Wireline channels were the first used for electrical
communications in the mid-nineteenth century for the telegraph.
Here, the channel is one of several wires connecting transmitter
to receiver. The transmitter simply creates a voltage related
to the message signal and applies it to the wire(s). We must
have a circuit—a closed path—that supports current flow. In the
case of single-wire communications, the earth is used as the
current's return path. In fact, the term ground
for the reference node in circuits originated in single-wire
telegraphs. You can imagine that the earth's electrical
characteristics are highly variable, and they are. Single-wire
metallic channels cannot support high-quality signal
transmission having a bandwidth beyond a few hundred Hertz over
any appreciable distance.
Consequently, most wireline channels today essentially consist
of pairs of conducting wires (Figure 1), and the transmitter applies a message-related
voltage across the pair. How these pairs of wires are
physically configured greatly affects their transmission
characteristics. One example is twisted pair,
wherein the wires are wrapped about each other. Telephone
cables are one example of a twisted pair channel. Another is
coaxial cable, where a concentric conductor
surrounds a central wire with a dielectric material in between.
Coaxial cable, fondly called "co-ax" by engineers, is what
Ethernet uses as its channel. In either case, wireline channels
form a dedicated circuit between transmitter and receiver. As
we shall find subsequently, several transmissions can share the
circuit by amplitude modulation techniques; commercial cable TV
is an example. These information-carrying circuits are designed
so that interference from nearby electromagnetic sources is
minimized. Thus, by the time signals arrive at the receiver,
they are relatively interference- and noise-free.
Both twisted pair and co-ax are examples of transmission
lines, which all have the circuit model shown in
Figure 2
for an infinitesimally small length. This circuit model arises
from solving Maxwell's equations for the particular
transmission line geometry.
The series resistance comes from the
conductor used in the wires and from the conductor's geometry.
The inductance and the capacitance derive from transmission line
geometry, and the parallel conductance from the medium between
the wire pair. Note that all the circuit elements have values
expressed by the product of a constant times a length; this
notation represents that element values here have
per-unit-length units. For example, the series resistance
R
∼
R
∼
has units of ohms/meter. For coaxial cable, the element values
depend on the inner conductor's radius
r
i
r
i
,
the outer radius of the dielectric
r
d
r
d
,
the conductivity of the conductors
σσ,
and the conductivity
σ
d
σ
d
,
dielectric constant
ε
d
ε
d
,
and magnetic permittivity
μ
d
μ
d
of the dielectric as
R
∼
=12πδσ(1
r
d
+1
r
i
)
R
∼
1
2
δ
σ
1
r
d
1
r
i
(1)
For twisted pair, having a separation
dd between the conductors that have
conductivity
σσ and common
radius
rr and that are immersed in
a medium having dielectric and magnetic properties, the element
values are then
R
∼
=1πrδσ
R
∼
1
r
δ
σ
(2)
The voltage between the two conductors and the current flowing
through them will depend on distance
xx along the transmission line as
well as time. We express this dependence as
vxt
v
x
t
and
ixt
i
x
t
.
When we place a sinusoidal source at one end of the transmission
line, these voltages and currents will also be sinusoidal because the
transmission line model consists of linear circuit elements. As
is customary in analyzing linear circuits, we express voltages
and currents as the real part of complex exponential signals,
and write circuit variables as a complex amplitude—here
dependent on distance—times a complex exponential:
vxt=ℜVxei2πft
v
x
t
V
x
2
f
t
and
ixt=ℜIxei2πft
i
x
t
I
x
2
f
t
.
Using the transmission line circuit model, we find from KCL,
KVL, and v-i relations the equations governing the complex
amplitudes.
Ix=Ix−Δx−Vx(
G
∼
+i2πf
C
∼
)Δx
I
x
I
x
Δ
x
V
x
G
∼
2
f
C
∼
Δ
x
(3)
Vx−Vx+Δx=Ix(
R
∼
+i2πf
L
∼
)Δx
V
x
V
x
Δ
x
I
x
R
∼
2
f
L
∼
Δ
x
(4)
Rearranging and taking the limit
Δx→0
Δ
x
0
yields the so-called
transmission line equations.
ddxIx=−((
G∼
+i2πf
C∼
)Vx)
x
Ix
G∼
2
f
C∼
V
x
(5)
By combining these equations, we can obtain a single equation
that governs how the voltage's or the current's complex
amplitude changes with position along the transmission line.
Taking the derivative of the second equation and plugging the
first equation into the result yields the equation governing the
voltage.
d
2
dx
2
Vx=(
G
∼
+i2πf
C
∼
)(
R
∼
+i2πf
L
∼
)Vx
x
2
V
x
G
∼
2
f
C
∼
R
∼
2
f
L
∼
V
x
(6)
This equation's solution is
Vx=
V
+
e−(γx)+
V
-
eγx
V
x
V
+
γ
x
V
-
γ
x
(7)
Calculating its second derivative and comparing the result with
our equation for the voltage can check this solution.
d
2
dx
2
Vx=γ2(
V
+
e−(γx)+
V
-
eγx)=γ2Vx
x
2
V
x
γ
2
V
+
γ
x
V
-
γ
x
γ
2
V
x
(8)
Our solution works so long as the quantity
γγ satisfies
γ=±(
G
∼
+i2πf
C
∼
)(
R
∼
+i2πf
L
∼
)=±af+ibf
γ
±
G
∼
2
f
C
∼
R
∼
2
f
L
∼
±
a
f
b
f
(9)
Thus,
γγ depends on
frequency, and we express it in terms of real and imaginary
parts as indicated. The quantities
V
+
V
+
and
V
-
V
-
are constants determined by the source and physical
considerations. For example, let the spatial origin be the
middle of the transmission line model
Figure 2.
Because the circuit model contains simple circuit elements,
physically possible solutions for voltage amplitude cannot
increase with distance along the transmission line. Expressing
γγ in terms of its real and
imaginary parts in our solution shows that such increases are a
(mathematical) possibility.
Vx=
V
+
e(−(a+ib))x+
V
-
e(a+ib)x
V
x
V
+
a
b
x
V
-
a
b
x
The voltage cannot increase without limit; because
af
a
f
is always positive, we must segregate the solution for negative
and positive
xx. The first term
will increase exponentially for
x<0
x
0
unless
V
+
=0
V
+
0
in this region; a similar result applies to
V
-
V
-
for
x>0
x
0
.
These physical constraints give us a cleaner solution.
Vx={
V
+
e(−(a+ib))x if x>0
V
-
e(a+ib)x if x<0
V
x
V
+
a
b
x
x
0
V
-
a
b
x
x
0
(10)
This solution suggests that voltages (and currents too) will
decrease
exponentially along a transmission
line. The
space constant, also known as the
attenuation constant, is the distance over which
the voltage decreases by a factor of
1e
1
.
It equals the reciprocal of
af
a
f
,
which depends on frequency, and is expressed by manufacturers in
units of dB/m.
The presence of the imaginary part of
γγ,
bf
b
f
,
also provides insight into how transmission lines work. Because
the solution for
x>0
x
0
is proportional to
e−(ibx)
b
x
,
we know that the voltage's complex amplitude will vary
sinusoidally in space. The complete solution for the
voltage has the form
vxt=ℜ
V
+
e−(ax)ei(2πft−bx)
v
x
t
V
+
a
x
2
f
t
b
x
(11)
The complex exponential portion has the form of a
propagating wave. If we could take a snapshot of
the voltage (take its picture at
t=
t
1
t
t
1
),
we would see a sinusoidally varying waveform along the
transmission line. One period of this variation, known as the
wavelength, equals
λ=2πb
λ
2
b
.
If we were to take a second picture at some later time
t=
t
2
t
t
2
,
we would also see a sinusoidal voltage. Because
2πf
t
2
−bx=2πf(
t
1
+
t
2
−
t
1
)−bx=2πf
t
1
−b(x−2πfb(
t
2
−
t
1
))
2
f
t
2
b
x
2
f
t
1
t
2
t
1
b
x
2
f
t
1
b
x
2
f
b
t
2
t
1
the second waveform appears to be the first one, but
delayed—shifted to the right—in space. Thus, the
voltage appeared to move to the right with a speed equal to
2πfb
2
f
b
(assuming
b>0
b
0
). We denote this
propagation speed by
cc, and it equals
c=|2πfℑ(
G
∼
+i2πf
C
∼
)(
R
∼
+i2πf
L
∼
)|
c
2
f
G
∼
2
f
C
∼
R
∼
2
f
L
∼
(12)
In the high-frequency region where
i2πf
L
∼
≫
R
∼
≫
2
f
L
∼
R
∼
and
i2πf
C
∼
≫
G
∼
≫
2
f
C
∼
G
∼
,
the quantity under the radical simplifies to
-4π2f2
L
∼
C
∼
-4
2
f
2
L
∼
C
∼
,
and we find the propagation speed to be
limit
f
→
∞
c=1
L
∼
C
∼
f
c
1
L
∼
C
∼
(13)
For typical coaxial cable, this propagation speed is a fraction
(one-third to two-thirds)
of the speed of light.
Find the propagation speed in terms of physical parameters
for both the coaxial cable and twisted pair examples.
In both cases, the answer depends less on geometry than on
material properties. For coaxial cable,
c=1
μ
d
ε
d
c
1
μ
d
ε
d
.
For twisted pair,
c=1μεarccoshd2rδ2r+arccoshd2r
c
1
μ
ε
d
2
r
δ
2
r
d
2
r
.
By using the second of the transmission line equation
Equation 5,
we can solve for the current's complex amplitude. Considering
the spatial region
x>0
x
0
,
for example, we find that
dd
x
Vx=−(γVx)=(−(
R
∼
+i2πf
L
∼
))Ix
x
V
x
γ
V
x
R
∼
2
f
L
∼
I
x
which means that the ratio of voltage and current complex
amplitudes does not depend on distance.
VxIx=
R
∼
+i2πf
L
∼
G
∼
+i2πf
C
∼
=
Z
0
V
x
I
x
R
∼
2
f
L
∼
G
∼
2
f
C
∼
Z
0
(14)
The quantity
Z
0
Z
0
is known as the transmission line's
characteristic
impedance. Note that when the signal frequency is
sufficiently high, the characteristic impedance is real, which
means the transmission line appears resistive in this
high-frequency regime.
limit f→∞
Z
0
=
L
∼
C
∼
f
Z
0
L
∼
C
∼
(15)
Typical values for characteristic impedance are 50 and
75 Ω.
A related transmission line is the optic fiber. Here, the
electromagnetic field is light, and it propagates down a
cylinder of glass. In this situation, we don't have two
conductors—in fact we have none—and the energy is
propagating in what corresponds to the dielectric material of
the coaxial cable. Optic fiber communication has exactly the
same properties as other transmission lines: Signal strength
decays exponentially according to the fiber's space constant and
propagates at some speed less than light would in free space.
From the encompassing view of Maxwell's equations, the only
difference is the electromagnetic signal's frequency. Because
no electric conductors are present and the fiber is protected by
an opaque “insulator,” optic fiber transmission is
interference-free.
From tables of physical constants, find the frequency of a
sinusoid in the middle of the visible light range. Compare
this frequency with that of a mid-frequency cable television
signal.
You can find these frequencies from
the spectrum allocation
chart.
Light in the middle of the visible band has a wavelength of
about 600 nm, which corresponds to a frequency of
5×1014Hz
514
Hz
.
Cable television transmits within the same frequency band as
broadcast television (about 200 MHz or
2×108Hz
28
Hz
).
Thus, the visible electromagnetic frequencies are over six
orders of magnitude higher!
To summarize, we use transmission lines for high-frequency
wireline signal communication. In wireline communication, we
have a direct, physical connection—a circuit—between
transmitter and receiver. When we select the transmission line
characteristics and the transmission frequency so that we
operate in the high-frequency regime, signals are not filtered
as they propagate along the transmission line: The
characteristic impedance is real-valued—the transmission
line's equivalent impedance is a resistor—and all the
signal's components at various frequencies propagate at the same
speed. Transmitted signal amplitude does decay exponentially
along the transmission line. Note that in the high-frequency
regime the space constant is approximately zero, which
means the attenuation is quite small.
What is the limiting value of the space constant in the high
frequency regime?
As frequency increases,
2πf
C
∼
≫
G
∼
≫
2
f
C
∼
G
∼
and
2πf
L
∼
≫
R
∼
≫
2
f
L
∼
R
∼
.
In this high-frequency region,
γ=i2πf
L
∼
C
∼
(1+
G
∼
i2πf
C
∼
)(1+
R
∼
i2πf
L
∼
)
γ
2
f
L
∼
C
∼
1
G
∼
2
f
C
∼
1
R
∼
2
f
L
∼
(16)
Thus, the attenuation (space) constant equals the real part
of this expression, and equals
af=
G
∼
Z
0
+
R
∼
Z
0
2
a
f
G
∼
Z
0
R
∼
Z
0
2
.
"Electrical Engineering Digital Processing Systems in Braille."