Chapter2_Notes

Line figure 2 13 17 input impedance weve learned a lot

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Unformatted text preview: jβd Z0 = Z0 ˜ 1 − Γe−j 2βd V0 [e − Γe−jβd ] I (d) (104) • Where we can define, Γd = Γe−j 2βd = |Γ|ej (θr −2βd) . Note the book also uses Γl which is just when d = l at the end of the line. • Do not confuse Z with Z0 ! The former is ratio of total voltage and current, while the latter is the ratio of the individual traveling wave components! Remember that any point total voltage (or current) is a sum of the traveling components! Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 57 • Note that, if d = l we are at a position at the input (at the source) of the transmission line. We can define that as the input impedance, Zin (l) = Z0 1 + Γl 1 + Γe−j 2βl = Z0 1 − Γe−j 2βl 1 − Γl (105) can be calculated for a known load impedance and properties of the transmission line. Γl = Γe−j 2βl = |Γ|ej (θr −2βl) Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (106) Electromagnetics I: Transmission lines 58 • Since we know how to calculate Γ, Γ= ZL − Z0 ZL + Z0 (107) We can use what we found earlier Z (d) = Z0 ejβd + Γe−jβd [ejβd − Γe−jβd ] (108) and we can substitute, ejβl = cos βl + j sin βl (109) e−jβl = cos βl − j sin βl (110) With some manipulations we can get to, Zin = Z0 ZL cos βl + jZ0 sin βl Z0 cos βl + jZL jZ0 sin βl Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (111) Electromagnetics I: Transmission lines Zin = Z0 59 ZL + jZ0 tan βl Z0 + jZL tan βl (112) We can also use the normalized load impedance, zL = Zin = Z0 ZL Z0 zL + j tan βl 1 + jzL tan βl (113) (114) We can finally solve for the voltage. This is where the “circuit” and “wave” approaches meet. • We have basically solved for the impedance of the transmission line and if we are at the source with the load ZL we have Zin (as in fig. 15) so we can replace everything to the right of the source with Zin . Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines Zg ~ Vg + ~ Ii 60 Transmission line + + ~ Vi Zin Z0 ~ VL ~ IL ZL - - Generator Load z = -l Zg ~ Vg + ~ Ii z=0 fl + ~ Vi Zin - Figure 2-14 Figure 15: Input impedance of transmission line + load. Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 61 • Once the transmission line and load are replaced by Zin , finding voltage at the input of transmission line is easy: ˜ Ii = ˜ Vg Zg + Zin (115) The voltage across Zin is then, ˜ ˜ Vi = Ii Zin = ˜ Vg Zin Zg + Zin (116) and this is the situation from the standpoint of the source (“circuit” picture) • From the transmission line, we solved for the voltage as the sum of the two waves: ˜ ˜ Vi = V (−l) = V0+ [ejβl + Γe−jβl ] Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for E...
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