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Display radius on the plot

5 vues (au cours des 30 derniers jours)
jeet Mehta
jeet Mehta le 11 Juin 2019
Commenté : Star Strider le 11 Juin 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

Réponse acceptée

Star Strider
Star Strider le 11 Juin 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 commentaires
jeet Mehta
jeet Mehta le 11 Juin 2019
Thank you i did relaised that and fixed it yesterday . Thanks heaps really appreciate
Star Strider
Star Strider le 11 Juin 2019
As always, my pleasure.

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Plus de réponses (1)

KSSV
KSSV le 11 Juin 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 commentaires
jeet Mehta
jeet Mehta le 11 Juin 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 le 11 Juin 2019
The above code doesn't show any error. It is wokring fine unless you make some changes.

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