## Transfer function vs. gains, differentiator, integrator = different, but I don't know why?

### Peter (view profile)

on 27 Dec 2011
Accepted Answer by Fangjun Jiang

### Fangjun Jiang (view profile)

Hi,
I'm really confused why the two systems shown in the picture below give different outputs. I thought they were mathematically equivalent but give different outputs.
Please could someone show me where I have gone wrong!
Many thanks,
Peter

### Fangjun Jiang (view profile)

Answer by Fangjun Jiang

### Fangjun Jiang (view profile)

on 27 Dec 2011

It must be due to the limitation of calculating du/dt. If you are using all the default setting, you'll notice some warning message in the Command Window regarding step size.
There is no perfect way of calculating the theoretical du/dt in numerical simulation. Doing the du/dt first and then the integration is the problem. If you move the two integrators ahead of the du/dt block (right after the negative feedback summation block), you'll find that the results match perfectly.

Peter

### Peter (view profile)

on 27 Dec 2011
Would there be any way that I could determine a mathematical equation of the first system then?
(I'm trying to match that system to a desired curve, i.e. a desired Y(s) where Y(s)=T(s)R(s) ).
Thanks!
Fangjun Jiang

### Fangjun Jiang (view profile)

on 28 Dec 2011
For canonical realizations, see http://en.wikipedia.org/wiki/Realization_(systems)
Fangjun Jiang

### Fangjun Jiang (view profile)

on 28 Dec 2011
Or a better reference at http://arri.uta.edu/acs/Lectures/realization.pdf