The Smith chart is a graphical aid or nomogram [1] designed as an aid in radio frequency engineering to assist in solving problems with transmission lines and matching circuits. The reflection coefficient is the relationship based on which the Smith chart is constructed:

The x-axis is the real part of the reflection coefficient (Γ_r) and the y-axis is the imaginary part of the reflection coefficient (Γ_i), where

The impedance plotted on the chart will be normalized with respect to the characteristic impedance Z₀:

The reflection coefficient can be represented as

and the load impedance as

In polar form, |Γ| and ϕ are, respectively the magnitude and angle of Γ as shown in Figure 2C.1.

Using Equations 2C.2 and 2C.3, and rearranging to get the real and imaginary parts of z_L, it can be seen that

After some further algebraic manipulation, it is found that

Equation 2C.8 represents a family of circles defined by the resistance r. These circles are centered at the point Γ_r = r/(1 + r), Γ_i = 0 (which is on the horizontal-axis), and have a radius equal to 1/(1 + r). When r = 0, this is a circle centered at the origin with radius = 1 (the unit circle). On the other hand, if r = ∞, the circle center is at (1, 0) with a radius of zero. Thus, for r = ∞, Equation 2C.8 represents a point on the Γ real axis at the far-right edge of the unit circle. If r = 1, Equation 2C.8 gives a circle centered at (½, 0) with a radius of ½. These circles along with those corresponding to r = ½ and 2 are shown in Figure 2C.2. Note that all the circles are centered on the positive Γ_r-axis and go through the point Γ_r = 1 and Γ_i = 0.

Equation 2C.9 also represents a family of circles, in this case centered at Γ_r = 1 and Γ_i = 1/x with a radius of 1/x. When x = ±∞, this represents a circle centered at (1, 0) with a radius of zero which is again the point at the right edge of the unit circle. If x = 1, the circle is centered at Γ = 1 + j1 with a radius of unity. Since |Γ| is always less than or equal to 1, only the portion of these circles within the unit circle are shown (Figure 2C.2). Figure 2C.3 shows a number of these circular sections for various values of x. Note that the circle representing x = 0 is along the horizontal-axis as the center of the circle moves toward ∞ along the Γ_r = 1 axis and the radius becomes infinite. As can be seen in Figure 2C.4, both sets of circles are plotted on the Smith chart. If Z_L is given, one may divide by Z₀ and find the intersection of the appropriate r and x circles and determine Γ from the plot.

To avoid unreadable clutter on the graph, the concentric circles for |Γ| and the radial lines for ϕ are not shown on the Smith chart. But rather graduated line segments at the bottom of the chart and circumference labels of angle around the outer circle are included as shown in Figure 2C.4. As an example, consider the point Z_L = 50 + j100 Ω on a 50-Ω line. Then z_L = Z_L/50 = 1 + j2, so that r = 1 and x = 2. This point is plotted as point P in Figure 2C.4 where it can be seen that |Γ| ≈ 0.7 and ϕ = 45°.

The final scale on the Smith chart, which is used to calculate the distance along a transmission line, is added on the circumference of the plot. The scale is in wavelength units with the zero point taken arbitrarily to the left. It is known that the voltage and the current at any point along a transmission line is given, respectively, by

and

Dividing Equation 2C.10 by Equation 2C.11 and normalizing gives the normalized input impedance:

Replacing z by −d and dividing the numerator and denominator by e^jβd, we obtain the general equation relating normalized input impedance, reflection coefficient, and line length (d):

Equation 2C.13 shows that the input impedance at any point z = −d can be obtained by replacing Γ, the reflection coefficient of the load, by Γ e^−j2βd. This corresponds to a decrease in the angle of Γ by 2βd radians as we go from the load to the line input. Only the angle of Γ changes, whereas the magnitude stays constant, thus the movement is along a constant radius arc. As we move from the load z_L to the input impedance z_in, we move toward the generator by a distance d on the transmission line and a clockwise angle 2βd on the Smith chart. One complete rotation around the Smith chart corresponds to a phase change of π radians or when d changes a half wavelength. The Smith chart is thus marked with a scale showing a change of 0.5λ for one circumnavigation of the unit circle. Two scales are normally given, one showing an increase in distance for clockwise movement and the other an increase for counterclockwise rotation. These two scales can be seen on the Smith chart of Figure 2C.5. One scale represents wavelengths toward generator (wtg) and increases for clockwise rotation. The other represents wavelengths toward load (wtl) and increases for counterclockwise rotation.

2C.1Reflection of Uniform Plane Waves

Note that the characters with tilde (~) represent a phasor quantity. A superscript plus sign indicates a wave propagating in the +z-direction, whereas a superscript minus sign represents a wave propagating in the −z-direction.

2C.1.1Free Space: Perfect Conductor Interface

Let us consider the following problem depicted in Figure 2C.8.

The space z > 0 is a perfect conductor and the space z < 0 is a free space. A time-harmonic electromagnetic field propagating in the +z-direction in the free space is normally incident on the interface from the left. Let the electric field of this wave be in the x-direction only, that is,

(2C.14)$${\tilde{E}}^{\text{I}}\left(z\right)={\tilde{E}}_0^{+}{\text{e}}^{-\text{j}\mathrm{\beta }\text{z}}\widehat{x},$$

where $\mathrm{\beta }=\mathrm{\omega }\text{}/\text{}c=\mathrm{\omega }\sqrt{{\mathrm{\mu }}_0{\mathrm{\epsilon }}_0}\text{and}\mathrm{\omega }=2\mathrm{\pi }f$. The magnetic field of this incident wave is

(2C.15)$${\tilde{H}}^{\text{I}}\left(z\right)=\frac{{\tilde{E}}_0^{+}}{{\mathrm{\eta }}_0}{\text{e}}^{-\text{j}\mathrm{\beta }z}\widehat{y},$$

where ${\mathrm{\eta }}_0=\sqrt{{\mathrm{\mu }}_0\text{/}{\mathrm{\epsilon }}_0}=120\mathrm{\pi }.$ The superscript I stands for incident wave. The electric field at z = 0 of this incident wave is ${\tilde{E}}_0^{+}$. But the boundary condition at z = 0 requires the electric field should be zero. Such a boundary condition can be satisfied if we assume that a reflected wave exists in the free space (propagating in the −z-direction) with the electric and magnetic fields as follows:

(2C.16)$${\tilde{E}}^{\text{R}}\left(z\right)={\tilde{E}}_0^{-}{\text{e}}^{+\text{j}\mathrm{\beta }z}\widehat{x},$$

(2C.17)$${\tilde{H}}^{\text{R}}\left(z\right)=-\frac{{\tilde{E}}_0^{-}}{{\mathrm{\eta }}_0}{\text{e}}^{+\text{j}\mathrm{\beta }z}\widehat{y},$$

and

(2C.18)$${\tilde{E}}_0^{+}+{\tilde{E}}_0^{-}=0,$$

so

(2C.19)$${\mathrm{\Gamma }}_0=\frac{{\tilde{E}}_0^{-}}{{\tilde{E}}_0^{+}}=-1.$$

2C.2Comparison with a Transmission Line Problem

A comparison of a normal plane wave reflection (from a free space to a perfect electric conductor) to the transmission line analogy is shown in Figure 2C.9 and summarized in Table 2C.1.

2C.2.1Dielectric–Dielectric Interface

Let us now consider the problem of reflection of a plane wave at the interface of two dielectrics as depicted in Figure 2C.10.

Let the incident wave be traveling in medium 1 in the positive z-direction and then for z < 0:

(2C.20)$${\tilde{E}}^{\text{I}}\left(z\right)={\tilde{E}}_0^{+}{\text{e}}^{-\text{j}{\mathrm{\beta }}_{1}z}\widehat{x},$$

(2C.21)$${\tilde{H}}^{\text{I}}\left(z\right)=\frac{{\tilde{E}}_0^{+}}{{\mathrm{\eta }}_1}{\text{e}}^{-\text{j}{\mathrm{\beta }}_{1}z},\hat{\mathbf{y}}={\tilde{H}}_0^{+}{\text{e}}^{-\text{j}{\mathrm{\beta }}_{1}z}\hat{\mathbf{y}},$$

Let this give rise to a reflected wave,

(2C.22)$${\tilde{E}}^{\text{R}}\left(z\right)={\tilde{E}}_0^{-}{\text{e}}^{+\text{j}{\mathrm{\beta }}_{1}z}\widehat{x},$$

(2C.23)$${\tilde{H}}^{\text{R}}\left(z\right)=-\frac{{\tilde{E}}_0^{-}}{{\mathrm{\eta }}_1}{\text{e}}^{+\text{j}{\mathrm{\beta }}_{1}z}\widehat{y}={\tilde{H}}_0^{-}{\text{e}}^{+\text{j}{\mathrm{\beta }}_{1}z}\widehat{y},$$

and a transmitted wave:

(2C.24)$${\tilde{E}}^{\text{T}}\left(z\right)={\tilde{E}}_0^{\left(2\right)}{\text{e}}^{-\text{j}{\mathrm{\beta }}_{2}z}\widehat{x},$$

(2C.25)$${\tilde{H}}^{\text{T}}\left(z\right)=\frac{{\tilde{E}}_0^{\left(2\right)}}{{\mathrm{\eta }}_2}{\text{e}}^{-\text{j}{\mathrm{\beta }}_{2}z}\widehat{y}={\tilde{H}}_0^{\left(2\right)}{\text{e}}^{-\text{j}{\mathrm{\beta }}_{2}z}\widehat{y}.$$

The problem is to determine the reflection coefficient ${\mathrm{\Gamma }}_0={\tilde{E}}_0^{-}\text{/}{\tilde{E}}_0^{+}$ and the transmission coefficient $T_0={\tilde{E}}_0^{\left(2\right)}\text{/}{\tilde{E}}_0^{+}$. The boundary conditions give us the required number of equations to solve for Γ₀ and T₀. The boundary conditions at a dielectric–dielectric interface are continuity of tangential components of $\overline{E}$ and $\overline{H}$:

(2C.26)$${\tilde{E}}_0^{+}+{\tilde{E}}_0^{-}={\tilde{E}}_0^{\left(2\right)},$$

(2C.27)$${\tilde{H}}_0^{+}+{\tilde{H}}_0^{-}={\tilde{H}}_0^{\left(2\right)},$$

or

(2C.28)$$\frac{{\tilde{E}}_0^{+}}{{\mathrm{\eta }}_1}-\frac{{\tilde{E}}_0^{-}}{{\mathrm{\eta }}_1}=\frac{{\tilde{E}}_0^{\left(2\right)}}{{\mathrm{\eta }}_2}.$$

Using Equations 2C.26 and 2C.28, we obtain

(2C.29)$${\mathrm{\Gamma }}_0=\frac{{\tilde{E}}_0^{-}}{{\tilde{E}}_0^{+}}=\frac{{\mathrm{\eta }}_2-{\mathrm{\eta }}_1}{{\mathrm{\eta }}_2+{\mathrm{\eta }}_1}$$

and

(2C.30)$$T_0=\frac{{\tilde{E}}_0^{\left(2\right)}}{{\tilde{E}}_0^{+}}=\frac{2{\mathrm{\eta }}_2}{{\mathrm{\eta }}_2+{\mathrm{\eta }}_1}.$$

2C.3Comparison with a Transmission Line Problem

Figure 2C.11a shows the normal incidence from the left of a plane wave on a dielectric–dielectric interface from dielectric 1 on the left to dielectric 2 on the right. Figure 2C.11b and c shows the transmission line analogy and the equivalent transmission line circuit, respectively. The transmission line 2 extends to infinity. On this line, the wave propagates in the +z-direction and there is no reflected wave. The reflection coefficient at z = ∞ is zero and Γ_∞ = 0, and therefore Z_in = Z₀₂ (see Figure 2C.11c). The reflection coefficient may now be obtained as follows:

(2C.31)$${\mathrm{\Gamma }}_0=\frac{Z_{02}-Z_{01}}{Z_{02}+Z_{01}}=\frac{{\mathrm{\eta }}_2-{\mathrm{\eta }}_1}{{\mathrm{\eta }}_2+{\mathrm{\eta }}_1}.$$

The purpose of developing the analogy with the transmission lines is to exploit the tools available for transmission line problems for solving the analogous boundary value problems involving plane waves. The Smith chart is one of them. Let us illustrate through a few examples.

Example 2C.2

A 4-GHz uniform plane wave is normally incident from region 1 (z < 0, ε₁ = 5, μ₁ = 1, σ₁ = 0) toward region 2 (z > 0, ε₂ = 2, μ₂ = 10, σ₂ = 0) (Figure 2C.12). Find s (Voltage Standing Wave Ratio) for regions 1 and 2 and the intrinsic impedance (η_in) at z = −0.6 cm.

For region 2, there is no reflected wave, Γ_∞ = 0, and s = (1 + Γ)/(1 − Γ) = 1. To determine various quantities in region 1, let us draw an equivalent transmission line (Figure 2C.13). The load is purely resistive and Z_L = R_L > Z₀. Therefore, s in region 1 is Z_L/Z₀ = 5. To determine η_in at −0.6 cm, we make use of the equivalent transmission line of Figure 2C.14 and the Smith chart shown in Figure 2C.15. In this case,

(2C.32)$${\mathrm{\beta }}_1=\mathrm{\omega }\sqrt{{\mathrm{\mu }}_1{\mathrm{\epsilon }}_1}=\frac{2\mathrm{\pi }\times 4\times {10}^9}{3\times {10}^8}\sqrt{5},$$

(2C.33)$${\mathrm{\lambda }}_1=\frac{2\mathrm{\pi }}{{\mathrm{\beta }}_1}=\frac{2\mathrm{\pi }\times 3\times {10}^8}{2\mathrm{\pi }\times 4\times {10}^9\sqrt{5}}=\frac{0.075\text{m}}{\sqrt{5}}=\frac{7.5\text{cm}}{\sqrt{5}},$$

(2C.34)$$\frac{d}{{\mathrm{\lambda }}_1}=\frac{0.6\sqrt{5}}{7.5}=\mathrm{0.1789.}$$

Load point A is located at r = 5, x = 0 on the Smith chart (0.25 wtg). To reach B on the |Γ| circle, move 0.1789 wtg from A on the |Γ| circle, that is, go to (0.25 + 0.1789) wtg = 0.4289 wtg. From the Smith chart read z_B = 0.25 − j0.45. Therefore,

(2C.35)$${\mathrm{\eta }}_{\text{in}}\left(z=-0.6\text{cm}\right)=\frac{120\mathrm{\pi }}{\sqrt{5}}\left(0.25-\text{j}0.45\right)=42.15-\text{j}75.87=86.8\angle -60.9°.$$

Example 2C.3

Design a radome. A radome is a dielectric material that covers an antenna and protects it from the weather. The thickness of the material should be such that there are no reflections (see Figure 2C.16).

For a transmission line analogy (see Figure 2C.17), it is obvious that η_input = η₀ if d = λ/2. If η_input = η₀, there is a perfect match at interface 1 and there will be no reflections at interface 1.

Consider a numerical example, where a uniform plane wave with λ = 3 cm in the free space is normally incident on a fiberglass (ε_r = 4.9, σ = 0) radome. (a) What thickness of fiberglass will produce no reflections? (b) What percentage of the incident energy will be transmitted through the fiberglass if the frequency of the incident wave is decreased by 10%?

In this problem, ε₁ = 4.9 ε₀, f = c/λ = 3 × 10⁸/3 × 10⁻² = 10 GHz.

$${\mathrm{\beta }}_1=\mathrm{\omega }\sqrt{{\mathrm{\mu }}_1{\mathrm{\epsilon }}_1}=\frac{2\mathrm{\pi }\times 10\times {10}^9}{3\times {10}^8}\sqrt{4.9}\text{and}{\mathrm{\lambda }}_1=\frac{2\mathrm{\pi }}{{\mathrm{\beta }}_1}=1.355\text{cm}.$$

1. d = λ₁/2 = 0.678 cm for no reflections.

2. If f_new = 0.9f, λ₀(new) = c/f_new = λ₀/0.9.

$${\mathrm{\beta }}_1\left(\text{new}\right)={\mathrm{\omega }}_{\text{new}}\sqrt{{\mathrm{\mu }}_1{\mathrm{\epsilon }}_1}=2\mathrm{\pi }{f}_{\text{new}}\sqrt{{\mathrm{\mu }}_1{\mathrm{\epsilon }}_1}=2\mathrm{\pi }\left(0.9\right)f\sqrt{{\mathrm{\mu }}_1{\mathrm{\epsilon }}_1}=0.9{\mathrm{\beta }}_1\text{and}{\mathrm{\lambda }}_1\left(\text{new}\right)=\frac{2\mathrm{\pi }}{{\mathrm{\beta }}_1\left(\text{new}\right)}=\frac{{\mathrm{\lambda }}_1}{0.9}.$$

Since the physical width d is the same,

$$\frac{d}{{\mathrm{\lambda }}_1\left(\text{new}\right)}=\frac{d\left(0.9\right)}{{\mathrm{\lambda }}_1}=\frac{\left({\mathrm{\lambda }}_1\text{/}2\right)\left(0.9\right)}{{\mathrm{\lambda }}_1}=\mathrm{0.45.}$$

Here η_input may now be obtained by solving the following transmission line problem (Figure 2C.18). To use the Smith chart (Figure 2C.19), let us use normalized values:

$$Z_0={\mathrm{\eta }}_1=\sqrt{\frac{{\mathrm{\mu }}_1}{{\mathrm{\epsilon }}_1}}=\frac{120\mathrm{\pi }}{\sqrt{4.9}}=\frac{{\mathrm{\eta }}_0}{\sqrt{4.9}}.$$

But Z_L = η₀, so $z_{\text{L}}=\sqrt{4.9}=\mathrm{2.214.}$ This is point A on the Smith chart (r = 2.214, x = 0). Now draw the |Γ| circle and move 0.45 wtg to reach B on the |Γ| circle. Read z_B = 1.6 + j0.85.

η_input = η₁ (1.6 + j0.85). The equivalent transmission line load = η₁ (1.6 + j0.85)/η₀ = 0.723 + j0.384. This normalized load of z_L = 0.723 + j0.384 is marked as point C on the Smith chart. To get the reflection coefficient magnitude measure OC = 0.26, |Γ|² = 0.0676. The transmitted energy = 1 − |Γ|² = 0.9324 = 93.24%.

2C.4Example of Use of the Smith Chart for Oblique Incidence [2]

The analogy between plane wave reflection problems for oblique incidence and transmission lines is summarized in Table 2C.2.

Consider the example shown in Figure 2C.20 where we have oblique incidence at an angle of 40° with perpendicular polarization from free space onto two dielectric layers situated within free space at a frequency of 10 GHz.

The incident wave freespace wavelength is

(2C.36)$${\mathrm{\lambda }}_0=\frac{c}{f}=\frac{3\times {10}^8}{10\times {10}^9}=3\text{cm}.$$

From Snell’s law,

(2C.37)$${\mathrm{\theta }}_2={sin}^{-1}\left(\frac{sin\left(40°\right)}{\sqrt{2.54}}\right)=23.79°,$$

(2C.38)$${\mathrm{\theta }}_3={sin}^{-1}\left(\frac{sin\left(40°\right)}{\sqrt{4}}\right)=18.73°,$$

(2C.39)$${\mathrm{\theta }}_4={sin}^{-1}\left(\frac{sin\left(40°\right)}{\sqrt{1}}\right)=40°.$$

For perpendicular polarization,

(2C.40)$$\begin{array}{ccc}\hfill Z_{01}& =\hfill & \frac{120\mathrm{\pi }}{cos\left(40°\right)}=492\mathrm{\Omega },\hfill \\ \hfill Z_{02}& =\hfill & \frac{120\mathrm{\pi }\text{/}\sqrt{2.54}}{cos\left({\mathrm{\theta }}_2\right)}=258.5\mathrm{\Omega },\hfill \\ \hfill Z_{03}& =\hfill & \frac{120\mathrm{\pi }\text{/}\sqrt{4}}{cos\left({\mathrm{\theta }}_3\right)}=199\mathrm{\Omega }.\hfill \end{array}$$

The length in wavelengths of each region

(2C.41)$${\mathrm{\lambda }}_{\text{m}z}=\frac{2\mathrm{\pi }}{q_{\text{m}}}=\frac{2\mathrm{\pi }}{k_{\text{m}}cos{\mathrm{\theta }}_{\text{m}}}=\frac{2\mathrm{\pi }}{\mathrm{\omega }\sqrt{{\mathrm{\mu }}_{\text{m}}{\mathrm{\epsilon }}_{\text{m}}}cos{\mathrm{\theta }}_{\text{m}}},$$

(2C.42)$${\mathrm{\lambda }}_{\text{mz}}=\frac{2\mathrm{\pi }}{\mathrm{\omega }\sqrt{{\mathrm{\mu }}_0{\mathrm{\epsilon }}_0}\sqrt{{\mathrm{\epsilon }}_{\text{r}}}cos{\mathrm{\theta }}_{\text{m}}}=\frac{{\mathrm{\lambda }}_0}{\sqrt{{\mathrm{\epsilon }}_{\text{r}}}cos{\mathrm{\theta }}_{\text{m}}},$$

(2C.43)$$\frac{{\text{d}}_2}{{\mathrm{\lambda }}_{2z}}=\frac{0.2\sqrt{2.54}cos\left(23.79\right)}{3}=0.097,$$

and therefore,

$$\frac{{\text{d}}_3}{{\mathrm{\lambda }}_{3z}}=\frac{0.3\sqrt{4}cos\left(18.73\right)}{3}=\mathrm{0.189.}$$

Now starting at interface AA, we have the equivalent transmission line shown in Figure 2C.21a that is point A on the Smith chart of Figure 2C.22:

(2C.44)$$Z_{\text{A}}=Z_{\text{L}3}=\frac{120\mathrm{\pi }}{cos\left(40°\right)}=492\mathrm{\Omega },$$

where

$$\begin{array}{c}\hfill Z_{03}=199\mathrm{\Omega },\hfill \\ \hfill z_{\text{A}}=\frac{492}{199}=\mathrm{2.47.}\hfill \end{array}$$

Also shown in the Smith chart of Figure 2C.22, moving from point A toward generator 0.189 wavelengths, we read (point B):

(2C.45)$$z_{\text{B}}=0.48-\text{j}0.325,$$

and therefore,

(2C.46)$$Z_{\text{B}}=\left(0.48-\text{j}0.325\right)199,$$

which can be seen as point B on the Smith chart and the transmission line of Figure 2C.21b.

Renormalizing,

(2C.47)$$z_{\text{B}\text{'}}=\left(0.48-\text{j}0.325\right)\frac{199}{258.5}=0.37-\text{j}\mathrm{0.25.}$$

This is represented by point B′ on the Smith chart and the transmission line of Figure 2C.21c. Now moving 0.097 wavelengths toward the generator, we have the transmission line analogy shown in Figure 2C.21d and point C on the Smith chart:

(2C.48)$$z_{\text{C}}=0.38+\text{j}0.3,$$

and therefore

(2C.49)$$Z_{\text{C}}=\left(0.38+\text{j}0.3\right)\mathrm{258.5.}$$

Renormalizing, we get

(2C.50)$$z_{\text{C}\text{'}}=\left(0.38+\text{j}0.3\right)\frac{258.5}{492}=0.2+\text{j}\mathrm{0.158.}$$

This is represented by point C′ on the Smith chart and the equivalent transmission line of Figure 2C.21e. The reflection coefficient can be read from the Smith chart as

(2C.51)$${\mathrm{\Gamma }}_{\text{s}}=0.66\angle 162°$$

and the power reflection coefficient is given as

(2C.52)$${\left|{\mathrm{\Gamma }}_{\text{s}}\right|}^2=\mathrm{0.435.}$$