Algebric Loop with FIR Filter in Simulink

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Muhammad Kamran
Muhammad Kamran el 11 de Abr. de 2016
Comentada: Ced el 12 de Abr. de 2016
Hi, I have FIR filter and I put it in a feedback. It has some feed foward coefficient so it gives an Algebric Loop error. How i can solve it?
Regards,

Respuestas (1)

Ced
Ced el 11 de Abr. de 2016
By feedforward, do you mean a feedthrough in your filter? Two possibilities I can think of:
1. Incorporate your feedforward as a gain, i.e. instead of having y = u + a*y, write y = u/(1-a). This is the exact solution.
2. As a quick fix, you can include a memory or delay element. This shifts everything by one time step though.
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Muhammad Kamran
Muhammad Kamran el 12 de Abr. de 2016
I am also facing another problem now.
The filter equation is y[n] = (a0 + a1*z^(-1)+ a2*z^(-1) +...) x[n]
If put it in a feedback loop with H =1; then it will be not easy to make the form you suggested. I have attached the picture of it.Now the path through a0 will give algebraic loop error.
your help is really appreciated. Thanks for looking into my problem
Ced
Ced el 12 de Abr. de 2016
I think it's still possible. Your system is something like y = (a0 + a1*z^(-1) + ... )*(r-y), where r is that signal coming from the left.
So, extracting the feedthrough term, taking it to the other side and then dividing the whole equation by (1+a0), you get
y = r*a0/(1+a0) + (r-y)*(a1*z^(-1) + ... )*a0/(1+a0)
Note that the a0/(1+a0) gain will be before the +- sign. The feedthrough of r doesn't have to be extracted if you don't want to, but this way, you only have one filter block. Otherwise, you need two blocks (one without feedthrough for y (in the feedback) and one with feedthrough for r).
Alternative: You could try and set an initial condition block on the feedback. Otherwise, even with discrete time, simulink arrives at your feedback point, but has no way of knowing what value to expect because nothing has been computed yet.

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