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Short-Circuited Lossless Line: For a line of length l with λ/4 < l < λ/2, the input impedance is predominantly of what nature?

Difficulty: Easy

Correct Answer: capacitive

Explanation:


Introduction / Context:
Input impedance of a short-circuited transmission line varies with electrical length. Recognizing whether it is inductive or capacitive in different length ranges is a core skill in RF design and matching network synthesis.


Given Data / Assumptions:

  • Lossless transmission line, short-circuited at the load end.
  • Electrical length satisfies λ/4 < l < λ/2.
  • Standard engineering sign convention: positive imaginary part → inductive, negative imaginary part → capacitive.


Concept / Approach:
For a short-circuited line, Zin = j Z0 tan(βl). Over l ∈ (λ/4, λ/2), βl ∈ (π/2, π). In this interval, tan(βl) is negative (approaches −∞ near λ/4 and 0⁻ near λ/2). Thus Zin has a negative imaginary part → capacitive behavior.


Step-by-Step Solution:

Zin = j Z0 tan(βl).For λ/4 < l < λ/2 ⇒ βl ∈ (π/2, π).In this interval, tan(βl) < 0.Therefore Im{Zin} = Z0 * tan(βl) < 0, so Zin is capacitive.


Verification / Alternative check:

At l = λ/3 (βl = 2π/3), tan(2π/3) = −√3 → Zin = jZ0(−√3), clearly capacitive.


Why Other Options Are Wrong:

Resistive: Occurs only at specific lengths (e.g., l = nλ/2), not in this interval.Inductive: Would require positive imaginary part, which does not occur here.None of the above: Incorrect since a clear capacitive nature is present.


Common Pitfalls:

Confusing tan behavior across quadrants; mixing up open- and short-circuited cases.


Final Answer:

capacitive
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