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0.8
0.9 0.5 0.7 0.9 1 Gr
q r=0
Opencircuit
load Unit circle 1 Figure 220 q r = 270° or 90° Figure 21: Complex Γ plane with some example points.
Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 85 Note that each impedance zL has a unique corresponding Γ, and
vice versa.
• We can write the reﬂection coeﬃcient in terms of the real and
imaginary parts,
Γ = Γr + j Γi
(168)
• We can also write the normalized impedance in a similar way,
zL = rL + jxL (169) • We can then write,
1+Γ
1−Γ
1 + Γ r + j Γi
rL + jxL =
1 − Γr − j Γi
zL = Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (170)
(171) Electromagnetics I: Transmission lines 86 • This can be manipulated to get,
rL = 1 − Γ2 − Γ2
r
i
(1 − Γr )2 + Γ2
i (172) • We can see from this that there are many possible values of Γr
and Γi that would give us the same rL (e.g., take Γr = 0.33, Γi =
0 which gives rL = 2 but Γr = 0.5, Γi = 0.29 also gives rL = 2).
• There are actually and inﬁnite number of possibilities for a given
rL .
• Likewise we can get,
xL = 2Γi
(1 − Γr )2 + Γ2
i Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (173) Electromagnetics I: Transmission lines 87 • In fact, we can further manipulate these equations to get the
following equation,
Γr − rL
1 + rL 2 1
1 + rL + Γ2 =
i 2 (174) • Recalling the standard equation for a circle,
2 2 (x − x0 ) + (y − y0 ) = a2 (175) • The above ﬁts this form where the x and y variables are the real
and imaginary parts of Γ.
• The circle is at the origin when x0 = 0 and y0 = 0 and oﬀset
when otherwise. So, in our case, we can see the circle will be
oﬀset on our xaxis (Γr ).
• The standard form shows the radius is a so our radius is 1
1+rL Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. . Electromagnetics I: Transmission lines 88 • Further, if we look at Γi = 0 we can see that,
Γr − rL
1 + rL = 1
1 + rL (176) taking just the positive root. We can then get,
Γr = 1
rL
+
=1
1 + rL
1 + rL (177) So, every circle will have a point at Γi = 0, Γr = 1.
• We can start by taking the minimum rL = 0 which gives us,
Γ2 + Γ2 = 1
r
i (178) which is a unit circle centered at the origin.
• For other constant values of rL we get other circles as shown in
the ﬁgure.
Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 89 • Similar manipulations yield a family of circles for the reactance,
2 (Γr − 1) + Γi − 1
xL 2 = 1
xL 2 (179) • So, in general, points of constant resistance and constant reactance lie on circles , which are called constant resistance and
constant reactance circles in the Γ complex plane, as shown
in Fig. 22
• Note, though some diﬀerences between the circles of constant
resistance...
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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.
 Fall '12
 MartinSiderious
 Electromagnet

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