How to find the coordinates of starting, end point and peak from graph?

How to find the peak value, end point and distance between starting and end point?

x = linespace(-5,5); y = sqrt((36-x.^2)/9); figure() plot (x,y)

Answer 1

[Star Strider]

Try this:

x = linspace(-5,5);
y = sqrt((36-x.^2)/9);
figure()
plot (x,y)

[onePeak,loc] = max(y)                          % Peak Value & Index
endPoint = y(end)                               % End Point Value
se_dist = max(x) - min(x)                       % Start To End Distance

producing:

onePeak =
    1.9999

loc =
    50

endPoint =
    1.1055

se_dist =
    10

[parag gupta]

Hi thank!

How to find a point when y = 0? I mean I want to know the point where y is zero except origin.

And could u pls tell how to find the distance between the stating point and the point when y= 0?

[Star Strider]

As always, my pleasure.

The value of ‘y’ at the origin is +2, not 0.

The easiest way to calculate the roots:

syms x 
y = sqrt((36-x.^2)/9)
ys = solve(y == 0)

producing:

ys =
 -6
  6

and the distance from the starting point to each root is:

sr_dist = -5 - ys
sr_dist =
   1
 -11

Note that for x<-6 and x>+6 ‘y’ is complex.

[parag gupta]

Is it possible to find the value of x when y = 0 from the graph except using solve command?

[parag gupta]

Can I use the following command?

yi = 0;

xDesired = interp1(y,x,yi)

[Star Strider]

To get it from the graph in the GUI, use the ‘Data Tips’ (from the ‘Tools’ drop-down menu, choose ‘Data Cursor’) to get the coordinates from the plot itself.

[parag gupta]

ok thanks star :)

[Star Strider]

As always, my pleasure.

‘Can I use the following command?’

Possibly, however it is likely not worth the effort, considering that ‘real(y)’ is not unique, and even forcing it to be unique will not return the desired values for ‘x’.

[Star Strider]

I can’t run this, since that syntax won’t work in R2019a. I also need values for ‘Z’ and ‘x’ to pass to ‘funtryp’:

sol = dde23(@(t,y)funtryp(t,y,Z,x),100,[0,0,0,0,0],[0,1500],options);

To find the point where ‘y1=0’ (from, the documentation):

Use the 'Events' option to specify a function that dde23 calls to find where functions

g(t,y(t),y(t−τ1),...,y(t−τk)) vanish. This function must be of the form

[value,isterminal,direction] = events(t,y,Z)

For more information, see the documentation section on ODE Event Location.

[Star Strider]

I’ve not written an event function for a delay differential equation solver before, largely because I rarely use delay differential equations (and have not in a while).

Try this:

function v = funtryp(t,y,Z);
r_v = 0.33; r_l = 0.33;b1 = 2*10^(-9);b3 = 0.5;t2 = 10;d_v = 0.5;d_l = 0.5;d_m = 0.1;c = 10^(2);       
C = 10^(8);gamma = 2.71;delta = 2.5*10^(-2);b2 = 0.15;
ylag1 = Z(:,1); x = 1;b2 = 0.15;
v = zeros(5,1);
v(1) = r_v*y(1)-b1*y(4)*y(1)-b2*y(1)-d_v*y(5)*y(1);
v(2) = r_l*y(2)+b1*y(4)*y(1)+b2*y(1)-gamma*y(2)-d_l*y(5)*y(2);
v(3) = gamma*y(2)-delta*y(3)-d_m*y(5)*y(3);
v(4) =y(1)+y(2)-b3*y(4);
v(5) = c*(((ylag1(1)+ylag1(2)+ylag1(3))/C)^x)*(1-y(5));
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% runfile
function [position,isterminal,direction] = trypevents(t,y,ydel)
position = y(1)-1E-2                            % Zero Value - Threshold
isterminal = 1;
direction = 0;
end

options = ddeset('RelTol',1e-5,'AbsTol',1e-8,'InitialY',[20,0,0,0,0], 'Events',@trypevents);
sol = dde23(@(t,y,Z)funtryp(t,y,Z),100,[0,0,0,0,0],[0,1500],options);
Q1 = sol.x
Q2 = sol.y
figure();
x1 = sol.x;
y1 = sol.y(1,:);
semilogy(x1,y1,'g','LineWidth',2); hold on;
xlabel('time t');
ylabel('v1(t)');

This runs without error in R2019a, and appears to give the correct result. The syntax you posted does not run in R2019a, so you may have to adapt my code to your version. I am using semilogy for the plot to be certain the event funciton returns the correct result. Change the event threshold value (that is here 1E-002, or eliminate it entirely) to produce the result you want.

[parag gupta]

Hey thank you so much @star :) thanksssssssss :)

[Star Strider]

As always, my pleasure!