Second-order filter — resonance, gain and cut-off: 22-Elec-A1 Circuits, May 2014, Q5

Question 5: Second-order filter — resonance, gain and cut-off [8 + 2 + 2 + 8]

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.

Element	Value
Series inductance $L$	2 mH
Shunt resistance $R$	5 Ω
Shunt capacitance $C$	0.5 µF
Source	$V_{in}=10\cos\omega t$

Find. $V_{out}$ and the gain at resonance, the filter type, and the cut-off frequency.

Figure 5 — Series $L$ feeding a parallel $R\,\|\,C$; $V_{out}$ is taken across the parallel combination. At DC the inductor is a short and $V_{out}=V_{in}$; at high frequency $C$ shorts the output — a low-pass shape.

Approach. Form the voltage-divider transfer function $H(\omega)=Z_{RC}/(Z_L+Z_{RC})$, reduce it to standard second-order form to read $\omega_0$ and $Q$, evaluate at $\omega_0=1/\sqrt{LC}$, then find the half-power frequency from $|H|=1/\sqrt2$.

Transfer function. With $Z_L=j\omega L$ and $Z_{RC}=R\,\|\,\tfrac{1}{j\omega C}=\tfrac{R}{1+j\omega RC}$, $$H(\omega)=\frac{Z_{RC}}{Z_L+Z_{RC}}=\frac{R}{R(1-\omega^{2}LC)+j\omega L} =\frac{1}{(1-\omega^{2}LC)+j\omega L/R}.$$ In standard form $H(s)=\dfrac{1/LC}{s^{2}+s/(RC)+1/LC}$, so $\omega_0=1/\sqrt{LC}$ and $Q=R\sqrt{C/L}$.
Resonance frequency. $$\omega_0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(2\times10^{-3})(0.5\times10^{-6})}}=\boxed{3.162\times10^{4}\ \text{rad/s}}\;\;(f_0=5.03\ \text{kHz}).$$
Output at resonance (a). At $\omega=\omega_0$ the real part of the denominator vanishes $(1-\omega_0^{2}LC=0)$, leaving $$H(\omega_0)=\frac{1}{j\omega_0 L/R}=\frac{R}{j\omega_0 L}=-j\,\frac{R}{\omega_0 L}=-j\,\frac{5}{63.25}=0.0791\angle{-}90^{\circ}.$$ With $|V_{in}|=10$, $$V_{out}(\omega_0)=0.0791\times 10=\boxed{0.791\ \text{V}}\ \ (\angle{-}90^{\circ}).$$
Gain at resonance (b). $$|H(\omega_0)|=\frac{R}{\omega_0 L}=Q=R\sqrt{\tfrac{C}{L}}=\boxed{0.0791}\;\;({-}22.0\ \text{dB}).$$
Filter type (c). $H(0)=1$ and $H\to0$ as $\omega\to\infty$, and $Q=0.079\ (<0.5)$ so the response is over-damped with its maximum at DC — this is a second-order low-pass filter. (The resonance $\omega_0$ lies well inside the stop-band, which is why the gain there is small.)
Cut-off (half-power) frequency (c). Set $|H|^{2}=\tfrac12$: $$(1-\omega^{2}LC)^{2}+\left(\tfrac{\omega L}{R}\right)^{2}=2.$$ Writing $x=\omega^{2}$, $(LC)^{2}x^{2}+\big[(L/R)^{2}-2LC\big]x-1=0$; the single positive root gives $$\omega_c=\boxed{2.516\times10^{3}\ \text{rad/s}}\;\;(f_c=400\ \text{Hz}).$$ (The quadratic’s other root is negative, i.e. non-physical, so there is one cut-off — as expected for a low-pass.)

Quantity	Result
Resonance frequency $\omega_0$	$3.162\times10^{4}$ rad/s ($5.03$ kHz)
$V_{out}$ at resonance	$\boxed{0.791\ \text{V}\ \angle{-}90^{\circ}}$
Gain at resonance	$\boxed{0.0791}$ ($-22.0$ dB)
Filter type	2nd-order low-pass
Cut-off frequency $\omega_c$	$\boxed{2.516\times10^{3}\ \text{rad/s}}$ ($400$ Hz)

Check: With $R=5\ \Omega$ the network is heavily over-damped ($Q\approx0.079$), so $V_{out}$ falls monotonically from $V_{in}$ at DC — there is no resonant rise, and only one half-power frequency exists. The two real poles are at $2.5\times10^{3}$ and $3.98\times10^{5}$ rad/s; the lower one sets the $-3$ dB corner.

Concept & Theory

Topic. Second-order $RLC$ transfer function: resonance frequency, quality factor, filter classification and the half-power (cut-off) frequency.

Key relations. $\omega_0=1/\sqrt{LC}$; $Q=R\sqrt{C/L}$ (for this series-$L$/shunt-$RC$ topology); half-power where $|H|=|H|_{max}/\sqrt2$; over-damped when $Q<\tfrac12$.

Why this works. The divider ratio is fixed by the impedance contrast between the series $L$ and the shunt $R\|C$. At DC the inductor vanishes and the output equals the input; as frequency rises the inductor’s impedance grows and the capacitor’s falls, both pulling the output down — a low-pass characteristic whose sharpness is set by $Q$.

Common pitfall. Assuming “resonance” must be a peak: with $Q<0.5$ the magnitude peaks at DC, not at $\omega_0$. Also, defining resonance as $\operatorname{Im}\{Z_{in}\}=0$ has no real solution for these values, so $\omega_0=1/\sqrt{LC}$ is the meaningful resonance measure.

Source. Sadiku §14.5–14.8 (series/parallel resonance, passive filters, half-power frequencies and bandwidth).

Question 5 of 6: Second-order filter — resonance, gain and cut-off

Question 5: Second-order filter — resonance, gain and cut-off [8 + 2 + 2 + 8]