Practical_Antenna_Handbook_0071639586

574 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
wave propagates along the line at 8 ⁄10 the speed of light, c. (c = 300,000,000 m/s.) The
length of the transmission line is 28 cm. The generator (a voltage source in this example)
V in is operated at a frequency of 4.5 GHz and produces a power output of 1.5 W. Let’s
see what we can glean from the Smith chart (Fig. 26.7).
First, normalize the load impedance by dividing it by the system impedance (in this
case, Z 0 = 50 W):
36 Ω + j40
Ω
Z′ =
50 Ω
= 0.72 + j0.8
Now locate the circle of constant resistive component that goes through the point
(0.72,0.0) in Fig. 26.7 and follow it until it intersects the +j0.8 constant reactance circle.
This point (identified with an arrow in Fig. 26.7) graphically represents the normalized
load impedance Z′ = 0.72 + j0.80.
Next, construct a VSWR circle centered on the point (1.0,0.0) and having an impedance
radius equal to the line connecting 1.0,0.0 (at the center of the chart) and the 0.72 +
j0.8 point.
At a frequency of 4.5 GHz, the length of a wave propagating in the transmission
line, assuming a velocity factor of 0.80, is
λ LINE
c υ
F
=
F (Hz)
8
(3 × 10 m/s) (0.80)
=
9
4.5 × 10 Hz
8
2.4 × 10 m/s
=
9
4.5 × 10 Hz
= 0.053 m ×
= 5.3 cm
100 cm
m
(26.15)
One wavelength in the transmission line is 5.3 cm, so a half-Âwavelength is 5.3/2, or
2.65 cm. The 28-Âcm line is 28 cm/5.3 cm, or 5.28 wavelengths long. With a straightedge,
draw a line from the center (1.0,0.0) through the load impedance and extend it to the
outermost circle; it should intersect that circle at 0.1325 (or 0.3675). Because one complete
revolution around this circle represents one half-Âwavelength, 5.28 wavelengths
from this point represents 10 revolutions plus 0.28 wavelengths (not revolutions) more.
The residual 0.28 wavelength figure is added to 0.1325 to form a value of (0.1325 + 0.28)
or 0.413. Now locate the point 0.413 on the circle, making sure to use the same markings
that the original line intersected at 0.1325, and place a mark there. Draw a line from
0.413 to the center of the circle, and note that it intersects the VSWR circle at 0.49 – j0.49,
which represents the input impedance Z′ in looking into the line.