Difficulty: Easy
Correct Answer: False
Explanation:
Introduction / Context:
RL filters complement RC filters. The placement of the output node determines whether the network is low-pass or high-pass. Misplacing the output node flips the filter type and changes the expected phase behavior.
Given Data / Assumptions:
Concept / Approach:
For RL low-pass (output across R), the transfer function is H(jω) = R / (R + jωL). Magnitude decreases as frequency increases due to the growing inductor reactance. The phase is ∠H = −arctan(ωL / R), a lagging phase that approaches −90° at very high frequency. Output across L instead yields H_HP(jω) = jωL / (R + jωL), a high-pass response.
Step-by-Step Solution:
Verification / Alternative check:
At ω → 0, H(jω) → 1 for output across R (low-pass). For output across L, H_HP(jω) → 0 at ω → 0, as expected for a high-pass shape, confirming the statement is incorrect.
Why Other Options Are Wrong:
Common Pitfalls:
Assuming the inductor’s voltage “lags” makes it low-pass. Voltage across L actually dominates at high frequency, which is characteristic of a high-pass output when measured at L.
Final Answer:
False
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