Smith Chart Basics: How to Read Impedance and Reflection

What the Smith chart plots

The Smith chart is a plot of the complex reflection coefficient, written here as Γ, inside a circle of radius 1. Every passive load has |Γ| of 1 or less, so every passive load lands somewhere inside that circle.

Two landmarks matter first. The center of the chart is Γ = 0, a perfect match with nothing reflected. The outer edge is |Γ| = 1, where all the power is reflected.

The impedance grid printed on the chart is a second set of coordinates drawn over the same circle. Each point has a Γ value and an impedance value at once. The chart is a map between the two, so you can read whichever is more useful.

From impedance to a point

First normalize the impedance by the system reference impedance:

Then the reflection coefficient follows from:

In a 50 ohm system, a load of 50 ohm gives z = 1 and Γ = 0. That is the center of the chart.

The horizontal axis holds the special cases:

• Left end: short circuit, z = 0, Γ = -1.
• Center: z = 1, Γ = 0.
• Right end: open circuit, z = infinity, Γ = +1.

The vertical direction carries the reactance. Points above the horizontal axis have positive reactance (inductive). Points below it have negative reactance (capacitive).

Two families of circles make up the grid. Circles of constant resistance all touch the right-hand end of the horizontal axis. Circles of constant reactance are arcs that also end at that same point. To place a load, find the resistance circle for its real part, then follow the reactance arc for its imaginary part. The point is where they cross.

Worked load: 100 + j50 ohm

Take a load of 100 + j50 ohm in a 50 ohm system.

Normalize it: z = (100 + j50) / 50 = 2 + j1.

Compute the reflection coefficient:

The magnitude is 0.447 and the angle is 26.6 degrees. On the chart, find the resistance circle for 2 and the reactance arc for +1. They cross in the upper half, so the load is inductive, as the +j50 ohm says.

The distance from the center is |Γ| = 0.447. Every point on a circle around the center has the same |Γ|, so that circle is also a constant-VSWR circle:

Two more numbers follow from |Γ|. The fraction of power reflected is |Γ|², which is 0.2, or 20 percent. The return loss is -20 log10(0.447), about 7.0 dB. The page on return loss and insertion loss explains how to use that number.

Moving along a line

Moving along a transmission line toward the generator turns the point clockwise around the center. The distance from the center does not change on a lossless line, so |Γ| and the VSWR stay the same. Only the angle changes.

One full turn corresponds to half a wavelength of line. A quarter wavelength is half a turn, which puts the point on the opposite side of the center.

Apply this to the worked load. A quarter wave toward the generator turns Γ = 0.4 + j0.2 into -0.4 - j0.2. The normalized impedance there is:

That is 20 - j10 ohm in a 50 ohm system, and it is the inverse of the original z. The VSWR is still 2.62, but the load now looks capacitive. This is why a line length changes what a generator sees even when nothing else changes.

Adding a series inductor or a shunt capacitor

Matching on the chart is a sequence of moves, and each part has a fixed direction.

• A series inductor adds positive reactance. It moves the point clockwise along a circle of constant resistance.
• A shunt capacitor adds positive susceptance. It moves the point clockwise along a circle of constant conductance.

Try both on the worked load. Add a series reactance of +j1 (normalized), and z becomes 2 + j2. The resistance stays at 2, and Γ becomes about 0.538 + j0.308. The point moves farther from the center, so this inductor makes the match worse.

For the shunt part, work in admittance. The inverse of 2 + j1 is y = 0.4 - j0.2. A shunt capacitor adds +jB to y. Adding +j0.2 gives y = 0.4, which is z = 2.5 and Γ of about 0.43. The reactance is gone, but the resistance has risen from 2 to 2.5. The conductance stayed at 0.4 the whole way, as the rule says.

Series parts move along resistance circles, and shunt parts move along conductance circles. A matching network combines the two types to bring the point to the center. The figure below shows such a path for 50 ohm to 200 ohm, and the L-network matching calculator gives the part values.

Reading checklist

1. Normalize the impedance by Z0.
2. Find the resistance circle and reactance arc, and mark the point.
3. Read the distance from the center as |Γ|, and the angle as the phase of Γ.
4. Convert |Γ| to VSWR, return loss or reflected power.
5. Move clockwise to represent line length toward the generator, or along the proper circle to represent a series or shunt part.

In plain words · for students and beginners

The Smith chart in plain words

An everyday analogy

Think of a target in a shooting range. The bullseye is a perfect match, where nothing bounces back. The farther your shot lands from the bullseye, the more signal bounces back toward the source.

The analogy breaks down in one place. A target has only one score for distance. The Smith chart gives you two readings for each point. One is how far it is from the center, which is how bad the mismatch is. The other is where it sits, which tells you the resistance and the type of reactance.

The same example in simple steps

The load is 100 + j50 ohm, and the system is 50 ohm.

1. Divide by 50. You get 2 + j1. This is the normalized impedance, which just means "relative to 50 ohm".
2. Find the circle marked 2 and the arc marked +1. Put a dot where they cross.
3. The dot is above the middle line, so the load is inductive. That means it behaves like a coil.
4. Measure how far the dot is from the center. The distance is 0.447. This is the reflection coefficient, the share of the wave amplitude that bounces back.
5. Turn that distance into VSWR, a number that says how uneven the standing wave on the line is. Here it is 2.62. A perfect match would be 1.

The figure below shows how VSWR relates to return loss, which is the same mismatch told in decibels.

Mistakes beginners make

• Forgetting to normalize. The chart works in ratios. If you plot 100 + j50 directly, the point is wrong. Divide by Z0 first.
• Mixing up the direction. Moving toward the generator is clockwise. Moving toward the load is counterclockwise.
• Using the wrong circles for the wrong part. A series part moves along a resistance circle. A shunt part moves along a conductance circle. Mixing these up sends the dot to the wrong place.
• Thinking the left side is bad. The left end is a short circuit, but the center is the goal. Distance from the center is what counts.

Remember

• The center is a perfect match, and the distance from the center is the size of the reflection.
• Above the middle line means inductive, and below it means capacitive.
• Clockwise means toward the generator, and also the direction a series inductor or shunt capacitor moves the dot.

Related pages on RF View

• L-Network Impedance Matching Calculator
• Return Loss vs Insertion Loss: What Each Number Tells You