If one day you have a completely different question from this topic, remember to open a new thread; partly because I don't have answers to everything, and partly because people are more willing to take a quick look at threads that are unanswered rather than those that someone has already responded to, and if the person who responded to you the first time is absent, you might wait a long time for an answer in the "old" thread you've bumped. :-D
That said, I'm probably not teaching you anything if you're used to forums.

There you go...

Looking forward to it, and maybe see you soon! ;-)
Have a great Sunday!
--
All the animals cried out loud
That he was the most beautiful toad when he played the banjo.

étudiant007 · Answer

Hello Pallas!

If it can help you (and if it's not too late since the milestone has passed...) here is what I would do:

use the find command to locate the negative elements of your vector z.
The first element of the vector that comes out is the index of your first negative z.
Then just truncate your vector z (and x as well) up to this index and then "plot(x,z)"

Fee Fay · Answer

Hello my friend!

We thank you from the bottom of our hearts for your incredibly useful participation.
It is indeed very interesting to repeat what was said two weeks ago in messages 8 and 9, it can always be useful... for people suffering from acute immediate memory issues... or for goldfish...
It is indeed well known that Matlab is the software of choice for these adorable aquatic vertebrates.

I wish you a very good rest of the afternoon!
--
All the animals cried out loud
That he was the most beautiful toad, when he played the banjo.

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