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The left part of the module screen shows a circuit where an inductor is inserted in parallel into a transmission line. The generator shows a voltage source, V sub g and an impedance, Z sub g, while the load impedance is Z sub L. The parameters of the transmission line are: Z sub 0 equals 50.0 ohms; epsilon sub r sub 1 equals 1.0. The frequency is 1.0 gigahertz, while length d equals lambda equals 0.3 meters, and load is Z sub L equals 100.0 plus j 80.0 ohms.

A text in the bottom-left part of the screen reads as follows: Step 4 First solution. To cancel the imaginary part of the line admittance we add a shunt inductance: Y sub 1 equals (j omega L sub 1) raised to negative 1 equals negative j 0.02668 S (actual); L sub 1 equals 5.96458 times 10 raised to negative 9 H; y sub 1 equals negative j 1.33417 (normalized)

The right part of the screen shows an interactive Smith chart, with an S W R circle passing through two points located in the upper and lower parts of the chart. The following parameters are indicated below the chart:

(Point in the upper part of the chart)

z sub L equals 2.0 plus j 1.6

gamma sub L equals 0.55494, angle 29.92213 degrees

(Point in the lower part of the chart)

y sub L equals 0.30488 minus j 0.2439

z of d equals 1.0 plus j 0.0

gamma of d equals 0.0, angle 0.0 degrees

y of d equals 1.0 plus j 0.0

d equals 0.2134 lambda; 2 beta d equals 2.6813 rad equals 153.6287 degrees

0.5 lambda minus d equals 0.2866 lambda; 2 beta (0.5 lambda minus d) equals 3.6019 rad equals 206.3713 degrees