.net barcode reader library VSWRdB = 20 log (VSWR) = 20 log (20) = (20) (0510) = 102 dB in Software

Printer Data Matrix ECC200 in Software VSWRdB = 20 log (VSWR) = 20 log (20) = (20) (0510) = 102 dB

VSWRdB = 20 log (VSWR) = 20 log (20) = (20) (0510) = 102 dB
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These points are plotted at C in Fig 4-5 Shortly, you will work an example to show how these factors are calculated in a transmission line problem from a known complex load impedance Transmission loss is a measure of the one-way loss of power in a transmission line because of reflection from the load Return loss represents the two-way loss, so it is exactly twice the transmission loss Return loss is found from Lossret = 10 log ( pwr ) and, for our example in which pwr = 028, Lossret = 10 log (028) = (10) ( 0553) = 553 dB [431] [432] [430]
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Smith chart applications 109 This point is shown as D in Fig 4-5 The transmission loss coefficient can be calculated from TLC = or, for our example, TLC = 1 + (028) 1 (028) 128 = 178 072 [434] 1 + pwr 1 pwr [433]
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The TLC is a correction factor that is used to calculate the attenuation caused by mismatched impedance in a lossy, as opposed to the ideal lossless, line The TLC is found from laying out the impedance radius on the loss coefficient scale on the radially scaled parameters at the bottom of the chart
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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
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Transmission line problems
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Figure 4-6 shows a 50- transmission line connected to a complex load impedance ZL of 36 + j40 The transmission line has a velocity factor v of 080, which means that the wave propagates along the line at 8 10 the speed of light (c = 300,000,000 m/s) The length of the transmission line is 28 cm The generator Vin is operated at a frequency of 45 GHz and produces a power output of 15 W See what you can glean from the Smith chart (Fig 4-7) First, normalize the load impedance This is done by dividing the load impedance by the systems impedance (in this case, Zo = 50 ): Z= 36 + j40 50 [436] [437]
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= 072 + j08
The resistive component of impedance Z is located along the 072 pure resistance circle (see Fig 4-7) Similarly, the reactive component of impedance Z is located by traversing the 072 constant resistance circle until the +j08 constant reactance circle is intersected This point graphically represents the normalized load impedance Z = 072 + j080 A VSWR circle is constructed with an impedance radius equal to the line between 10 (in the center of the chart) and the 072 + j08 point
110 The Smith chart
Zo =50 R ZS =50 I f Zo ZS ZL P 08 28 cm 45 GHz 50 50 36 j40 15 W ZL X j40 36
Vin F 45 GHz
109 Hz
4-6 Transmission line and load circuit
At a frequency of 45 GHz, the length of a wave propagating in the transmission line, assuming a velocity factor of 080, is line = = cv FHz (3 108 m/s) (080) 45 109 Hz 24 108 m/s 45 109 Hz 100 cm = 53 cm m [438]
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= 0053 m
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One wavelength is 53 cm, so a half-wavelength is 53 cm/2, or 265 cm The 28-cm line is 28 cm/53 cm, or 528 wavelengths long A line drawn from the center (10) to the load impedance is extended to the outer circle, and it intersects the circle at 01325 Because one complete revolution around this circle represents onehalf wavelength, 528 wavelengths from this point represents 10 revolutions plus 028 more The residual 028 wavelengths is added to 01325 to form a value of (01325 + 028), or 0413 The point 0413 is located on the circle, and is marked A line is then drawn from 0413 to the center of the circle, and it intersects the VSWR circle at 049 j049, which represents the input impedance Zin looking into the line To find the actual impedance represented by the normalized input impedance, you have to denormalize the Smith chart impedance by multiplying the result by Zo:
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