The Smith Chart and Impedance Matching
The input-impedance formula of the last chapter is exact but unfriendly — full of complex tangents that hide what is going on. The Smith chart turns that algebra into geometry. Because the reflection coefficient always lives inside the unit circle, you can plot it there, overlay the impedance grid, and read impedance, reflection, and standing-wave ratio off a single picture. Moving along the line becomes a simple rotation; matching a load becomes a path to the centre. Built in 1939 and still printed on every microwave engineer's notepad, it is the most enduring graphical tool in electrical engineering.
- Why the reflection coefficient maps the whole right-half impedance plane into the unit circle.
- The bilinear relation \(\Gamma=(z-1)/(z+1)\) and its inverse \(z=(1+\Gamma)/(1-\Gamma)\).
- The constant-resistance circles and constant-reactance arcs that form the chart.
- How to read \(z\), \(\Gamma\), and the SWR off one diagram.
- That moving toward the generator is a clockwise rotation, one revolution per \(\lambda/2\).
- The SWR circle, the admittance chart, and matching by quarter-wave and single-stub tuners.
Why a Chart?
Impedance is unbounded: a resistance can run from zero to infinity, and a reactance from minus to plus infinity. You cannot plot that on a finite page. The reflection coefficient, by contrast, is bounded — for any passive load \(|\Gamma|\le1\), so it always sits inside a circle of radius one. The Smith chart is built on this single observation: work in the \(\Gamma\)-plane, where everything fits, and carry the impedance information along as a printed grid.
The bridge between the two is the relation from Chapter 24, written in normalized form by dividing every impedance by the line's \(Z_0\). That normalization is what makes one chart serve every line, whatever its characteristic impedance.
Normalized Impedance and Γ
Define the normalized impedance \(z=Z/Z_0=r+jx\). The load reflection coefficient and its inverse are then a matched pair of bilinear (Möbius) transformations — each a smooth, one-to-one map between the impedance plane and the \(\Gamma\)-plane:
Three anchor points fix the whole picture. A matched load \(z=1\) maps to \(\Gamma=0\), the centre of the chart. A short \(z=0\) maps to \(\Gamma=-1\), the far left. An open \(z\to\infty\) maps to \(\Gamma=+1\), the far right. The entire right-half impedance plane (all passive loads, \(r\ge0\)) folds neatly into the unit disc, and the rim \(|\Gamma|=1\) is the locus of pure reactances — loads that reflect everything.
Resistance and Reactance Circles
The magic of the map is that the rectangular grid of constant \(r\) and constant \(x\) in the impedance plane becomes a grid of circles in the \(\Gamma\)-plane. Substituting \(\Gamma=u+jv\) and separating real and imaginary parts gives two families:
Every constant-\(r\) circle passes through the right-hand point \(\Gamma=+1\); larger resistances are smaller circles bunched toward the right. Every constant-\(x\) arc also springs from \(\Gamma=+1\), curving up for inductive (\(+x\)) loads and down for capacitive (\(-x\)) ones, with the real axis itself being \(x=0\). Where an \(r\)-circle crosses an \(x\)-arc is the unique point for that impedance.
Reading the Chart
To place a load: normalize it to \(z_L=Z_L/Z_0\), find the constant-\(r\) circle for its real part and the constant-\(x\) arc for its imaginary part, and mark where they meet. That point is \(\Gamma_L\): its distance from the centre (as a fraction of the chart radius) is \(|\Gamma_L|\), and the angle it makes from the positive real axis is \(\theta_\Gamma\). Reading backward — from a plotted \(\Gamma\) to the impedance — you just note which \(r\)-circle and \(x\)-arc the point sits on.
The grid gives \(z\); the radius and angle give \(\Gamma\); the circle through the point crosses the right-hand axis at \(s\). No formula needed.
Moving Along the Line
Here the chart earns its keep. From Chapter 24, the reflection coefficient at distance \(\ell\) from the load is \(\Gamma(\ell)=\Gamma_L e^{-2j\beta\ell}\). On a lossless line the magnitude is fixed, so the point simply rotates on a circle of constant radius about the centre. Moving toward the generator decreases the phase, which is a clockwise turn; moving toward the load is counter-clockwise:
So the input impedance at any distance is found by spinning the load point clockwise through an angle \(2\beta\ell\) and reading the new grid location — the whole \(\tan\beta\ell\) calculation, done with a compass. The rim of the chart even carries a "wavelengths toward generator" scale so the rotation can be measured directly.
The SWR Circle and Admittance
The constant-radius circle the point travels on is the SWR circle. Because it crosses the positive real axis where the impedance is purely resistive and maximal, that crossing reads the standing-wave ratio directly: there \(z=s\). The opposite crossing gives \(z=1/s\), the voltage-minimum point. Drawing this one circle therefore captures the match quality and both reference points at a glance:
Many matching problems are easier in admittance, because shunt elements add admittances. The normalized admittance is \(y=1/z=(1-\Gamma)/(1+\Gamma)\), and on the chart taking the reciprocal is just a 180° rotation through the centre — the point diametrically opposite \(z\) on the same SWR circle. Read with the same grid, the chart now gives conductance circles and susceptance arcs for free.
Impedance Matching
Matching means driving the operating point to the centre of the chart (\(z=1\), \(\Gamma=0\)), where there is no reflection and all the power reaches the load. Two methods dominate, and both are natural on the chart.
The quarter-wave transformer of Chapter 24 handles a real load: a \(\lambda/4\) section of impedance \(Z_0'=\sqrt{Z_0 R_L}\) inverts the load onto \(Z_0\). On the chart this is the half-turn from a real load on one side to \(z=1\) on the other.
The single-stub tuner matches any load, on the admittance chart, in two moves. First, run along the line (rotate toward the generator) until the point lands on the \(g=1\) circle — at that distance \(d\) the admittance is \(y=1+jb\), the right conductance but a leftover susceptance. Second, connect a shunt stub — a short- or open-circuited length of line — chosen to add exactly \(-jb\), cancelling the susceptance and dropping the point onto the centre. The stub contributes only susceptance because, terminated in a short or open, it is purely reactive (Chapter 24). Two lengths, no lumped parts, perfect match.
Worked Examples
Problem. A \(50\,\Omega\) line carries \(Z_L=100+j50\,\Omega\). Find the normalized \(z_L\), the reflection coefficient, and the SWR.
Solution. Normalize, then \(\Gamma_L=(z_L-1)/(z_L+1)\) and \(s=(1+|\Gamma|)/(1-|\Gamma|)\):
Problem. A point sits at \(\Gamma=0.5\angle60^\circ\) on a \(50\,\Omega\) chart. What impedance is it?
Solution. Write \(\Gamma=0.25+j0.433\), then \(z=(1+\Gamma)/(1-\Gamma)\) and \(Z=Z_0 z\):
Problem. A load plots at radius \(|\Gamma|=0.4\). Find the SWR and the impedance at the voltage maximum.
Solution. Use \(s=(1+|\Gamma|)/(1-|\Gamma|)\); the right-axis crossing gives \(z=s\):
Problem. On a \(50\,\Omega\) line, \(z_L=1+j1\). Find the input impedance \(\lambda/8\) toward the generator (a \(90^\circ\) clockwise rotation).
Solution. \(\Gamma_L=0.2+j0.4=0.447\angle63.4^\circ\); rotate \(-90^\circ\) to \(0.447\angle-26.6^\circ=0.4-j0.2\), then convert back:
Problem. An SWR of \(s=3\) is measured on a \(50\,\Omega\) line. Find \(|\Gamma|\) and the impedances at the voltage maximum and minimum.
Solution. \(|\Gamma|=(s-1)/(s+1)\); maxima see \(sZ_0\), minima \(Z_0/s\):
Problem. Find the normalized admittance of \(z=0.5-j0.5\) — the point a half-turn across the chart.
Solution. Take \(y=1/z\):
Chapter Summary
\(z=Z/Z_0\); \(\Gamma=\dfrac{z-1}{z+1}\), \(z=\dfrac{1+\Gamma}{1-\Gamma}\); all loads fit in \(|\Gamma|\le1\).
Centre \(z=1\); left \(z=0\) (short); right \(z=\infty\) (open); rim \(|\Gamma|=1\).
Constant-\(r\) circles and constant-\(x\) arcs; their crossing is the impedance.
Toward generator = clockwise; one revolution = \(\lambda/2\); \(\lambda/4=180^\circ\).
SWR circle crosses axis at \(s\) and \(1/s\); \(y=1/z\) is a 180° rotation.
Drive the point to centre — quarter-wave \(\sqrt{Z_0 R_L}\) or single-stub tuner.
Problems
For each item, normalize first (\(z=Z/Z_0\)); the chart only knows normalized values. Remember that distance along the line is an angle (\(2\beta\ell\)), clockwise toward the generator. Difficulty rises down the list.
- Normalize \(Z_L=25-j50\,\Omega\) on a \(50\,\Omega\) line and find \(\Gamma_L\).
- Find the impedance at \(\Gamma=0.6\angle0^\circ\) on a \(75\,\Omega\) chart.
- A load plots at \(|\Gamma|=0.6\). Find the SWR.
- An SWR of \(s=4\) is measured. Find \(|\Gamma|\) and the radius of its SWR circle as a fraction of the chart radius.
- Where on the chart do all constant-\(r\) circles meet? Where do all constant-\(x\) arcs meet?
- For \(z_L=0.4+j0.5\), find \(\Gamma_L\) and \(s\).
- On a \(50\,\Omega\) line with \(z_L=1+j1\), find \(z_{\text{in}}\) a quarter wavelength toward the generator (a \(180^\circ\) rotation).
- Find the normalized admittance of \(z=2+j1\), and state the rotation that produced it.
- A load \(Z_L=200\,\Omega\) is matched to a \(50\,\Omega\) line with a quarter-wave transformer. Find \(Z_0'\), and describe the move on the chart.
- At what two normalized resistances does the \(s=2\) SWR circle cross the real axis?
- Explain why a lossy line spirals inward on the chart as you move toward the generator, rather than tracing a closed circle.
- In single-stub matching, explain why the first step targets the \(g=1\) circle specifically, and what the stub does in the second step.