This preview shows page 1. Sign up to view the full content.
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
Vg 107 +
Z0 Generator Matching
network Zin - A M'
Transmission line ZL
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
Yin = Yd + Ys
• Generally, Yd is complex but Ys will be purely imaginary because we will attach something purely reactive like a capacitor
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
gd = 1, bd = −bs
• 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
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)...
View Full Document
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.
- Fall '12