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Display radius on the plot

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jeet Mehta
jeet Mehta el 11 de Jun. de 2019
Comentada: Star Strider el 11 de Jun. de 2019
%% Plot Region of Attraction
for k=1:360
xra(k)=2*cos(2*pi*k/360);
yra(k)=2*sin(2*pi*k/360);
end
figure(1)
p1=plot(xra,yra,'--r')
title('Initial Condition with radius ')
xlabel('x1(t)')
ylabel('x2(t)')
grid
axis(2.5*[-1 1 -1 1])
hold on
%% Plot Initial Circle
for k=1:360
xc(k)=ro*cos(2*pi*k/360); %Projection on x1
yc(k)=ro*sin(2*pi*k/360); %Projection on x2
end
figure(1)
p2=plot(xc,yc,'--k')
%% Run Simulation for N different initial conditions
for k =1:N
x1_o=ro*cos(2*pi*k/10+pi/6); %Projection on x1
x2_o=ro*sin(2*pi*k/10+pi/6); %Projection on x2
%% Simulation Linear Model
sim('sim_nonlinear_new.slx')
%states
x1=x(:,1);
x2=x(:,2);
scatter(x1_o,x2_o,100,'ok','filled') %Plot initial condition
plot(x1,x2) %Plot trajectory
drawnow;
legend([p1 p2],{'First','Second'});
end
hold off
I have above code for region of attraction circle and intial circle. I want to be able to display with a line drawing out and radius of the two circles. Please help

Respuesta aceptada

Star Strider
Star Strider el 11 de Jun. de 2019
You did not define ‘ro’.
When you do, you can plot the radius easily as:
plot([0 xra(45)], [0 yra(45)], '-b', [0 xc(135)],[0 yc(135)],'-g')
after the second loop. Choose whatever index you want to define the radius lines.
I would also use:
axis equal
so your circles don’t look like elipses.
  5 comentarios
jeet Mehta
jeet Mehta el 11 de Jun. de 2019
Thank you i did relaised that and fixed it yesterday . Thanks heaps really appreciate
Star Strider
Star Strider el 11 de Jun. de 2019
As always, my pleasure.

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Más respuestas (1)

KSSV
KSSV el 11 de Jun. de 2019
YOu can display radius of the circle like below:
th = linspace(0,2*pi) ;
R = 1. ; % REadius of Circle
C = [0. 0.] ; % cneter of circle
x = C(1)+R*cos(th) ;
y = C(2)+R*sin(th) ;
plot(x,y)
hold on
plot([C(1) x(1)],[C(2) y(1)])
  2 comentarios
jeet Mehta
jeet Mehta el 11 de Jun. de 2019
It giving me error saying its bound to index 1 . Can u please have a look at the code above and suggest
KSSV
KSSV el 11 de Jun. de 2019
The above code doesn't show any error. It is wokring fine unless you make some changes.

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