Smith chart analysis of 75+j100 Ω load

Given: Z0 = 50 Ω, ZL = 75 + j100 Ω. The values below were computed exactly; Smith chart readings agree to about ±0.005λ.

zL = ZL/Z0 = (75 + j100)/50 = 1.5 + j2.0

Plot zL at the intersection of the r = 1.5 circle and the x = 2.0 arc. It reads 0.198λ on the WTG scale.

(a) Reflection coefficient ΓL

ΓL = (zL − 1)/(zL + 1) = (0.5 + j2)/(2.5 + j2)
   = 0.644 ∠ 37.3°

On the chart: |Γ| = OP/OR (distance from the centre to the point, divided by the chart radius) = 0.644. The angle is read on the "angle of reflection coefficient" scale: 37.3°.

(b) VSWR

VSWR = (1 + |Γ|)/(1 − |Γ|) = 1.644/0.356 = 4.62

On the chart: draw the constant-|Γ| circle through zL. It cuts the right-hand real axis at r = 4.62.

(c) Zin at 0.375λ from the load

Move 0.375λ toward the generator (clockwise) on the |Γ| circle:

0.198λ + 0.375λ = 0.573λ → 0.073λ WTG
zin = (zL + j tan βl)/(1 + j zL tan βl),  βl = 135°, tan βl = −1
    = 0.267 + j0.467
Zin = 50 × zin = 13.33 + j23.33 Ω

(d) Shortest length for a purely resistive impedance

Moving clockwise from 0.198λ, the |Γ| circle first meets the real axis at the voltage maximum (right side, 0.25λ):

dmax = 0.25λ − 0.198λ = 0.0518λ
(= θΓ/(4π) = 37.3°/720° × λ)

The next resistive point is the voltage minimum at dmin = 0.0518λ + 0.25λ = 0.3018λ.

(e) Value of this resistance

At dmax: R = Z0 × VSWR = 50 × 4.617 = 230.8 Ω
(at dmin: R = Z0/VSWR = 10.83 Ω)

Answer: ΓL = 0.644∠37.3°, VSWR = 4.62, Zin(0.375λ) = 13.33 + j23.33 Ω, shortest length = 0.0518λ, R = 230.8 Ω.