Question 9.11: A 50[Ω] -lossless line is terminated in a load impedance ZL ......
A 50[Ω] -lossless line is terminated in a load impedance Z_{L} = 17.5 – j 55 [Ω]. Find the location and length of the short-circuited stub to be connected in parallel for the impedance matching.
Normalized load impedance is
z_{L} = (17.5 – j 55) / 50 = 0.35 – j1.1.
The intersection between the r = 0.35 circle and the x = –1.1 circle is marked as A on the Smith chart in Fig. 9.18. It is located at 0.363λ on the wtg scale.
To obtain the normalized load admittance, at the inversion of point A through the origin, or at point B, we read r = 0.26 and x = 0.83 . Point B is located at 0.113λ on the wtg scale.
Thus, y_{L} = 0.26 + j0.83.
We next move along the circle of constant |Γ| from point B to point C, which is a point on the r = 1 circle. At point C, we read x = 2.2 and wtg = 0.191 λ . The difference in wtg between points C and B is (0.191 – 0.113) λ = 0.078λ.
Thus, we have y_{\ell} = 1 + j2.2 at a distance \ell = 0.078λ from the load.
A short circuit has Z_{L} = 0 and Y_{L} = ∞ , which is marked as D on the admittance chart in Fig. 9.18. We move from point D to point E, which is the intersection between the circle of |Γ| = 1 and the circle of b = −2.2 . At point E, we read 0.318λ on the wtg scale.
The difference in wtg between points E and D is the length of the stub, that is, d = (0.318 – 0.25)λ = 0.068λ .
