Question 9.10: A 100[Ω] -lossless line is 0.24λ long, and is terminated in ......

(a) Normalized load impedance is

Z_{L} = (140 + j130) / 100 = 1.4 + j1.3.

The intersection of the r = 1.4 circle and the x = 1.3 circle is marked as A on the Smith chart in Fig. 9.16.

From the Smith chart we find that

(1)  \frac{\overline{OA} }{\overline{OD} } = 0.5 = \left|\Gamma \right|, and

(2) point A is located at 0.188λ on the wtg scale.

Noting that the wtg of 1λ corresponds to 4π[rad]  , the phase angle of point A is calculated as follows:

Φ = (0.25 – 0.188) × 4π = 0.78[rad].

Thus, \Gamma = 0.5 e^{j 0.78}.

(b) Move from point A to point C along the circle of |Γ| = 0.5, in the clockwise
direction, by increasing the wtg by 0.24λ .

At point C, we read r = 0.40 and x = −0.42 .
The input impedance is therefore

Z_{in} = 100 (0.40 – j0.42) = 40 – j42[\Omega ].
(c) The circle of |Γ| = 0.5 crosses the positive \Gamma_{R}-axis at point B, where we read r = 3.0 and x = 0 .
Thus, the standing wave ratio is S = 3.0 .
(d) A voltage maximum occurs at a distance \ell ^{\prime} . from the load, at which

z_{in}(\ell^{\prime }) = r^{\prime } + j0  (r^{\prime } \gt 1) . This point is marked as B. The difference in wtg between points B and A is equal to \ell ^{\prime} , that is,
\ell ^{\prime} = (0.25 − 0.188)λ = 0.062λ.