Chapter2_Notes

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Unformatted text preview: nding wave circle transforms impedance z to admittance y (and vice versa). That means these points are diametrically opposite each other. • The Smith chart can be used for this conversion as well. • All it takes is a rotation by λ/4 (how many degrees is that?). • If admittances are used, constant resistance circle becomes constant conductance circle and constant reactance circle becomes constant susceptance circle. Values are read off directly from the chart. Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines M Zg ~ Vg 107 + Z0 Generator Matching network Zin - A M' Transmission line ZL A' Load Figure 2-28 Figure 28: Illustration of matching to transmission line. 1.11. Impedance matching Transmission lines are not just a nuisance that has to be taken into account. We can use them for transforming impedance. Why? Typically, to ensure the best power transfer. • See Fig. 28 for circuit setup. • Best case: have a load such that ZL = Z0 . Usually not possible. • What else? Place impedance matching network between the load and transmission line. Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 108 • Work with admittances (why?). • The load admittance is YL , the line admittance Y0 and the shunt admittance is Ys . • A the point M M , the admittance of the line to the right is Yd (includes the length d of line and the load). • So, the input impedance at a point just to the left of M M is the sum, Yin = Yd + Ys (196) • Generally, Yd is complex but Ys will be purely imaginary because we will attach something purely reactive like a capacitor or inductor. • So, Yin = (Gd + jBd ) + jBs = Gd + j (Bd + Bs ) Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (197) Electromagnetics I: Transmission lines 109 • Or, the normalized form, yin = gd + j (bd + bs ) (198) • To have the line matched, we want zin = 1 (because Γ = 0 for matched line so we have Zin = Z0 ) and therefore yin = 1. So that requires, gd = 1 and bd + bs = 0. So, we have two conditions, gd = 1, bd = −bs (199) • We have two conditions to satisfy so we have two degrees of freedom. We will use the parameter d (length of line from the load where we attach shunt element) to satisfy the gd = 1 condition. We will use an appropriate capacitor of inductor to satisfy the bd = −bs condition. • To see how to get gd = 1, we have to do some more manipulations. 1 − yL (200) Γ= 1 + yL Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 110 and, yd = 1 − Γej 2βd 1 − |Γ|ej (θr −2βd) = 1 + Γej 2βd 1 + |Γ|ej (θr −2βd) (201) • We can use Eulers and do some further manipulations to get the real and imaginary parts, gd = 1 − |Γ|2 1 + |Γ|2 + 2|Γ| cos(θr − 2βd) (202) bd = −2|Γ| sin(θr − 2βd)...
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This note was uploaded on 09/25/2013 for the course ECE 331 taught by Professor Martinsiderious during the Fall '12 term at Portland State.

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