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184 Impedance Matching
chart locus from P2 to P, is plotted with a radius that decays with the angle by
the - 2NP exponential term in (6.26); i.e., it spirals inward.
Example 6.5. Consider the lossless (NP= 0) transmission line in Figure 6.11,
terminated by Z2 = 40 + j30 ohms. Suppose that the characteristic impedance is
50 ohms. By (6.24), P2=0.3333 190°; this is plotted in Figure 6.12. By (6.26),
a 45-degree line rotates P2 clockwise by 90 chart degrees to locate input Pia; a
90-degree line rotates P2 clockwise by 180 chart degrees to locate P,b' By
(6.24), Pia = 0.3333 10° corresponds to Z, = 100+jO ohms; similarly, P,b=
0.3333 1_90° corresponds to Z, =40-j30 ohms. If dissipation loss NP had
not been zero, the radius of 0.3333 would have decreased with rotation.
Further considerations will involve only lossless transmission lines; so it is
useful to equate "If = jO in (6.23). This reduces to the input impedance
expression for a lossless transmission line:
Z2+jyZO
Z,=ZOZ . Z '
o+JY 2
where definition (4.22) is repeated:
y=tanO.
(6.27)
(6.28)
X=" +0.5
x = -0.5
-1
I I I I I I I I I I I
1.0 0.8 0,6 0.4 0.2 0 0.2' 0.4 0.6 0.8 1.0
_I
Figure 6.12. LossJess transmission line rotation on a Smith chart for Example 6.5.