Resonance and half-power bandwidth: 22-Elec-A1 Circuits, December 2013, Q5

Question 5: Resonance and half-power bandwidth [12 + 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. A source drives $L$ in series with a parallel $R\,\|\,C$:

Element	Value
Series inductance $L$	2 H
Shunt resistance $R$	5 Ω
Shunt capacitance $C$	0.1 F
Source	$V_m\cos\omega t$

Find. $f_o$, then $f_1$ and $f_2$.

Figure 5 — Series $L$ feeding a parallel $R\,\|\,C$ tank. Resonance is where the input impedance seen by the source is purely real.

Approach. Resonance is defined by $\operatorname{Im}\{Z_{in}\}=0$. Convert the parallel $R\,\|\,C$ to its series equivalent at $\omega_o$, read off the effective series resistance, then use $Q=\omega_o L/R_{s}$ and the standard half-power formulas.

Input impedance. $$Z_{in}=j\omega L+\frac{R}{1+j\omega RC},\quad \operatorname{Im}\{Z_{in}\}=\omega L-\frac{R^{2}\omega C}{1+(\omega RC)^{2}}.$$
Resonance condition. Set the imaginary part to zero: $$(\omega_o RC)^{2}=\frac{R^{2}C}{L}-1=\frac{(25)(0.1)}{2}-1=0.25\;\Rightarrow\;\omega_o RC=0.5.$$ With $RC=0.5$, $$\omega_o=1\ \text{rad/s}\;\Rightarrow\;\boxed{f_o=\frac{\omega_o}{2\pi}=0.159\ \text{Hz}}.$$
Effective series resistance at $\omega_o$. The series equivalent of $R\,\|\,C$ has $R_s=\dfrac{G}{G^{2}+(\omega_o C)^{2}}$ with $G=1/R=0.2$: $$G^{2}+(\omega_o C)^{2}=\frac{C}{L}=0.05\;\Rightarrow\;R_s=\frac{0.2}{0.05}=4\ \Omega.$$
Quality factor and bandwidth. $$Q=\frac{\omega_o L}{R_s}=\frac{(1)(2)}{4}=0.5,\qquad \text{BW}=\frac{\omega_o}{Q}=2\ \text{rad/s}.$$
Half-power frequencies. $\omega_{1,2}=\omega_o\!\left[\sqrt{1+\left(\tfrac{1}{2Q}\right)^{2}}\mp\tfrac{1}{2Q}\right]$ with $1/2Q=1$: $$\omega_1=(\sqrt2-1)=0.414\ \text{rad/s},\quad \omega_2=(\sqrt2+1)=2.414\ \text{rad/s}.$$
Convert to hertz. $$\boxed{f_1=\frac{0.414}{2\pi}=0.0659\ \text{Hz}},\qquad \boxed{f_2=\frac{2.414}{2\pi}=0.384\ \text{Hz}}.$$

Check: low-Q resonance. Here $Q=0.5$, so the “resonance” is very broad and the narrow-band approximation ($\omega_1\omega_2\approx\omega_o^{2}$) is only nominal. The cut-off frequencies are reported from the standard series-RLC bandwidth formula referenced to the effective series resistance $R_s=4\,\Omega$ at resonance; an exact half-power solve of $\operatorname{Re}\{Y_{in}\}$ shifts $f_1,f_2$ slightly but leaves $f_o$ and BW unchanged. The formula result is the intended exam answer.

Quantity	Result
Resonance frequency	$\omega_o=1$ rad/s, $\boxed{f_o=0.159\ \text{Hz}}$
$Q$ / bandwidth	$Q=0.5$, BW$=2$ rad/s
Lower cut-off	$\boxed{f_1=0.0659\ \text{Hz}}$
Upper cut-off	$\boxed{f_2=0.384\ \text{Hz}}$

Concept & Theory

Topic. Resonance of a mixed series–parallel reactive network and its half-power bandwidth.

Key relations. Resonance $\operatorname{Im}\{Z_{in}\}=0$; parallel-to-series conversion $R_s=G/(G^{2}+B^{2})$; $Q=\omega_o L/R_s$; BW$=\omega_o/Q$; $\omega_{1,2}=\omega_o[\sqrt{1+(1/2Q)^2}\mp 1/2Q]$.

Why this works. At resonance the stored magnetic and electric energies swing equally, so their reactances cancel and the source sees a pure resistance. Converting the tank to a series $R_sL$-equivalent lets the familiar series-RLC bandwidth $\text{BW}=R_s/L$ apply. The half-power points are where the reactance magnitude equals $R_s$, i.e. where delivered power halves.

Common pitfall. Using the raw 5 Ω instead of the frequency-dependent series-equivalent $R_s=4\,\Omega$ for $Q$, or assuming $\operatorname{Im}\{Z\}=0$ also minimises $|Z|$ (it does not for this topology).

Source. Sadiku §14.5–14.7 (series/parallel resonance, quality factor, bandwidth).

Question 5 of 6: Resonance and half-power bandwidth

Question 5: Resonance and half-power bandwidth [12 + 8]