We will now go on and look at what happens when
we excite the line. Let's take a DC voltage source with a source
internal impedance
R
s
R
s
and connect it to our semiinfinite line. The sketch
in Figure 1 is sort of awkward looking, and will
be hard to analyze, so let's make a more "schematic like"
drawing Figure 2, keeping in mind that it is a
situation such as Figure 1 which we trying to
represent.
Why have we shown an
I
+
I
+
and a
V
+
V
+
but not
V

V

or
I

I

? The answer is, that if the line is semiinfinite,
then the "other" end is at infinity, and we know there
are no sources at infinity. The current flowing through the
source resistor is just
I
1
+
I
1
+
, so we can do a KVL around the loop
V
s
−
I
1
+
0t
R
s
−
V
1
+
0t=0
V
s
I
1
+
0
t
R
s
V
1
+
0
t
0
(1)
Substituting for
I
1
+
I
1
+
in terms of
V
1
+
V
1
+
using
this
equation:
V
s
−
V
1
+
0t
Z
0
R
s
−
V
1
+
0t=0
V
s
V
1
+
0
t
Z
0
R
s
V
1
+
0
t
0
(2)
Which we rewrite as
V
1
+
0t(1+
R
s
Z
0
)=
V
s
V
1
+
0
t
1
R
s
Z
0
V
s
(3)
Or, on solving for
V
1
+
0t
V
1
+
0
t
:
V
1
+
0t=
Z
0
Z
0
+
R
s
V
s
V
1
+
0
t
Z
0
Z
0
R
s
V
s
(4)
This
should look both reasonable and
familiar to you. The line and the source resistance are acting
as a
voltage
divider. In fact,
Equation 4 is just the usual
voltage divider equation for two resistors in series. Thus, the
generator can not tell the difference between a semiinfinite
transmission line of characteristic impedance
Z
0
Z
0
and a resistor with a resistance of the same value
Figure 3.
Have you ever heard of "
300Ω
300
Ω
twinlead" or maybe "
75Ω
75
Ω
coax" and wondered why people would want to use wires
with such a high resistance value to bring a TV signal to their
set? Now you know. The
300Ω
300
Ω
characterization is not a measure of the resistance of
the wire, rather it is a specification of the transmission
line's impedance. Thus, if a TV signal coming from your antenna
has a value of, say,
30μV
30
μV
, and it is being brought down from the roof with
300Ω
300
Ω
twinlead, then the current flowing in the wires is
I=30μV300Ω=100nA
I
30
μV
300
Ω
100
nA
, which is a very small current indeed!
Why then, did people decide on
300Ω
300
Ω
? An antenna which is just a halfwavelength long
(Which turns out to be both a convenient and efficient choice
for signals in the 100 MHz (
λ≃3m
λ
3
m
) range) acts like a voltage source with a source
resistance of about
300Ω
300
Ω
. If you remember from ELEC 242, when we have a source
with a source resistance
R
s
R
s
and a load resistor with load resistance value
R
L
R
L
Figure 4, you calculate the power
delivered to the load using the following method.
P
L
P
L
, the power in the load, is just product of the voltage
across the load times the current through the load. We can use
the voltage divider law to find the voltage across
R
L
R
L
and the resistor sum law to find the current through
it.
P
L
=
V
L
I
L
=
R
L
R
L
+
R
s
V
s
V
s
R
L
+
R
s
=
R
L
R
L
+
R
s
2
V
s
2
P
L
V
L
I
L
R
L
R
L
R
s
V
s
V
s
R
L
R
s
R
L
R
L
R
s
2
V
s
2
(5)
If we take the derivative of
Equation 5 with respect
to
R
L
R
L
, the load resistor (which we assume we can pick, given
some predetermined
R
s
R
s
) we have (ignoring the
V
s
2
V
s
2
),
dd
R
L
P
L
=1
R
L
+
R
s
22
R
L
R
L
+
R
s
3=0
R
L
P
L
1
R
L
R
s
2
2
R
L
R
L
R
s
3
0
(6)
Putting everything on
R
L
+
R
s
3
R
L
R
s
3
and then just looking at the numerator:
R
L
+
R
s
−2
R
L
=0
R
L
R
s
2
R
L
0
(7)
Which obviously says that for maximum power transfer, you want
your load resistor
R
L
R
L
to have the same value as your source resistor
R
s
R
s
! Thus, people came up with
300Ω
300
Ω
twin lead so that they could maximize the energy
transfer between the TV antenna and the transmission line
bringing the signal to the TV receiver set. It turns out that
for a coaxial transmission line (such as your TV cable)
75Ω
75
Ω
minimizes the signal loss, which is why that value was
chosen for CATV.
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