Page 357 - Engineering Electromagnetics, 8th Edition

CHAPTER 10 Transmission Lines 339

line length,

1 + e − j2βl 1 +| |e j(φ−2βl)
z in = = (114)
1 − e − j2βl 1 −| |e j(φ−2βl)
Note that when l = 0, we are located at the load, and z in = (1 + )/(l − ) = z L ,as
shown by (107).
Equation (114) shows that the input impedance at any point z =−l can be
obtained by replacing , the reflection coefficient of the load, by e − j2βl . That is, we
decrease the angle of by 2βl radians as we move from the load to the line input.
Only the angle of is changed; the magnitude remains constant.
Thus, as we proceed from the load z L to the input impedance z in ,wemove
toward the generator a distance l on the transmission line, but we move through a
clockwise angle of 2βl on the Smith chart. Since the magnitude of stays constant, the
movement toward the source is made along a constant-radius circle. One lap around
the chart is accomplished whenever βl changes by π rad, or when l changes by one-
half wavelength. This agrees with our earlier discovery that the input impedance of a
half-wavelength lossless line is equal to the load impedance.
The Smith chart is thus completed by the addition of a scale showing a change
of 0.5λ for one circumnavigation of the unit circle. For convenience, two scales are
usually given, one showing an increase in distance for clockwise movement and
the other an increase for counterclockwise travel. These two scales are shown in
Figure 10.13. Note that the one marked “wavelengths toward generator” (wtg) shows
increasing values of l/λ for clockwise travel, as described previously. The zero point
of the wtg scale is rather arbitrarily located to the left. This corresponds to input
impedances having phase angles of 0 and R L < Z 0 .Wehave also seen that voltage
◦
minima are always located here.
EXAMPLE 10.10
The use of the transmission line chart is best shown by example. Let us again consider
a load impedance, Z L = 25 + j50 , terminating a 50- line. The line length is
60 cm and the operating frequency is such that the wavelength on the line is 2 m. We
desire the input impedance.
Solution. We have z L = 0.5 + j1, which is marked as A on Figure 10.14, and we
read = 0.62 82 .By drawing a straight line from the origin through A to the
◦
circumference, we note a reading of 0.135 on the wtg scale. We have l/λ = 0.6/2 =
0.3, and it is, therefore, 0.3λ from the load to the input. We therefore find z in on the
| |= 0.62 circle opposite a wtg reading of 0.135 + 0.300 = 0.435. This construction
is shown in Figure 10.14, and the point locating the input impedance is marked B.
The normalized input impedance is read as 0.28 − j0.40, and thus Z in = 14 − j20.
A more accurate analytical calculation gives Z in = 13.7 − j20.2.

Information concerning the location of the voltage maxima and minima is also
readily obtained on the Smith chart. We already know that a maximum or minimum