Question 3 of 7: Inverting Active Filter with T-Network Feedback

Question 3: Inverting Active Filter with T-Network Feedback (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $R_1=10\text{k}\Omega$, $C_1=0.1\,\mu\text{F}$ (series input impedance); $R_2=R_3=R_4=40\text{k}\Omega$ (T-network feedback); ideal op-amp; part (c) $V_i(t)=10\sin(120\pi t)\text{V}$ ($f=60\text{Hz}$).

Find. (a) $V_o(j\omega)/V_i(j\omega)$; (b) $3\text{dB}$ corner(s), sketch; (c) $V_o(t)$ at $60\text{Hz}$.

Approach. Reduce the resistive T-network (a standard trick for synthesizing a large effective feedback resistance from moderate resistor values) to a single equivalent feedback resistance $R_{f,\text{eq}}$ by nodal analysis at its mid-node, then treat the circuit as an ordinary inverting amplifier with input impedance $Z_{in}=R_1+1/(j\omega C_1)$ and feedback $R_{f,\text{eq}}$.

1. Part (a) — T-network reduction and transfer function. Let $I$ be the current forced into the T-network at the (virtual-ground) inverting node; with $V_o$ the far end and the mid-node voltage $V_M$, KCL at $V_M$ (no current into the ideal op-amp input) gives the standard series-series/shunt T-to-equivalent-resistor result $$R_{f,\text{eq}}=R_2+R_4+\frac{R_2R_4}{R_3}=40+40+\frac{40\times40}{40}=\boxed{120\text{ k}\Omega}$$ (three $40\text{k}\Omega$ resistors synthesizing an effective $120\text{k}\Omega$ feedback path without one large physical resistor). The circuit is then an ordinary inverting amplifier: $$\frac{V_o(j\omega)}{V_i(j\omega)}=-\frac{R_{f,\text{eq}}}{Z_{in}(j\omega)}=-\frac{R_{f,\text{eq}}}{R_1+\dfrac{1}{j\omega C_1}}=\boxed{-\dfrac{j\omega R_{f,\text{eq}}C_1}{1+j\omega R_1C_1}}$$ — a single-pole high-pass shape (zero at DC, pole at $\omega_p=1/(R_1C_1)$).
2. Part (b) — frequency response and $3\text{dB}$ corner. The passband (high-frequency) gain is $$A_{v,\text{HF}}=-\frac{R_{f,\text{eq}}}{R_1}=-\frac{120\text{k}\Omega}{10\text{k}\Omega}=\boxed{-12\text{ V/V}}\ (21.6\text{ dB})$$ and the single $3\text{dB}$ corner (where the response is $3\text{dB}$ below the flat passband gain, rising at $+20\text{dB/dec}$ below it) is $$f_{3\text{dB}}=\frac{1}{2\pi R_1C_1}=\frac{1}{2\pi(10\text{k}\Omega)(0.1\,\mu\text{F})}=\boxed{159.2\text{ Hz}}$$
3. Part (c) — response at $60\text{Hz}$. With $\omega=120\pi\text{ rad/s}$ ($f=60\text{Hz}$), $\omega R_1C_1=120\pi\times10^{-3}=0.377$: $$H(j\omega)=-\frac{12\times j0.377}{1+j0.377}\ \Rightarrow\ |H|=4.23,\ \angle H=-110.7^{\circ}$$ $$V_o(t)=10\times4.23\,\sin(120\pi t-110.7^{\circ})=\boxed{42.3\sin(120\pi t-110.7^{\circ})\text{ V}}$$