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 $1/RC \ll 2\pi/T_w$ ?

  • $\begingroup$ 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.