 Solution

ID: 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^{iω t}

Z_{1} = R_{1} + 1/ (iω C)

V_{in} / Z_{1} = - V_{out} / R_{2}

, hence

H (ω) = \frac{V _{out}}{V _{in}} = - \frac{iω R _{2} C}{1 + iω R _{1} C}, ∣ H ∣ = \frac{ω R _{2} C}{1 + ω ^{2} R _{1}^{2} C ^{2}} .

This is an inverting high-pass filter. Its phase can be chosen continuously as

ar g H = - \frac{π}{2} - tan^{- 1} (ω R_{1} C) .

Thus

V_{out} (t) = V_{0} ∣ H ∣ cos (ω t + ar g 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

9 0^{\circ} + tan^{- 1} (ω R_{1} C)

, between

9 0^{\circ}

and

18 0^{\circ}

. At

ω R_{1} C = 1

the lag is

13 5^{\circ}

; only for

ω R_{1} C ≫ 1

does it approach

18 0^{\circ}

. Being within the amplifier's own bandwidth does not specify the external RC corner frequency.

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