A band-pass RLC circuit resonates at f0 = 150 kHz and the inductor has a coil quality factor Q = 30. Using the narrowband approximation, what are the −3 dB cutoff frequencies (F1 and F2) that define the passband?

Introduction:
Band-pass filters around a resonant frequency f0 are characterized by a bandwidth BW and a quality factor Q = f0 / BW. Determining the lower and upper cutoff frequencies (F1 and F2) from Q and f0 is a routine but vital task in RF design, selectivity analysis, and tuning stages.

Given Data / Assumptions:

• Resonant (center) frequency: f0 = 150 kHz.
• Coil quality factor: Q = 30 (assumed representative of the filter Q).
• Narrowband assumption applies (Q ≫ 1).

Concept / Approach:
For narrowband resonators, BW = f0 / Q. The cutoff frequencies are approximately symmetric around f0: F1 ≈ f0 − BW/2 and F2 ≈ f0 + BW/2. This symmetry gives a quick, accurate estimate when BW is small compared with f0.

Step-by-Step Solution:
Compute bandwidth: BW = f0 / Q = 150 kHz / 30 = 5 kHzHalf-bandwidth: BW/2 = 2.5 kHzLower cutoff: F1 ≈ 150 kHz − 2.5 kHz = 147.5 kHzUpper cutoff: F2 ≈ 150 kHz + 2.5 kHz = 152.5 kHzCheck: F2 − F1 = 5.0 kHz = BW (consistent)

Verification / Alternative check:
Using the geometric-mean identity for narrowband filters, f0 ≈ sqrt(F1 * F2). With F1 = 147.5 kHz and F2 = 152.5 kHz, sqrt(147.5 * 152.5) kHz ≈ 150.0 kHz, validating the approximation.

Why Other Options Are Wrong:

• 100.0–155.0 kHz: Far too wide; BW would be 55 kHz, not 5 kHz.
• 295.5 kHz–4500 kHz: Nonsensical relative to f0.
• 149,970–150,030 Hz: Extremely narrow (BW 60 Hz) inconsistent with Q = 30.
• 130–170 kHz: BW 40 kHz; contradicts Q = 30.

Common Pitfalls:

• Using BW itself as the offset from f0 instead of BW/2.
• Confusing coil Q with loaded Q; here we assume they are approximately equal.

Final Answer:
147.5 kHz to 152.5 kHz