Smith chart, load 100+j150 on 75 Ω line

Given: Z_L = 100 + j150 Ω, Z₀ = 75 Ω. Normalised load:

z_L = Z_L/Z₀ = 1.333 + j2.0

Plot z_L (r = 1.33 circle, x = +2 arc). With the compass centred on the chart centre, draw the circle through z_L.

(a) Reflection coefficient

Read |Γ| from the radial scale and the angle from the angle-of-reflection scale:

Γ = (z_L − 1)/(z_L + 1) = (0.333 + j2)/(2.333 + j2)
  = 2.028∠80.5° / 3.073∠40.6°
  = 0.660 ∠39.9°

(b) VSWR

Where the |Γ| circle cuts the right side of the real axis:

VSWR = (1 + 0.660)/(1 − 0.660) = 4.88 ≈ 4.9

(c) Load admittance

Draw the diameter through z_L and read the opposite point on the same circle (rotate 180° = λ/4):

y_L = 1/z_L = 0.231 − j0.346
Y_L = y_L/Z₀ = (0.231 − j0.346)/75 = 3.08 − j4.62 mS

(d) Z_in at 0.4λ from the load

Move 0.4λ toward the generator (clockwise); the angle of Γ becomes 39.9° − 2×360°×0.4 = 39.9° − 288° = −248.1° ≡ 111.9°. Reading the chart at this point:

z_in ≈ 0.29 + j0.64  →  Z_in = 75 × z_in ≈ 22 + j47.6 Ω   (52.4∠65.2°)

Check by formula with βl = 144°: Z_in = 21.97 + j47.61 Ω.

(e) Z_in at the generator, line 0.6λ long

0.6λ is more than a half wavelength; the chart repeats every 0.5λ, so it is the same as 0.1λ:

Angle of Γ = 39.9° − 72° = −32.1°.

z_in ≈ 1.78 − j2.21  →  Z_in = 75 × z_in ≈ 133.6 − j165.7 Ω   (212.9∠−51.1° Ω)

Answer: (a) 0.66∠39.9°; (b) 4.88; (c) y_L = 0.231 − j0.346 (Y_L = 3.08 − j4.62 mS); (d) 22 + j47.6 Ω; (e) 133.6 − j165.7 Ω.