# 2.10 The Smith Chart

The Smith chart, developed by P. H. Smith in 1939, is a widely used graphical tool for analyzing and designing transmission-line circuits. Even though it was originally intended to facilitate calculations involving complex impedances, the Smith chart has become an important avenue for comparing and characterizing the performance of microwave circuits. As the material in this and the next section demonstrates, use of the Smith chart not only avoids tedious manipulations of complex numbers, but it also allows an engineer to design impedance-matching circuits with relative ease.

# 2.10.1 Parametric Equations

The reflection coefficient Γ is, in general, a complex quantity composed of a magnitude |Γ| and a phase angle θ r or, equivalently, a real part Γ r and an imaginary part Γ i,

Γ=|Γ|e j θ r=Γ r+j Γ i,(2.109)

where

Γ r=|Γ|cos θ r,(2.110a)
Γ i=|Γ|sin θ r.(2.110b)

The Smith chart lies in the complex Γ plane. In Fig. 2-24, point A represents a reflection coefficient Γ A=0.3+j 0.4 or, equivalently,

|Γ A|=[(0.3)2+(0.4)2]1/2=0.5

and

θ r A=tan−1(0.4/0.3)=53°.

Similarly, point B represents Γ B=−0.5−j 0.2, or |Γ B|=0.54 and θ r B=202° or, equivalently,

θ r B=(360°−202°)=−158°.

# Figure 2-24 The complex Γ plane. Point A is at Γ A=0.3+j 0.4=0.5 e j 53°, and point B is at Γ B=−0.5−j 0.2=0.54 e j 202°. The unit circle corresponds to |Γ|=1. At point C, Γ=1, corresponding to an open-circuit load, and at point D, Γ=−1, corresponding to a short circuit.

Figure 2 24 Full Alternative Text

The unit circle shown in Fig. 2-24 corresponds to |Γ|=1. Because |Γ|≤1 for a transmission line terminated with a passive load, only that part of the Γ r–Γ i plane that lies within the unit circle is useful to us; hence, future drawings will be limited to the domain contained within the unit circle.

Impedances on a Smith chart are represented by their values normalized to the line’s characteristic impedance, Z 0. From

Γ=Z L/Z 0−1 Z L/Z 0+1=z L−1 z L+1,(2.111)

the inverse relation is

The normalized load impedance z L is, in general, a complex quantity composed of a normalized load resistance r L and a normalized load reactance x L:

z L=r L+j x L.(2.113)

Using Eqs. (2.109) and (2.113) in Eq. (2.112), we have

r L+j x L=(1+Γ r)+j Γ i(1−Γ r)−j Γ i,(2.114)

which can be manipulated to obtain explicit expressions for each of r L and x L in terms of Γ r and Γ i. This is accomplished by multiplying the numerator and denominator of the right-hand side of Eq. (2.114) by the complex conjugate of the denominator and then separating the result into real and imaginary parts. These steps lead to

r L=1−Γ r 2−Γ i 2(1−Γ r)2+Γ i 2,(2.115a)
x L=2 Γ i(1−Γ r)2+Γ i 2.(2.115b)

Equation (2.115a) implies that there exist many combinations of values for Γ r and Γ i that yield the same value for the normalized load resistance r L. For example, (Γ r,Γ i)=(0.33,0) gives r L=2, as does (Γ r,Γ i)=(0.5,0.29), as well as an infinite number of other combinations. In fact, if we were to plot in the Γ r–Γ i plane all possible combinations of Γ r and Γ i corresponding to r L=2, we would obtain the circle labeled r L=2 in Fig. 2-25. Similar circles can be obtained for other values of r L. After some algebraic manipulations, Eq. (2.115a) can be rearranged into the parametric equation for the circle in the Γ r–Γ i plane corresponding to a given value of r L:

Γ r−r L 1+r L 2+Γ i 2=1 1+r L 2.(2.116)

The standard equation for a circle in the x–y plane with center at (x 0,y 0) and radius a is

(x−x 0)2+(y−y 0)2=a 2.(2.117)

Comparison of Eq. (2.116) with Eq. (2.117) shows that the r L circle is centered at Γ r=r L/(1+r L) and Γ i=0, and its radius is 1/(1+r L). It therefore follows that all r L-circles pass through the point (Γ r,Γ i)=(1,0). The largest circle shown in Fig. 2-25 is for r L=0, which also is the unit circle corresponding to |Γ|=1. This is to be expected, because when r L=0, |Γ|=1 regardless of the magnitude of x L.

# Figure 2-25 Families of r L and x L circles within the domain |Γ|≤1. The r L circles are concentric with the outermost circle corresponding to r L=0 and the innermost circle (of zero radius) corresponding to r L=∞. The parts of the x L circles contained in the Smith chart are in the upper half of the chart if x L is positive and in the lower half of the chart if x L is negative.

Figure 2 25 Full Alternative Text

A similar manipulation of the expression for x L given by Eq. (2.115b) leads to

(Γ r−1)2+Γ i−1 x L 2=1 x L 2,(2.118)

which is the equation of a circle of radius (1/x L) centered at (Γ r,Γ i)=(1,1/x L). The x L circles in the Γ r–Γ i plane are quite different from those for constant r L. To start with, the normalized reactance x L may assume both positive and negative values, whereas the normalized resistance cannot be negative (negative resistances cannot be realized in passive circuits). Hence, Eq. (2.118) represents two families of circles: one for positive values of x L and another for negative ones. Furthermore, as shown in Fig. 2-25, only part of a given circle falls within the bounds of the |Γ|=1 unit circle.

The families of circles of the two parametric equations given by Eqs. (2.116) and (2.118) plotted for selected values of r L and x L constitute the Smith chart shown in Fig. 2-26. The Smith chart provides a graphical evaluation of Eqs. (2.115a and b) and their inverses.

# Figure 2-26 Point P represents a normalized load impedance z L=2−j 1. The reflection coefficient has a magnitude |Γ|=O P‾/O R‾=0.45 and an angle θ r=−26.6°. Point R is an arbitrary point on the r L=0 circle (which also is the |Γ|=1 circle).

Figure 2 26 Full Alternative Text

For example, point P in Fig. 2-26 represents a normalized load impedance z L=2−j 1, which corresponds to a voltage reflection coefficient Γ=0.45 exp(−j 26.6°). The magnitude |Γ|=0.45 is obtained by dividing the length of the line between the center of the Smith chart and point P (designated O P‾ in Fig. 2-26), by O R‾, which is the length of the line between the center of the Smith chart and the edge of the unit circle (the radius of the unit circle corresponds to |Γ|=1). The perimeter of the Smith chart contains three concentric scales. The innermost scale is labeled angle of reflection coefficient in degrees. This is the scale for θ r. As indicated in Fig. 2-26, θ r=−26.6° (−0.46 rad) for point P. The meanings and uses of the other two scales are discussed next.

# 2.10.2 SWR Circle

Consider point A in the Smith chart of Fig. 2-27. At point A, the normalized load impedance is z L=2−j 1. The magnitude of the corresponding reflection coefficient is

|Γ|=z L−1 z L+1=2−j 1−1 2−j 1+1=1−j 1 3−j 1=2 10=0.45.

Let us construct a circle centered at (Γ r,Γ i)=(0,0) and passing through point A. Every point on this circle has the same value for |Γ|, namely 0.45. This constant-|Γ| circle is also a constant-SWR circle. This follows from the relation between the voltage standing-wave ratio (SWR) and |Γ|, namely

S=1+|Γ|1−|Γ|.(2.119)

# Figure 2-27 Point A represents a normalized load z L=2−j 1 at 0.287 λ on the WTG scale. Point B represents the line input at d=0.1 λ from the load. At B, z(d)=0.6−j 0.66.

Figure 2 27 Full Alternative Text

The utility of the SWR circle will become evident shortly.

# 2.10.3 Wave Impedance

From Eq. (2.74), the normalized wave impedance looking toward the load at a distance d from the load is

z(d)=Z(d)Z 0=1+Γ d 1−Γ d,(2.120)

where

Γ d=Γ e−j 2 β d=|Γ|e j(θ r−2 β d)(2.121)

is the phase-shifted voltage reflection coefficient. The form of Eq. (2.120) is identical with that for z L given by Eq. (2.112):

z L=1+Γ 1−Γ.(2.122)

This similarity in form suggests that, if Γ is transformed into Γ d, z L gets transformed into z(d). On the Smith chart, the transformation from Γ to Γ d is achieved by maintaining |Γ| constant and decreasing its phase θ r by 2 β d, which corresponds to a clockwise rotation (on the Smith chart) over an angle of 2 β d radians. A complete rotation around the Smith chart corresponds to a phase change of 2 π in Γ. The length d corresponding to this phase change satisfies

2 β d=2 2 π λ d=2 π,(2.123)

from which it follows that d=λ/2.

In some transmission-line problems, it may be necessary to move from some point on the transmission line toward a point closer to the load, in which case the phase of Γ must be increased, which corresponds to rotation in the counterclockwise direction. For convenience, the Smith chart contains a third scale around its perimeter (in between the θ r scale and the WTG scale) for accommodating such an operation. It is called the wavelengths toward load (WTL) scale.

# 2.10.4 SWR, Voltage Maxima, and Minima

Consider a load with z L=2+j 1. Figure 2-28 shows a Smith chart with a SWR circle drawn through point A, representing z L. The SWR circle intersects the real (Γ r) axis at two points, labeled P m a x and P min. At both points, Γ i=0 and Γ=Γ r. Also, on the real axis, the imaginary part of the load impedance x L=0. From the definition of Γ,

Γ=z L−1 z L+1,(2.124)

It follows that points P m a x and P min correspond to

Γ=Γ r=r 0−1 r 0+1(for Γ i=0),(2.125)

where r 0 is the value of r L where the SWR circle intersects the Γ r axis. Point P min corresponds to r 0<1, and P m a x corresponds to r 0>1. Rewriting Eq. (2.119) for |Γ| in terms of S, we have

|Γ|=S−1 S+1.(2.126)

For point P m a x, |Γ|=Γ r; hence

Γ r=S−1 S+1.(2.127)

# Figure 2-28 Point A represents a normalized load with z L=2+j 1. The standing-wave ratio is S=2.6 (at P m a x), the distance between the load and the first voltage maximum is d m a x=(0.25−0.213)λ=0.037 λ, and the distance between the load and the first voltage minimum is d min=(0.037+0.25)λ=0.287 λ.

Figure 2 28 Full Alternative Text

The similarity in form of Eqs. (2.125) and (2.127) suggests that S equals the value of the normalized resistance r 0. By definition, S≥1, and at point P m a x, r 0>1, which further satisfies the similarity condition. In Fig. 2-28, r 0=2.6 at P m a x; hence S=2.6.

Points P m i n and P m a x also represent locations on the line where the magnitude of the voltage |V˜| is a minimum and a maximum, respectively. This is easily demonstrated by considering Eq. (2.121) for Γ d. At point P m a x, the total phase of Γ d, that is, (θ r−2 β d), equals zero or −2 n π (with n being a positive integer), which is the condition corresponding to |V˜|m a x, as indicated by Eq. (2.69). Similarly, at P min the total phase of Γ d equals −(2 n+1)π, which is the condition for |V˜|min. Thus, for the transmission line represented by the SWR circle shown in Fig. 2-28, the distance between the load and the nearest voltage maximum is d m a x=0.037 λ, which is obtained by moving clockwise from the load at point A to point P m a x, and the distance to the nearest voltage minimum is d min=0.287 λ, corresponding to the clockwise rotation from A to P min. Since the location of |V˜|m a x corresponds to that of |I˜|min and the location of |V˜|min corresponds to that of |I˜|m a x, the Smith chart provides a convenient way to determine the distances from the load to all maxima and minima on the line (recall that the standing-wave pattern has a repetition period of λ/2).

# 2.10.5 Impedance to Admittance Transformations

In solving certain types of transmission line problems, it is often more convenient to work with admittances than with impedances. Any impedance Z is in general a complex quantity consisting of a resistance R and a reactance X:

Z=R+j X(Ω).(2.128)

The admittance Y is the reciprocal of Z:

Y=1 Z=1 R+j X=R−j X R 2+X 2(S).(2.129)

The real part of Y is called the conductance G, and the imaginary part of Y is called the susceptance B. That is,

Y=G+j B(S).(2.130)

Comparison of Eq. (2.130) with Eq. (2.129) reveals that

G=R R 2+X 2 (S),(2.131a)
B=−X R 2+X 2 (S).(2.131b)

A normalized impedance z is defined as the ratio of Z to the characteristic impedance of the line, Z 0. The same concept applies to the definition of the normalized admittance y; that is,

y=Y Y 0=G Y 0+j B Y 0=g+j b(dimensionless),(2.132)

where Y 0=1/Z 0 is the characteristic admittance of the line and

g=G Y 0=G Z 0 (dimensionless),(2.133a)
b=B Y 0=B Z 0 (dimensionless).(2.133b)

The lowercase quantities g and b represent the normalized conductance and normalized susceptance of y, respectively. Of course, the normalized admittance y is the reciprocal of the normalized impedance z,

y=Y Y 0=Z 0 Z=1 z.(2.134)

Accordingly, using Eq. (2.122), the normalized load admittance y L is given by

Now let us consider the normalized wave impedance z(d) at a distance d=λ/4 from the load. Using Eq. (2.120) with 2 β d=4 π d/λ=4 π λ/4 λ=π gives

z(d=λ/4)=1+Γ e−j π 1−Γ e−j π=1−Γ 1+Γ=y L.(2.136)

In Fig. 2-29, the points representing z L and y L are diametrically opposite to each other on the SWR circle. In fact, such a transformation on the Smith chart can be used to determine any normalized admittance from its corresponding normalized impedance, and vice versa.

The Smith chart can be used with normalized impedances or with normalized admittances. As an impedance chart, the Smith chart consists of r L and x L circles: the resistance and reactance of a normalized load impedance z L, respectively.

# Figure 2-29 Point A represents a normalized load z L=0.6+j 1.4. Its corresponding normalized admittance is y L=0.25−j 0.6, and it is at point B.

Figure 2 29 Full Alternative Text

# Example 2-11: Smith Chart Calculations

A 50 Ω lossless transmission line of length 3.3 λ is terminated by a load impedance Z L=(25+j 50) Ω. Use the Smith chart to find (a) the voltage reflection coefficient, (b) the voltage standing-wave ratio, (c) the distances of the first voltage maximum and first voltage minimum from the load, (d) the input impedance of the line, and (e) the input admittance of the line.

Solution: (a) The normalized load impedance is

z L=Z L Z 0=25+j 50 50=0.5+j 1,

which is marked as point A on the Smith chart in Fig. 2-30. A radial line is drawn from the center of the chart at point O through point A to the outer perimeter of the chart. The line crosses the scale labeled “angle of reflection coefficient in degrees” at θ r=83°. Next, measurements are made to determine lengths O A‾ and O O′‾, of the lines between O and A and between points O and O′, respectively, where O′ is an arbitrary point on the r L=0 circle. The length O O′‾ is equal to the radius of the |Γ|=1 circle. The magnitude of Γ is then obtained from |Γ|=O A‾/O O′‾=0.62. Hence,

Γ=0.62 83°.(2.137)

# Figure 2-30 Solution for Example 2-11. Point A represents a normalized load z L=0.5+j 1 at 0.135 λ on the WTG scale. At A, θ r=83° and |Γ|=O A‾/O O′‾=0.62. At B, the standing-wave ratio is S=4.26. The distance from A to B gives d m a x=0.115 λ and from A to C gives d min=0.365 λ. Point D represents the normalized input impedance Z in=0.28−j 0.40, and point E represents the normalized input admittance y in=1.15+j 1.7.

Figure 2 30 Full Alternative Text

(b) The SWR circle passing through point A crosses the Γ r axis at points B and C. The value of r L at point B is 4.26, from which it follows that

S=4.26.

(c) The first voltage maximum is at point B on the SWR circle, which is at location 0.25 λ on the WTG scale. The load, represented by point A, is at 0.135 λ on the WTG scale. Hence, the distance between the load and the first voltage maximum is

d m a x=(0.25−0.135)λ=0.115 λ.

The first voltage minimum is at point C. Moving on the WTG scale between points A and C gives

d min=(0.5−0.135)λ=0.365 λ,

which is 0.25 λ past d m a x.

(d) The line is 3.3 λ long; subtracting multiples of 0.5 λ leaves 0.3 λ. From the load at 0.135 λ on the WTG scale, the input of the line is at (0.135+0.3)λ=0.435 λ. This is labeled as point D on the SWR circle, and the normalized impedance is

z in=0.28−j 0.40,

which yields

Z in=z in Z 0=(0.28−j 0.40)50=(14−j 20) Ω.

(e) The normalized input admittance y in is found by moving 0.25 λ on the Smith chart to the image point of z in across the circle, labeled point E on the SWR circle. The coordinates of point E give

y in=1.15+j 1.7,

and the corresponding input admittance is

Y in=y in Y 0=y in Z 0=1.15+j 1.7 50=(0.023+j 0.034) S.

# Example 2-12: Determining ZL Using the Smith Chart

This problem is similar to Example 2-6, except that now we demonstrate its solution using the Smith chart.

Given that the voltage standing-wave ratio S=3 on a 50 Ω line, that the first voltage minimum occurs at 5 cm from the load, and that the distance between successive minima is 20 cm, find the load impedance.

Solution: The distance between successive minima equals λ/2. Hence, λ=40 cm. In wavelength units, the first voltage minimum is at

d min=5 40=0.125 λ.

Point A on the Smith chart in Fig. 2-31 corresponds to S=3. Using a compass, the constant S circle is drawn through point A. Point B corresponds to locations of voltage minima. Upon moving 0.125 λ from point B toward the load on the WTL scale (counterclockwise), we arrive at point C, which represents the location of the load. The normalized load impedance at point C is

z L=0.6−j 0.8.

Multiplying by Z 0=50 Ω, we obtain

Z L=50(0.6−j 0.8)=(30−j 40) Ω.

# Figure 2-31 Solution for Example 2-12. Point A denotes that S=3, point B represents the location of the voltage minimum, and point C represents the load at 0.125 λ on the WTL scale from point B. At C, z L=0.6−j 0.8.

Figure 2 31 Full Alternative Text

Concept Question 2-19: The outer perimeter of the Smith chart represents what value of |Γ|? Which point on the Smith chart represents a matched load?

Concept Question 2-20: What is an SWR circle? What quantities are constant for all points on an SWR circle?

Concept Question 2-21: What line length corresponds to one complete rotation around the Smith chart? Why?

Concept Question 2-22: Which points on the SWR circle correspond to locations of voltage maxima and minima on the line and why?

Concept Question 2-23: Given a normalized impedance z L, how do you use the Smith chart to find the corresponding normalized admittance y L=1/z L?