The term "Smith Chart" is a trademark of Analog Instruments Company of Providence, NJ, which distributes printed charts. Blank charts are also available from the American Radio Relay League, of Newington, CT, http://www.arrl.org
This tutorial gives a procedure for making simple Smith charts. It can be applied to many different drawing packages, including Microsoft Office Shapes, Inkscape, or even to drawing them by hand.
Drawing the charts by hand takes patience, but it can be done. I recommend using a set of dividers to accurately position the circles and arcs on their respective axes.
Mathematics
The basis for the chart is the reflection coefficient, \(\rho\), defined by the equation
$$\rho = \frac{\frac{Z}{Z_0} - 1}{\frac{Z}{Z_0} + 1}$$
Where \(Z\) is the complex impedance, \(R+jX\), and \(Z_0\) is a reference impedance to normalize it. \(Z_0\) is often chosen to be the characteristic impedance of a transmission line we are working with, like 50 or 450 ohms.
\(\frac{Z}{Z_0}\) is itself a complex value. If we let A be the real (resistance) part, and B the imaginary (reactance) part, the expression for \(\rho\) becomes
$$\rho = \frac{A^2 + B^2 - 1 + j2B}{\left(A + 1\right)^2 + B^2}$$
The lines on the chart are reflection coefficient points where either the resistance is held constant and the reactance varies from minus to plus infinity or vice versa.
Because of the logarithmic nature of the chart, choosing indices of 0.2, 0.5, 1.0, 2.0, 5.0 and 10 yields nice, even divisions.\
Zero Resistance, Zero Reactance
Zero resistance is represented by the circle that defines the outer edge of the chart. The left side of the circle is \(A=0\), \(B=0\). Looking at the equation above, you can see that as \(A\) gets very large, \(\rho\) approaches \(1\). This is the right side of the circle.
Zero reactance is a line through the middle of the chart. You can also think of this as another circle of infinite radius, centered at infinity.
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| Zero Resistance, Zero Reactance |
It helps to draw a line perpendicular to the zero reactance line when constructing the chart with a compass and ruler. This line is the \(A=\infty\) axis, and is where the constant reactance circles are centered.
Constant Resistance Circles
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| Constant Resistance Circles |
Now the chart gets interesting. From the reflection coefficient equation, we can show\footnote{See notes at the end of the tutorial.} that the radius of each constant resistance circle is
$$r = \frac{1}{A+1}$$
and that the center is at
$$x = \frac{A}{A+1}$$
where \(x\) is the linear value along the \(B=0\) axis, \(-1\) to \(1). These values are shown below.
| A | Center | Radius |
|---|---|---|
| 0 | 0 | 1 |
| 0.2 | 0.166667 | 0.833333 |
| 0.5 | 0.333333 | 0.666667 |
| 1.0 | 0.5 | 0.5 |
| 2.0 | 0.666667 | 0.333333 |
| 5.0 | 0.833333 | 0.166667 |
| 10 | 0.909091 | 0.090909 |
Constant Reactance Circles
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| Adding Constant Reactance Circles |
It can also be shown that the center of each constant reactance circle along the \(A=\infty\) axis is
$$y = \frac{1}{B}$$
The radius is
$$r = \left|\frac{1}{B}\right|$$
The centers and radii for the circles are shown in the following table.
| B | Center | Radius |
| -5.0 | -0.2 | 0.2 |
| -2.0 | -0.5 | 0.5 |
| -1.0 | -1.0 | 1.0 |
| -0.5 | -2.0 | 2.0 |
| -0.2 | -5.0 | 5.0 |
| 0.2 | 5.0 | 5.0 |
| 0.5 | 2.0 | 2.0 |
| 1.0 | 1.0 | 1.0 |
| 2.0 | 0.5 | 0.5 |
| 5.0 | 0.2 | 0.2 |
By the way, setting\(A\)equal to zero in the reflection coefficient equation gives the point where the constant reactance and zero resistance circles intersect.
$$\rho = \frac{B^2 + j2B - 1}{B^2 + 1} = -\frac{1-jB}{1+jB}$$
Numbering and Formatting
I like to number the resistance circles along the zero reactance line, and number the reactance circles around the zero resistance circle.
Instead of complete constant resistance circles, I suggest stopping them at strategic points. Since everything on a Smith chart is centered on the right side at \(B=0\), it tends to get a little cluttered there.
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| The Decluttered Chart |
To create the chart above, select the desired circle that is being drawn, and then the conjugate circle where it stops. Enter these as A and B.
$$\left(x - \frac{A}{A+1}\right)^2 + y^2 = \left(\frac{1}{A+1}\right)^2$$
$$\left(x - 1\right)^2 + \left(y - \frac{1}{B}\right)^2 = \frac{1}{B^2}$$
Simplifying,
$$y^2 = \left(\frac{1}{A+1}\right)^2 - \left(x - \frac{A}{A+1}\right)^2 \\
>= x^2 - \frac{2A}{A+1}x + \frac{A^2+1}{\left(A+1\right)^2}$$
$$y = x - \frac{\frac{2A}{A+1} \pm \sqrt{\frac{4A^2}{\left(A+1\right)^2}-\frac{4A^2+4}{\left(A+1\right)^2}}}{2} = x - \frac{A\pm2}{A+1}$$
$$x^2 - 2x + 1 + y^2 - \frac{2y}{B} + \frac{1}{B^2} = \frac{1}{B^2} \\
x^2 - 2x + y^2 - \frac{2y}{B} = -1$$
Of course, to draw the chart by hand, simply use the centers and radii found in the tables above.
Notes
Derivation Radius and Position of the Constant Resistance Circles
Setting \(B\) to zero in the reflection coefficient equation gives the point where the right side of the constant resistance circle intersects the zero reactance axis.
$$\rho = \frac{A^2-1}{\left(A+1\right)^2} = \frac{A-1}{A+1}$$
The radius of the circle will be the distance from there to the right side of the zero resistance circle (\(x=1\)).
$$\frac{1 - \left(A - 1\right)/\left(A + 1\right)}{2} = \frac{1}{A+1}$$
And the center will be
$$1 - \frac{1}{A+1} = \frac{A}{A+1}$$
This gives a standard form for the equation of the circles as
$$\left(x - \frac{A}{A+1}\right)^2 + y^2 = \left(\frac{1}{A+1}\right)^2$$
Derivation of the Height of the Constant Reactance Circles
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| Constant Reactance Circle Construction |
Right triangle DEF has angle \(\alpha\) at the point where the constant reactance circle intersects the zero resistance circle.
Since sides D and H are parallel, angles \(alpha\) and \(rho\) are equal. This means that triangles DEF and FGH are similar, because they share a side and each have two equal angles (angle-angle-side). Therefore, the ratios of side F to side D for triangle DEF is the same as the ratio of triangle FGH side H to side F. We can express side H in terms of sides F and D
$$H=\frac{F}{D} F = \frac{F^2}{D}$$
From Pythagorean theorem,
$$F=\sqrt{D^2 + E^2}$$
\(E\) is the radius of the zero resistance circle, \(1\) minus the real part of the zero-resistance - constant reactance intersection, \(C\).
This gives us the expression for \(H\).
$$H = \frac{D^2 + \left(1-C\right)^2}{D}$$
\(C\) and \(D\) come from the reflection coefficient equation.
$$H=\frac{\left(\frac{2B}{B^2+1}\right)^2+\left(1-\frac{B^2-1}{B^2+1}\right)^2}{\frac{2B}{B^2+1}}=\frac{4B^2+4}{{2B}\left(B^2+1\right)}=\frac{2}{B}$$
The center and radius of the constant reactance circle will be\(1/2H = 1/B\).
The standard form for the circle becomes
$$\left(x - 1\right)^2 + \left(y - \frac{1}{B}\right)^2 = \frac{1}{B^2}$$
