Past exam of the physics 2 paper of the University of Oxford / 2017 / 7 / Solution

The original circuit has $R_1$ and $C$ in series between the input and the inverting node, with $R_2$ providing feedback and the noninverting node grounded. An ideal operational amplifier in stable linear negative feedback draws no input electric current and holds the inverting node at virtual ground. For phasors proportional to $e^{i\omega t}$, the input complex impedance is $Z_1=R_1+1/(i\omega C)$. Kirchhoff's current law gives ${\tilde{V}}_{\mathrm{in}}/Z_1=-{\tilde{V}}_{\mathrm{out}}/R_2$, henceThis is an inverting high-pass filter. Its phase can be chosen continuously asThus $V_{\mathrm{out}}(t)=V_0|H|\cos (\omega t+\arg H)$. The ratio above is a phasor ratio, not an instantaneous ratio at input zero crossings. Setting $R_1=R_2$ does not remove the capacitor's phase: the output lags by $90^{\circ }+{\tan }^{-1}(\omega R_{1}C)$, between $90^{\circ }$ and $180^{\circ }$. At $\omega R_{1}C=1$ the lag is $135^{\circ }$; only for $\omega R_{1}C\gg 1$ does it approach $180^{\circ }$. Being within the amplifier's own bandwidth does not specify the external RC corner frequency.