Bandwidth from quality factor: A tuned RLC circuit resonates at f0 = 150 kHz and has quality factor Q = 30. Determine the -3 dB bandwidth BW and the approximate half-power frequencies f1 and f2 (centered around resonance).

Introduction:
Quality factor (Q) characterizes the sharpness of resonance in RLC circuits. For narrowband resonant circuits, Q links the center frequency f0, the -3 dB bandwidth BW, and the half-power (cutoff) frequencies f1 and f2 that bound the passband of a bandpass behavior.

Given Data / Assumptions:

• Resonant frequency f0 = 150 kHz
• Quality factor Q = 30
• Narrowband assumption valid (Q ≫ 1)

Concept / Approach:

For a bandpass resonant network, BW = f2 - f1 and Q = f0 / BW. Hence BW = f0 / Q. For high Q, the half-power frequencies are approximately symmetric about f0, so f1 ≈ f0 - BW/2 and f2 ≈ f0 + BW/2.

Step-by-Step Solution:

Verification / Alternative check:

Since BW is much less than f0 (5 kHz ≪ 150 kHz), the symmetry assumption holds well. Also, Q = f0 / BW = 150/5 = 30, matching the given Q.

Why Other Options Are Wrong:

• 100.0 kHz to 155.0 kHz: BW = 55 kHz ⇒ Q ≈ 2.7, not 30.
• 4500 kHz to 295.5 kHz: Nonsensical ordering and values.
• 149,970 Hz to 150,030 Hz: BW = 60 Hz ⇒ Q = 2500, not 30.
• 140.0 kHz to 160.0 kHz: BW = 20 kHz ⇒ Q = 7.5, not 30.

Common Pitfalls:

• Confusing BW = f2 - f1 with fractional bandwidth (BW / f0).
• Forgetting that higher Q means narrower bandwidth.

Final Answer:

147.5 kHz to 152.5 kHz