In the series RLC circuit in the figure, L = 1 mH, C = 0.1 µF and R = 50 Ω. Which of the following is closest to the frequency of the damped oscillation that appears in the step response?

Answer: A. About 15.4 kHz

Method 1 (roots of the characteristic equation): the roots of Ls² + Rs + 1/C = 0 are s = −α ± jωd, with α = R/(2L) = 50/(2 × 10⁻³) = 2.5×10⁴ s⁻¹ and ωn = 1/√(LC) = 1×10⁵ rad/s. ωd = √(ωn² − α²) = √(10¹⁰ − 6.25×10⁸) ≈ 9.68×10⁴ rad/s, so fd = ωd/(2π) ≈ 15.4 kHz. Method 2 (from the damping ratio): ζ = α/ωn = 0.25, so ωd = ωn√(1 − ζ²) = 10⁵ × √0.9375 ≈ 9.68×10⁴ rad/s, the same result. The response oscillates at this frequency while decaying inside the envelope e^(−αt).

• A. Correct. fd = ωn√(1 − ζ²)/(2π) ≈ 15.4 kHz.
• B. About 15.9 kHz is the natural frequency fn = ωn/(2π); it ignores the drop in frequency due to damping.
• C. 100 kHz mistakes ωn = 1×10⁵ rad/s for a value in Hz.