RSMTC Foundations Practice Exam

Compute the magnitude of Γ for Z0 = 50 Ω and ZL = 25 + j40 Ω.

|Γ| ≈ 0.40 |Γ| ≈ 0.56 |Γ| ≈ 0.75 |Γ| ≈ 0.20

Correct Answer:

|Γ| ≈ 0.20

Explanation:

The quantity being tested is the magnitude of the reflection coefficient on a transmission line, which is |Γ| = |ZL − Z0| / |ZL + Z0|. With Z0 = 50 and ZL = 25 + j40, compute the two magnitudes: ZL − Z0 = (25 − 50) + j40 = −25 + j40, whose magnitude is √((-25)² + 40²) = √(625 + 1600) = √2225 ≈ 47.17. ZL + Z0 = (25 + 50) + j40 = 75 + j40, whose magnitude is √(75² + 40²) = √(5625 + 1600) = √7225 = 85. Thus |Γ| ≈ 47.17 / 85 ≈ 0.554, about 0.56. So the correct magnitude is around 0.56, not 0.20; the listed option near 0.56 is the match, indicating a possible misprint in the provided answer.

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