Air line 400 Ω, SWR and ZL

Given: air line, Z₀ = 400 Ω, f = 200 MHz, Z_in = 200 − j200 Ω.

λ = c/f = 3×10⁸ / 2×10⁸ = 1.5 m
β = 2π/λ = 4.189 rad/m

(a) SWR

Reflection coefficient at the input end:

Γ_in = (Z_in − Z₀)/(Z_in + Z₀) = (−200 − j200)/(600 − j200)
     = 282.8∠−135° / 632.5∠−18.43°
     = 0.447 ∠−116.6°

|Γ| is the same everywhere on a lossless line, so

S = (1 + 0.447)/(1 − 0.447) = 1.447/0.553 = 2.62

SWR = 2.62

(b) Load impedance for l = 1 m

βl = 4.189 rad = 240°, tan 240° = tan 60° = 1.732.

Moving back to the load (inverse of the input-impedance formula):

Z_L = Z₀ (Z_in − jZ₀ tanβl)/(Z₀ − jZ_in tanβl)

Numerator:   200 − j200 − j400(1.732) = 200 − j892.8
Denominator: 400 − j(200 − j200)(1.732) = 400 − (346.4 + j346.4)
           = 53.6 − j346.4

Z_L = 400 (200 − j892.8)/(53.6 − j346.4)

Equivalent check by Γ: Γ_L = Γ_in e^{+j2βl} = 0.447∠(−116.6° + 480°) = 0.447∠3.4°, and Z_L = Z₀(1 + Γ_L)/(1 − Γ_L).

Z_L ≈ 1042 + j70 Ω

Z_L ≈ 1.04 kΩ + j70 Ω

(c) Distance from load to the nearest voltage maximum

A voltage maximum occurs where the phase of Γ(d) = Γ_L e^{−j2βd} is zero:

2βd = θ_L = 3.43° = 0.0599 rad
d = 0.0599 / (2 × 4.189) = 0.00716 m ≈ 0.72 cm  (0.0048 λ)

d ≈ 0.7 cm from the load (the load is nearly at a voltage maximum because Z_L is almost real and larger than Z₀).