VSWR → ∞: What Does It Say About the Load Termination? If the voltage standing-wave ratio (VSWR) on a line is infinite, which termination(s) are implied?

Difficulty: Easy

Correct Answer: either (a) or (c)

Explanation:


Introduction / Context:
VSWR quantifies mismatch on a transmission line. It is related to the reflection coefficient magnitude |Γ| by VSWR = (1 + |Γ|) / (1 − |Γ|). Understanding extreme values of VSWR helps identify the nature of the terminating load.


Given Data / Assumptions:

  • Lossless (or low-loss) line near the termination.
  • Standard definition of VSWR and reflection coefficient.


Concept / Approach:

VSWR tends to infinity when |Γ| → 1. An open circuit (ZL → ∞) gives Γ = +1; a short circuit (ZL = 0) gives Γ = −1; both have |Γ| = 1. Thus either open or short produces an infinite VSWR. Finite complex impedances produce |Γ| < 1 and hence finite VSWR.


Step-by-Step Solution:

1) Start with VSWR relation: VSWR = (1 + |Γ|) / (1 − |Γ|).2) For open: Γ = +1 → VSWR → ∞.3) For short: Γ = −1 → VSWR → ∞.4) Any finite ZL ≠ 0, ∞ results in |Γ| < 1 (for passive lossy loads), giving finite VSWR.


Verification / Alternative check:

Smith chart positions at the extreme right (open) and left (short) lie on the |Γ| = 1 circle, which corresponds to infinite VSWR.


Why Other Options Are Wrong:

Short only (a) or open only (c) are incomplete since both cause VSWR → ∞. “Complex impedance” is vague and not necessarily |Γ| = 1. A purely reactive load of finite magnitude typically still yields |Γ| < 1 unless it is open/short.


Common Pitfalls:

Assuming all reactive loads give infinite VSWR; forgetting that only |Γ| = 1 causes infinite ratio.


Final Answer:

either (a) or (c)

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