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

Codex  0 2026-10-25

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} = 2 R

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 _{out}}{2 R}, V_{out} = - 2 (V_{a} + 2 V_{b}) .

Figure 1.
An inverting summing amplifier with input resistances $R, R /2$ and feedback resistance $2 R$
. Details.

Replacing the feedback resistor by a capacitor

C

makes a weighted op amp integrator:

\frac{d V _{out}}{d t} = - \frac{V _{a} + 2 V _{b}}{RC}, V_{out} (t) = V_{out} (t_{0}) - \frac{1}{RC} \int_{t_{0}}^{t} [V_{a} (u) + 2 V_{b} (u)] d u .

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.

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