Question 11.4: A lossless transmission line with Zo = 50 Ω is 30 m long and...

This problem will be solved with and without using the Smith chart.

Method 1 (without the Smith chart):

(a)

\Gamma=\frac{Z_{L}-Z_{o}}{Z_{L}+Z_{o}}=\frac{60+j40-50}{50+j40+50}=\frac{10+j40}{110+j40}=0.3523\angle56^{\circ}

(b)

s=\frac{1+|\Gamma|}{1-|\Gamma|}=\frac{1+0.3523}{1-0.3523}=2.088

(c) Since u=\omega/\beta, or \beta=\omega/u

\beta\ell=\frac{\omega\ell}{u}=\frac{2\pi(2\times10^{6})(30)}{0.6(3\times10^{8})}=\frac{2\pi}{3}=120^{\circ}

Note that \beta\ell is the electrical length of the line.

Z_{in}=Z_{o}\left[\frac{Z_{L}+jZ_{o}\tan\beta\ell}{Z_{o}+jZ_{L}\tan\beta\ell}\right]=\frac{50(60+j40+j50\tan120^{\circ})}{[50+j(60+j40)\tan120^{\circ}]}

=\frac{50(6+j4-j5\sqrt{3})}{(5+4\sqrt{3}-j6\sqrt{3})}=24.01\angle3.22^{\circ}=23.97+j1.35\Omega

Method 2 (using the Smith chart):

(a) Calculate the normalized load impedance

z_{L}=\frac{Z_{L}}{Z_{o}}=\frac{60+j40}{50}=1.2+j0.8

Locate z_{L} on the Smith chart of Figure 11.15 at point P, where the r = 1.2 circle and the x = 0.8 circle meet.

To get \Gamma at z_{L}, extend OP to meet the r = 0 circle at Q and measure OP and OQ. Since OQ corresponds to |\Gamma|=1, then at P

|\Gamma|=\frac{OP}{OQ}=\frac{3.2cm}{9.1cm}=0.3516

Note that OP = 3.2 cm and OQ = 9.1 cm were taken from the Smith chart used by the author; the Smith chart in Figure 11.15 is reduced, but the ratio of OP/OQ remains the same.

Angle \theta_{\Gamma} is read directly on the chart as the angle between OS and OP; that is

\theta_{\Gamma}=angle  POS=56^{\circ}

Thus

\Gamma =0.3516\angle56^{\circ}

(b) To obtain the standing wave ratio s, draw a circle with radius OP and center at O. This is the constant s or |\Gamma| circle. Locate point S where the s-circle meets the \Gamma_{r}-axis. [This is easily shown by setting \Gamma_{i}=0 in eq. (11.49a).]:

r=\frac{1-\Gamma_{r}^{2}-\Gamma_{i}^{2}}{(1-\Gamma_{r})^{2}+\Gamma_{i}^{2}}

The value of r at this point is s; that is

s=r(for   r\geq1)=2.1

(c) To obtain Z_{in}, first express \ell in terms of \lambda or in degrees:

\lambda =\frac{u}{f}=\frac{0.6(3\times 10^{8})}{2\times 10^{6}}=90m

\ell=30m=\frac{30}{90}\lambda=\frac{\lambda}{3}\rightarrow \frac{720^{\circ}}{3}=240^{\circ}

Since \lambda corresponds to an angular movement of 720° on the chart, the length of the line corresponds to an angular movement of 240°. That means we move toward the generator (or away from the load, in the clockwise direction) 240° on the s-circle from point P to point G. At G, we obtain

z_{in}=0.47+j0.03

Hence

Z_{in}=Z_{o}z_{in}=50(0.47+j0.03)=23.5+j1.5\Omega

Although the results obtained using the Smith chart are only approximate, for engineering purposes they are close enough to the exact ones obtained by Method 1. However, an inexpensive modern calculator can handle the complex algebra in less time and with much less effort than are needed to use the Smith chart. The value of the Smith chart is that it allows us to observe the variation of Z_{in} with \ell.