Output tapping and filter type (repaired) — for a series RLC network driven by a source, if the output is taken across the resistor (R), does the resulting two-port behave as a band-pass filter?

Difficulty: Easy

Correct Answer: True

Explanation:


Introduction / Context:
Filter behavior in simple RLC networks depends on where you probe or take the output. A single series RLC path can realize different transfer functions depending on whether the output is measured across L, C, or R. This question clarifies the case when output is taken across R.


Given Data / Assumptions:

  • Single series RLC driven by a voltage source.
  • Output is the voltage across the resistor, VR.
  • Linear time-invariant conditions; moderate Q.


Concept / Approach:
The magnitude of VR is low at both very low and very high frequencies: far below resonance, current is limited by large XC; far above resonance, current is limited by large XL. At resonance (where XL = XC), impedance is minimum and current peaks, so VR = I * R peaks. This “low at extremes, high at center” behavior is characteristic of a band-pass filter centered at f0.


Step-by-Step Solution:

1) Write |I(jω)| = |V_source| / |Z_series|, with Z_series minimum at resonance.2) Compute VR = I * R; since I peaks at f0, so does VR.3) Observe that VR diminishes away from f0 in both directions.4) Conclude that the transfer |VR/Vs| exhibits a band-pass profile.


Verification / Alternative check:
Bode plots for |VR/Vs| in series RLC show a peak at f0 with attenuation on both sides, confirming band-pass behavior.


Why Other Options Are Wrong:

“False” or “always low-pass”: contradicts the current-peaking nature at resonance when measuring across R.“Only if Q>5/ideal components”: while Q affects sharpness and peak gain, the qualitative band-pass shape persists for finite Q with real components.


Common Pitfalls:
Assuming the network type (LPF/HPF/BPF) is intrinsic to the parts rather than the measurement node.


Final Answer:
True

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