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 µs 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 µΩ 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.
|
