Chapter2_Notes

Use them for transforming impedance why typically to

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Unformatted text preview: 1 + |Γ|2 + 2|Γ| cos(θr − 2βd) (203) and, • We can see from gd that if we want to make that 1 we have to have, cos(θr − 2βd) = −|Γ| (204) • That will give us a denominator, 1 + |Γ|2 − 2|Γ|2 = 1 − |Γ|2 Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (205) Electromagnetics I: Transmission lines 111 • So that, we would have, gd = 1+ |Γ|2 1 − |Γ|2 1 − |Γ|2 =1 = + 2|Γ| cos(θr − 2βd) 1 + |Γ|2 − 2|Γ|2 (206) • So, for the condition, cos(θr − 2βd) = −|Γ| (207) we have to adjust d since that is all we have control over. So, we select a d to do the above. • Once we fix d, then we have to calculate what bd is and then add our shunt element so that bs = −bd . • Many different ways to accomplish matching. We could add a capacitor or inductor, or we canl use transmission lines in an arrangement called “single stub”, illustrated in Fig. 29. (Remember a transmission line can be the same as an inductor or capacitor). Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 112 • To satisfy both our two degrees of freedom we have two length of d and the length of a “stub” l placed in shunt. Stub is either S-C or O-C. • The procedure: 1. Convert ZL to YL 2. Select a distance d so as to transform the load admittance into YL into Yd = Y0 + jB (i.e. so that yL = 1 looking into M M before adding the stub). 3. Add shunt stub (O-C or S-C) with Ys = −jB so that total admittance looking into M M (with stub) is Yin = Y0 + jB − jB = Y0 , as needed! Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines Feedline Y0 Yin 113 d M Yd M' Ys Y0 YL Load Y0 l Feedline Shorted stub M Yd Yin Ys M' Figure 2-29 Figure 29: Impedance matching using shorted-stub. Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 114 1.12. Transients on Transmission lines Everything we’ve done so far was frequency-centric. Also, largely applicable only to narrowband signals. What do we do with wideband applications? For that we need the transient response of the transmission line network. • Start with a single pulse of amplitude V0 and duration τ . • Decompose it into two step functions: V (t) = V1 (t) + V2 (t) = V0 U (t) − V0 U (t − τ ) (208) where U (t) is a unit step function. Illustrated in fig. 30. • Why bother? If we know what happens to a step response we can figure out pulse responses! Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 115 V(t) V(t) V0 V1(t) = V0 U(t) V0 τ t τ t V2(t) = V0 U(t - τ) (a) Pulse of duration τ (b) V(t) = V1(t) + V2(t) Figure 30: Rectangular pulse as a sum of two step functions. • Transient response Figure 2-32 For a transient response we need to include...
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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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