Asked by George Sterpu
on 4 Dec 2012

Hi.

I need to implement the following behavior :

The Integrator and my_Integrator blocks have to be equivalent I/O.

How should I write the Matlab Function ?

Thanks for any reply.

Answer by Ryan G
on 4 Dec 2012

Accepted answer

As this looks like a homework problem, I can't answer directly. However I will point you in the direction of persistent variables.

Answer by Azzi Abdelmalek
on 7 Dec 2012

Edited by Azzi Abdelmalek
on 8 Dec 2012

I don't know why do you need this, maybe if you explain exactly what you need, there is better way

Show 6 older comments

George Sterpu
on 8 Dec 2012

The solution it's already posted above. But since you and Guy disapprove this, I am waiting for the better solution.

I didn't need any clock input , please see http://www.mathworks.com/matlabcentral/answers/55705-represent-simulink-integrator-block-as-matlab-function#comment_115977

And even if I did, I guess it could have been fetched using the get_param command (never tried this though)

Azzi Abdelmalek
on 8 Dec 2012

Ok, I see, If T is constant, you must then set, in model configuration parameters your fixed step time to T, and also your step block sample time to T. In this case you don't need a clock.

function y = fcn(u) persistent uold yold T=0.01; if isempty(uold) uold=0;yold=0; end y = u*T+yold-(u-uold)*T/2 yold=y;uold=u;

George Sterpu
on 8 Dec 2012

Changing the sample time of the Step block to 0.01 removed the previous offset. Thanks

Answer by Guy Rouleau
on 5 Dec 2012

This is not a good idea. The MATLAB function is not designed for this purpose.

Answer by George Sterpu
on 5 Dec 2012

Edited by George Sterpu
on 6 Dec 2012

My main goal is to implement the differential equations of a physical system using a single Matlab Function. As the sums and gains were easy to represent, I couldn't find any alternative for the integration.

Show 1 older comment

George Sterpu
on 5 Dec 2012

Your idea would be to declare a persistent variable for the numeric integration ?

Y(s) = U(s) / s => y(z) = yOld + u(z)

so you declare persistent x = y to stand for yOld ?

Can I avoid this low-level arithmetic and call a predefined method instead (ode45 for example) ?

Ryan G
on 5 Dec 2012

What you have written is close it would be more like:

y(z) = yOld+u(z)/SampleTime

You cannot use the ODE solver in the MATLAB function block.

George Sterpu
on 7 Dec 2012

Any idea on how to get rid of this offset ?

Code looks like:

function y = fcn(u) %#codegen T=0.01;

persistent yOld; persistent uOld;

if (isempty(yOld)) yOld = 0; end

if (isempty(uOld)) uOld = 0; end

y = yOld + (T/2)* (u + uOld);

%y=yOld + u*T; yOld = y; uOld = u;

Answer by George Sterpu
on 5 Dec 2012

Can anybody suggest a better way of implementing the numerical integration ? The code has to be written inside the Matlab Function Block though.

Related Content

MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi

Learn moreOpportunities for recent engineering grads.

Apply Today
## 2 Comments

## Azzi Abdelmalek (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/55705#comment_115365

What is your goal?

## George Sterpu (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/55705#comment_115533

Please see http://www.mathworks.com/matlabcentral/answers/55705#answer_67551