Short-Circuited Lossless Line: For a line of length l with λ/4

Question

Solve this multiple-choice question and choose the correct option.

Curioustab · Accepted Answer

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

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

Answer

Explanation