- Define constant delays.
understanding DDE23 function format
4 views (last 30 days)
Show older comments
I look at this example https://www.mathworks.com/help/matlab/math/dde-with-constant-delays.html
DDE23 function has a format of
ddex1de(t,y,Z)
How should one understand Z?
0 Comments
Accepted Answer
Guru Mohanty
on 4 May 2020
I understand you are trying to solve system of differential equation using ‘dde23’. To do this you need to do following steps.
In this system of equation there are two lags i.e. (t-1) and (t-0.2).
lags = [1 0.2];
2. Define Solution History.
It Defines the first solution from which the solver starts iterations.
function s = history(t)
s = ones(3,1);
end
3. Form the equation.
function dydt = ddex1de(t,y,Z)
ylag1 = Z(:,1);
ylag2 = Z(:,2);
dydt = [ylag1(1);
ylag1(1)+ylag2(2);
y(2)];
end
Here In the call to ddex1de,
‘t’ is a scalar indicates the current ‘t’ in the equation
‘y’ is a column vector approximates y(t)
‘Z’ is a column vector approximates y(t – αj) for delay αj= lags(J).
In the following example Solution history is [1;1;1]. So Approximate values of lag ie ‘Z’ will be a matrix of size 3x2.In ‘Z’ Row defines the number of equations and column defines Number of lags.
4. Solve using ‘dde23’.
tspan = [0 5];
sol = dde23(@ddefun, lags, @history, tspan);
2 Comments
Guru Mohanty
on 4 May 2020
As it is a System of 3 Equations, 'y', 'dydt', 'ylag1' 'ylag2' have dimension 3x1. There are two constant Delay present, So The dimention of 'Z' is 3x2.
More Answers (0)
See Also
Products
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!