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 573 These points are plotted at C in Fig. 26.5. Shortly, we will work an example to show how these factors are determined graphically in a transmission line problem when we are given the complex load impedance. Transmission loss is a measure of the one-Âway loss of power in a transmission line because of reflection from the load. Return loss represents the two-Âway loss, so it is exactly twice the transmission loss. Return loss is found from Loss = 10 log ( G ) (26.13) RET PWR and, for our example in which G PWR = 0.28, Loss = 10 log (0.28) RET = (10) ( – 0.553) = – 5.53 dB This point is shown as D in Fig. 26.5. The transmission loss coefficient (TLC) can be calculated from TLC = 1 + Γ 1− Γ PWR PWR (26.14) or, for our example, TLC = 1 + (0.28) 1 − (0.28) 1.28 = 0.72 = 1.78 The TLC is a correction factor that is used to calculate the attenuation caused by mismatched impedance in a lossy, as opposed to an ideal or “lossless”, line. The TLC is found from laying out the impedance radius on the loss coefficient scale of the radially scaled parameters at the bottom of the chart. Smith Chart Applications One of the best ways to demonstrate the usefulness of the Smith chart is by practical example. The following sections look at two general cases: transmission line problems and stub matching systems. Transmission Line Calculations Figure 26.6 shows a 50-ÂW transmission line connected to a complex load impedance Z L of 36 + j40 W. The transmission line has a velocity factor v F of 0.80, which means that the
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.
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C h a p t e r 2 6 : T h e S m i t h C h a r t 573<br />
These points are plotted at C in Fig. 26.5. Shortly, we will work an example to show<br />
how these factors are determined graphically in a transmission line problem when we<br />
are given the complex load impedance.<br />
Transmission loss is a measure of the one-Âway loss of power in a transmission line<br />
because of reflection from the load. Return loss represents the two-Âway loss, so it is exactly<br />
twice the transmission loss. Return loss is found from<br />
Loss = 10 log ( G )<br />
(26.13)<br />
RET<br />
PWR<br />
and, for our example in which G PWR = 0.28,<br />
Loss = 10 log (0.28)<br />
RET<br />
= (10) ( – 0.553)<br />
= – 5.53 dB<br />
This point is shown as D in Fig. 26.5.<br />
The transmission loss coefficient (TLC) can be calculated from<br />
TLC = 1 + Γ<br />
1− Γ<br />
PWR<br />
PWR<br />
(26.14)<br />
or, for our example,<br />
TLC = 1 + (0.28)<br />
1 − (0.28)<br />
1.28<br />
=<br />
0.72<br />
= 1.78<br />
The TLC is a correction factor that is used to calculate the attenuation caused by<br />
mismatched impedance in a lossy, as opposed to an ideal or “lossless”, line. The TLC is<br />
found from laying out the impedance radius on the loss coefficient scale of the radially<br />
scaled parameters at the bottom of the chart.<br />
Smith Chart Applications<br />
One of the best ways to demonstrate the usefulness of the Smith chart is by practical<br />
example. The following sections look at two general cases: transmission line problems and<br />
stub matching systems.<br />
Transmission Line Calculations<br />
Figure 26.6 shows a 50-ÂW transmission line connected to a complex load impedance Z L<br />
of 36 + j40 W. The transmission line has a velocity factor v F of 0.80, which means that the
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