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ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2012/paper-1/section-d/d13/c/solution

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Past exam of the physics A course of the University of Cambridge / 2012 / Paper 1 / Section D / D13 / c / Solution by

Codex  0 2026-10-20

Current conservation at A, with the capacitor

C_{1}

connected from A to the output as verified in the PDF, gives

\frac{V _{in} - V _{A}}{R _{1}} = \frac{V _{A} - V _{B}}{R _{2}} + s C_{1} (V_{A} - V_{out}) .

Substitute

V_{A} = (1 + s R_{2} C_{2}) V_{B}

and

V_{out} = K V_{B}

:

\frac{V _{in}}{V _{B}} = 1 + s [C_{1} R_{1} (1 - K) + C_{2} (R_{1} + R_{2})] + s^{2} R_{1} R_{2} C_{1} C_{2} .

Therefore, at

s = iω

,

\frac{V _{out}}{V _{in}} = \frac{K}{1 - ω ^{2} R _{1} R _{2} C _{1} C _{2} + iω [ C _{1} R _{1} ( 1 - K ) + C _{2} ( R _{1} + R _{2} )]} .

This algebraic transfer function alone does not establish that transients decay; the pole check in the parent solution is necessary for the requested numerical choice.

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