Chapter 7 Practice Problems
Practice Problems
Problem 1 — Series RLC Parameters
A series RLC circuit has R = 10 Ω, L = 100 mH, C = 10 μF.
(a) Calculate the undamped natural frequency \(\omega_0\) (in rad/s) and \(f_0\) (in Hz).
(b) Calculate the damping coefficient \(\alpha\).
(c) Classify the response as overdamped, critically damped, or underdamped.
(d) Calculate the quality factor Q and bandwidth.
Solution
(a) Natural frequency:
[\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.1 \times 10 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-6}}} = 1000\text{ rad/s}] [f_0 = \frac{\omega_0}{2\pi} = \frac{1000}{6.283} = 159.2\text{ Hz}]
(b) Damping coefficient for series RLC:
(c) Since \(\alpha = 50 < \omega_0 = 1000\), the circuit is underdamped. Damping ratio: \(\zeta = \alpha/\omega_0 = 50/1000 = 0.05\).
(d) Quality factor and bandwidth:
[Q = \frac{\omega_0 L}{R} = \frac{1000 \times 0.1}{10} = 10] [\text{BW} = \frac{f_0}{Q} = \frac{159.2}{10} = 15.9\text{ Hz}]
Problem 2 — Critical Damping Design
A series RLC circuit has L = 50 mH and C = 100 nF.
(a) What value of R produces critical damping?
(b) What is the natural frequency of this circuit?
(c) With R found in (a), write the step response form (leaving constants A1 and A2 to be determined by initial conditions).
Solution
(a) Natural frequency first: \(\omega_0 = 1/\sqrt{LC} = 1/\sqrt{0.05 \times 100\times10^{-9}} = 1/\sqrt{5\times10^{-9}} = 14{,}142\text{ rad/s}\)
Critical damping requires \(\alpha = \omega_0\):
(b) \(f_0 = 14{,}142 / (2\pi) = 2{,}251\text{ Hz} \approx 2.25\text{ kHz}\)
(c) For critical damping, the step response form is:
where \(\alpha = \omega_0 = 14{,}142\text{ rad/s}\) and \(A_1, A_2\) are determined from initial conditions \(x(0^+)\) and \(\dot{x}(0^+)\).
Problem 3 — Underdamped Step Response
A series RLC circuit has R = 5 Ω, L = 100 mH, C = 25 μF. A 10 V step is applied at t = 0 (zero initial conditions).
(a) Calculate \(\omega_0\), \(\alpha\), and \(\omega_d\).
(b) Determine the damping ratio \(\zeta\) and percent overshoot.
(c) Estimate the settling time (±2% criterion: \(t_s \approx 4/\alpha\)).
Solution
(a) Circuit parameters:
[\omega_0 = \frac{1}{\sqrt{0.1 \times 25\times10^{-6}}} = \frac{1}{\sqrt{2.5\times10^{-6}}} = 632.5\text{ rad/s}] [\alpha = \frac{R}{2L} = \frac{5}{0.2} = 25\text{ rad/s}] [\omega_d = \sqrt{\omega_0^2 - \alpha^2} = \sqrt{632.5^2 - 25^2} = \sqrt{400{,}006 - 625} = \sqrt{399{,}381} = 631.9\text{ rad/s}]
(b) Damping ratio and overshoot:
[\zeta = \frac{\alpha}{\omega_0} = \frac{25}{632.5} = 0.0395] [\%\text{OS} = 100\,e^{-\pi \times 0.0395/\sqrt{1-0.0395^2}} = 100\,e^{-0.124} = 100 \times 0.883 = 88.3\%]
This very low damping ratio produces extreme overshoot — the circuit nearly oscillates without decay.
(c) Settling time:
Problem 4 — Resonance in Series RLC
A series RLC band-pass filter has L = 10 mH, C = 250 pF, R = 20 Ω.
(a) Calculate the resonant frequency in Hz.
(b) Calculate the quality factor Q.
(c) Calculate the 3 dB bandwidth.
(d) At resonance, what is the impedance of the circuit?
Solution
(a) Resonant frequency:
[\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10\times10^{-3} \times 250\times10^{-12}}} = \frac{1}{\sqrt{2.5\times10^{-12}}} = 632{,}456\text{ rad/s}] [f_0 = \frac{632{,}456}{2\pi} = 100{,}658\text{ Hz} \approx 100.7\text{ kHz}]
(b) Quality factor:
(c) Bandwidth:
(d) At resonance, \(Z_L = j\omega_0 L\) and \(Z_C = 1/(j\omega_0 C)\) are equal and opposite, so they cancel. Total impedance = R = 20 Ω (purely resistive).
Problem 5 — Parallel RLC Circuit
A parallel RLC circuit has R = 1 kΩ, L = 5 mH, C = 200 pF.
(a) Calculate the resonant frequency and damping coefficient for the parallel configuration (\(\alpha = 1/(2RC)\)).
(b) Calculate Q and bandwidth.
(c) At resonance, what is the admittance of the circuit? What current flows from the source if 10 V is applied?
Solution
(a) Resonant frequency:
[\omega_0 = \frac{1}{\sqrt{5\times10^{-3} \times 200\times10^{-12}}} = \frac{1}{\sqrt{10^{-12}}} = 10^6\text{ rad/s}] [f_0 = \frac{10^6}{2\pi} = 159.2\text{ kHz}]
Parallel RLC damping coefficient:
(b) Since \(\alpha > \omega_0\), this is overdamped. \(Q = \omega_0/(2\alpha) = 10^6/(5\times10^6) = 0.2\).
(c) At resonance, \(Z_L\) and \(Z_C\) cancel in the parallel combination (they present a very high combined impedance). The circuit reduces to R. Admittance \(Y = 1/R = 1/1000 = 1\text{ mS}\). Current: \(I = V \times Y = 10 \times 10^{-3} = 10\text{ mA}\).