Table of Contents

**The Smith Chart**

**Smith Chart:** The mathematics required to design and analyze transmission lines is complex, whether the line is a physical cable connecting a transceiver to an antenna or is being used as a filter or impedance-matching network. This is so because the impedances involved are complex ones, involving both resistive and reactive elements. The impedances are in the familiar rectangular form R + jX. Computations with complex numbers such as this are long and time-consuming.

Furthermore, many calculations involve trigonometric relationships. Although no individual calculation is difficult, the sheer volume of the calculations can lead to errors.

In the 1930s, one clever engineer decided to do something to reduce the chance of error in transmission line calculations. The engineer’s name was Philip H. Smith, and in January 1939 he published the Smith chart, a sophisticated graph that permits visual solutions to transmission line calculations.

Today, of course, the mathematics of transmission line calculations is not a problem, because of the wide availability of electronic computing options. Transmission line equations can be easily programmed into a scientific and engineering calculator for fast, easy solutions. Personal computers provide an ideal way to make these calculations, either by using special mathematics software packages or by using BASIC, Fortran, C, or another language to write the specific programs. The math software packages commonly available today for engineering and scientific computation also provide graphical outputs, if desired.

Despite the availability of all the computing options today, the Smith chart is still used. Its unique format provides a more or less standardized way of viewing and solving transmission line and related problems. Furthermore, a graphical representation of an equation conveys more information than is gained by a simple inspection of the equation. For these reasons, it is desirable to become familiar with the Smith chart.

equation conveys more information than is gained by a simple inspection of the equation. For these reasons, it is desirable to become familiar with the Smith chart.

Fig. 13-30 is a Smith chart. This imposing graph is created by plotting two sets of orthogonal (at right angles) circles on a third circle. Smith chart graph paper is available from the original publisher, Analog Instruments Company, and from the American Radio Relay League. Smith chart graph paper is also available in some college and university bookstores. The first step in creating a Smith chart is to plot a set of eccentric circles along a horizontal line, as shown in Fig. 13-31. The horizontal axis is the pure resistance or zero reactance line. The point at the far left end of the line represents zero resistance, and the point at the far right represents infinite resistance. The resistance circles are centered on and pass-through this pure resistance line. The circles are all tangent to one another at the infinite resistance point, and the centers of all the circles fall on the resistance line.

Each circle represents all the points of some fixed resistance value. Any point on the outer circle represents a resistance of 0 Ω. The other circles have other values of resistance. The R = 1 circle passes through the exact center of the resistance line and is known as the prime center. Values of pure resistance and the characteristic impedance of the transmission line are plotted on this line.

The most common transmission line impedance in use today is 50 Ω. For that reason, most of the impedances and reactances are in the 50-Ωrange. It is convenient, then,

to have the value of 50 V located at the prime center of the chart. This means that all points on the R = 1 circle represent 50 Ω, all points on the R = 0.5 circle represent 25 Ω, all points on the R = 2 circle represent 100 Ω, and so on.

Smith charts are in what is called normalized form, with R = 1 at the prime center. Users customize the Smith chart for specific applications by assigning a different value to the prime center.

The remainder of the Smith chart is completed by adding reactance circles, as shown in Fig. 13-32. Like the resistive circles, these are eccentric, with all circles meeting at the infinite resistance point.

Each circle represents a constant reactance point, with the inductive-reactance circles at the top and the capacitive-reactance circles at the bottom. Note that the reactance circles on the chart are incomplete. Only those segments of the circles within the R = 0 line are included on the chart. Like the resistive circles, the reactive circles are presented in normalized form. Compare Figs. 13-31 and 13-32 to the complete Smith chart in Fig. 13-30 before you proceed.

**Plotting and Reading Impedance Values**

The Smith chart in Fig. 13-33 shows several examples of plotted impedance values:

Z1 = 1.5 + j0.5

Z2 = 5 – j1.6

Z3 = 0.2 + j3

Z4 = 0.4 – j0.36

Locate each of those points on the Smith chart, and be sure that you understand how each was obtained.

The impedance values plotted in Fig. 13-33 are normalized values. Scaling the Smith chart to a particular impedance range requires multiplying those values by some common factor. For example, many Smith charts are plotted with 50 **Ω** at the prime center. The above-listed impedances for a prime center value of 50 **Ω** would be as follows:

Z1 = 75 + j25

Z2 = 250 – j80

Z3 = 25 + j150

Z4 = 20 – j18

To solve problems with real impedance values, put them into normalized form by dividing the resistive and reactive numbers by a factor equal to the resistance value at the prime center. Then plot the numbers.

When you are reading values from a normalized Smith chart, convert them to standard impedance form by multiplying the resistance and reactance by a factor equal to the resistance of the prime center.

**Wavelength Scales**

The three scales on the outer perimeter of the Smith chart in Fig. 13-30 are wavelength toward the generator, wavelength toward the load, and angle of the reflection coefficient in degrees. The scale labeled “toward generator” begins at the zero-resistance and zero-reactance line and moves clockwise around to the infinite resistance position. One-half of a circular rotation is 90º, and on the Smith chart scale, it represents the one-quarter wavelength. One full rotation is one-half wavelength. The transmission line patterns of voltage and current distribution along a line repeat every one-half wavelength.

The scale labeled “toward load” also begins at the zero-resistance, zero-reactance point and goes in a counterclockwise direction for one complete rotation or one-half wavelength. The infinite resistance point is the one-quarter wavelength mark. The reflection coefficient (the ratio of the reflected voltage to the incident voltage) has a range of 0 to 1, but it can also be expressed as an angle from 08 to 360º, from 08 to positive 180º, or from 08 to minus 180º. The zero marker is at the infinite resistance point on the right hand of the resistance line.

**SWR Circle**

The SWR of a transmission line plotted on the Smith chart is a circle. If the load is resistive and matched to the characteristic impedance of the line, the standing wave ratio is 1. This is plotted as a single point at the prime center of the Smith chart. The impedance of the line is flat 50 V or any other normalized value. However, if the load is not perfectly matched to the prime impedance, standing waves will exist. The SWR in such cases is represented by a circle whose center is the prime center.

To draw an SWR circle, first, calculate the SWR by using one of the previously given formulas. For this example, assume an SWR of 2. Starting at the prime center, move to the right on the resistance line until the value 2 is encountered. Then, using a drawing compass, place the point at the center and draw a circle through the 2 marks to the right of the prime center. The circle should also pass through the 0.5 marks to the left of the prime center. The red circle drawn in Fig. 13-33 represents a plot of the impedance variations along an unmatched or resonant transmission line. The variations in the voltage and current standing waves mean that there is a continuous variation in the impedance along the line. In other words, the impedance at one point on the unmatched line is different from the impedance at all other points on the line. All the impedance values appear on the SWR circle.

The SWR circle can also be used to determine the maximum and minimum voltage points along the line. For example, the point at which the SWR circle crosses the resistance line to the right of the prime center indicates the point of maximum or peak voltage of the standing wave in wavelengths from the load. This is also the maximum impedance point on the line. The point at which the SWR circle passes through the resistance line to the left of the prime center indicates the minimum voltage and impedance points.

The linear scales printed at the bottom of Smith’s charts are used to find the SWR, dB loss, and reflection coefficient. For example, to use the linear SWR scale, simply draw a straight line tangent to the SWR circle and perpendicular to the resistance line on the left side of the Smith chart. Make the line long enough that it intersects the SWR scale at the bottom of the chart. This has been done in Fig. 13-33. Note that the SWR circle for a value of 2 can be read from the linear SWR scale.

**Using the Smith Chart: Examples**

As discussed above, when the load does not match the characteristic impedance in a given application, the length of the line becomes a part of the total impedance seen by the generator. The Smith chart provides a way to find this impedance. Once the impedance is known, an impedance-matching circuit can be added to compensate for the conditions and make the line flat and the SWR as close to 1 as possible.

**Example 1 for Fig. 13-34** The operating frequency for a 24-ft piece of RG-58A/U coaxial cable is 140 MHz. The load is resistive, with a resistance of 93 Ω. What is the impedance seen by a transmitter?

The first step is to find the number of wavelengths represented by 24 ft of cable. A 140-MHz signal has a wavelength of

λ = 984/f = 984/140 = 7.02 ft

Remember that coaxial cable has a velocity factor less than 1 due to the slowdown of RF signals in a cable. The velocity factor of RG-58A/U is 0.66. Therefore, one wavelength at 140 MHz is

λ = 7.02 x 0.66 = 4.64 ft

The number of wavelengths represented by the 24-ft cable is 24/4.64 = 5.17 λ. As indicated earlier, the impedance variations along a line repeat every one-half wavelength and therefore every full wavelength; thus for the purposes of calculation, we need only the 0.17 λ of the above value.

Next, we normalize the Smith chart to the characteristic impedance of the coaxial cable, which is 53.5 Ω. The value of the prime center is 53.5Ω. The SWR is then computed:

SWR = Z_{l}/Z_{0} = 93/53.5 = 1.74

This SWR is now plotted on the Smith chart. See Fig. 13-35, where point X represents the resistive load of 93 Ω.

To find the impedance at the transmitter end of the coaxial cable, we move along line 5.17 λ from the load back to the transmitter or generator. Remember that one full rotation around the Smith chart is one-half wavelength, because the values repeat every one-half wavelength. Starting at point X, then, move in a clockwise direction (toward the generator) around the SWR circle for 10 complete revolutions, which represents 5 λ. This brings you back to point X.

Continue the clockwise rotation for an additional 0.17 λ, stop there, and mark that point on the SWR circle. Draw a line from the prime center to the location that represents 0.17 λ. The marking on the right-hand side of point X is the 0.25-λ mark. You need to go 0.17 λ from that, or 0.25+0.17 = 0.42. Stop at the 0.42 mark on the lower part of the circle at Y. Draw a line from there to the prime center. Refer again to Fig. 13-35.

The point at which the line cuts the SWR circle is the impedance that the generator sees. Reading the values from the chart, you have R ≈ 0.68 and XC ≈ 0.35. Since the point is in the lower half of the chart, the reactance is capacitive. Thus the impedance at this point is 0.68 -j0.35.

To fi nd the actual value, convert from the normalized value by multiplying by the impedance of the prime center, or 53.5:

Z = 53.5(0.68 – j0.35) = 36.4 – j18.725 Ω

The transmitter is seeing what appears to be a 36.4-Ω resistor in series with a capacitor with a reactance of 18.725** **Ω. The equivalent capacitance value is [using X_{C} = 1/(2πf C) or C = 1/(2πf X_{C})]

C = 1/6.28(140 x 10^{6} )(18.725) = 60.74 pF

The equivalent circuit is shown in Fig. 13-34(b).

**Example 2 for Fig. 13.36** An antenna is connected to the 24-ft 53.5-Ω RG-58A/U line described in Example 1. The load is 40 + j30 Ω. What impedance does the transmitter see?

The prime center is 53.5 Ω, as before. Before plotting the load impedance on the chart, you must normalize it by dividing by 53.5:

Zl = 40 + j30/53.5 = 0.75 + j0.56 Ω

The plot for this value is shown in Fig. 13-36. Remember that all impedances fall on the SWR circle. You can draw the SWR circle for this example simply by placing the compass on the prime center with a radius out to the load impedance point and rotating 360°.

The SWR is obtained from the chart at the bottom of the figure by extending a line perpendicular to the resistance axis down to the SWR line. The SWR is about 2:1.

The next step is to draw a line from the prime center through the plotted load impedance so that it intersects the wavelength scales on the perimeter of the curve. Refer again to Fig. 13-36. Because our starting point of reference is the load impedance, we use the “toward generator” scale to determine the impedance at the input to the line. At point X is the wavelength value of 0.116.

Now, because the generator is 5.17λ away from the load and because the readings

repeat every one-half wavelength, you move toward the generator in the clockwise direction 0.17λ: 0.116 and 0.17 = 0.286. This point is marked Y in Fig. 13-36.

Draw a line from point Y through the prime center. The point at which it intersects the SWR circle represents the impedance seen by the generator. The normalized value is 1.75 – j0.55. Correcting this for the 53.5 Ω prime center, we get

Z = 53.5(1.75 – j0.55) = 93.6 – j29.4 Ω

This is the impedance that the generator sees.

**Example 3 for Fig. 13.37**

In many cases, the antenna or other load impedance is not known. If it is not matched to the line, the line modifies this impedance so that the transmitter sees a different impedance. One way to find the overall impedance, as well as the antenna or other load impedance, is to measure the combined impedance of the load and the transmission line at the transmitter end, using an impedance bridge. Then the Smith chart can be used to find the individual impedance values.

A 50-ft RG-11/U foam dielectric coaxial cable with a characteristic impedance of 75 Ω and a velocity factor of 0.8 has an operating frequency of 72 MHz. The load is an antenna whose actual impedance is unknown. Measurement at the transmitter end of the cable gives a complex impedance of 82 + j43. What is the impedance of the antenna?

One wavelength at 72 MHz is 984/72 =13.67 ft. Taking the velocity factor into account yields

λ = 13.67 x 0.8 = 10.93 ft

The length of the 50-ft cable, in wavelengths, is 50/10.93 = 4.57. As before, it is the 0.57 λ value that is of practical use.

First, the measured impedance is normalized and plotted. The prime center is 75 Ω. In Fig. 13-37 Z=(82 + j43)/75 = 1.09 + j0.52 is plotted on the Smith chart. The SWR circle is plotted through this point from the prime center. A tangent line is then drawn from the circle to the linear SWR chart on the bottom. The SWR is 1.67.

Next, a line is drawn from the prime center to the normalized impedance point and through the wavelength plots on the perimeter of the graphs. It intersects the “toward load” scale at point X, or 0.346 λ. Refer to the figure.

The plotted impedance is what is seen at the generator end, so it is necessary to move around the graph to the load in the counterclockwise direction to find the load impedance. We move a distance equal to the length of the cable, which is 4.57 λ. Nine full rotations from point X represent 4.5 λ, which brings us back to point X. We then rotate 0.07 λ more, counterclockwise, to complete the length. This puts us at the 0.346 + 0.07 = 0.416λ point, which is designated point Y in Fig. 13-37. We then draw a line from Y to the prime center. The point at which this line intersects the SWR circle is the actual antenna impedance. The normalized value is 0.72 + j0.33. Multiplying by

75 gives the actual value of the antenna impedance:

Z = 75(0.72 + j0.33) = 54+ j24.75 Ω

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