Determining bandwidth of a resonant filter Is the bandwidth of a resonant (RLC) filter governed by both the quality factor Q and the resonant frequency f0 (for example, BW = f0 / Q)?
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ATrue
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BFalse
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CTrue only for series RLC, not parallel RLC
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DTrue only for active filters with op-amps
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EFalse unless Q ≥ 10
Answer
Correct Answer: True
Explanation
Introduction / Context:Bandwidth quantifies the width of the passband (or stopband) around resonance for band-pass or band-stop filters. The quality factor Q reflects how underdamped the resonance is. Their relationship guides component selection for target selectivity.
Given Data / Assumptions:
- Resonant frequency f0 for an RLC circuit.
- Quality factor Q defined as 2 * pi * (energy stored / energy dissipated per cycle) or, equivalently, f0 / BW for many resonant responses.
- Small-signal linear behavior.
Concept / Approach:
For classical second-order band-pass responses, BW = fH − fL and Q = f0 / BW, with f0 commonly near sqrt(fL * fH). This relation holds for both series and parallel RLC configurations when defined with standard half-power (−3 dB) points.
Step-by-Step Solution:
Identify fL and fH as the half-power frequencies.Compute BW = fH − fL.Relate to Q using Q = f0 / BW, so BW = f0 / Q.Thus, bandwidth depends on both f0 and Q; increasing Q narrows BW for fixed f0.Verification / Alternative check:
Component-level: for a series RLC with R, L, C, Q ≈ (1/R) * sqrt(L/C) and f0 = 1 / (2 * pi * sqrt(L * C)). Substituting shows that BW depends on R through Q and on f0 via L and C, reinforcing the stated dependency.
Why Other Options Are Wrong:
- It is not limited to series or active filters; the relationship is generic for second-order resonances.
- No minimum Q such as 10 is required for the formula to hold; it is definitional.
Common Pitfalls:
Confusing absolute bandwidth with fractional bandwidth (BW / f0). Also, mixing peak voltage bandwidth with power bandwidth; standard practice uses half-power points for BW.
Final Answer:
True