Line with 50+j50 load, first maximum, power

Given: Z₀ = 50 Ω, l = 1.5 m, V_g = 60∠0° V, Z_g = 50 Ω, Z_L = 50 + j50 Ω, f = 100 MHz, v = c.

λ = c/f = 3×10⁸/10⁸ = 3 m
β = 2π/3 = 2.094 rad/m
l = 1.5 m = 0.5λ   (half-wave line)

(a) Distance of first voltage maximum from the load

Γ_L = (Z_L − Z₀)/(Z_L + Z₀) = j50/(100 + j50)
    = 50∠90° / 111.8∠26.57°
    = 0.447 ∠63.43°
S = (1 + 0.447)/(1 − 0.447) = 2.618

A voltage maximum occurs where the phase of Γ is zero: θ_Γ − 2βd = 0.

d_max = θ_Γ / (2β) = (63.43° × π/180) / (2 × 2.094)
      = 1.1071 / 4.189 = 0.264 m  (= 0.088 λ)

First voltage maximum at 0.264 m from the load.

(b) Power delivered to the load

A half-wave line repeats the load impedance: tan βl = tan 180° = 0, so Z_in = Z_L = 50 + j50 Ω.

I_in = V_g/(Z_g + Z_in) = 60/(100 + j50) = 0.5367 ∠−26.57° A
P_in = ½ |I_in|² Re(Z_in) = ½ (0.5367)² × 50 = 7.2 W

The line is lossless, so P_L = P_in.

Check: V_in = V_g Z_in/(Z_g + Z_in) = 60(50 + j50)/(100 + j50) = 37.9∠18.4° V, and P = ½ Re(V_in I_in*) = ½ × 37.9 × 0.5367 × cos(45°) = 7.2 W. ✓

Answer: d_max = 0.264 m (0.088λ); P_L = 7.2 W.