If the load impedance is much larger than the characteristic impedance (ZL » Z0), what is the approximate relationship between ZL, Z0, and VSWR?
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AZ0 = ZL (VSWR)
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BZL = Z0 (VSWR)
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CZL = Z0 (VSWR - 1)
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DZL ≈ Z0 * VSWR (for ZL » Z0)
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EZL ≈ Z0 / VSWR
Answer
Correct Answer: ZL ≈ Z0 * VSWR (for ZL » Z0)
Explanation
Introduction:This item checks your facility with the relationship between load mismatch and standing wave ratio (VSWR) on a line terminated in a real load that is much larger than the line impedance. In that limit, the VSWR connects directly to the impedance ratio.
Given Data / Assumptions:
- Purely resistive load ZL » Z0.
- Lossless or low-loss line for simplicity.
- Standard definitions: Γ = (ZL − Z0) / (ZL + Z0), VSWR = (1 + |Γ|) / (1 − |Γ|).
Concept / Approach:
When ZL » Z0, |Γ| ≈ (ZL − Z0) / (ZL + Z0) → (ZL / Z0 − 1) / (ZL / Z0 + 1) ≈ 1 − 2 Z0 / ZL. The corresponding VSWR ≈ (1 + |Γ|) / (1 − |Γ|) ≈ (2 − 2 Z0 / ZL) / (2 Z0 / ZL) ≈ ZL / Z0. Hence, ZL ≈ Z0 * VSWR.
Step-by-Step Solution:
1) Start from Γ = (ZL − Z0) / (ZL + Z0).2) Assume ZL / Z0 ≫ 1 ⇒ |Γ| ≈ 1 − 2 Z0 / ZL.3) Compute VSWR = (1 + |Γ|) / (1 − |Γ|) ≈ (2 − 2 Z0 / ZL) / (2 Z0 / ZL) = (ZL / Z0) − 1 ≈ ZL / Z0.4) Rearranged: ZL ≈ Z0 * VSWR.Verification / Alternative check:
Numerical example: Z0 = 50 Ω, ZL = 1000 Ω ⇒ exact VSWR ≈ 1000/50 = 20 (very close), confirming the approximation.
Why Other Options Are Wrong:
- Proportionalities using VSWR − 1 or inverses do not match the high-ratio limit.
- Equations setting Z0 = ZL(VSWR) invert the relationship incorrectly.
Common Pitfalls:
Applying the approximation outside its range; for modest mismatches, use exact formulas rather than the ZL/Z0 ≈ VSWR shortcut.
Final Answer:
ZL ≈ Z0 * VSWR (for ZL » Z0)