How can I implement the following blocks and code in function block ?

2020 4 8
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更新时间：2020 4 10
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I would use the following integral equation as function block in my simulation. I know I can do it as what I attached but I need a way to implement it by code inside function block.

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Willim
Willim 2020-4-8
I have read it but you know the simlink will run the blocks at each instant but in the link the result will give you over one period. Sorry I=the link cannot help me.
Just consider M=4 ; N=2 and I have a clock start from t=0:0.1:3, I need the y(t) to have all value at each t=0:0.1:3;

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Mehmed Saad
Mehmed Saad 2020-4-8
编辑：Mehmed Saad 2020-4-8

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This is an example of matlab given in integral help
% Without For loop
fun = @(x,c) 1./(x.^3-2*x-c);
q_f = integral(@(x)fun(x,5),0,2);
q_f = -0.4605
Now i just change the time of integration and sum them
% With For loop
t = 0:0.1:2;
q =0;
for i =2:length(t)
q =q+ integral(@(x)fun(x,5),t(i-1),t(i));
end
q = -0.4605

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Willim
Willim 2020-4-9
Thank you , but you still do not answer the question. This result will be over the whole period of time, but what I need is that integrate the equation and evaluate it as each step of the time. That means if we have 3000-time steps, we are going to integrate 3000 times for each step.
1- integrate
2- evaluate the integration at each step of the time
the result should match the size of t
Mehmed Saad
Mehmed Saad 2020-4-9
编辑：Mehmed Saad 2020-4-9
Yes, so what you have to do is divide the integral interval in small intervals
suppose your time starts from 0 and ends at 10s with 1s second step (so there will ber 11 steps of time)
0 1 2 3 4 5 6 7 8 9 10
Iteration 1:
y=integral(your_function,0,1);
Iteration 2:
y = integral(your_function,1,2);
Iteration 3:
y = integral(your_function,2,3);
.
.
.
Iteration 10:
y = integral(your_function,9,10);
So you have 10 different outputs
Similarly you can do it for 3000 time steps, what you need is current time step and previous time step (feedback).
I am not much familiar with simulink, if it doesnot work, try setting the xmin =0 for every iteration
(integral(fun,xmin,xmax))
Willim
Willim 2020-4-9
Inside the simulation when I use clock as my time I cannot control the time to divide it into periods
Walter Roberson
Walter Roberson 2020-4-9
Is this for continuous or discrete work?
What is the expected output for these three cases:
  • current time is "close enough" to the end of one of the defined intervals
  • previous time and current time are both in the middle of one of the defined intervals
  • previous time was in the previous defined interval compared to the current time
Willim
Willim 2020-4-10
I am using solver type ( Variable-step ) solver ode45. I am using this to have current time results
Walter Roberson
Walter Roberson 2020-4-10
With variable-step solvers, the block will be called at irregular intervals that might not match the interval that you want solutions at. The variable-step solvers can also go backwards in time (I encountered a case in which this really happened within the last few days.) ode45() invokes the function 6 times per step, using slightly different times and slightly different boundary conditions in order to make projections and check the error measurement.
In order to get results for a fixed time interval, because the function could be called for times spanning multiple intervals, you would have to emit a variable number of results -- and you would have to be prepared to have the values "revoked" if ode45 decided that the step was a failure. ode45 does not always go backwards in time when it decides that a step is a failure.
What is the reason that you are trying to replace those blocks with a MATLAB Function Block ? Are you trying to move to a discrete time system, using MATLAB Function Block to replace the ode45() work over a known interval?

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