This preview shows page 1. Sign up to view the full content.
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 = Γej (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 = Γej (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 214
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...
View
Full
Document
 Fall '12
 MartinSiderious
 Electromagnet

Click to edit the document details