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A lossless transmission line with characteristic impedance Z0 is terminated in a pure reactance equal to −j·Z0. What is the resulting Voltage Standing Wave Ratio (VSWR) on the line?

Difficulty: Easy

Correct Answer: infinity

Explanation:


Introduction / Context:
This question tests standing-wave behavior on a lossless transmission line when the load is purely reactive. Understanding how the reflection coefficient magnitude relates to VSWR is core to RF and microwave engineering.


Given Data / Assumptions:

  • Lossless line with real Z0.
  • Load ZL = −j·Z0 (pure reactance, magnitude equal to Z0).
  • VSWR is defined using the magnitude of the reflection coefficient |Γ|.


Concept / Approach:
The reflection coefficient at the load is Γ = (ZL − Z0) / (ZL + Z0). For a purely reactive load, |Γ| = 1 because the load cannot absorb average power; it only stores and returns energy. VSWR = (1 + |Γ|) / (1 − |Γ|).


Step-by-Step Solution:

Let ZL = −j·Z0.Γ = (−j·Z0 − Z0) / (−j·Z0 + Z0) = (−1 − j) / (1 − j).|−1 − j| = √(1 + 1) = √2 and |1 − j| = √2 ⇒ |Γ| = 1.VSWR = (1 + 1) / (1 − 1) = 2 / 0 → infinity.


Verification / Alternative check:
If a load is purely reactive, average real power at the load is zero; thus, all incident power is reflected (|Γ| = 1), which forces VSWR to be unbounded (infinite).


Why Other Options Are Wrong:

2 or 10 or 5: These correspond to |Γ| strictly less than 1; not possible with a pure reactance on a lossless line.1: VSWR = 1 only for a perfect match (ZL = Z0), not for a reactive load.


Common Pitfalls:

Confusing the sign or value of the reactance with resistive matching; any purely reactive termination on a lossless line yields |Γ| = 1.


Final Answer:

infinity
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