= Solution
A non-ideal amplifier is not specified by infinite <input impedance> alone: its <open-loop gain> and <output impedance> must also be modeled. Take its internal controlled output to be $u=-A(s)v$, where $v$ is the inverting-input potential, and place a resistance $r_o$ between $u$ and the actual output $y=V_{\mathrm{out}}$. The non-inverting input remains grounded; assume no external load. This gives an explicit <finite-gain operational-amplifier integrator> model.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2019/paper-1-nonideal-integrator.png]
{title=Non-ideal integrator model and its gain and phase for a specified dominant-pole amplifier}
Writing <Kirchhoff's current law> at the input and output nodes gives
$$
\frac{V_{\mathrm{in}}-v}{R}=sC(v-y),\qquad \frac{u-y}{r_o}=sC(y-v),\qquad u=-A(s)v.
$$
The second equation gives $y=(sCr_o-A)v/(1+sCr_o)$. Substitution in the first gives the general <transfer function>
$$
\boxed{H(s)=\frac{V_{\mathrm{out}}}{V_{\mathrm{in}}}=\frac{sCr_o-A(s)}{1+sC\{r_o+R[1+A(s)]\}}.}
$$
For sinusoidal <phasors>, set $s=i\omega$. The numerator and denominator are complex, so the magnitude gives the <amplitude> ratio and their arguments give the relative phase.
The simplest model takes $r_o=0$ and a constant finite gain $A_0>0$. It gives
$$
H(i\omega)=-\frac{A_0}{1+i\omega RC(1+A_0)},\qquad |H|=\frac{A_0}{\sqrt{1+\omega^2R^2C^2(1+A_0)^2}},\qquad \arg H=\pi-\tan^{-1}[\omega RC(1+A_0)].
$$
Thus below $\omega_l=[(1+A_0)RC]^{-1}$ the amplifier has gain magnitude $A_0$ and inversion. Above that <frequency> it approaches an integrator with magnitude $A_0/[(1+A_0)\omega RC]\simeq1/(\omega RC)$ and phase $+90^\circ$. Constant finite gain by itself predicts no further high-frequency turnover.
For a more realistic sketch, use the standard dominant-pole <open-loop gain> $A(s)=A_0/(1+s/\omega_a)$, still with $r_o=0$. Then
$$
\boxed{H(s)=\frac{-A_0}{1+bs+cs^2},\quad b=\omega_a^{-1}+(1+A_0)RC,\quad c=RC/\omega_a.}
$$
The exact <frequency> response is
$$
|H(i\omega)|=\frac{A_0}{\sqrt{(1-c\omega^2)^2+b^2\omega^2}},\qquad \phi(\omega)=\pi-\operatorname{atan2}(b\omega,1-c\omega^2).
$$
Assuming $A_0\gg1$ and $A_0\omega_aRC\gg1$, the turnover frequencies are well separated: $\omega_l\simeq1/(A_0RC)$ and $\omega_h\simeq A_0\omega_a$, the approximate unity-gain angular <frequency> of the amplifier. The regimes are
$$
\boxed{\begin{array}{c|c|c}
\text{frequency}&|H|&\text{phase}\\ \hline
\omega\ll\omega_l&A_0&180^\circ\\
\omega_l\ll\omega\ll\omega_h&(\omega RC)^{-1}&90^\circ\\
\omega\gg\omega_h&A_0\omega_a/(\omega^2RC)&0^\circ
\end{array}}
$$
with magnitude slopes $0,-20,-40\,\mathrm{dB}$ per decade and smooth phase transitions. This is the solid curve in the figure; $180^\circ$ and $-180^\circ$ denote the same phase.
If finite $r_o$ is retained and $A(i\omega)\to0$, the general expression instead approaches $r_o/(R+r_o)$ at very high <frequency>: the <capacitor> couples the input to the output through a passive <voltage> divider. The dashed curve illustrates that eventual feedthrough. Other amplifier poles or an external load would alter the response again. No particular values of $A_0,\omega_a,r_o$ are given in the paper, so the numerical sketch is illustrative and the general expression states the model dependence explicitly.
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