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Two-wire 40 Ω line, input and output voltage

Short answer

Compute Z_in = 40(60 + j40 tanβl)/(40 + j60 tanβl) = 37.84 − j15.73 Ω, then use source voltage division and line propagation. Results: V_in ≈ 65.1∠−4.4° V and V_L ≈ 75.7∠−36.4° V.

Consider a two-wire 40 Ω line (Z0 = 40 Ω) connecting the source of 80 V, 400 kHz with series resistance 10 Ω to the load of Z_L = 60 Ω. The line is 75 m long and the velocity on the line is 2.5×10⁸ m/s. Find the voltage V_in at input end and V_L at output end of the transmission line.

Given: Z₀ = 40 Ω, V_g = 80 V, R_g = 10 Ω, Z_L = 60 Ω, f = 400 kHz, l = 75 m, v = 2.5×10⁸ m/s (lossless).

λ = v/f = 2.5×10⁸/4×10⁵ = 625 m
β = 2π/λ = 0.010053 rad/m
βl = 0.010053 × 75 = 0.7540 rad = 43.2°   (l = 0.12λ)
tan βl = 0.9391
Γ_L = (60 − 40)/(60 + 40) = 0.2

Step 1: Input impedance

Z_in = Z₀ (Z_L + jZ₀ tanβl)/(Z₀ + jZ_L tanβl)
     = 40 (60 + j37.56)/(40 + j56.35)
     = 37.84 − j15.73 Ω   (= 40.98∠−22.6°)

Step 2: Input voltage (voltage divider with the source)

V_in = V_g Z_in/(R_g + Z_in) = 80 (37.84 − j15.73)/(47.84 − j15.73)
     = 80 × 40.98∠−22.58° / 50.36∠−18.20°
     = 65.1 ∠−4.4° V

Step 3: Output voltage

On the line, V(d) = V⁺_L (e^{jβd} + Γ_L e^{−jβd}), where d is the distance from the load. At d = l:

V_in = V⁺_L (e^{j43.2°} + 0.2 e^{−j43.2°})
     = V⁺_L (0.729 + j0.685 + 0.146 − j0.137)
     = V⁺_L (0.875 + j0.548) = V⁺_L (1.0325∠32.07°)

V⁺_L = 65.1∠−4.4° / 1.0325∠32.07° = 63.08 ∠−36.4° V

V_L = V⁺_L (1 + Γ_L) = 1.2 × 63.08∠−36.4° = 75.7 ∠−36.4° V

Answer: V_in ≈ 65.1∠−4.4° V; V_L ≈ 75.7∠−36.4° V.

Check: P_in = ½ |V_in|² Re(Z_in)/|Z_in|² = ½ × 65.1² × 37.84/40.98² = 47.7 W; P_L = ½ × 75.7²/60 = 47.7 W. ✓ (lossless line).

✓ Answer checked by Claude Opus 5.5 (high effort)

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