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Now check what you have learned from the Smith chart Recall that 15 W of 45GHz microwave RF signal was input to a 50- transmission line that was 28 cm long The load connected to the transmission line has an impedance of 36 + j40 From the Smith chart: Admittance (load) VSWR VSWR (dB): Reflection coefficient (E) Reflection coefficient (P) Reflection coefficient angle Return loss Reflection loss Transmission loss coefficient 062 j069 26:1 83 dB 044 02 84 degrees 7 dB 105 dB 15
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A properly designed matching system will provide a conjugate match to a complex impedance Some sort of matching system or network is needed any time the load impedance ZL is not equal to the characteristic impedance Zo of the transmission line In a transmission line system, it is possible to use a shorted stub connected in
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Smith chart applications 115 parallel with the line, at a critical distance back from the mismatched load, in order to effect a match The stub is merely a section of transmission line that is shorted at the end not connected to the main transmission line The reactance (hence also susceptance) of a shorted line can vary from to + , depending upon length, so you can use a line of critical length L2 to cancel the reactive component of the load impedance Because the stub is connected in parallel with the line it is a bit easier to work with admittance parameters, rather than impedance Consider the example of Fig 4-8, in which the load impedance is Z = 100 + j60, which is normalized to 20 + j12 This impedance is plotted on the Smith chart in Fig 4-9, and a VSWR circle is constructed The admittance is found on the chart at point Y = 037 j022 In order to provide a properly designed matching stub, you need to find two lengths L1 is the length (relative to wavelength) from the load toward the generator (see L1 in Fig 4-8); L2 is the length of the stub itself The first step in finding a solution to the problem is to find the points where the unit conductance line (10 at the chart center) intersects the VSWR circle; there are two such points shown in Fig 4-9: 10 + j11 and 10 j11 Select one of these (choose 10 + j11) and extend a line from the center 10 point through the 10 + j11 point to the outer circle (WAVELENGTHS TOWARD GENERATOR) Similarly, a line is drawn from the center through the admittance point 037 022 to the outer circle These two lines intersect the outer circle at the points 0165 and 0461 The distance of the stub back toward the generator is found from: L1 = 0165 + (0500 0461) [470]
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Transmission line
50 Vin
Matching stub Conditions: ZS 50 50 Zo ZL 100
4-8 Matching stub length and position
116 The Smith chart
j11
j022
Y 0461
j11
Y 0368
j11
4-9 Solution to problem (Courtesy of Kay Elementrics)
Smith chart applications 117 = 0165 + 0039 = 0204 [471] [472]
The next step is to find the length of the stub required This is done by finding two points on the Smith chart First, locate the point where admittance is infinite (far right side of the pure conductance line); second, locate the point where the admittance is 0 j11 (Note that the susceptance portion is the same as that found where the unit conductance circle crossed the VSWR circle) Because the conductance component of this new point is 0, the point will lay on the j11 circle at the intersection with the outer circle Now draw lines from the center of the chart through each of these points to the outer circle These lines intersect the outer circle at 0368 and 0250 The length of the stub is found from L2 = (0368 0250) = 0118 [473] [474]
From this analysis you can see that the impedance, Z = 100 + j60, can be matched by placing a stub of a length 0118 at a distance 0204 back from the load
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