 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2015/paper-1/section-a/5/solution

negative feedback and the ideal operational amplifier's virtual short hold the inverting node at zero potential, with no current entering either input. The parallel feedback resistor and capacitor have electrical impedance

Z_{f} = (R_{f}^{- 1} + iω C)^{- 1} = \frac{R _{f}}{1 + iω R _{f} C} .

Kirchhoff's current law at the inverting node gives

V_{in} / R_{i} = - V_{out} / Z_{f}

. For the component values read from the circuit,

G (f) = \frac{V _{out}}{V _{in}} = - \frac{1000}{1 + i 2 π f ( 1 s )}, ∣ G (f) ∣ = \frac{1000}{1 + ( f /0.15915 Hz ) ^{2}} .

f_{c} = (2 π R_{f} C)^{- 1} = 0.15915

Hz. At the corner,

∣ G ∣ = 707.1

. Above it the gain approaches

1/ (2 π f R_{i} C) = 159.15/ f

, with

f

in Hz: a slope of

- 1

on logarithmic axes, or

- 20

0.01, 1, 10, 100

Hz the magnitudes are approximately

998, 157.2, 15.91, 1.592

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