Chapter2_Notes

# I transmission lines 83 110 smith chart the smith

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Unformatted text preview: 0.7 -0.8 -0.9 0.5 0.7 0.9 1 Gr q r=0 Open-circuit load Unit circle -1 Figure 2-20 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 x-axis (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.

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