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Past exam of the physics 2 paper of the University of Oxford

2018

Use an ideal inverting <summing amplifier>, grounding the non-inverting input and connecting the two sources through resistances $R$ and $R/2$. A feedback resistance $R_f=2R$ gives the desired weights. <Operational-amplifier negative feedback> makes the inverting node a <virtual ground>, and negligible input current lets <Kirchhoff's current law> give
$$
\frac{V_a}{R}+\frac{V_b}{R/2}=-\frac{V_{\mathrm{out}}}{2R},\qquad
\boxed{V_{\mathrm{out}}=-2(V_a+2V_b).}
$$

\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2018-weighted-summing-amplifier.png]
{title=An inverting <summing amplifier> with input resistances $R,R/2$ and feedback resistance $2R$}

Replacing the feedback <resistor> by a <capacitor> $C$ makes a weighted <op amp integrator>:
$$
\boxed{\frac{dV_{\mathrm{out}}}{dt}=-\frac{V_a+2V_b}{RC},\qquad
V_{\mathrm{out}}(t)=V_{\mathrm{out}}(t_0)-\frac1{RC}\int_{t_0}^t[V_a(u)+2V_b(u)]\,du.}
$$
The initial capacitor charge determines the integration constant. A nonzero DC weighted input produces a ramp; a real <operational amplifier> eventually reaches its output-voltage limit. A parallel feedback resistance can limit drift while preserving integration over an appropriate frequency range.

Solution

Past exam of the physics 2 paper of the University of Oxford / 2018 / 6 / Solution

Past exam of the physics 2 paper of the University of Oxford / 2018 / 6 / Solution

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Past exam of the physics 2 paper of the University of Oxford / 2018 / 6 / Solution

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