= Solution
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2021-operational-amplifier-stages.png]
{title=An inverting amplifier followed by an integrator, with an optional offset-correction branch}
Use the following arrangement: an <inverting operational amplifier> with input <resistor> $R_a$ and feedback <resistor> $R_f$, followed by an <op amp integrator> with input <resistor> $R_i$ and feedback <capacitor> $C_i$. Ground both non-inverting inputs. Each <ideal operational amplifier> has infinite <input impedance>, so its inputs draw no <electric current>; <negative feedback> makes the inverting input a <virtual ground>.
<Kirchhoff's current law> at the first summing junction gives $V_1=-(R_f/R_a)V_{\rm in}$. At the second summing junction it gives $V_1/R_i+C_i\,dV_{\rm out}/dt=0$. Consequently the two inversions cancel:
$$
\boxed{V_{\rm out}(t)=V_{\rm out}(t_0)+\frac{R_f}{R_aR_iC_i}\int_{t_0}^t V_{\rm in}(t')\,dt'.}
$$
The initial output <voltage> records the initial <capacitor> charge. A reset switch across $C_i$ sets it to zero. The formula holds while both stages remain within their supply rails; a persistent input <voltage> otherwise produces <amplifier saturation>.
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