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30 m line at 2 MHz, Γ SWR Zin

Short answer

Γ = (Z_L − Z₀)/(Z_L + Z₀) = 0.352∠56°. SWR = 2.09. Input impedance is 39.8 − j32.0 Ω, or 51.1∠−38.8° Ω.

A lossless transmission line with Z0 = 50 Ω is 30 m long and operates at 2 MHz. The line is terminated with a load Z_L = (60 + j40) Ω. If velocity (v) = 3×10⁸ m/s on the line, find (a) the reflection coefficient, (b) the standing wave ratio and the input impedance.

Given: Z₀ = 50 Ω, l = 30 m, f = 2 MHz, Z_L = 60 + j40 Ω, v = 3×10⁸ m/s.

λ = v/f = 3×10⁸/2×10⁶ = 150 m
β = 2π/150 = 0.04189 rad/m
βl = 0.04189 × 30 = 1.2566 rad = 72°   (l = 0.2λ)

(a) Reflection coefficient

Γ = (Z_L − Z₀)/(Z_L + Z₀) = (10 + j40)/(110 + j40)
  = 41.23∠75.96° / 117.05∠19.98°
  = 0.352 ∠55.98°

(b) Standing wave ratio

S = (1 + |Γ|)/(1 − |Γ|) = 1.352/0.648 = 2.09

Input impedance

tan 72° = 3.0777.

Z_in = Z₀ (Z_L + jZ₀ tanβl)/(Z₀ + jZ_L tanβl)

Numerator:   60 + j40 + j50(3.0777) = 60 + j193.9
Denominator: 50 + j(60 + j40)(3.0777) = 50 + j184.7 − 123.1 = −73.1 + j184.7

Z_in = 50 (60 + j193.9)/(−73.1 + j184.7)
     = 50 × 202.9∠72.8° / 198.6∠111.6°
     = 51.1 ∠−38.8° Ω
     = 39.8 − j32.0 Ω

Answer: Γ = 0.352∠56°; S = 2.09; Z_in = 39.8 − j32.0 Ω (51.1∠−38.8° Ω).

✓ Answer checked by Claude Opus 5.5 (high effort)

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