Is there any possibility to solve
x = L*y*s + 2*y , without a du/dt block. ..
(x:output ; y:input)
Just e.g. integrators.
My system is very dynamic and du/dt would cause errors.

6 comentarios

Walter Roberson
Walter Roberson el 28 de Jun. de 2015
What you posted looks like it only needs multiplication and addition, not integration or any du/dt block.
Marcel Büttner
Marcel Büttner el 29 de Jun. de 2015
Editada: Marcel Büttner el 29 de Jun. de 2015
It looks like the following.
What I do not really know is how to prevent my system of errors, when I use the du/dt
Walter Roberson
Walter Roberson el 29 de Jun. de 2015
I think you were intending to attach an image?
Marcel Büttner
Marcel Büttner el 29 de Jun. de 2015
yes sorry I attached it now
Walter Roberson
Walter Roberson el 29 de Jun. de 2015
Under what circumstances would you get an error from the du/dt block?
Marcel Büttner
Marcel Büttner el 29 de Jun. de 2015
The values would rise too high. The simulation stops cause of infinity.

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 Respuesta aceptada

Stefan
Stefan el 29 de Jun. de 2015
Editada: Stefan el 29 de Jun. de 2015

1 voto

Two remarks here:
1. If you draw the Bode plot of the transfer function of the derivator, it looks like a line that ramps up with 20 dB/decade. In other words, for high frequencies it amplifies the signal (most likely it is a noise in this region). So, what you want is to avoid this phenomena, however keeping the derivator behavior for lower frequencies. In order to achieve this, replace your derivative block with a transfer function block:
where you have to choose tau sufficiently low. Notice that if tau = 0, then you end up with the derivator H(s) = s.
2. You say something about using integrators. This is possible if you solve your equation the other way around, namely you know x and want to calculate y. The dynamics of y is described by:
dy/dt = (1/L) (2y-x)
For this, you need to know the initial condition y(0) of your system, together with x(t). You code this equation in matlab or Simulink using an integrator block.

2 comentarios

Marcel Büttner
Marcel Büttner el 29 de Jun. de 2015
Perfectly explained. Thanks Stefan
Marcel Büttner
Marcel Büttner el 29 de Jun. de 2015
The second remark is not right for this question, cause your input is y and your output is x . As you explained I could only get y as an output value.

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