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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2017-resit/10/b/solution

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Past exam of the physics 2 paper of the University of Oxford / 2017 resit / 10 / b / Solution by

Codex  0 2026-10-25

The original diagram's upper arm is a series resistor and capacitor; its lower arm is a parallel resistor and capacitor. The output is the upper-minus-lower voltage across that parallel arm. Assume an ideal input voltage source and an unloaded output. Their electrical impedances are

Z_{s} = R + \frac{1}{iω C}, Z_{p} = \frac{R}{1 + iω RC} .

(1)

For this series-parallel RC band-pass filter, the voltage divider gives, with

w = ω RC

H (ω) = \frac{V _{out}}{V _{in}} = \frac{Z _{p}}{Z _{s} + Z _{p}} = \frac{1}{3 + i ( w - w ^{- 1} )} = \frac{i w}{1 - w ^{2} + 3 i w} .

(2)

Taking the real part of

V_{0} H e^{iω t}

yields the steady output

V_{out} (t) = \frac{V _{0}}{9 + ( w - w ^{- 1} ) ^{2}} cos [ω t - arctan (\frac{w - w ^{- 1}}{3})] .

(3)

Initial charging contributes decaying transients if the source is switched on at a finite time; the displayed expression is the sinusoidal steady state. The output polarity matters: reversing the indicated output voltage would multiply

H

- 1

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