A transmission line is open-circuited at its load end. For a general (possibly lossy) line with propagation constant γ and characteristic impedance Z0, is the input impedance Z_in = Z0 * tanh(γ l)?

Introduction / Context:
Transmission-line input impedances depend on termination. Knowing the correct expressions for open and short circuits prevents design mistakes in matching networks and stubs.

Given Data / Assumptions:

• Characteristic impedance Z0.
• Propagation constant γ = α + jβ.
• Open-circuited termination (Z_L → ∞).

Concept / Approach:

The general input-impedance formula is Z_in = Z0 * (Z_L + Z0 tanh(γ l)) / (Z0 + Z_L tanh(γ l)). Taking the limit Z_L → ∞ gives Z_in(open) = Z0 * coth(γ l). For a short circuit (Z_L = 0), Z_in(short) = Z0 * tanh(γ l). Thus the statement equating the open-circuit input impedance to Z0 tanh(γ l) is incorrect.

Step-by-Step Solution:

Verification / Alternative check:

In the lossless limit (α → 0), coth(jβl) = −j cot(βl), which matches the well-known result Z_in(open) = −j Z0 cot(βl).

Why Other Options Are Wrong:

• True / length-specific claims: Confuse open- and short-circuit formulas.
• Only true for lossless lines: Still wrong; even lossless case gives −j Z0 cot(βl).

Common Pitfalls:

Memorizing tanh for both terminations; forgetting to apply limiting forms for Z_L → ∞ and Z_L → 0.

Final Answer:

False