Standing waves and resonance
Whenever there is a mismatch of impedance
between transmission line and load, reflections will occur.
If the incident signal is a continuous AC waveform, these
reflections will mix with more of the oncoming incident
waveform to produce stationary waveforms called standing
waves.
The following illustration shows how a
triangle-shaped incident waveform turns into a mirror-image
reflection upon reaching the line's unterminated end. The
transmission line in this illustrative sequence is shown as
a single, thick line rather than a pair of wires, for
simplicity's sake. The incident wave is shown traveling from
left to right, while the reflected wave travels from right
to left:
If we add the two waveforms together, we
find that a third, stationary waveform is created along the
line's length:
This third, "standing" wave, in fact,
represents the only voltage along the line, being the
representative sum of incident and reflected voltage waves.
It oscillates in instantaneous magnitude, but does not
propagate down the cable's length like the incident or
reflected waveforms causing it. Note the dots along the line
length marking the "zero" points of the standing wave (where
the incident and reflected waves cancel each other), and how
those points never change position:
Standing waves are quite abundant in the
physical world. Consider a string or rope, shaken at one
end, and tied down at the other (only one half-cycle of hand
motion shown, moving downward):
Both the nodes (points of little or no
vibration) and the antinodes (points of maximum vibration)
remain fixed along the length of the string or rope. The
effect is most pronounced when the free end is shaken at
just the right frequency. Plucked strings exhibit the same
"standing wave" behavior, with "nodes" of maximum and
minimum vibration along their length. The major difference
between a plucked string and a shaken string is that the
plucked string supplies its own "correct" frequency of
vibration to maximize the standing-wave effect:
Wind blowing across an open-ended tube also
produces standing waves; this time, the waves are vibrations
of air molecules (sound) within the tube rather than
vibrations of a solid object. Whether the standing wave
terminates in a node (minimum amplitude) or an antinode
(maximum amplitude) depends on whether the other end of the
tube is open or closed:
A closed tube end must be a wave node, while
an open tube end must be an antinode. By analogy, the
anchored end of a vibrating string must be a node, while the
free end (if there is any) must be an antinode.
Note how there is more than one wavelength
suitable for producing standing waves of vibrating air
within a tube that precisely match the tube's end points.
This is true for all standing-wave systems: standing waves
will resonate with the system for any frequency (wavelength)
correlating to the node/antinode points of the system.
Another way of saying this is that there are multiple
resonant frequencies for any system supporting standing
waves.
All higher frequencies are integer-multiples
of the lowest (fundamental) frequency for the system. The
sequential progression of harmonics from one resonant
frequency to the next defines the overtone
frequencies for the system:
The actual frequencies (measured in Hertz)
for any of these harmonics or overtones depends on the
physical length of the tube and the waves' propagation
velocity, which is the speed of sound in air.
Because transmission lines support standing
waves, and force these waves to possess nodes and antinodes
according to the type of termination impedance at the load
end, they also exhibit resonance at frequencies determined
by physical length and propagation velocity. Transmission
line resonance, though, is a bit more complex than resonance
of strings or of air in tubes, because we must consider both
voltage waves and current waves.
This complexity is made easier to understand
by way of computer simulation. To begin, let's examine a
perfectly matched source, transmission line, and load. All
components have an impedance of 75 Ω:
Using SPICE to simulate the circuit, we'll
specify the transmission line (t1) with a 75 Ω
characteristic impedance (z0=75) and a propagation
delay of 1 microsecond (td=1u). This is a
convenient method for expressing the physical length of a
transmission line: the amount of time it takes a wave to
propagate down its entire length. If this were a real 75 Ω
cable -- perhaps a type "RG-59B/U" coaxial cable, the type
commonly used for cable television distribution -- with a
velocity factor of 0.66, it would be about 648 feet long.
Since 1 �s is the period of a 1 MHz signal, I'll choose to
sweep the frequency of the AC source from (nearly) zero to
that figure, to see how the system reacts when exposed to
signals ranging from DC to 1 wavelength.
Here is the SPICE netlist for the circuit
shown above:
Transmission line
v1 1 0 ac 1 sin
rsource 1 2 75
t1 2 0 3 0 z0=75 td=1u
rload 3 0 75
.ac lin 101 1m 1meg
* Using "Nutmeg" program to plot analysis
.end
Running this simulation and plotting the source impedance
drop (as an indication of current), the source voltage, the
line's source-end voltage, and the load voltage, we see that
the source voltage -- shown as vm(1) (voltage
magnitude between node 1 and the implied ground point of
node 0) on the graphic plot -- registers a steady 1 volt,
while every other voltage registers a steady 0.5 volts:
In a system where all impedances are
perfectly matched, there can be no standing waves, and
therefore no resonant "peaks" or "valleys" in the Bode plot.
Now, let's change the load impedance to 999
MΩ, to simulate an open-ended transmission line. We should
definitely see some reflections on the line now as the
frequency is swept from 1 mHz to 1 MHz:
Transmission line
v1 1 0 ac 1 sin
rsource 1 2 75
t1 2 0 3 0 z0=75 td=1u
rload 3 0 999meg
.ac lin 101 1m 1meg
* Using "Nutmeg" program to plot analysis
.end
Here, both the supply voltage vm(1)
and the line's load-end voltage vm(3) remain steady
at 1 volt. The other voltages dip and peak at different
frequencies along the sweep range of 1 mHz to 1 MHz. There
are five points of interest along the horizontal axis of the
analysis: 0 Hz, 250 kHz, 500 kHz, 750 kHz, and 1 MHz. We
will investigate each one with regard to voltage and current
at different points of the circuit.
At 0 Hz (actually 1 mHz), the signal is
practically DC, and the circuit behaves much as it would
given a 1-volt DC battery source. There is no circuit
current, as indicated by zero voltage drop across the source
impedance (Zsource: vm(1,2)), and full
source voltage present at the source-end of the transmission
line (voltage measured between node 2 and node 0: vm(2)).
At 250 kHz, we see zero voltage and maximum
current at the source-end of the transmission line, yet
still full voltage at the load-end:
You might be wondering, how can this be? How
can we get full source voltage at the line's open end while
there is zero voltage at its entrance? The answer is found
in the paradox of the standing wave. With a source frequency
of 250 kHz, the line's length is precisely right for 1/4
wavelength to fit from end to end. With the line's load end
open-circuited, there can be no current, but there will be
voltage. Therefore, the load-end of an open-circuited
transmission line is a current node (zero point) and a
voltage antinode (maximum amplitude):
At 500 kHz, exactly one-half of a standing
wave rests on the transmission line, and here we see another
point in the analysis where the source current drops off to
nothing and the source-end voltage of the transmission line
rises again to full voltage:
At 750 kHz, the plot looks a lot like it was
at 250 kHz: zero source-end voltage (vm(2)) and
maximum current (vm(1,2)). This is due to 3/4 of a
wave poised along the transmission line, resulting in the
source "seeing" a short-circuit where it connects to the
transmission line, even though the other end of the line is
open-circuited:
When the supply frequency sweeps up to 1
MHz, a full standing wave exists on the transmission line.
At this point, the source-end of the line experiences the
same voltage and current amplitudes as the load-end: full
voltage and zero current. In essence, the source "sees" an
open circuit at the point where it connects to the
transmission line.
In a similar fashion, a short-circuited
transmission line generates standing waves, although the
node and antinode assignments for voltage and current are
reversed: at the shorted end of the line, there will be zero
voltage (node) and maximum current (antinode). What follows
is the SPICE simulation and illustrations of what happens at
all the interesting frequencies: 0 Hz, 250 kHz, 500 kHz, 750
kHz, and 1 MHz. The short-circuit jumper is simulated by a 1
�Ω load impedance:
Transmission line
v1 1 0 ac 1 sin
rsource 1 2 75
t1 2 0 3 0 z0=75 td=1u
rload 3 0 1u
.ac lin 101 1m 1meg
* Using "Nutmeg" program to plot analysis
.end
In both these circuit examples, an
open-circuited line and a short-circuited line, the energy
reflection is total: 100% of the incident wave reaching the
line's end gets reflected back toward the source. If,
however, the transmission line is terminated in some
impedance other than an open or a short, the reflections
will be less intense, as will be the difference between
minimum and maximum values of voltage and current along the
line.
Suppose we were to terminate our example
line with a 100 Ω resistor instead of a 75 Ω resistor.
Examine the results of the corresponding SPICE analysis to
see the effects of impedance mismatch at different source
frequencies:
Transmission line
v1 1 0 ac 1 sin
rsource 1 2 75
t1 2 0 3 0 z0=75 td=1u
rload 3 0 100
.ac lin 101 1m 1meg
* Using "Nutmeg" program to plot analysis
.end
If we run another SPICE analysis, this time
printing numerical results rather than plotting them, we can
discover exactly what is happening at all the interesting
frequencies (DC, 250 kHz, 500 kHz, 750 kHz, and 1 MHz):
Transmission line
v1 1 0 ac 1 sin
rsource 1 2 75
t1 2 0 3 0 z0=75 td=1u
rload 3 0 100
.ac lin 5 1m 1meg
.print ac v(1,2) v(1) v(2) v(3)
.end
freq v(1,2) v(1) v(2) v(3)
1.000E-03 4.286E-01 1.000E+00 5.714E-01 5.714E-01
2.500E+05 5.714E-01 1.000E+00 4.286E-01 5.714E-01
5.000E+05 4.286E-01 1.000E+00 5.714E-01 5.714E-01
7.500E+05 5.714E-01 1.000E+00 4.286E-01 5.714E-01
1.000E+06 4.286E-01 1.000E+00 5.714E-01 5.714E-01
At all frequencies, the source voltage,
v(1), remains steady at 1 volt, as it should. The load
voltage, v(3), also remains steady, but at a lesser
voltage: 0.5714 volts. However, both the line input voltage
(v(2)) and the voltage dropped across the source's
75 Ω impedance (v(1,2), indicating current drawn
from the source) vary with frequency.
At odd harmonics of the fundamental
frequency (250 kHz and 750 kHz), we see differing levels of
voltage at each end of the transmission line, because at
those frequencies the standing waves terminate at one end in
a node and at the other end in an antinode. Unlike the
open-circuited and short-circuited transmission line
examples, the maximum and minimum voltage levels along this
transmission line do not reach the same extreme values of 0%
and 100% source voltage, but we still have points of
"minimum" and "maximum" voltage. The same holds true for
current: if the line's terminating impedance is mismatched
to the line's characteristic impedance, we will have points
of minimum and maximum current at certain fixed locations on
the line, corresponding to the standing current wave's nodes
and antinodes, respectively.
One way of expressing the severity of
standing waves is as a ratio of maximum amplitude (antinode)
to minimum amplitude (node), for voltage or for current.
When a line is terminated by an open or a short, this
standing wave ratio, or SWR is valued at
infinity, since the minimum amplitude will be zero, and any
finite value divided by zero results in an infinite
(actually, "undefined") quotient. In this example, with a 75
Ω line terminated by a 100 Ω impedance, the SWR will be
finite: 1.333, calculated by taking the maximum line voltage
at either 250 kHz or 750 kHz (0.5714 volts) and dividing by
the minimum line voltage (0.4286 volts).
Standing wave ratio may also be calculated
by taking the line's terminating impedance and the line's
characteristic impedance, and dividing the larger of the two
values by the smaller. In this example, the terminating
impedance of 100 Ω divided by the characteristic impedance
of 75 Ω yields a quotient of exactly 1.333, matching the
previous calculation very closely.
A perfectly terminated transmission line
will have an SWR of 1, since voltage at any location along
the line's length will be the same, and likewise for
current. Again, this is usually considered ideal, not only
because reflected waves constitute energy not delivered to
the load, but because the high values of voltage and current
created by the antinodes of standing waves may over-stress
the transmission line's insulation (high voltage) and
conductors (high current), respectively.
Also, a transmission line with a high SWR
tends to act as an antenna, radiating electromagnetic energy
away from the line, rather than channeling all of it to the
load. This is usually undesirable, as the radiated energy
may "couple" with nearby conductors, producing signal
interference. An interesting footnote to this point is that
antenna structures -- which typically resemble open- or
short-circuited transmission lines -- are often designed to
operate at high standing wave ratios, for the very
reason of maximizing signal radiation and reception.
The following photograph shows a set of
transmission lines at a junction point in a radio
transmitter system. The large, copper tubes with ceramic
insulator caps at the ends are rigid coaxial transmission
lines of 50 Ω characteristic impedance. These lines carry RF
power from the radio transmitter circuit to a small, wooden
shelter at the base of an antenna structure, and from that
shelter on to other shelters with other antenna structures:
Flexible coaxial cable connected to the
rigid lines (also of 50 Ω characteristic impedance) conduct
the RF power to capacitive and inductive "phasing" networks
inside the shelter. The white, plastic tube joining two of
the rigid lines together carries "filling" gas from one
sealed line to the other. The lines are gas-filled to avoid
collecting moisture inside them, which would be a definite
problem for a coaxial line. Note the flat, copper "straps"
used as jumper wires to connect the conductors of the
flexible coaxial cables to the conductors of the rigid
lines. Why flat straps of copper and not round wires?
Because of the skin effect, which renders most of the
cross-sectional area of a round conductor useless at radio
frequencies.
Like many transmission lines, these are
operated at low SWR conditions. As we will see in the next
section, though, the phenomenon of standing waves in
transmission lines is not always undesirable, as it may be
exploited to perform a useful function: impedance
transformation.
-
REVIEW:
-
Standing waves are waves of voltage
and current which do not propagate (i.e. they are
stationary), but are the result of interference between
incident and reflected waves along a transmission line.
-
A node is a point on a standing
wave of minimum amplitude.
-
An antinode is a point on a
standing wave of maximum amplitude.
-
Standing waves can only exist in a
transmission line when the terminating impedance does not
match the line's characteristic impedance. In a perfectly
terminated line, there are no reflected waves, and
therefore no standing waves at all.
-
At certain frequencies, the nodes and
antinodes of standing waves will correlate with the ends
of a transmission line, resulting in resonance.
-
The lowest-frequency resonant point on a
transmission line is where the line is one
quarter-wavelength long. Resonant points exist at every
harmonic (integer-multiple) frequency of the fundamental
(quarter-wavelength).
-
Standing wave ratio, or SWR,
is the ratio of maximum standing wave amplitude to minimum
standing wave amplitude. It may also be calculated by
dividing termination impedance by characteristic
impedance, or visa-versa, which ever yields the greatest
quotient. A line with no standing waves (perfectly
matched: Zload to Z0) has an SWR
equal to 1.
-
Transmission lines may be damaged by the
high maximum amplitudes of standing waves. Voltage
antinodes may break down insulation between conductors,
and current antinodes may overheat conductors.
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