SWR → ∞ conditions on a transmission line For a lossless line with characteristic impedance Z0, the standing-wave ratio (SWR) becomes infinite when |Γ| = 1. Which load terminations guarantee |Γ| = 1?
Select the correct combination:

Introduction / Context:SWR quantifies how severe standing waves are due to reflection from a load. Infinite SWR corresponds to complete reflection (|Γ| = 1). This occurs for specific classes of terminations and has direct implications for power handling and node voltages on lines.

Given Data / Assumptions:

• Lossless line with characteristic impedance Z0 (real > 0).
• Loads considered: (1) short circuit, (2) arbitrary complex impedance, (3) open circuit, (4) purely reactive load jX.
• Standard definitions: Γ = (ZL − Z0) / (ZL + Z0), VSWR = (1 + |Γ|) / (1 − |Γ|).

Concept / Approach:

|Γ| = 1 whenever the load is lossless and not equal to Z0 in magnitude/phase such that the denominator magnitude equals the numerator magnitude. Shorts (ZL = 0) and opens (ZL → ∞) obviously yield |Γ| = 1. Purely reactive loads ZL = jX also give |Γ| = 1 because |jX − Z0| = |jX + Z0| for real Z0, leading to unit magnitude reflection (phase depends on X). A general “complex impedance” may have |Γ| < 1 if it possesses a resistive part not equal to 0 and not infinite; therefore it does not guarantee |Γ| = 1.

Step-by-Step Solution:

Verification / Alternative check:

Compute VSWR from |Γ|; |Γ| = 1 implies denominator 0 → SWR → ∞, matching intuition of complete reflection.

Why Other Options Are Wrong:

Any option including “complex impedance” as a guarantee is overbroad; many complex impedances (with resistive parts) yield |Γ| < 1. Options excluding short/open/pure reactance miss guaranteed |Γ| = 1 cases.

Common Pitfalls:

Believing any non-matched impedance leads to |Γ| = 1; in reality, |Γ| spans (0, 1] depending on mismatch severity.

Final Answer:

1, 3, and 4 (short circuit; open circuit; pure reactance).