Question 3 of 7

Question 3 (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.

Approach. (a) general active-filter advantages. (b) apply the ideal-op-amp idealizations directly. (c) write nodal (KCL) equations for Stage 1 in the Laplace domain to get $H_1(s)=V_2/V_1$, multiply by Stage 2's non-inverting gain $H_2=V_3/V_2$. (d) substitute $s=j(0.5\omega_o)$ into the derived $F(s)$ and evaluate magnitude/phase as a complex number.

1. Part (a) — advantages of active filters. (i) They can provide gain (and even a steep, sharp roll-off) using only R, C and op-amps — no inductors are needed, which are bulky, lossy and hard to fabricate in IC form at audio/instrumentation frequencies. (ii) The op-amp's low output impedance buffers each stage from the next, so cascaded stages do not load one another and the overall transfer function is simply the product of each stage's individual transfer function — a property passive RLC filters do not have.
2. Part (b) — $R_{in}$, $R_{out}$. With ideal op-amps, each op-amp's input terminals draw zero current (infinite input impedance) and each op-amp's output behaves as an ideal voltage source (zero output impedance) even after feedback is applied. Since $R_{in}$ and $R_{out}$ are marked at op-amp input/output nodes in Figure P3: $$R_{in}=\boxed{\infty}\qquad R_{out}=\boxed{0\ \Omega}$$
3. Part (c) — deriving $F(s)$. Stage 1: in Figure P3, $V_1$ feeds a capacitor $C$ into node $V_x$; from $V_x$ a second $C$ goes to the op-amp’s $+$ input (node $V_+$), which returns to ground through $R$ (current $I_3$), and a resistor $R$ runs from $V_x$ to the output $V_2$ (current $I_2$). The op-amp is a unity-gain follower, so $V_+=V_2$, and its $+$ input draws no current. KCL at $V_+$ (current through the second $C$ = current $I_3$ in $R$ to ground): $$sC(V_x-V_2)=\frac{V_2}{R}\ \Rightarrow\ V_x=V_2\,\frac{1+sRC}{sRC}$$ KCL at $V_x$ (current $I_1$ in from $V_1$ = current on through the second $C$ + current $I_2$ through $R$ to the output): $$sC(V_1-V_x)=sC(V_x-V_2)+\frac{V_x-V_2}{R}$$ With $x=sRC$, the first equation gives $V_x-V_2=V_2/x$; multiplying the second by $R$ gives $x(V_1-V_x)=(1+x)V_2/x$, so $$xV_1=xV_x+\frac{(1+x)V_2}{x}=(1+x)V_2+\frac{(1+x)V_2}{x}=\frac{(1+x)^2}{x}\,V_2\ \Rightarrow\ \frac{V_2}{V_1}=\frac{x^2}{(1+x)^2}$$ Substituting $x=sRC=s/\omega_o$ with $\omega_o=1/RC$: $$H_1(s)=\frac{V_2(s)}{V_1(s)}=\frac{s^2}{s^2+\dfrac{2}{RC}s+\dfrac{1}{(RC)^2}}=\frac{s^2}{s^2+2\omega_os+\omega_o^2},\qquad\omega_o=\frac{1}{RC}$$ Stage 2: a standard non-inverting amplifier with gain-setting resistors $R$ (to ground) and $9R$ (feedback): $$H_2(s)=\frac{V_3(s)}{V_2(s)}=1+\frac{9R}{R}=10$$ Multiplying the two stages (they do not load each other, since the op-amp output driving Stage 2 is ideal): $$F(s)=H_1(s)H_2(s)=\boxed{\dfrac{10s^2}{s^2+2\omega_os+\omega_o^2}}\qquad\checkmark$$
4. Part (d) — magnitude and phase at $\omega=0.5\omega_o$. Let $x=s/\omega_o=j0.5$ (normalizing frequency to $\omega_o$): $$\frac{F(j\omega)}{10}=\frac{x^2}{x^2+2x+1}=\frac{(j0.5)^2}{(j0.5)^2+2(j0.5)+1}=\frac{-0.25}{0.75+j1.0}$$ The denominator has magnitude $\sqrt{0.75^2+1^2}=1.25$ and phase $\arctan(1/0.75)=53.13^\circ$; the numerator is real and negative (magnitude 0.25, phase $180^\circ$): $$\left|\frac{F}{10}\right|=\frac{0.25}{1.25}=0.2\ \Rightarrow\ |F|=10(0.2)=\boxed{2.0}\quad\Rightarrow\quad 20\log_{10}(2.0)=\boxed{6.02\ \text{dB}}$$ $$\angle F=180^\circ-53.13^\circ=\boxed{126.87^\circ}$$ (the $\times10$ gain factor is real and positive, so it does not shift the phase.)