Practical_Antenna_Handbook_0071639586

564 P a r t V I I : T u n i n g , T r o u b l e s h o o t i n g , a n d D e s i g n A i d
center of the chart, so it represents all points on the chart with a normalized resistance
of 1.0. This particular circle is sometimes called the unity resistance circle.
Constant Reactance Circles
Constant reactance circles are highlighted in Fig. 26.4. Circles (or circle segments) above
the pure resistance line (Fig. 26.4A) represent inductive reactance (+X), while circles (or
segments) below the pure resistance line (Fig. 26.4B) represent capacitive reactance (–X). In
both graphs, circle A corresponds to a normalized reactance of 0.80.
One of the outer circles (i.e., circle A in Fig. 26.4C) is called the pure reactance circle.
Points along circle A represent reactance only; in other words, R = 0 everywhere on this
outer circle, and all impedances along it are of the form Z′ = 0 ± jX.
Plotting Impedance and Admittance
Let’s use Fig. 26.4D to see how to plot any impedance and admittance on the Smith
chart. Consider an example in which system impedance Z 0 is 50 W, and the load impedance
is Z L = 95 + j55 W. This load impedance normalizes to
Z′ Z L
=
Z
L
0
=
95 + j55
Ω
50 Ω
(26.4)
= 1.9 + j1.1
An impedance radius is constructed by drawing a line from the point represented
by the normalized load impedance, 1.9 + j1.1, to the point represented by the normalized
system impedance (1.0, 0.0) in the center of the chart. A circle, called the VSWR
circle, is constructed by drawing a circle with this radius, centered on the location
(1.0,0.0).
Admittance is the reciprocal of impedance, so in normalized form it is found from
Y
Z
1
′ = ′
(26.5)
On the Smith chart, any admittance value is located exactly opposite, or 180 degrees
from, its corresponding impedance. That is, if the radius line connecting the center of
the Smith chart to the impedance point is doubled to become a diameter, the coordinates
of the point at the opposite end of the line from the original impedance point describe
the normalized admittance corresponding to the reciprocal of the original normalized
impedance. In other words, by extending a simple radius to a diameter, we have solved
Eq. (26.5) with graphical techniques, rather than with mathematical ones!
Example 26.1 To show the value of the Smith chart in easily solving real problems, let’s
first find the complex admittance manually: Obtain the reciprocal of the complex impedance
by multiplying the simple reciprocal by the complex conjugate (see App. A) of the