How can I compute the Area and the Centroid of the following shape?

How can I compute the Area and the Centroid of the following shape?

I have used the following code to construct it:

xA = 0;
xB = 1;
xf = 1

xf = 1

x  = linspace(0, xf, xf*1e4 + 1);
a  = 9.2;   % 
c  = 0.5;     % center of function
Y  = sigmf(x, [a c]).*((0 <= (x - xA)) & ((x - xA) < (xB - xA)));
plot(x, Y, 'linewidth', 1.5), grid on, xlim ([0 1.1])

[Torsten]

You have the functional equation. So you can use "int" or "integral".

The integral formula for the centroid is given here:

https://en.wikipedia.org/wiki/Centroid

Section:

Locating by integral formula of a bounded region.

[M]

@Torsten Thanks. Is there a MATLAB function can help to copmute the centroid in this case?

[Torsten]

For functional areas: not that I know of.

But the formulae given in Wikipedia are not that complicated, are they ?

[M]

@Torsten does the g(x) in my case is the x axis?

and the x is the length of the x?

[M]

@Torsten If that the case, the x_Centroid will be the length of the x always?

[M]

Hi @Star Strider, Do you have any idea how to compute the centroid here. Thanks

Answer 1

[Star Strider]

The polyshape approach is the easiest way to go on this.

However it is possible to calculate this using the trapz function and the centroid definition —

xA = 0;
xB = 1;
xf = 1

xf = 1

x  = linspace(0, xf, xf*1e4 + 1);
a  = 9.2;   % 
c  = 0.5;     % center of function
Y  = sigmf(x, [a c]).*((0 <= (x - xA)) & ((x - xA) < (xB - xA)));

cx = trapz(x,x.*Y) ./ trapz(x,Y)

cx = 0.7138

cy = trapz(Y,x.*Y) ./ trapz(Y,x)

cy = 0.3934

p = polyshape(x,Y);

Warning: Polyshape has duplicate vertices, intersections, or other inconsistencies that may produce inaccurate or unexpected results. Input data has been modified to create a well-defined polyshape.

[xc,yc] = centroid(p)

xc = 0.7176

yc = 0.3974

figure
plot(x, Y, 'linewidth', 1.5), grid on, xlim ([0 1.1])

The values from the two approaches are not exactly the same, however they are reasonably close.

.

[M]

@Star Strider Thank you so much.

What do you think of the approach that Torsten suggested (https://en.wikipedia.org/wiki/Centroid

Section:

Locating by integral formula of a bounded region)

Do you think it gives more accurate answer?

Because any difference it causes effects in my application

Thanks

[Star Strider]

As always, my pleasure!

That is esentially the approach I used, with trapz.

The polyshape approach may be more accurate, however the difference is slight —

xA = 0;
xB = 1;
xf = 1;
x  = linspace(0, xf, xf*1e4 + 1);
a  = 9.2;   % 
c  = 0.5;     % center of function
Y  = sigmf(x, [a c]).*((0 <= (x - xA)) & ((x - xA) < (xB - xA)));

cx = trapz(x,x.*Y) ./ trapz(x,Y)

cx = 0.7138

cy = trapz(Y,x.*Y) ./ trapz(Y,x)

cy = 0.3934

p = polyshape(x,Y);

Warning: Polyshape has duplicate vertices, intersections, or other inconsistencies that may produce inaccurate or unexpected results. Input data has been modified to create a well-defined polyshape.

[xc,yc] = centroid(p)

xc = 0.7176

yc = 0.3974

x_accuracy = abs(xc-cx)/mean([xc cx])

x_accuracy = 0.0053

y_accuracy = abs(yc-cy)/mean([yc cy])

y_accuracy = 0.0100

figure
plot(x, Y, 'linewidth', 1.5, 'DisplayName','Data'), grid on, xlim ([0 1.1])
hold on
plot(cx,cy,'r+', 'DisplayName','trapz')
plot(xc,yc,'rx', 'DisplayName','polyshape+centroid')
hold off
legend('Location','best')

The relative deviation of the individual values from the mean of each set is about 0.5% for the x-coordinate and about 1% for the y-coordinate. I am not certain how accurate they can be.

.

Answer 2

[Image Analyst]

centroid

Centroid of polyshape

collapse all in page

Syntax

[x,y] = centroid(polyin)

[x,y] = centroid(polyin,I)

Description

example

[x,y] = centroid(polyin) returns the x-coordinates and the y-coordinates of the centroid of a polyshape.

example

[x,y] = centroid(polyin,I) returns the coordinates of the centroid of the Ith boundary of polyin.

This syntax is only supported when polyin is a scalar polyshape object.

[M]

@Image Analyst, This is not a ploy shape!

[Image Analyst]

So convert x, Y into one!

[M]

@Image Analyst I didnt get you. Can you explain please!

[Image Analyst]

You didn't look up the help for polyshape, did you? It's just as trivial as @Star Strider showed:

p = polyshape(x,Y);
[x,y] = centroid(p)

Answer 3

[Bruno Luong]

xA = 0;
xB = 1;
a  = 9.2;
c  = 0.5; 

sfun  = @(x)sigmf(x, [a c]);
Area = integral(@(x)sfun(x),xA,xB);
xc = integral(@(x)sfun(x).*x,xA,xB)/Area

xc = 0.7138

yc = integral(@(x)sfun(x).^2,xA,xB)/(2*Area)

yc = 0.3935

Answer 4

[Sam Chak]

Hi @M

If your intention is to find the defuzzified output value for membership function mf at the interval in x using the centroid method, then you can try this method:

xA = 0;
xB = 1;
x  = xA:0.0001:xB;
mf = sigmf(x, [9.2 0.5]);
plot(x, mf), grid on, xlim([-0.2 1.2]), ylim([0 1.2]), xlabel('\it{x}'), ylabel('\mu(\it{x})')
xc = defuzz(x, mf, 'centroid')

xc = 0.7138

xline(xc, '--', sprintf('%.4f', xc), 'LabelVerticalAlignment', 'middle');

Note that there should be no significant difference between

mf = sigmf(x, [9.2 0.5]).*((0 <= (x - xA)) & ((x - xA) < (xB - xA)));

and

mf = sigmf(x, [9.2 0.5]);