![]() ![]() Points with suffix P are in the Z plane and points with suffix Q are in the Y plane. Shunt line with smith chart series#The first transformation is OP 1 along the line of constant normalized resistance in this case the addition of a normalized reactance of - j0.80, corresponding to a series capacitor of 40 pF. As is the same as the system impedance, this is represented by a point at the centre of the Smith chart. The analysis starts with a Z Smith chart looking into R 1 only with no other components present. If there were very different values of resistance present a value closer to these might be a better choice. Despite there being no transmission line as such, a system impedance must still be defined to enable normalization and de-normalization calculations and is a good choice here as. The component dimensions themselves will be in the order of millimetres so the assumption of lumped components will be valid. Shunt line with smith chart free#At this frequency the free space wavelength is 3 m. The following circuit will be analyzed using a Smith chart at an operating frequency of 100 MHz. In this case the wavelength scaling on the Smith chart circumference is not used. The Smith chart may be used to analyze such circuits in which case the movements around the chart are generated by the (normalized) impedances and admittances of the components at the frequency of operation. The analysis of lumped element components assumes that the wavelength at the frequency of operation is much greater than the dimensions of the components themselves. If ZLpassband response of a binomial matching transformer is optimum to have as flat as possible near the design frequency, and is known as maximally flat. The important characteristicsÄ¡4 Example5.8: Design a quarter-wave matching transformer to match a 10ï load to a 50ï line? Determine the percent bandwidth for SWRï£1.5? Solution ï The length l= ï¬/4 at design frequency f0. 6.ZL = 60ï-j 80ï at 2 GHz can find R= 60ï, C=0.995pF.Ä¡2 Problem 3: Repeat example 5.7 using analytic solution. The susceptance of the second stubs should beÄ¥. Rotating ï¬/8 toward the load (WTL) to construct 1+jb circle can find two values of first stub b1 = 1.314 bâ1 = 3. Two possible solutions b1,b2 and b1â,b2â with the same distance d.Ä© Example5.7: Design a double-stub (open circuit) shunt tuning networks to match this load ZL = 60ï-j 80ï to a 50ï line, at a frequency of 2 GHz? Solution 1. The shaded region forms a forbidden range of load admittances. Original circuit Equivalent circuitĨ Disadvantage is the double-stub tuner cannot match all load impedancesÄisadvantage is the double-stub tuner cannot match all load impedances. ![]() In practice, stub spacing is chosen as ï¬/8 or 3ï¬/8 and far away 0 or ï¬/2 to reduce frequency sensitive. Shunt stubs are easier to implement in practice than series stubs. ZL = 100ï+j 80ï at 2 GHz can find R= 100ï,L=6.37nH.Ħ Problem 2: Repeat example 5.6 using analytic solution.ħ Shunt stubs are easier to implement in practice than series stubs.Äouble-Stub Matching adjustable tuning Variable length of length d between load and stub to have adjustable tuning between load and the first stub. The stub length for tuning z1 to 1 requires l1 = 0.397ï¬, and for tuning z1 to 1 needs l2 = 0.103ï¬.Ĥ 1. SWR circle intersects the 1+jx circle at both points z1 = 1.0-j1.33 z2 = 1.0+j1.33. The normalized load impedance ZL= 2-j1.6. 2 Problem 1: Repeat example 5.5 using analytic solution.Ä£ Example5.6: Design two single-stub (open circuit) series tuning networks to match this load ZL = 100ï+j 80ï to a 50ï line, at a frequency of 2 GHz? Solution 1. ![]()
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