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

The ideal operational-amplifier rules are zero input currents and, when stable negative feedback maintains unsaturated linear operation, equal input voltages. Infinite open-loop gain motivates the latter; it must not be assumed to guarantee asymptotic stability for every external network.

For the numerical sketch and usefulness discussion, use the transfer function derived in part (c). The chosen values give $K=5$ and $\tau =RC=10^{-4}\mathrm{s}$. With $x=\omega \tau$,The formal amplitude falls from 5 to zero, with high-frequency slope $-40$ dB per decade. Its unwrapped phase rises from zero through $+90^{\circ }$ at $\omega =10^4{\mathrm{s}}^{-1}$ to $+180^{\circ }$; the corresponding characteristic frequency is 1.59 kHz. At that frequency the amplitude is 2.5.

These are formal sinusoidal particular-solution curves, not a stable settled filter response. For the formal sinusoidal response of an unstable linear system, replacing $i\omega$ by the Laplace variable $s$ gives $G(s)=5/(1-s\tau {)}^2$, with a repeated pole at $s=+1/\tau$. A homogeneous output contains $(A+Bt)e^{t/\tau }$ and grows, so real perturbations lead to supply-limited nonlinear behaviour. With equal positive resistors and capacitors in the frequency-setting network, the general denominator is $1+(3-K)s\tau +s^2{\tau }^2$: the Sallen-Key equal-component stability bound is $K<3$. For example $K=3-\sqrt{2}$ gives a Butterworth filter response. Thus the Sallen-Key low-pass filter architecture is useful for active filtering with suitable gain and damping, but the stated $K=5$ component set is unsuitable for linear filtering. The component values are retained in this diagnosis.^[ref]

Zero capacitance removes the capacitor branches. No current enters the non-inverting input, so no current flows through $R_1$ or $R_2$ and $V_{+}=V_{\mathrm{in}}$. The inverting divider gives $V_{-}=V_{\mathrm{out}}{R}_3/(R_3+R_4)$. Equating the two input voltages in the linear solution gives

Let $V_B$ be the non-inverting input voltage. The feedback divider gives $V_B=V_{\mathrm{out}}/K$. With $s=i\omega$, current conservation at this wavefunction node gives $(V_A-V_B)/R_2=s{C}_2{V}_B$, because the amplifier input draws no current. Hence

Current conservation at A, with the capacitor $C_1$ connected from A to the output as verified in the PDF, givesSubstitute $V_A=(1+s{R}_2{C}_2)V_B$ and $V_{\mathrm{out}}=K{V}_B$:Therefore, at $s=i\omega$,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.