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One way to describe a practical integrator ("leaky integrator") is

$$ H(s) = \frac{g R}{1 + sRC} $$

I am trying to understand how to choose the values $g$, $R$ and $C$ because in practice, I will have constraints on them.

Here is my attempt: Suppose for a moment, that $R \rightarrow \infty$, then it becomes an ideal integrator

$$ H(s) = \frac{g}{sC} $$

When I just look at this equation (or look it its time domain version) it seems that g and C do not matter because it's just a scalar. Do I understand it correctly that $g/C$ relates to saturation (practically, there is a maximum output value). Suppose I want to implement a finite integration:

$$ x_i = \frac{g}{C} \int_0^{T_w} x(t) \, dt $$

and I know that $x(t)$ and $x_i$ can have a maximum value of $K$. Then I could derive (to avoid saturation in the worst case):

$$ K < \frac{g}{C} T_w K \Rightarrow \frac{g}{C} > T_w $$

This would restrict the ratio between $g$ and $C$. Does this make sense?

Back to the non-ideal integrator used for a finite integration interval. In this case, $1/RC$ can be interpreted as the first pole and after this frequency, the ideal integrator starts.

Since I integrate over a finite period $T_w$, is it valid to say that the lowest integrateable frequency is $1/T_w$ and hence $1/RC \ll 2\pi/T_w$ ?

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  • $\begingroup$ Your formula for $x(t)$ does not make any sense because the definite integral of $y(t)$ over a fixed interval is just a constant, not a function of time. $\endgroup$
    – Matt L.
    Aug 24, 2014 at 10:25
  • $\begingroup$ Of course, that was not intended. I corrected it to $x_i$. Thanks $\endgroup$
    – divB
    Aug 24, 2014 at 18:02

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