= Finite-gain operational-amplifier integrator
{title2=$H(s)=\frac{sCr_o-A(s)}{1+sC\{r_o+R[1+A(s)]\}}$}
Consider an <op amp integrator> with infinite <input impedance>, an internal controlled voltage $u=-A(s)v$ where $v$ is the inverting-input voltage, and a series <output impedance> equal to a resistance $r_o$. Assume no external output load. Writing $y$ for the output, <Kirchhoff's current law> gives
$$
\frac{v_{\mathrm{in}}-v}{R}=sC(v-y),\qquad \frac{u-y}{r_o}=sC(y-v).
$$
Eliminating $u$ and $v$ yields
$$
\boxed{H(s)=\frac{y}{v_{\mathrm{in}}}=\frac{sCr_o-A(s)}{1+sC\{r_o+R[1+A(s)]\}}.}
$$
The expression at $r_o=0$ is obtained by its limit. In particular, constant gain $A_0$ and $r_o=0$ give $H(s)=-A_0/[1+sRC(1+A_0)]$: a finite direct-current gain followed by an integrating regime. For the dominant-pole model $A(s)=A_0/(1+s/\omega_a)$ and $r_o=0$,
$$
H(s)=\frac{-A_0}{1+s[\omega_a^{-1}+(1+A_0)RC]+s^2RC/\omega_a}.
$$
If $A_0\gg1$ and $A_0\omega_aRC\gg1$, there are separated turnover frequencies near $1/(A_0RC)$ and $A_0\omega_a$. The gain magnitude is respectively constant, proportional to $\omega^{-1}$ and proportional to $\omega^{-2}$. With the convention $e^{i\omega t}$ a continuous phase choice passes from $180^\circ$ through $90^\circ$ to $0^\circ$. If $r_o>0$ and $A(i\omega)\to0$, capacitive feedthrough instead gives the ultimate limit $r_o/(R+r_o)$. Specifying finite gain alone therefore does not specify a unique high-frequency response.
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