What does this plot mean? Can an equation be made of it?

This is the plot I am trying to get an equation for:

plot((1:N_rel), [e_mean; mean(lambdaSO(1:end-1,:),2)]./e_total, 'ko');

This is what each value is:

N_rel = 20
e_mean = 4.6600e+04

lambdaSO = 20x30 double

e_total = 1x30 double

The first coordinate on the plot is (1, 0.996058).

Answer 1

[Image Analyst]

With this:

N_rel = 20;
e_mean = 4.6600e+04;
lambdaSO = rand(20,30);
e_total = rand(1,30);
x = 1 : N_rel;
y = [e_mean; mean(lambdaSO(1:end-1,:),2)]./e_total

y = 20×30

1.0e+05 * 1.2372 0.4987 2.1508 0.6137 1.1839 1.1303 2.2412 0.5112 1.3909 0.5290 0.5288 1.0551 0.6578 0.5781 1.2955 0.9397 2.7138 0.6341 4.5755 0.9004 0.8226 0.8600 0.8630 0.9650 5.6663 0.9112 2.0831 1.7258 0.5786 0.9998 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

plot(x, y, '.-', 'MarkerSize', 20);
grid on;
legend

whos x

  Name      Size            Bytes  Class     Attributes

  x         1x20              160  double              

whos y

  Name       Size            Bytes  Class     Attributes

  y         20x30             4800  double              

y is a matrix not a vector like x is. So it will plot columns. Each of the 30 columns of y is like a separate curve. What exactly are you hoping to see?

[Blakeley Ficenec]

I do currently get a plot with MATLAB, which is an energy fraction graph, but I wanted to understand what that line of code is doing to see if I can try to plot it in excel as well. I'm doing some research on Proper Orthogonal Decomposition.

[Image Analyst]

Well, like I said, it plots each column in your matrix as a separate curve.

Answer 2

[John D'Errico]

Can you get an "equation" for it? No. At least not any more than what you already have. Anyway, an equation involves an equality. You might hope to get an expression that approximates the relation you showed. But you aready have that. And there is no reason you have suggested that anything more meaningful exists, than what you already have.

[Blakeley Ficenec]

Thank you. I wanted to try and plot this in excel. I don't understand what the plot is currently doing in MATLAB.

Answer 3

[Cris LaPierre]

There are likely many equations you could generate that could satisfy the one point you have shared. However, you must take all the points into consideration when coming up with the equation, and you have not shared that.

As John D'Errico said, that process most likely will be an approximation of the data. If that is what you want to do, then you might consider exploring the Curve Fitter app (this does require having the curve fitting toolbox installed).