non-zero threshold for event function

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9times6
9times6 el 10 de Sept. de 2019
Comentada: Star Strider el 10 de Sept. de 2019
I have written the following event function and I want to detect an event which is non-zero. Therefore, I am using the logical operator since event function will only work for value=0. In this code, y is 4x1 vector.
function [value,isterminal,direction] = myevent12d1(t,y)
% Locate the time when height passes through zero in a decreasing direction
% and stop integration.
z=y(1)<y(3);
value = z; % detect height = 0
isterminal = 1; % stop the integration
direction = -1; % negative
However, upon running the code, I am getting the following error message:
Undefined function 'sign' for input arguments of type 'logical'.
Error in odezero (line 46)
indzc = find((sign(vL) ~= sign(vR)) & (direction .* (vR - vL) >= 0));
Error in ode45 (line 352)
[te,ye,ie,valt,stop] = ...
Error in onedof2dof (line 19)
[t1,y1,te1,ye1,ie1]=ode45(@(t,y) sp121(t,y,k1,k2,m1,m2),[tstart tfinal],y0,options1);
Any help in this regard would be appreciated.

Respuesta aceptada

Star Strider
Star Strider el 10 de Sept. de 2019
Editada: Star Strider el 10 de Sept. de 2019
I am not certain what you are doing. If you want to evaluate ‘z’ as numeric rather than logical, just do an arithmetic operation on it, here that being uplus (unary plus):
y = 1:4; % Create ‘y’
z1 = y(1)<y(3) % Logical Result
z2 = +(y(1)<y(3)) % Double Result
Note the parentheses, so the logical operation is done first, then the arithmetic operation.
Using the ‘z2’ approach will at least avoid the problem with logical variables.
EDIT —
Also:
z3 = y(3) - y(1);
Note that the documentation on ODE Event Location states: ‘Event functions take an expression that you specify, and detect an event when that expression is equal to zero.’ So a variable in your ODE evalutation does not itself have to be zero, it simply needs for the ‘value’ output to be zero to trigger the event.
  2 comentarios
9times6
9times6 el 10 de Sept. de 2019
Thanks a lot, Star Strider! It worked :)
Star Strider
Star Strider el 10 de Sept. de 2019
As always, my pleasure!

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