Smith chart, load 80+j100 on 50 Ω line

Given: Z_L = 80 + j100 Ω, Z₀ = 50 Ω, f = 200 MHz, λ = 2 m, l = 0.8 m = 0.4λ.

Step 1: Normalise and plot

z_L = Z_L/Z₀ = 1.6 + j2.0. Mark this point on the chart (intersection of the r = 1.6 circle and the x = +2 arc, upper half). Draw the constant-|Γ| circle through it with centre at the chart centre.

Reading the chart: |Γ| = 0.637, angle θ_Γ = 35.7°. (Check: Γ = (z_L − 1)/(z_L + 1) = (0.6 + j2)/(2.6 + j2) = 0.637∠35.7°.)

(a) Input impedance at 0.4λ

Move 0.4λ from the load toward the generator (clockwise on the WTG scale). The Γ angle changes by −2βl = −2 × 360° × 0.4 = −288°:

Γ_in = 0.637 ∠(35.7° − 288°) = 0.637 ∠107.7°   (≡ −252.3°)

Reading z_in at this point on the chart: z_in ≈ 0.33 + j0.68. Using the exact formula Z_in = Z₀(Z_L + jZ₀ tanβl)/(Z₀ + jZ_L tanβl), with βl = 144°, tan = −0.7265:

Numerator:   80 + j100 + j50(−0.7265) = 80 + j63.7      (102.2∠38.5°)
Denominator: 50 + j(80 + j100)(−0.7265) = 122.7 − j58.1  (135.7∠−25.3°)
Z_in = 50 × 0.753∠63.9° = 37.7∠63.9° = 16.6 + j33.8 Ω

Result: Z_in = 16.6 + j33.8 Ω (z_in = 0.33 + j0.68), which the chart reading confirms.

(b) Standing wave ratio

At the point where the |Γ| circle crosses the real axis on the right side (r = s):

s = (1 + |Γ|)/(1 − |Γ|) = 1.637/0.363 = 4.5

s ≈ 4.5

(c) Distance from load to nearest voltage maximum

The voltage maximum is where Γ is real positive, i.e. on the right of the real axis. From the load angle 35.7°, rotate clockwise to 0°, a change of 35.7° = 2βd.

d = 35.7° / 720° × λ = 0.0496 λ = 0.0496 × 2 m ≈ 0.099 m ≈ 0.1 m

On the wavelengths-toward-generator scale this is about 0.0496λ. Distance ≈ 0.1 m (0.05λ).