We put the load at \(z = 0\) on the line and measure distance \(d = -z\) back toward the source. Lesson 6 showed that the ratio of the reflected wave to the incident wave at that point is the load’s reflection coefficient multiplied by \(e^{-j2\beta d}\): the incident wave travels \(d\) to reach the load and the reflected wave travels \(d\) back. Lesson 10 showed that every point of the \(\Gamma\) plane carries one normalized impedance. We now put the two results together.

Writing \(\Gamma_L\) in polar form shows what the factor \(e^{-j2\beta d}\) does. It leaves the magnitude alone, so the point stays on the circle of constant \(\vert\Gamma\vert\) that Lesson 10 called the VSWR circle; on a lossless line the VSWR is the same everywhere. It subtracts \(2\beta d\) from the angle, and a falling angle is a clockwise turn. A full turn needs \(2\beta d = 2\pi\), which is \(d = \lambda/2\), so the impedance on any lossless line repeats every half wavelength. Moving toward the source, or generator, is clockwise; moving toward the load is counterclockwise.

The rotation gives the input impedance formula of Lesson 7 without the voltage and current waves. We substitute \(\Gamma(d)\) into the mapping from Lesson 10 and write \(\Gamma_L\) in terms of \(z_L\). Multiplying the numerator and denominator by \(\left(z_L + 1\right)e^{j\beta d}\) clears both fractions. We then group the terms with \(z_L\) and the terms without it and use \(e^{j\beta d} + e^{-j\beta d} = 2\cos\beta d\) and \(e^{j\beta d} - e^{-j\beta d} = j2\sin\beta d\). Dividing the top and bottom by \(2\cos\beta d\) gives the normalized form of Lesson 7’s result. The algebra and the rotation are the same calculation; the chart does it with a compass.

The printed chart carries two scales around its rim so that nobody has to convert angles by hand. The outer one, wavelengths toward generator, starts at zero at the short circuit on the left edge and counts clockwise to \(0.5\lambda\) in one full turn. Because \(720^\circ\) of \(\Gamma\) phase is one wavelength of travel, a point at angle \(\theta_\Gamma\) reads \((180^\circ - \theta_\Gamma)/720^\circ\) wavelengths on that scale. To move a distance \(d\) toward the source, we draw a ray from the center through the load, read its WTG value, add \(d\), and draw a second ray at the new value. The input impedance is where that ray crosses the VSWR circle.

A quarter wavelength is half a turn, so it carries \(\Gamma_L\) to the point diametrically opposite, \(-\Gamma_L\). Substituting \(-\Gamma_L\) into the mapping flips the fraction, so the normalized input impedance is the reciprocal of the normalized load. Undoing the normalization gives \(Z_{in} = Z_0^2/Z_L\), the quarter-wave transformer of Lesson 8. The right panel of the worked example figure shows it: half a turn from \(z_L = 1.5 + j1.2\) lands on \(1/z_L = 0.407 - j0.325\).

Shunt elements add as admittances, so Lessons 12 and 13 need \(y\) as often as \(z\). The normalized admittance \(y = Y/Y_0 = 1/z\) has conductance \(g\) and susceptance \(b\). Writing \(1/z\) in terms of \(\Gamma\) gives the same mapping as before with \(-\Gamma\) in place of \(\Gamma\). So we find \(y\) by reflecting the point through the center and reading the same grid, now labeled as \(g\) and \(b\); that is the quarter-wave half turn of the previous frame. On the grid used for admittance, the upper half holds capacitive susceptance and the lower half inductive, and the left edge is now the open circuit. Some charts print a second, mirrored grid for admittance; this course uses one grid and the reflection.

Lesson 6 wrote the voltage magnitude as \(\vert V\vert = \vert V_0^+\vert \vert 1 + \Gamma(d)\vert\). It is largest when \(\Gamma(d)\) points along the positive real axis, so the two terms add, and smallest when \(\Gamma(d)\) points the other way. On the chart, those are the two crossings of the VSWR circle with the real axis: the maximum at \(r = \text{VSWR}\) on the right and the minimum at \(r = 1/\text{VSWR}\) on the left, a quarter wavelength apart. For our load, the first maximum is the clockwise turn of \(41.7^\circ\), which is \(0.058\lambda\); on the WTG scale the right axis reads \(0.25\lambda\), and \(0.25 - 0.192 = 0.058\). The first minimum is \(0.25\lambda\) farther, where the impedance is \(0.362\), or \(18.1\ \Omega\).

We can now move a load anywhere around its VSWR circle. That circle passes through the \(r = 1\) circle twice, and on the \(r = 1\) circle only a reactance separates the input from a match. Lesson 12 uses that fact to design matching networks with one capacitor or inductor, in series or in shunt, and shows how each element moves the point on the chart.

Before Lesson 12, finish the Smith chart walkthrough on the materials page, and bring a printed blank Smith chart to class.