tutorial
Transmission lines
as antennas
The bane of transmission line leakage
can also be a surprise benefit.
characteristic (surge) impedance, Zo. This article will
quantify when a particular open-wire transmission
line becomes an antenna. Conventional wisdom
asserts that when equal magnitude but opposite
sign currents exist on the opposite sides of the twinlead, radiation does not occur. But this rule is valid
only when the two wires are electrically close
together. In fact a loop antenna is a degenerate form
of twin-lead, and it is interesting to discover what
physical line dimensions encourage radiation.
The surge impedance of balanced twin-lead in
free space with an air dielectric is:
Zo = 120 ln (2S/d) Ω
By Grant Bingeman
W
hen does an open-wire transmission line
begin to lose its ideal characteristics and
start to behave as an antenna? Is there such a
thing as a balanced, but radiating transmission line
in free space? If so, can we call this a line antenna?
This article correlates twin-lead transmission line
input impedance with antenna radiation resistance.
A discussion of transmission
lines and radiation
Electromagnetic radiation from an open-wire
length of twin-lead transmission line in free space
can exist even when its standing wave ratio (SWR)
is 1.00; that is, when the line is terminated in its
Termination
Impedance
(1)
where:
S = spacing between wire centers
d = diameter of wire
ln = natural log
*note that the dimension units of S and d do not
matter as long as they are the same.
In theory, when 1000 W are delivered to the input
of a lossless and perfect transmission line, then
1000 W should appear at the termination impedance, Zt, when Zt = Zo. However, if the transmission
line behaves as an antenna, then the input resistance to the line also includes a radiation component, Rr such that Rin = Rline + Rr. An ideal transmission line without any leaks would have no radiation
resistance. An ideal antenna, on the other hand
would radiate all power delivered to it from a generator, such that Rline would be zero. A leaky transmission line would exhibit both transmission line
and antenna characteristics. Normally, a physical
termination impedance, Zt is not added to an antenna except in special cases where bandwidth needs to
be augmented, etc. Keep in mind that this article is
not about a conventional transmission line feeding a
conventional antenna, but rather about a hybrid
transmission-line/antenna mixture.
Modeling an example
using twin-lead line
Source
Figure 1. Model of 10 meter twin-lead transmission line.
74
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As an example, a 10 meter length of twin-lead is
used as a pair of 5 mm diameter wires spaced one
meter (see Figure 1). Equation 1 suggests that Zo
should be about 720 Ω in this case. Method of
moments (MoM) analysis using three current segments per meter predicts a 10 MHz input impedance to the twin-lead of 731 - j9.4 Ω when the line is
terminated in a pure resistance of 720 Ω. If we use a
730 Ω termination resistor instead, then the 10
MHz input impedance is exactly 730 Ω, leading one
to believe that Zo is actually closer to 730 Ω in this
case. But because of the accuracy limits of the MoM
model, the Zo can only be said to be about 720 Ω.
However, keep in mind that what is being delt
with is not quite a perfect transmission line. At 10
MHz the electrical length of the line is 120°, (1/3 of a
wavelength), and the width of the line is 12° (about
3% of a wavelength). This width is actually not a ter-
January 2001
0 dB
resistance can be determined from:
Rr = Rin Pr/P
(2)
-10
where:
-20
Rin = input resistance to transmission
line.
-30
Pr = radiated power.
Pr = P – Pt = input power – load power.
P = input power to transmission line.
The average value for R r is 3.3 Ω.
Remember that accuracy is constrained
by the MoM model, but the results are
close enough to deduce some valid generalizations.
Table 3, where the electrical wire
spacing is less than three degrees, conveys that the radiation from the line is
not so much a function of mismatch or
SWR, but more a function of wire spacing. However, this conclusion may not
be true of unbalanced lines, and it is
sometimes difficult to obtain perfectly
balanced transmission lines in the real
world anyway.
Figure 2. 10 MHz radiation pattern.
ribly small value electrically, so it
should be suspected that there is some
radiation from this open-wire line at frequencies above 10 MHz. According to
Table 1, for an input power of 1000 W,
the radiated power, Pr, and the radiation
resistance, Rr, increase with increasing
frequency even when the line is properly
terminated in Zt = Zo. One percent radiated power occurs when the electrical
width of the example twin-lead is about
20°, which is about 6% of a wavelength
at 17 MHz. Recall that the one meter
spacing between conductors is rather
large for typical open-wire line. At low
RF power levels, this large spacing
would not be required. But the point of
this exercise is to show that such a line
MHz
2
5
7
10
15
20
L
24°
60
84
120
180
240
can degenerate into an antenna when
the frequency is high enough. And it
was thought that transmission line
surge impedance was not a function of
frequency? Think again. The assumptions upon which many of our RF formulas are designed to keep life simple, but
are based beyond those assumptions,
then calculation accuracy begins to
degrade.
Assuming an ideal, lossless transmission line, a 1000 W is delivered to the
input of that line, but only 990 W is
received in the load resistance, then the
missing 10 W must appear as radiation.
Table 2 demonstrates what to expect at
10 MHz for high SWR on the ten meter
length of 720 Ω twin-lead. The radiation
Transmission line
MoM segment Pr
W
Zin
2.4°
717 – j6.2 Ω
2.4°
0W
6.0
711 – j4.2
6.0
1
8.4
715 + j0.2
8.4
2
12.0
731 + j9.4
12
4
18.0
705 – j45.9
18
8
24.0
689 – j4.0
24
13
Table 1. Zin when Zt = 720 Ω (L = length, W = width)
76
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Rr
0Ω
0.7
1.4
2.9
5.6
9.0
The transmission line
as an antenna
Consider the case where Zt = 0 Ω.
Referring to Figure 1, the antenna
looks like two parallel horizontal
dipoles, one of which is driven directly
by a generator. The second dipole is
driven by the ten meter length of
transmission line connected to the
first dipole. Thus, the phase of the current in the second dipole, hence the
radiation pattern, are affected by this
transmission line. There is no physical
termination impedance placed in the
second dipole, although it does have
an operating impedance that is reflected back down the transmission line to
the first dipole, which influences the
value of Zin.
The 10, 15 and 20 MHz radiation
patterns in the free-space horizontal
plane are described in Figures 2
through 4. Clearly the transmission
line can behave as an antenna, especially when it is terminated in some
impedance other than Z0. The operating particulars are listed in table 4.
Note that all these pattern shapes
and gains are very different. As expected, the greatest power leak from the
open-wire line occurs at the highest frequency, where the greatest electrical
January 2001
0 dB
wavelengths long at 20 MHz. Thus
the free-space coupling or mutual
impedance between these two dipoles
is small, and the current in dipole 2 is
mostly contributed by the direct
transmission line connection.
Referring to Table 4 and Figure 5, one
can see that most of the current exists
in dipole 2.
The current in dipole 2 can be
manipulated with a terminating reactance in order to reduce the radiation
effects, if desired. In any case the
physical wiring at the source and termination ends of the transmission line
is a necessary part of the complete
model. When the spacing between
transmission line conductors becomes
large electrically, the same is true of
the wiring at the input and output
ends of the transmission line, which
can be considered as radiating dipoles.
-10
-20
-30
Summary
Figure 3. 15 MHz radiation pattern.
spacing between conductors exists
(Figure 4). It is also important to note
that the electrical size of the dipoles at
Rt
10 Ω
50
500
1k
5k
10 k
50 k
Zin = Rin +jXin
SWR
27.3 – j845Ω
72
120 – j836
14.4
719 – j274
1.4
665 + j210
1.4
185 + j606
6.9
95.2 – j623
13.9
21.3 – j629
69.4
Pt
Pr
860 W 140 W
968
32
995
5
995
5
985
15
971
29
872
128
Table 2. 10 MHz radiation vs terminating resistance, Rt
each end of the transmission line is
greatest at the highest frequency.
The current distribution along the
transmission line is described in
Figure 5. This is a standing wave,
MHz
10
15
20
Zin
3.8 – j846
3.1 + j183
212 + j4710
where the node repeats every half
wavelength, which is 7.5 meters at 20
MHz. The transmission line currents
dipole 1
16.2 A/0°
18.0 at 0
2.2 at 0
Rr
3.8 W
3.8
3.6
3.3
2.8
2.8
2.7
Zt
Zin
SWR
Pt
Pr
Rr
10 Ω 12.4 + j350 Ω 72.0 999.5 W 0.5 W 0.006 W
50
62.0 + j348
14.4 999.9
0.1
0.006
500
555 + j159
1.4 1000
0
0.0
Table 3. 2 MHz radiation vs. terminating impedance.
are
equal
magnitude,
but opposite
in phase, and
do not contribute too much to the radiated field
intensity. It is the dipole currents that
primarily affect the radiation pattern.
The dipole currents exist at each end
of the transmission line, which is 1.5
dipole 2
24.8 A/0.2°
18.7 at –0.1
14.7 at –2.6
max gain (dBi)
–2.1
–12.1
+1.3
Table 4. Antenna operating parameters, 1000 Watts (dipole 1 and 2 numbers are Amps @degree).
78
The article has shown that electromagnetic radiation exists from a balanced open-wire line when the electrical spacing between wires is large
(dipole size of source and termination
wires is large), regardless of the value
of terminating impedance.
It has also shown that a transmission
line can behave as an antenna, in which
case its input resistance has separate
line and radiation components.
www.rfdesign.com
Further, termination impedance does
not significantly affect radiation from a
balanced open-wire line unless the spacing between conductors (dipole size) is
electrically large. And source and termination dipoles are a necessary part of
the radiating transmission line model.
In today’s age of computers it is
advantageous to use one of the several
available modeling programs to aid in
the development of such anaylses. In
this case EZNAC 1 was used as the
application tool. The appendix provides
a technique and the equations used to
develop a lossy transmission line model.
(continued on page 80)
January 2001
0 dB
-10
-20
-30
Figure 4. 20 MHz radiation pattern.
References
1) EZNEC version 3.0 is available
from Roy Lewallen, www.eznec.com.
2) For further information about
R F N e t w o r k D e s i g n e r , go to
www.qsl.net/km5kg
that of air, which determines the
velocity of propagation of the electromagnetic signal through the line, and
also determines the capacitance of
the line. Sometimes the insulation is
a foam, or a series of stand-offs, or
some other mixture of various
dielectrics. In such special cases, the
effective dielectric constant may have
to be estimated or measured.
Since the velocity of propagation of
the electromagnetic energy along a
transmission line is less than that of
air, a wavelength in a transmission
line is shorter than its free-space
value. Normally assign 360 electrical
degrees are assigned to a wavelength.
Thus a physical quarter-wave section
of transmission line with air insulation (velocity factor = 1.0) would be
90° long. However the same physical
length of line with solid polyethylene
insulation would be about 136° long,
since its velocity factor is about 66%.
The phase shift across this line is
denoted as negative 136° to convey
the fact that a time lag or delay is
associated with the propagation of the
energy from the input to the output.
Note that the phase shift across a
transmission line is dependent on Zt,
and the value of –136° in our example
assumes that Z t = Z 0 . Also keep in
mind that the voltage phase shift and
the current phase shift are the same
only when Zt = Zo.
Losses in a transmission line consist
of dielectric heating (the dissipation factor of the insulation determines this),
DIPOLE 2
Appendix – a lossy transmission
line model
A transmission line conveys an
electromagnetic signal from its input
to its output. In other words, a transmission line can transport RF power
from a generator to a load. The power
is developed in the resistive portion
o f th e l o ad i m p e d a n ce , Z t . So me
power is lost to heating of the transmission line. A transmission line can
be as a simple two-port using either
a lumped-parameter tee network or a
p i n e tw o r k. Lu m p e d -p a r a m e t e r
i n c l u d e r e si s t o rs , i n d u c t o rs a n d
capacitors. A transmission line is
defined by its length, surge impedance (Z0 or characteristic impedance),
dielectric constant and loss. All other
parameters can be calculated from
this information.
The insulating material between
the conductors in a transmission line
has a dielectric constant greater than
80
DIPOLE 1
Figure 5. Distribution along the transmission line.
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January 2001
and conductor heating (I2R) caused by
resistance to the RF current.
Sometimes when a transmission line
is mis-terminated additional loss can
appear as electromagnetic radiation
from currents on the outside of a coaxial line, but this effect is not accommodated in the following equations.
The maximum rated power-handling capability of a transmission line
is based on a fixed temperature
increase, and assumes the line is terminated in its characteristic impedance (Zt = Z0). In general, the larger
the transmission line, the lower the
losses, and the higher the frequency,
the higher the losses.
A Tee network model of a lossy
transmission line:
A π network model of a lossy
transmission line:
Distributed circuit model components of a transmission line:
π network shunt arms: Z a = Z b =
Z0/tanh (y/2)
π network series arm: Zc = Z0sinh (y)
Tee network series arms: Z1 = Z2 = Zo
tanh (y/2)
Tee network shunt arm: Z3 = Z0/sinh (y)
Input impedance to line: Zin = [Zt + Z0
tanh (y)]/[1 + (Zt/Z0) tanh (y )] Ω
note that Z0 has a reactance term when
loss is substantial, but the line is still
terminated in R0 for a perfect match.
and:
y = a + jb:
a = line loss (Nepers)
b = electrical length of line
hyperbolic tangent:
where:
the dielectric constant, e = 1/vel2
velocity factor relative to the speed of
light, vel < 1.0
C = 1016 sqrt (e)/Z0, pF / foot
L = 0.001016 Z0 sqrt (e), uH/foot
Z0 = characteristic impedance of line
tanh (y) = [sinh (2a) + jsin (2b)]/[cosh
(2a) + cos (2b)]
(1 Neper = 8.686 dB)
About the author
Grant Bingeman is a registered
professional engineer in Texas,
and a principal engineer with
Continental Electronics in Dallas.
His e-mail address is DrBingo@compuserve.com and his RF web site is
www.qsl.net/km5kg
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January 2001