EE 311  Lecture 7
•
Impedance on lossless lines
•
Reflection coefficient
•
Impedance equation
•
Shorted line example
Assigned reading: Sec 2.5.1 of Ulaby
1
Impedance on lossless lines
•
For lossless lines,
γ
=
jω
√
L
0
C
0
=
jβ
; we will focus on lossless
lines almost always from here on out
•
Thus
˜
V
(
z
) =
V
+
0
e

jβz
+
V

0
e
jβz
(1)
˜
I
(
z
) =
1
Z
0
(
V
+
0
e

jβz

V

0
e
jβz
)
(2)
where
V
+
0
and
V

0
are two phasor constants and
Z
0
=
p
L/C
for a lossless line
•
The impedance on a lossless line is then
Z
(
z
) =
˜
V
(
z
)
˜
I
(
z
)
=
Z
0
•
V
+
0
e

jβz
+
V

0
e
jβz
V
+
0
e

jβz

V

0
e
jβz
‚
which is a function of
z
!
•
Thus, the impedance on a transmission line changes depending
where you measure it on the line
2
Alternate form
•
Note that we can factor out
V
+
0
e

jβz
on the top and bottom to
get:
Z
(
z
) =
Z
0
V
+
0
e

jβz
V
+
0
e

jβz
1 +
V

0
V
+
0
e
2
jβz
1

V

0
V
+
0
e
2
jβz
=
Z
0
1 + Γ
e
2
jβz
1

Γ
e
2
jβz
¶
where we have defined Γ =
V

0
V
+
0
as a “voltage reflection
coefficient”
•
Note that if Γ = 0 (we have no
V

0
wave),
Z
(
z
) =
Z
0
meaning
that the impedance is no longer a function of z
•
This is usually the case when we talk about infinitely long
lines, where + propagating waves can propagate forever
without generating a

traveling wave
•
Finite length lines terminated in an impedance, however, will
typically have Γ
6
= 0 as we will see next...
3
Finding gamma
•
Consider a Xmission line of length
l
terminated in an
impedance
Z
L
. A picture is below
•
The two parallel lines in this picture represent a transmission
line. It could be any type of line (parallel plate, coax, etc.) but
we are told the
Z
0
and the
μ, ²
inside the line which is all we
need to know to talk about voltages, currents, and impedances
•
Since we’ve drawn a circuit impedance
Z
L
at
z
= 0, the
impedance on the transmission line at
z
= 0 must be
Z
L
V
g
I
i
Z
g
Z
in
Z
0
Z
L
~
V
i
~
~
+
+


V
L
~
I
L
~
+

Transmission line
Generator
Load
z
= 
l
z
= 0
⇓
4
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•
Using our previous formula for
Z
(
z
) and plugging in
z
= 0 we
get
Z
(0) =
Z
0
1 + Γ
1

Γ
=
Z
L
which can be solved to obtain
Γ =
Z
L

Z
0
Z
L
+
Z
0
•
Note Γ in general is complex; Ulaby writes Γ =

Γ

e
jθ
r
•
Knowing Γ we can find the impedance anywhere else on the
line, for example at
z
=

l
, by using
Z
(
z
) =
Z
0
1 + Γ
e
2
jβz
1

Γ
e
2
jβz
¶
since all quantities in this equation are now known (note
β
=
ω
√
LC
=
ω
√
μ²
)
•
Another way of writing this equation is
Z
(
z
) =
Z
0
1 + Γ(
z
)
1

Γ(
z
)
¶
where Γ(
z
) = Γ
e
2
jβz
5
Impedance equation without
Γ
•
We can also plug in Γ =
Z
L

Z
0
Z
L
+
Z
0
and simplify some to get
Z
(
z
) =
Z
0
Z
L

jZ
0
tan(
βz
)
Z
0

jZ
L
tan(
βz
)
¶
so that
Z
(

l
) =
Z
0
Z
L
+
jZ
0
tan(
βl
)
Z
0
+
jZ
L
tan(
βl
)
¶
•
This last formula is probably the most useful of all the
equations since it directly gives the impedance at
z
=

l
if we
know
Z
0
,
Z
L
,
β
, and
l
•
Notice the dependence on
l
occurs only in the tan(
βl
) terms,
which are periodic in
βl
with period
π
. Since
β
= 2
π/λ
, the
impedance is periodic in
l
with period
λ/
2.
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 Fall '08
 Johnson,J
 Electromagnet, Impedance, Transmission line, Impedance matching, ZL, Ulaby, Gamma Plane

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