Page 347 - Engineering Electromagnetics, 8th Edition

CHAPTER 10 Transmission Lines 329

Z L cos(βl) + jZ 0 sin(βl)

Z in = Z 0 (98)
Z 0 cos(βl) + jZ L sin(βl)
This is the quantity that we need in order to create the equivalent circuit in Figure 10.7.
One special case is that in which the line length is a half-wavelength, or an integer
multiple thereof. In that case,

2π mλ
βl = = mπ (m = 0, 1, 2,...)
λ 2
Using this result in (98), we find

Z in (l = mλ/2) = Z L (99)
Fora half-wave line, the equivalent circuit can be constructed simply by removing
the line completely and placing the load impedance at the input. This simplification
works, of course, provided the line length is indeed an integer multiple of a half-
wavelength. Once the frequency begins to vary, the condition is no longer satisfied,
and (98) must be used in its general form to find Z in .
Another important special case is that in which the line length is an odd multiple
of a quarter wavelength:

2π λ π
βl = (2m + 1) = (2m + 1) (m = 0, 1, 2,...)
λ 4 2
Using this result in (98) leads to

Z 2
Z in (l = λ/4) = 0 (100)
Z L

An immediate application of (100) is to the problem of joining two lines having
different characteristic impedances. Suppose the impedances are (from left to right)
Z 01 and Z 03 .At the joint, we may insert an additional line whose characteristic
impedance is Z 02 and whose length is λ/4. We thus have a sequence of joined lines
whose impedances progress as Z 01 , Z 02 , and Z 03 ,in that order. A voltage wave is
now incident from line 1 onto the joint between Z 01 and Z 02 .Now the effective load
at the far end of line 2 is Z 03 . The input impedance to line 2 at any frequency is now
Z 03 cos β 2 l + jZ 02 sin β 2 l
Z in = Z 02 (101)
Z 02 cos β 2 l + jZ 03 sin β 2 l
Then, since the length of line 2 is λ/4,
Z 2
Z in (line 2) = 02 (102)
Z 03
Reflections at the Z 01 –Z 02 interface will not occur if Z in = Z 01 . Therefore, we can
match the junction (allowing complete transmission through the three-line sequence)