Lesson 8 moved a distance \(d\) from the load to a point where the normalized impedance is \(z = 1 + jx\), then cancelled \(jx\) with a series capacitor or inductor. The reactance \(x\) came from the magnitude of \(\Gamma\), which does not change along a lossless line, and the distance came from how far the phase of \(\Gamma\) has to rotate. For the \(33.9 + j17.9\ \Omega\) load, the two answers were a \(5.44\) pF capacitor at \(0.064\lambda\) and a \(4.65\) nH inductor at \(0.269\lambda\).

The normalized admittance \(y\) is the reciprocal of the normalized impedance. Its real part \(g\) is the normalized conductance and its imaginary part \(b\) is the normalized susceptance. We substitute \(z = 1/y\) into the reflection coefficient and multiply the numerator and denominator by \(y\), which gives \(\Gamma\) in terms of \(y\). The result has the same form as before with the sign of the numerator flipped. Admittances of elements in parallel add, just as impedances in series add, and that is why admittance is the natural variable for a parallel component. A match is still \(y = 1\).

The design follows Lesson 8 step for step. Moving away from the load rotates \(\Gamma\), and the normalized conductance passes through \(g = 1\) twice in every half wavelength. At either point the admittance is \(y = 1 + jb\). A component connected across the line at that point adds its own admittance, so a susceptance of \(-jb\) leaves \(y = 1\), and the source sees \(Z_0\). The figure shows that component built as a shorted stub, which we come to after the lumped version.

We substitute \(y = 1 + jb\) into the admittance form of the reflection coefficient. The magnitude of \(-jb\) is \(\vert b\vert\), the same as the magnitude of \(jx\) in Lesson 8, so the magnitude equation is identical with \(b\) in place of \(x\). Since \(\vert\Gamma\vert\) is fixed along the line, we solve for \(b\) exactly as before. For the same load, the susceptance at the parallel matching points has the same size as the reactance at the series matching points.

The angles differ from Lesson 8 because of the minus sign. For \(b > 0\) the numerator \(-jb\) sits at \(-90^\circ\) and the denominator at \(\tan^{-1}(b/2)\), so the angle of the quotient is \(-90^\circ - \tan^{-1}(b/2)\). For \(b < 0\) every angle changes sign. The rest of the derivation is unchanged: moving a distance \(d\) rotates the load’s phase by \(-2\beta d\), so we set \(\theta_\Gamma - 2\beta d = \theta_d\) and solve for \(d\), at \(720^\circ\) of rotation per wavelength.

The load and its reflection coefficient are the ones from Lesson 8, so \(\vert b\vert = 0.585\) and \(\tan^{-1}(0.585/2) = 16.3^\circ\). For \(b = -0.585\) the matching phase is \(+106.3^\circ\), and the load’s phase must rotate from \(119.9^\circ\) down to \(106.3^\circ\), which takes only \(13.6/720 = 0.019\lambda\). For \(b = +0.585\) the matching phase is \(-106.3^\circ\), a rotation of \(226.2^\circ\), which takes \(0.314\lambda\). In RG-58 at 1 GHz these distances are 0.38 cm and 6.28 cm.

The susceptance to cancel is \(b/Z_0 = 0.585/50 = 11.7\) mS. Solution 1 has \(-j11.7\) mS at its matching point, so it needs a parallel element with \(+j11.7\) mS. A capacitor has admittance \(j\omega C\):

Solution 2 has \(+j11.7\) mS, so it needs an inductor, whose admittance is \(\frac{1}{j\omega L} = -\frac{j}{\omega L}\). Setting \(\frac{1}{\omega L} = 11.7\) mS gives \(L = 13.6\) nH.

Lesson 7 showed that a shorted or open stub has a purely reactive input impedance, so its admittance is purely susceptive. We take the reciprocal of each input impedance and multiply by \(Z_0\) to normalize it, using \(\frac{1}{j} = -j\) and \(\frac{1}{\tan} = \cot\). A stub of the same line, cut to the right length \(\ell_s\) and connected across the line at the matching point, supplies the cancelling susceptance with no lumped component at all. Every susceptance from \(-\infty\) to \(+\infty\) is available from a stub shorter than half a wavelength.

Solution 1 needs a stub with normalized susceptance \(b_s = +0.585\). An open stub gives it when \(\tan\beta\ell_s = 0.585\), so \(\beta\ell_s = 30.3^\circ\) and \(\ell_s = 30.3/360 = 0.084\lambda\). A shorted stub gives it when \(\cot\beta\ell_s = -0.585\), which first happens at \(\beta\ell_s = 120.3^\circ\). Solution 2 needs \(b_s = -0.585\): a shorted stub at \(59.7^\circ\), or an open stub at \(149.7^\circ\). A stub’s length is a one-way trip, so it is \(360^\circ\) per wavelength, not \(720^\circ\).

Solution 1 with an open stub is the most compact design here: \(0.38\) cm of line and a \(1.68\) cm stub. Shorter lengths usually hold the match over a wider band, for the same reason as in Lesson 8. The open end of a stub is not a perfect open circuit; it radiates a little and has fringing capacitance, so at high frequency a shorted stub is often more predictable. A shorted stub also puts a DC short across the line, which is harmless for a signal path but matters when the line carries bias current to an amplifier.

These steps match any load with one parallel element, and they differ from the series procedure only in the angle \(\theta_d\) and in what the last step adds. The reflection coefficient can be found from either \(z_L\) or \(y_L\); the two forms give the same \(\Gamma\). Lessons 10 through 13 do this design on the Smith chart, where finding \(d\) becomes a rotation on a circle and finding the stub length becomes reading a second point off the chart’s edge.

Every matching design in Lessons 8 and 9 came down to the same move: rotate \(\Gamma\) at constant magnitude until the real part of \(z\) or \(y\) equals 1. The Smith chart draws the circles of constant \(r\) and constant \(g\) on the plane of \(\Gamma\), so that move becomes a rotation you can see and measure with a compass. Lesson 10 builds the chart from \(z = \frac{1 + \Gamma}{1 - \Gamma}\), the relation we derived in Lesson 8.

Before Lesson 10, print a blank Smith chart from the handouts on the Materials page and bring it to class.