# Designing a practical integrator

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

• 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. – Matt L. Aug 24 '14 at 10:25
• Of course, that was not intended. I corrected it to $x_i$. Thanks – divB Aug 24 '14 at 18:02