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🧭 The Smith Chart

A single sheet for reading impedances, getting the reflection coefficient and the SWR, seeing what a length of line does to a load and designing the match, without solving a complex equation at every step.

Transmission Media · Electronic Engineering · In the syllabus it appears as “the Smith chart and its use.” Lossless line.

The Γ plane with an impedance grid on top

The chart is the radius-1 circle of the reflection coefficient Γ. Every point inside is a possible load. On top of it is drawn the grid that tells you which impedance corresponds to that Γ: constant-resistance circles and constant-reactance arcs. Move the cursor over the chart: the circle and the arc that cross at that point are highlighted. Click to pin it.

constant resistance r inductive reactance (+jx) capacitive reactance (−jx)

The three equations behind the chart

\( z = \frac{Z}{Z_0} = r + j x \) \( \Gamma = \frac{z -1}{z + 1} \) \( z = \frac{1 + \Gamma}{1 -\Gamma} \) The impedance is divided by the characteristic impedance of the line, so the same chart works for a 50 Ω coax, a 75 Ω coax or a 300 Ω open-wire line.

All the r circles pass through the right-hand end and have their centers on the horizontal axis. The x arcs also start at the right-hand end: above the axis the load is inductive, below it is capacitive, and on the axis it is purely resistive.

\( \text{circle of }r : \; \text{center }\left(\frac{r}{1 + r}, 0 \right) , \; \text{radius }\frac{1}{1 + r} \) \( \text{arc of }x : \; \text{center }\left(1, \frac{1}{x} \right) , \; \text{radius }\frac{1}{\left| x \right|} \) They come from substituting z = r + jx into the formula for Γ and separating the real and imaginary parts.

A load on the chart

Choose the load impedance and the line. The point sits where its r circle and its x arc cross. The distance to the center is |Γ| and the angle is the phase of the reflected wave. The dotted circle is the constant-|Γ| circle: that is where the load lives as seen from any point on the line.

load zL constant-|Γ| circle (SWR)

What you read from that point

\( \mathrm{SWR} = \frac{1 + | \Gamma |}{1 -| \Gamma |} \) \( L_{\operatorname{R}} = -20 \, \log_{10} | \Gamma | \, \operatorname{dB} \) \( \frac{P_{\operatorname{refl}}}{P_{\operatorname{inc}}} = {\left| \Gamma \right|}^2 \) SWR is the standing wave ratio (also written VSWR). LR is the return loss.

A shortcut the chart gives you for free: the |Γ| circle crosses the horizontal axis, to the right of the center, exactly at r = SWR. There Γ is real and positive, and the formula for z gives exactly the one for the SWR. You read it without doing any arithmetic.

Moving away from the load means rotating

On a lossless line, as you move away from the load toward the generator, |Γ| does not change: only its phase does. The point rotates around its circle clockwise and makes a full turn every half wavelength. The scale on the edge measures that distance in wavelengths.

load zL impedance seen at distance d path toward the generator

Voltage along the line

The incident and reflected waves add: where they arrive in phase there is a voltage maximum, and where they arrive in antiphase, a minimum. The ratio between the two is the SWR.

The calculations the chart saves you

\( \Gamma (d) = \Gamma_L \, e^{-j 2 \beta d} \) \( \beta = \frac{2 \pi}{\lambda} \) Multiplying by e−j2βd rotates the point by an angle 2βd clockwise: 720° per wavelength.
\[ Z_{\operatorname{in}} = Z_0 \frac{Z_L + j Z_0 \tan (\beta d)}{Z_0 + j Z_L \tan (\beta d)} \] The same answer, without the chart: the input impedance of a line of length d loaded with ZL.
Two lengths worth recognizing. With λ/2 the point goes all the way around and returns to the load: the line is “not there.” With λ/4 the point ends up on the opposite side and the impedance is inverted, \( Z_{\operatorname{in}} = Z_0^2 / Z_L \): this is the quarter-wave transformer.

Moving the load to the center

Each component you add moves the point along a fixed path. An element in series adds reactance: the point moves along its constant-r circle. An element in parallel adds susceptance: it moves along its constant-conductance g circle, which is the one on the admittance grid. A length of line makes it rotate around the center. Try to bring the point to the center, which is the matched load.

impedance z admittance y = 1/z inductor capacitor line section

Impedance and admittance on the same sheet

\( y = \frac{1}{z} = g + j b \) \( \Gamma_y = -\Gamma_z \) The admittance of a point is at the diametrically opposite location, at the same distance from the center. That is why the admittance grid is the impedance grid turned around.

For elements in series it is convenient to think in impedances, because they add; for elements in parallel, in admittances, for the same reason. The chart lets you switch from one to the other simply by crossing to the other side of the center. Watch the signs: an inductor adds +jx in series but subtracts susceptance, −jb, in parallel; with a capacitor it is the other way around.

Matching with a parallel stub

The classic method for matching a load to the same line: moving away from the load, you look for the point where the conductance is exactly 1, and there you connect in parallel a short-circuited length of line, the stub, whose length is chosen to cancel the leftover susceptance. Follow the steps.

load g = 1 circle path along the line (d) stub (l)

The resulting circuit