Smith chart from the NEXT ANTENNAS calculator with the load 25 + j50 ohms plotted at the intersection of the r = 0.5 circle and the x = +1 arc
Fig 1 — A Smith chart for a 50 Ω system with the load Z_L = 25 + j50 Ω plotted. Green circles are constant resistance, orange arcs constant reactance; the highlighted circle and arc pass through the load. Try it in the Smith Chart Calculator →

What Is a Smith Chart?

The Smith chart is a graphical calculator for impedance. It is a plot of the complex reflection coefficient Γ (gamma) inside a circle of radius one, overlaid with a grid of circles and arcs that let you read the corresponding impedance directly. Every passive impedance — any combination of resistance, inductance and capacitance — lands on exactly one point inside that circle.

In Part 3 we defined the reflection coefficient as the ratio of the reflected to the incident voltage wave at a load. In Part 5 we saw that moving along a transmission line changes the impedance in a complicated way, but changes Γ in a very simple way: its magnitude stays the same and only its angle rotates. The Smith chart exploits exactly this. By drawing impedance on the Γ plane, line-length problems become rotations, matching problems become moves along circles, and VSWR becomes a radius you can see.

📌 In one sentence

A Smith chart is a polar plot of Γ with the impedance grid z = (1 + Γ)/(1 − Γ) drawn on top, so you can switch between reflection coefficient and impedance just by looking.

History and Development of the Smith Chart

The chart is named after Phillip H. Smith (1905–1987), an engineer at Bell Telephone Laboratories. In the early 1930s Smith was designing antenna feeds and transmission-line matching for radio transmitters, and every calculation of impedance along a line meant laborious work with hyperbolic functions, slide rules and tables. He first drew rectangular charts that plotted impedance against line position; they worked but needed huge sheets, because impedance runs from zero to infinity.

His key insight was to plot the reflection coefficient instead. Because |Γ| ≤ 1 for every passive load, the entire range of impedances fits inside a single circle, and moving along a lossless line is simply a rotation. He published the chart as a "Transmission Line Calculator" in Electronics magazine in January 1939, followed by an improved version in 1944 that added the scales still printed on paper charts today. A very similar chart was published independently in Japan by Tosaku Mizuhashi in 1937, and in Japan it is sometimes called the Mizuhashi–Smith chart.

YearMilestone
Early 1930sSmith, at Bell Labs, uses rectangular impedance charts for antenna-feed design
1937Tosaku Mizuhashi publishes a circular impedance chart in Japan
1939"Transmission Line Calculator" published in Electronics (January)
1944"An Improved Transmission Line Calculator" adds the outer wavelength and angle scales
1969Smith's book Electronic Applications of the Smith Chart
1970s – todayThe chart becomes the standard display of vector network analyzers and RF CAD tools

Nearly ninety years later, the computer has replaced the paper chart as a calculator, but not as a way of thinking. Every vector network analyzer (VNA) and every RF simulator still displays impedance on a Smith chart, because no other picture shows so much about a match at a glance.

Why Is the Smith Chart Used?

  • It fits every impedance on one page. Impedances from 0 to ∞ ohms, inductive or capacitive, all lie inside the unit circle.
  • Transmission lines become rotations. Moving a distance d along a lossless line rotates the point by an angle 2βd around the centre. A half wavelength is one full turn.
  • Matching becomes geometry. Adding a series reactance moves the point along a circle of constant resistance; adding a shunt reactance moves it along a circle of constant conductance. A matching network is a path from the load to the centre.
  • Mismatch is visible. Distance from the centre is |Γ|. Circles around the centre are circles of constant VSWR and return loss.
  • Impedance and admittance live on one chart. Rotating a point by 180° converts impedance into admittance, which is what you need for shunt components.
  • Frequency behaviour is a curve. A swept measurement traces a curve across the chart, showing at once where a device is matched, how fast it drifts and whether it resonates.

Relationship Between Impedance and Reflection Coefficient

For a load Z_L on a line of characteristic impedance Z₀, the reflection coefficient is fixed by the mismatch between the two. Dividing by Z₀ gives the normalised impedance z = Z_L/Z₀ (the subject of Part 7), and the relationship becomes independent of the system impedance:

Reflection coefficient from normalised impedance — and back
Γ=ZL−Z0ZL+Z0 =z−1z+1 z=1+Γ1−Γ
Γ = u + jv is a complex number with |Γ| ≤ 1 for any passive load (r ≥ 0). The Smith chart is drawn on the u–v plane.

This is a bilinear (Möbius) transformation. It has a remarkable property: it maps circles and straight lines into circles. The whole right half of the impedance plane (all r ≥ 0) maps into the unit disk, every vertical line of constant resistance becomes a circle, and every horizontal line of constant reactance becomes an arc of a circle. That is why the Smith chart is made entirely of circles.

Fig 2 — The same lines in the impedance plane (left) and on the Smith chart (right): r = 0 becomes the rim, r = 1 becomes the circle through the centre, x = +1 becomes an arc, and the real axis stays straight
z = 0.5 + j1 r = 0 r = 1 x = +1 x = 0 resistance r → reactance x →
Impedance plane (r ≥ 0)
z = 0.5 + j1
Smith chart (Γ plane)
📘 Go deeper: reflection coefficient on transmission lines

The reflection coefficient, incident and reflected waves and how Γ varies along a line are covered in detail in the transmission line series. Open the Transmission Line Theory series →

Understanding the Smith Chart Geometry

Substituting Γ = u + jv into z = (1 + Γ)/(1 − Γ) and separating real and imaginary parts gives the normalised resistance and reactance at any point on the chart:

Normalised resistance and reactance at a point (u, v) of the Γ plane
r=1−u2−v2(1−u)2+v2 x=2v(1−u)2+v2
Holding r constant and rearranging gives the equation of a circle; holding x constant gives another family of circles. These two families are the grid of the chart.

The chart is therefore three things laid on top of each other: a polar plot of Γ (magnitude = distance from the centre, angle = direction from the positive u axis), a family of constant-resistance circles, and a family of constant-reactance arcs. Paper charts add outer scales for wavelengths and reflection angle, covered in Part 8.

Resistance and Reactance Circles

Constant-resistance circles

Circle of constant normalised resistance r
(u−r1+r)2+v2=(11+r)2
Centre at (r/(1+r), 0), radius 1/(1+r). Every r circle touches the point Γ = +1 on the right edge. r = 0 is the outer rim; r = 1 passes through the centre; as r → ∞ the circle shrinks to the point Γ = +1.

Constant-reactance arcs

Circle of constant normalised reactance x
(u−1)2+(v−1x)2=(1x)2
Centre at (1, 1/x), radius 1/|x|. Only the part inside the unit circle is used, so they appear as arcs. Positive x (inductive) arcs curve into the upper half; negative x (capacitive) arcs into the lower half. x = 0 is the horizontal axis.
Fig 3 — Constant-resistance circles (left) and constant-reactance arcs (right). All of them pass through the open-circuit point on the right edge.
00.20.5125+j0.2−j0.2+j0.5−j0.5+j1−j1+j2−j2+j5−j5∞
r = 0r = 0.5r = 1r = 2r = 5
00.20.5125+j0.2−j0.2+j0.5−j0.5+j1−j1+j2−j2+j5−j5∞
x = ±0.5x = ±1x = ±2

A point on the chart is the intersection of one r circle and one x arc. To read an impedance you find which resistance circle and which reactance arc pass through it; to plot an impedance you find the intersection of the circle and arc you want. Part 9 turns this into a step-by-step procedure.

Center Point and Outer Boundary

Fig 4 — Landmarks on the Smith chart: the matched centre, the short and open circuits at the ends of the real axis, the inductive upper half and the capacitive lower half
0.20.525+j0.2−j0.2+j0.5−j0.5+j1−j1+j2−j2+j5−j5Inductive (+jx)Capacitive (−jx)Short (z = 0)Open (z = ∞)Matched (z = 1)+j50 Ω (pure L)−j50 Ω (pure C)
LocationΓzMeaning
Centre01 + j0Perfect match: Z_L = Z₀, no reflection, VSWR = 1
Left end of the axis−1 (1∠180°)0Short circuit
Right end of the axis+1 (1∠0°)∞Open circuit
Horizontal axisrealr + j0Pure resistance: left of centre r < 1, right of centre r > 1
Outer rim|Γ| = 10 + jxPure reactance (lossless L or C): all power reflected
Upper halfv > 0x > 0Inductive impedance
Lower halfv < 0x < 0Capacitive impedance
Outside the rim|Γ| > 1r < 0Negative resistance (oscillators, unstable amplifiers)

The distance from the centre is the magnitude of the mismatch. Points close to the centre are well matched; points near the rim reflect almost all the incident power. Any circle centred on the chart is a circle of constant |Γ| — and therefore of constant VSWR and return loss — which is why the centre is the target of every matching problem.

Applications in RF Engineering

Impedance matchingDesign L-, π- and T-networks and stub matches by moving the load point to the centre (Part 10, Part 11).
Transmission-line analysisFind the input impedance of a line, voltage maxima and minima and stub positions by rotation (Part 5).
Antenna measurementA VNA shows an antenna's S11 as a curve on the chart; the loop near the centre shows the matched band (Part 12).
Amplifier designStability, constant-gain and noise-figure circles are drawn on the chart to choose source and load impedances.
Filters and resonatorsResonances appear as loops and crossings of the real axis; coupling and Q can be read from the trace.
OscillatorsNegative-resistance devices plot outside the unit circle on a compressed chart.
Smith chart from the NEXT ANTENNAS calculator with both the impedance grid and the purple admittance grid shown
Fig 5 — The combined impedance–admittance (ZY) chart. The purple admittance grid is the impedance grid rotated by 180°; it is used whenever shunt components are added (Parts 8 and 10). Try it in the Smith Chart Calculator →

Worked Examples

All examples use a 50 Ω reference impedance. Each one can be reproduced by entering the load in the Smith Chart Calculator.

01

Plotting a 100 Ω Resistive Load

Problem: Where does a 100 Ω resistor terminating a 50 Ω line appear on the Smith chart? Find Γ and the VSWR.

z = 100/50 = 2; Γ = (z − 1)/(z + 1)
Γ=2−12+1=0.333∠0° VSWR=1+0.3331−0.333=2.0
Smith chart with a 100 ohm load plotted on the horizontal axis where the r = 2 circle crosses it
The 100 Ω load sits on the horizontal axis, on the r = 2 circle, one third of the way from the centre to the right edge. The calculator reports Γ = 0.333 ∠ 0°, VSWR 2.00 and S11 = −9.54 dB. Try it in the Smith Chart Calculator →
Answer: z = 2, Γ = 0.333 ∠ 0°, VSWR = 2.0

Any pure resistance lies on the horizontal axis. For R > Z₀ the point is right of centre and the VSWR equals r itself (here 2); for R < Z₀ it is left of centre and VSWR = 1/r.

02

Locating 25 + j50 Ω

Problem: Find Γ for Z_L = 25 + j50 Ω on a 50 Ω line, and show that the point lies on both the r = 0.5 circle and the x = +1 arc.

z = 0.5 + j1
Γ=−0.5+j11.5+j1 =(−0.5+j)(1.5−j)3.25 =0.0769+j0.6154=0.620∠82.9°
Check against the circle equations
r = 0.5: (0.0769−0.3333)2+0.61542=0.444=(0.667)2 ✓ x = 1: (0.0769−1)2+(0.6154−1)2=1.000 ✓
The load lies where the r = 0.5 circle and the x = +1 arc intersect; the arrow is the Γ vector
00.20.5125+j0.2−j0.2+j0.5−j0.5+j1−j1+j2−j2+j5−j5∞z = 0.5 + j1
r = 0.5x = +1|Γ| = 0.62
Answer: Γ = 0.620 ∠ 82.9° — exactly what the calculator in Fig 1 reports

The point is in the upper half (inductive) and 62% of the way to the rim, so the load is badly mismatched: VSWR = 4.27 and 38% of the incident power is reflected.

03

From Γ Back to Impedance

Problem: A network analyzer shows Γ = 0.5 ∠ −90° at the end of a 50 Ω cable. What is the load impedance?

Γ = −j0.5; z = (1 + Γ)/(1 − Γ)
z=1−j0.51+j0.5 =(1−j0.5)21.25 =0.6−j0.8 ZL=50z=30−j40 Ω
Answer: Z_L = 30 − j40 Ω (capacitive)

An angle of −90° points straight down, into the capacitive lower half — consistent with the negative reactance. At 1 GHz, −j40 Ω corresponds to a series capacitance of 1/(2π × 10⁹ × 40) ≈ 4.0 pF.

04

Short, Open, Matched and Pure Reactance

Problem: Find Γ for a short circuit, an open circuit, a 50 Ω load, +j50 Ω and −j50 Ω on a 50 Ω line, and locate each on the chart.

LoadzΓPosition (see Fig 4)
Short, 0 Ω0−1 = 1∠180°Left end of the axis
Open, ∞∞+1 = 1∠0°Right end of the axis
50 Ω10Centre
+j50 Ω+j1(j − 1)/(j + 1) = +j = 1∠90°Top of the rim
−j50 Ω−j1−j = 1∠−90°Bottom of the rim
Answer: |Γ| = 1 for the short, open and both pure reactances; Γ = 0 only for 50 Ω

Lossless reactances reflect all the incident power — they only shift its phase — so they sit on the rim. This is why an ideal inductor or capacitor alone can never be a matched load, but can be used to move a lossy load toward the centre.

05

Reading r and x From a Point

Problem: A point on the chart has Γ = 0.2 + j0.4. Use the geometry formulas to find r and x, then the impedance in a 50 Ω system.

Denominator (1 − u)² + v² = 0.64 + 0.16 = 0.80
r=1−0.04−0.160.80=1.0 x=2×0.40.80=1.0 Z=50+j50 Ω
Answer: z = 1 + j1, Z = 50 + j50 Ω

The point lies on the r = 1 circle — the circle through the centre. That makes it special: adding a series −j50 Ω (a capacitor) would move it along that circle straight to the centre, a one-component match. Points on the r = 1 or g = 1 circles are the stepping stones of every matching design in Parts 10 and 11.

Summary

ConceptExpression / locationKey point
Chart coordinatesΓ = u + jv, |Γ| ≤ 1Polar plot of the reflection coefficient
Impedance ↔ ΓΓ = (z − 1)/(z + 1), z = (1 + Γ)/(1 − Γ)Bilinear map: lines become circles
Resistance circlecentre (r/(1+r), 0), radius 1/(1+r)All touch the open-circuit point
Reactance arccentre (1, 1/x), radius 1/|x|Upper half inductive, lower half capacitive
CentreΓ = 0, z = 1Perfect match
Rim|Γ| = 1Pure reactance; total reflection
Circles about the centre|Γ| = constantConstant VSWR and return loss
  • The Smith chart plots the reflection coefficient and overlays the impedance grid, so every passive impedance fits inside one circle.
  • It was published by Phillip Smith of Bell Labs in 1939 and remains the standard display of network analyzers and RF CAD tools.
  • Constant-resistance lines become circles touching the right edge; constant-reactance lines become arcs.
  • The centre is a perfect match, the rim is pure reactance, the upper half is inductive and the lower half capacitive.

The chart is drawn in normalised units. In Part 7: Understanding Normalized Impedance on the Smith Chart we look at why, and how to convert between the chart and real ohms in 50 Ω and 75 Ω systems.

Put It Into Practice

Use the NEXT ANTENNAS calculators to apply the ideas from this article to real designs.