Practical_Antenna_Handbook_0071639586

C h a p t e r 2 6 : T h e S m i t h C h a r t 575
I
Z o =50
R 36
Z S =50
F 4.5 GHz
V in
v F 0.8
I 28 cm
f 4.5 GHz 4.5 10 9 Hz
Z o 50
Z S 50
Z L 36 j40
P 1.5 W
Z L
X j40
Figure 26.6 Transmission line and load circuit.
Finally, to find the actual impedance represented by the normalized input impedance
we have just read off the chart, “denormalize” the Smith chart impedance by multiplying
the result by Z 0 :
Z
in
= (0.49 − j0.49) (50 Ω)
= 24.5 − j24.5
Ω
This is the impedance that must be matched at the generator end of the transmission
line by a conjugate matching network.
The admittance of the load is the reciprocal of the load impedance and is found by
extending the impedance radius through the center of the VSWR circle until it intersects
the circle again on the other side (as an impedance diameter). This point is found to be Y′ =
0.62 – j0.69; it is the normalized input admittance. Confirming the solution mathematically:
1
Y′ =
Z′
1 0.72 − j0.80
=
×
0.72 + j0.80
0.72 − j0.80
0.72 − j0.80
=
1.16
= 0.62 − j0.69
The VSWR at the generator end of the line is found by transferring the “impedance
radius” of the VSWR circle to the radial scales below the circle chart. At the leftmost