An RLC circuit resonates at f0 = 2000 Hz and exhibits a bandwidth BW = 250 Hz. Using the narrowband approximation, what is the high cutoff (upper half-power) frequency f2?

Introduction:For band-pass responses around resonance, the two half-power (−3 dB) frequencies f1 and f2 bound the passband. The bandwidth BW is defined as f2 − f1. When the bandwidth is small compared with the center frequency (narrowband), simple arithmetic approximations around f0 are accurate and widely used in design and quick checks.

Given Data / Assumptions:

• Resonance (center) frequency: f0 = 2000 Hz.
• Bandwidth: BW = 250 Hz.
• Narrowband approximation (BW ≪ f0), so f1 ≈ f0 − BW/2 and f2 ≈ f0 + BW/2.

Concept / Approach:Use the identity BW = f2 − f1 and the narrowband symmetry around f0. Thus f2 ≈ f0 + (BW / 2). This method is accurate enough for filters with modest Q where BW/f0 is small.

Step-by-Step Solution:Compute half-bandwidth: BW/2 = 250 / 2 = 125 HzUpper cutoff: f2 ≈ f0 + BW/2 = 2000 + 125 = 2125 HzLower cutoff (for reference): f1 ≈ 2000 − 125 = 1875 HzCheck: f2 − f1 = 2125 − 1875 = 250 Hz (matches given BW)

Verification / Alternative check:The exact relationship for narrowband resonators also satisfies f0 ≈ sqrt(f1 * f2). With the approximated f1 and f2 above, sqrt(1875 * 2125) ≈ 1999.999 Hz, essentially 2000 Hz, validating the arithmetic method.

Why Other Options Are Wrong:

• 2250 Hz: This would imply BW/2 = 250 Hz, doubling the bandwidth incorrectly.
• 1750 Hz: That is below f0 and cannot be the upper cutoff.
• 8.0 Hz: Irrelevant magnitude; not connected to the given data.
• 2000 Hz: The resonant (center) frequency, not the high cutoff.

Common Pitfalls:

• Confusing center frequency f0 with cutoff frequencies f1 and f2.
• Forgetting that BW is the difference f2 − f1, not f0 − f1 or f2 − f0 alone.
• Misapplying the geometric-mean identity outside the narrowband regime.

Final Answer:2125 Hz