A 50-ohm cable?
Early in my explorations of electricity,
I came across a length of coaxial cable with the label "50 ohms"
printed along its outer sheath. Now, coaxial cable is a two-conductor
cable made of a single conductor surrounded by a braided wire jacket,
with a plastic insulating material separating the two. As such, the
outer (braided) conductor completely surrounds the inner (single wire)
conductor, the two conductors insulated from each other for the entire
length of the cable. This type of cabling is often used to conduct weak
(low-amplitude) voltage signals, due to its excellent ability to shield
such signals from external interference.
I was mystified by the "50 ohms" label on
this coaxial cable. How could two conductors, insulated from each other
by a relatively thick layer of plastic, have 50 ohms of resistance
between them? Measuring resistance between the outer and inner
conductors with my ohmmeter, I found it to be infinite (open-circuit),
just as I would have expected from two insulated conductors. Measuring
each of the two conductors' resistances from one end of the cable to the
other indicated nearly zero ohms of resistance: again, exactly what I
would have expected from continuous, unbroken lengths of wire. Nowhere
was I able to measure 50 Ω of resistance on this cable, regardless of
which points I connected my ohmmeter between.
What I didn't understand at the time was
the cable's response to short-duration voltage "pulses" and
high-frequency AC signals. Continuous direct current (DC) -- such as
that used by my ohmmeter to check the cable's resistance -- shows the
two conductors to be completely insulated from each other, with nearly
infinite resistance between the two. However, due to the effects of
capacitance and inductance distributed along the length of the cable,
the cable's response to rapidly-changing voltages is such that it acts
as a finite impedance, drawing current proportional to an applied
voltage. What we would normally dismiss as being just a pair of wires
becomes an important circuit element in the presence of transient and
high-frequency AC signals, with characteristic properties all its own.
When expressing such properties, we refer to the wire pair as a
transmission line.
This chapter explores transmission line
behavior. Many transmission line effects do not appear in significant
measure in AC circuits of powerline frequency (50 or 60 Hz), or in
continuous DC circuits, and so we haven't had to concern ourselves with
them in our study of electric circuits thus far. However, in circuits
involving high frequencies and/or extremely long cable lengths, the
effects are very significant. Practical applications of transmission
line effects abound in radio-frequency ("RF") communication circuitry,
including computer networks, and in low-frequency circuits subject to
voltage transients ("surges") such as lightning strikes on power lines.
Circuits and the speed of light
Suppose we had a simple one-battery,
one-lamp circuit controlled by a switch. When the switch is closed, the
lamp immediately lights. When the switch is opened, the lamp immediately
darkens:
Actually, an incandescent lamp takes a
short time for its filament to warm up and emit light after receiving an
electric current of sufficient magnitude to power it, so the effect is
not instant. However, what I'd like to focus on is the immediacy of the
electric current itself, not the response time of the lamp filament. For
all practical purposes, the effect of switch action is instant at the
lamp's location. Although electrons move through wires very slowly, the
overall effect of electrons pushing against each other happens at the
speed of light (approximately 186,000 miles per second!).
What would happen, though, if the wires
carrying power to the lamp were 186,000 miles long? Since we know the
effects of electricity do have a finite speed (albeit very fast), a set
of very long wires should introduce a time delay into the circuit,
delaying the switch's action on the lamp:
Assuming no warm-up time for the lamp
filament, and no resistance along the 372,000 mile length of both wires,
the lamp would light up approximately one second after the switch
closure. Although the construction and operation of superconducting
wires 372,000 miles in length would pose enormous practical problems, it
is theoretically possible, and so this "thought experiment" is valid.
When the switch is opened again, the lamp will continue to receive power
for one second of time after the switch opens, then it will de-energize.
One way of envisioning this is to imagine
the electrons within a conductor as rail cars in a train: linked
together with a small amount of "slack" or "play" in the couplings. When
one rail car (electron) begins to move, it pushes on the one ahead of it
and pulls on the one behind it, but not before the slack is relieved
from the couplings. Thus, motion is transferred from car to car (from
electron to electron) at a maximum velocity limited by the coupling
slack, resulting in a much faster transfer of motion from the left end
of the train (circuit) to the right end than the actual speed of the
cars (electrons):
Another analogy, perhaps more fitting for
the subject of transmission lines, is that of waves in water. Suppose a
flat, wall-shaped object is suddenly moved horizontally along the
surface of water, so as to produce a wave ahead of it. The wave will
travel as water molecules bump into each other, transferring wave motion
along the water's surface far faster than the water molecules themselves
are actually traveling:
Likewise, electron motion "coupling"
travels approximately at the speed of light, although the electrons
themselves don't move that quickly. In a very long circuit, this
"coupling" speed would become noticeable to a human observer in the form
of a short time delay between switch action and lamp action.
- REVIEW:
- In an electric circuit, the effects of
electron motion travel approximately at the speed of light, although
electrons within the conductors do not travel anywhere near that
velocity.
Characteristic impedance
Suppose, though, that we had a set of
parallel wires of infinite length, with no lamp at the end. What
would happen when we close the switch? Being that there is no longer a
load at the end of the wires, this circuit is open. Would there be no
current at all?
Despite being able to avoid wire
resistance through the use of superconductors in this "thought
experiment," we cannot eliminate capacitance along the wires' lengths.
Any pair of conductors separated by an insulating medium creates
capacitance between those conductors:
Voltage applied between two conductors
creates an electric field between those conductors. Energy is stored in
this electric field, and this storage of energy results in an opposition
to change in voltage. The reaction of a capacitance against changes in
voltage is described by the equation i = C(de/dt), which tells us that
current will be drawn proportional to the voltage's rate of change over
time. Thus, when the switch is closed, the capacitance between
conductors will react against the sudden voltage increase by charging up
and drawing current from the source. According to the equation, an
instant rise in applied voltage (as produced by perfect switch closure)
gives rise to an infinite charging current.
However, the current drawn by a pair of
parallel wires will not be infinite, because there exists series
impedance along the wires due to inductance. Remember that current
through any conductor develops a magnetic field of proportional
magnitude. Energy is stored in this magnetic field, and this storage of
energy results in an opposition to change in current. Each wire develops
a magnetic field as it carries charging current for the capacitance
between the wires, and in so doing drops voltage according to the
inductance equation e = L(di/dt). This voltage drop limits the voltage
rate-of-change across the distributed capacitance, preventing the
current from ever reaching an infinite magnitude:
Because the electrons in the two wires
transfer motion to and from each other at nearly the speed of light, the
"wave front" of voltage and current change will propagate down the
length of the wires at that same velocity, resulting in the distributed
capacitance and inductance progressively charging to full voltage and
current, respectively, like this:
The end result of these interactions is a
constant current of limited magnitude through the battery source. Since
the wires are infinitely long, their distributed capacitance will never
fully charge to the source voltage, and their distributed inductance
will never allow unlimited charging current. In other words, this pair
of wires will draw current from the source so long as the switch is
closed, behaving as a constant load. No longer are the wires merely
conductors of electrical current and carriers of voltage, but now
constitute a circuit component in themselves, with unique
characteristics. No longer are the two wires merely a pair of
conductors, but rather a transmission line.
As a constant load, the transmission
line's response to applied voltage is resistive rather than reactive,
despite being comprised purely of inductance and capacitance (assuming
superconducting wires with zero resistance). We can say this because
there is no difference from the battery's perspective between a resistor
eternally dissipating energy and an infinite transmission line eternally
absorbing energy. The impedance (resistance) of this line in ohms is
called the characteristic impedance, and it is fixed by the
geometry of the two conductors. For a parallel-wire line with air
insulation, the characteristic impedance may be calculated as such:
If the transmission line is coaxial in
construction, the characteristic impedance follows a different equation:
In both equations, identical units of
measurement must be used in both terms of the fraction. If the
insulating material is other than air (or a vacuum), both the
characteristic impedance and the propagation velocity will be affected.
The ratio of a transmission line's true propagation velocity and the
speed of light in a vacuum is called the velocity factor of that
line.
Velocity factor is purely a factor of the
insulating material's relative permittivity (otherwise known as its
dielectric constant), defined as the ratio of a material's electric
field permittivity to that of a pure vacuum. The velocity factor of any
cable type -- coaxial or otherwise -- may be calculated quite simply by
the following formula:
Characteristic impedance is also known as
natural impedance, and it refers to the equivalent resistance of
a transmission line if it were infinitely long, owing to distributed
capacitance and inductance as the voltage and current "waves" propagate
along its length at a propagation velocity equal to some large fraction
of light speed.
It can be seen in either of the first two
equations that a transmission line's characteristic impedance (Z0)
increases as the conductor spacing increases. If the conductors are
moved away from each other, the distributed capacitance will decrease
(greater spacing between capacitor "plates"), and the distributed
inductance will increase (less cancellation of the two opposing magnetic
fields). Less parallel capacitance and more series inductance results in
a smaller current drawn by the line for any given amount of applied
voltage, which by definition is a greater impedance. Conversely,
bringing the two conductors closer together increases the parallel
capacitance and decreases the series inductance. Both changes result in
a larger current drawn for a given applied voltage, equating to a lesser
impedance.
Barring any dissipative effects such as
dielectric "leakage" and conductor resistance, the characteristic
impedance of a transmission line is equal to the square root of the
ratio of the line's inductance per unit length divided by the line's
capacitance per unit length:
- REVIEW:
- A transmission line is a pair
of parallel conductors exhibiting certain characteristics due to
distributed capacitance and inductance along its length.
- When a voltage is suddenly applied to
one end of a transmission line, both a voltage "wave" and a current
"wave" propagate along the line at nearly light speed.
- If a DC voltage is applied to one end
of an infinitely long transmission line, the line will draw current
from the DC source as though it were a constant resistance.
- The characteristic impedance (Z0)
of a transmission line is the resistance it would exhibit if it were
infinite in length. This is entirely different from leakage resistance
of the dielectric separating the two conductors, and the metallic
resistance of the wires themselves. Characteristic impedance is purely
a function of the capacitance and inductance distributed along the
line's length, and would exist even if the dielectric were perfect
(infinite parallel resistance) and the wires superconducting (zero
series resistance).
- Velocity factor
is a fractional value relating a transmission line's propagation speed
to the speed of light in a vacuum. Values range between 0.66 and 0.80
for typical two-wire lines and coaxial cables. For any cable type, it
is equal to the reciprocal (1/x) of the square root of the relative
permittivity of the cable's insulation.
Finite-length transmission lines
A transmission line of infinite length is
an interesting abstraction, but physically impossible. All transmission
lines have some finite length, and as such do not behave precisely the
same as an infinite line. If that piece of 50 Ω "RG-58/U" cable I
measured with an ohmmeter years ago had been infinitely long, I actually
would have been able to measure 50 Ω worth of resistance between the
inner and outer conductors. But it was not infinite in length, and so it
measured as "open" (infinite resistance).
Nonetheless, the characteristic impedance
rating of a transmission line is important even when dealing with
limited lengths. An older term for characteristic impedance, which I
like for its descriptive value, is surge impedance. If a
transient voltage (a "surge") is applied to the end of a transmission
line, the line will draw a current proportional to the surge voltage
magnitude divided by the line's surge impedance (I=E/Z). This simple,
Ohm's Law relationship between current and voltage will hold true for a
limited period of time, but not indefinitely.
If the end of a transmission line is
open-circuited -- that is, left unconnected -- the current "wave"
propagating down the line's length will have to stop at the end, since
electrons cannot flow where there is no continuing path. This abrupt
cessation of current at the line's end causes a "pile-up" to occur along
the length of the transmission line, as the electrons successively find
no place to go. Imagine a train traveling down the track with slack
between the rail car couplings: if the lead car suddenly crashes into an
immovable barricade, it will come to a stop, causing the one behind it
to come to a stop as soon as the first coupling slack is taken up, which
causes the next rail car to stop as soon as the next coupling's slack is
taken up, and so on until the last rail car stops. The train does not
come to a halt together, but rather in sequence from first car to last:
A signal propagating from the source-end
of a transmission line to the load-end is called an incident wave.
The propagation of a signal from load-end to source-end (such as what
happened in this example with current encountering the end of an
open-circuited transmission line) is called a reflected wave.
When this electron "pile-up" propagates
back to the battery, current at the battery ceases, and the line acts as
a simple open circuit. All this happens very quickly for transmission
lines of reasonable length, and so an ohmmeter measurement of the line
never reveals the brief time period where the line actually behaves as a
resistor. For a mile-long cable with a velocity factor of 0.66 (signal
propagation velocity is 66% of light speed, or 122,760 miles per
second), it takes only 1/122,760 of a second (8.146 microseconds) for a
signal to travel from one end to the other. For the current signal to
reach the line's end and "reflect" back to the source, the round-trip
time is twice this figure, or 16.292 µs.
High-speed measurement instruments are
able to detect this transit time from source to line-end and back to
source again, and may be used for the purpose of determining a cable's
length. This technique may also be used for determining the presence
and location of a break in one or both of the cable's conductors,
since a current will "reflect" off the wire break just as it will off
the end of an open-circuited cable. Instruments designed for such
purposes are called time-domain reflectometers (TDRs). The basic
principle is identical to that of sonar range-finding: generating a
sound pulse and measuring the time it takes for the echo to return.
A similar phenomenon takes place if the
end of a transmission line is short-circuited: when the voltage
wave-front reaches the end of the line, it is reflected back to the
source, because voltage cannot exist between two electrically common
points. When this reflected wave reaches the source, the source sees the
entire transmission line as a short-circuit. Again, this happens as
quickly as the signal can propagate round-trip down and up the
transmission line at whatever velocity allowed by the dielectric
material between the line's conductors.
A simple experiment illustrates the
phenomenon of wave reflection in transmission lines. Take a length of
rope by one end and "whip" it with a rapid up-and-down motion of the
wrist. A wave may be seen traveling down the rope's length until it
dissipates entirely due to friction:
This is analogous to a long transmission
line with internal loss: the signal steadily grows weaker as it
propagates down the line's length, never reflecting back to the source.
However, if the far end of the rope is secured to a solid object at a
point prior to the incident wave's total dissipation, a second wave will
be reflected back to your hand:
Usually, the purpose of a transmission
line is to convey electrical energy from one point to another. Even if
the signals are intended for information only, and not to power some
significant load device, the ideal situation would be for all of the
original signal energy to travel from the source to the load, and then
be completely absorbed or dissipated by the load for maximum
signal-to-noise ratio. Thus, "loss" along the length of a transmission
line is undesirable, as are reflected waves, since reflected energy is
energy not delivered to the end device.
Reflections may be eliminated from the
transmission line if the load's impedance exactly equals the
characteristic ("surge") impedance of the line. For example, a 50 Ω
coaxial cable that is either open-circuited or short-circuited will
reflect all of the incident energy back to the source. However, if a 50
Ω resistor is connected at the end of the cable, there will be no
reflected energy, all signal energy being dissipated by the resistor.
This makes perfect sense if we return to
our hypothetical, infinite-length transmission line example. A
transmission line of 50 Ω characteristic impedance and infinite length
behaves exactly like a 50 Ω resistance as measured from one end. If we
cut this line to some finite length, it will behave as a 50 Ω resistor
to a constant source of DC voltage for a brief time, but then behave
like an open- or a short-circuit, depending on what condition we leave
the cut end of the line: open or shorted. However, if we terminate
the line with a 50 Ω resistor, the line will once again behave as a 50 Ω
resistor, indefinitely: the same as if it were of infinite length again:
In essence, a terminating resistor
matching the natural impedance of the transmission line makes the line
"appear" infinitely long from the perspective of the source, because a
resistor has the ability to eternally dissipate energy in the same way a
transmission line of infinite length is able to eternally absorb energy.
Reflected waves will also manifest if the
terminating resistance isn't precisely equal to the characteristic
impedance of the transmission line, not just if the line is left
unconnected (open) or jumpered (shorted). Though the energy reflection
will not be total with a terminating impedance of slight mismatch, it
will be partial. This happens whether or not the terminating resistance
is greater or less than the line's characteristic
impedance.
Re-reflections of a reflected wave may
also occur at the source end of a transmission line, if the
source's internal impedance (Thevenin equivalent impedance) is not
exactly equal to the line's characteristic impedance. A reflected wave
returning back to the source will be dissipated entirely if the source
impedance matches the line's, but will be reflected back toward the line
end like another incident wave, at least partially, if the source
impedance does not match the line. This type of reflection may be
particularly troublesome, as it makes it appear that the source has
transmitted another pulse.
- REVIEW:
- Characteristic impedance is also known
as surge impedance, due to the temporarily resistive behavior
of any length transmission line.
- A finite-length transmission line will
appear to a DC voltage source as a constant resistance for some short
time, then as whatever impedance the line is terminated with.
Therefore, an open-ended cable simply reads "open" when measured with
an ohmmeter, and "shorted" when its end is short-circuited.
- A transient ("surge") signal applied
to one end of an open-ended or short-circuited transmission line will
"reflect" off the far end of the line as a secondary wave. A signal
traveling on a transmission line from source to load is called an
incident wave; a signal "bounced" off the end of a transmission
line, traveling from load to source, is called a reflected wave.
- Reflected waves will also appear in
transmission lines terminated by resistors not precisely matching the
characteristic impedance.
- A finite-length transmission line may
be made to appear infinite in length if terminated by a resistor of
equal value to the line's characteristic impedance. This eliminates
all signal reflections.
- A reflected wave may become
re-reflected off the source-end of a transmission line if the source's
internal impedance does not match the line's characteristic impedance.
This re-reflected wave will appear, of course, like another pulse
signal transmitted from the source.
"Long" and "short" transmission lines
In DC and low-frequency AC circuits, the
characteristic impedance of parallel wires is usually ignored. This
includes the use of coaxial cables in instrument circuits, often
employed to protect weak voltage signals from being corrupted by induced
"noise" caused by stray electric and magnetic fields. This is due to the
relatively short timespans in which reflections take place in the line,
as compared to the period of the waveforms or pulses of the significant
signals in the circuit. As we saw in the last section, if a transmission
line is connected to a DC voltage source, it will behave as a resistor
equal in value to the line's characteristic impedance only for as long
as it takes the incident pulse to reach the end of the line and return
as a reflected pulse, back to the source. After that time (a brief
16.292 µs for the mile-long coaxial cable of the last example), the
source "sees" only the terminating impedance, whatever that may be.
If the circuit in question handles
low-frequency AC power, such short time delays introduced by a
transmission line between when the AC source outputs a voltage peak and
when the source "sees" that peak loaded by the terminating impedance
(round-trip time for the incident wave to reach the line's end and
reflect back to the source) are of little consequence. Even though we
know that signal magnitudes along the line's length are not equal at any
given time due to signal propagation at (nearly) the speed of light, the
actual phase difference between start-of-line and end-of-line signals is
negligible, because line-length propagations occur within a very small
fraction of the AC waveform's period. For all practical purposes, we can
say that voltage along all respective points on a low-frequency,
two-conductor line are equal and in-phase with each other at any given
point in time.
In these cases, we can say that the
transmission lines in question are electrically short, because
their propagation effects are much quicker than the periods of the
conducted signals. By contrast, an electrically long line is one
where the propagation time is a large fraction or even a multiple of the
signal period. A "long" line is generally considered to be one where the
source's signal waveform completes at least a quarter-cycle (90o
of "rotation") before the incident signal reaches line's end. Up until
this chapter in the Lessons In Electric Circuits book series, all
connecting lines were assumed to be electrically short.
To put this into perspective, we need to
express the distance traveled by a voltage or current signal along a
transmission line in relation to its source frequency. An AC waveform
with a frequency of 60 Hz completes one cycle in 16.66 ms. At light
speed (186,000 m/s), this equates to a distance of 3100 miles that a
voltage or current signal will propagate in that time. If the velocity
factor of the transmission line is less than 1, the propagation velocity
will be less than 186,000 miles per second, and the distance less by the
same factor. But even if we used the coaxial cable's velocity factor
from the last example (0.66), the distance is still a very long 2046
miles! Whatever distance we calculate for a given frequency is called
the wavelength of the signal.
A simple formula for calculating
wavelength is as follows:
The lower-case Greek letter "lambda" (λ)
represents wavelength, in whatever unit of length used in the velocity
figure (if miles per second, then wavelength in miles; if meters per
second, then wavelength in meters). Velocity of propagation is usually
the speed of light when calculating signal wavelength in open air or in
a vacuum, but will be less if the transmission line has a velocity
factor less than 1.
If a "long" line is considered to be one
at least 1/4 wavelength in length, you can see why all connecting lines
in the circuits discussed thusfar have been assumed "short." For a 60 Hz
AC power system, power lines would have to exceed 775 miles in length
before the effects of propagation time became significant. Cables
connecting an audio amplifier to speakers would have to be over 4.65
miles in length before line reflections would significantly impact a 10
kHz audio signal!
When dealing with radio-frequency
systems, though, transmission line length is far from trivial. Consider
a 100 MHz radio signal: its wavelength is a mere 9.8202 feet, even at
the full propagation velocity of light (186,000 m/s). A transmission
line carrying this signal would not have to be more than about 2-1/2
feet in length to be considered "long!" With a cable velocity factor of
0.66, this critical length shrinks to 1.62 feet.
When an electrical source is connected to
a load via a "short" transmission line, the load's impedance dominates
the circuit. This is to say, when the line is short, its own
characteristic impedance is of little consequence to the circuit's
behavior. We see this when testing a coaxial cable with an ohmmeter: the
cable reads "open" from center conductor to outer conductor if the cable
end is left unterminated. Though the line acts as a resistor for a very
brief period of time after the meter is connected (about 50 Ω for an
RG-58/U cable), it immediately thereafter behaves as a simple "open
circuit:" the impedance of the line's open end. Since the combined
response time of an ohmmeter and the human being using it greatly
exceeds the round-trip propagation time up and down the cable, it is
"electrically short" for this application, and we only register the
terminating (load) impedance. It is the extreme speed of the propagated
signal that makes us unable to detect the cable's 50 Ω transient
impedance with an ohmmeter.
If we use a coaxial cable to conduct a DC
voltage or current to a load, and no component in the circuit is capable
of measuring or responding quickly enough to "notice" a reflected wave,
the cable is considered "electrically short" and its impedance is
irrelevant to circuit function. Note how the electrical "shortness" of a
cable is relative to the application: in a DC circuit where voltage and
current values change slowly, nearly any physical length of cable would
be considered "short" from the standpoint of characteristic impedance
and reflected waves. Taking the same length of cable, though, and using
it to conduct a high-frequency AC signal could result in a vastly
different assessment of that cable's "shortness!"
When a source is connected to a load via
a "long" transmission line, the line's own characteristic impedance
dominates over load impedance in determining circuit behavior. In other
words, an electrically "long" line acts as the principal component in
the circuit, its own characteristics overshadowing the load's. With a
source connected to one end of the cable and a load to the other,
current drawn from the source is a function primarily of the line and
not the load. This is increasingly true the longer the transmission line
is. Consider our hypothetical 50 Ω cable of infinite length, surely the
ultimate example of a "long" transmission line: no matter what kind of
load we connect to one end of this line, the source (connected to the
other end) will only see 50 Ω of impedance, because the line's infinite
length prevents the signal from ever reaching the end where the
load is connected. In this scenario, line impedance exclusively defines
circuit behavior, rendering the load completely irrelevant.
The most effective way to minimize the
impact of transmission line length on circuit behavior is to match the
line's characteristic impedance to the load impedance. If the load
impedance is equal to the line impedance, then any signal source
connected to the other end of the line will "see" the exact same
impedance, and will have the exact same amount of current drawn from it,
regardless of line length. In this condition of perfect impedance
matching, line length only affects the amount of time delay from signal
departure at the source to signal arrival at the load. However, perfect
matching of line and load impedances is not always practical or
possible.
The next section discusses the effects of
"long" transmission lines, especially when line length happens to match
specific fractions or multiples of signal wavelength.
- REVIEW:
- Coaxial cabling is sometimes used in
DC and low-frequency AC circuits as well as in high-frequency
circuits, for the excellent immunity to induced "noise" that it
provides for signals.
- When the period of a transmitted
voltage or current signal greatly exceeds the propagation time for a
transmission line, the line is considered electrically short.
Conversely, when the propagation time is a large fraction or multiple
of the signal's period, the line is considered electrically long.
- A signal's wavelength is the
physical distance it will propagate in the timespan of one period.
Wavelength is calculated by the formula λ=v/f, where "λ" is the
wavelength, "v" is the propagation velocity, and "f" is the signal
frequency.
- A rule-of-thumb for transmission line
"shortness" is that the line must be at least 1/4 wavelength before it
is considered "long."
- In a circuit with a "short" line, the
terminating (load) impedance dominates circuit behavior. The source
effectively sees nothing but the load's impedance, barring any
resistive losses in the transmission line.
- In a circuit with a "long" line, the
line's own characteristic impedance dominates circuit behavior. The
ultimate example of this is a transmission line of infinite length:
since the signal will never reach the load impedance, the
source only "sees" the cable's characteristic impedance.
- When a transmission line is terminated
by a load precisely matching its impedance, there are no reflected
waves and thus no problems with line length.
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.
Impedance transformation
Standing waves at the resonant frequency
points of an open- or short-circuited transmission line produce unusual
effects. When the signal frequency is such that exactly 1/2 wave or some
multiple thereof matches the line's length, the source "sees" the load
impedance as it is. The following pair of illustrations shows an
open-circuited line operating at 1/2 and 1 wavelength frequencies:
In either case, the line has voltage
antinodes at both ends, and current nodes at both ends. That is to say,
there is maximum voltage and minimum current at either end of the line,
which corresponds to the condition of an open circuit. The fact that
this condition exists at both ends of the line tells us that the
line faithfully reproduces its terminating impedance at the source end,
so that the source "sees" an open circuit where it connects to the
transmission line, just as if it were directly open-circuited.
The same is true if the transmission line
is terminated by a short: at signal frequencies corresponding to 1/2
wavelength or some multiple thereof, the source "sees" a short circuit,
with minimum voltage and maximum current present at the connection
points between source and transmission line:
However, if the signal frequency is such
that the line resonates at 1/4 wavelength or some multiple
thereof, the source will "see" the exact opposite of the termination
impedance. That is, if the line is open-circuited, the source will "see"
a short-circuit at the point where it connects to the line; and if the
line is short-circuited, the source will "see" an open circuit:
Line open-circuited; source
"sees" a short circuit:
Line short-circuited; source
"sees" an open circuit:
At these frequencies, the transmission
line is actually functioning as an impedance transformer,
transforming an infinite impedance into zero impedance, or visa-versa.
Of course, this only occurs at resonant points resulting in a standing
wave of 1/4 cycle (the line's fundamental, resonant frequency) or some
odd multiple (3/4, 5/4, 7/4, 9/4 . . .), but if the signal frequency is
known and unchanging, this phenomenon may be used to match otherwise
unmatched impedances to each other.
Take for instance the example circuit
from the last section where a 75 Ω source connects to a 75 Ω
transmission line, terminating in a 100 Ω load impedance. From the
numerical figures obtained via SPICE, let's determine what impedance the
source "sees" at its end of the transmission line at the line's resonant
frequencies:
A simple equation relates line impedance
(Z0), load impedance (Zload), and input impedance
(Zinput) for an unmatched transmission line operating at an
odd harmonic of its fundamental frequency:
One practical application of this
principle would be to match a 300 Ω load to a 75 Ω signal source at a
frequency of 50 MHz. All we need to do is calculate the proper
transmission line impedance (Z0), and length so that exactly
1/4 of a wave will "stand" on the line at a frequency of 50 MHz.
First, calculating the line impedance:
taking the 75 Ω we desire the source to "see" at the source-end of the
transmission line, and multiplying by the 300 Ω load resistance, we
obtain a figure of 22,500. Taking the square root of 22,500 yields 150 Ω
for a characteristic line impedance.
Now, to calculate the necessary line
length: assuming that our cable has a velocity factor of 0.85, and using
a speed-of-light figure of 186,000 miles per second, the velocity of
propagation will be 158,100 miles per second. Taking this velocity and
dividing by the signal frequency gives us a wavelength of 0.003162
miles, or 16.695 feet. Since we only need one-quarter of this length for
the cable to support a quarter-wave, the requisite cable length is
4.1738 feet.
Here is a schematic diagram for the
circuit, showing node numbers for the SPICE analysis we're about to run:
We can specify the cable length in SPICE
in terms of time delay from beginning to end. Since the frequency is 50
MHz, the signal period will be the reciprocal of that, or 20 nano-seconds
(20 ns). One-quarter of that time (5 ns) will be the time delay of a
transmission line one-quarter wavelength long:
Transmission line
v1 1 0 ac 1 sin
rsource 1 2 75
t1 2 0 3 0 z0=150 td=5n
rload 3 0 300
.ac lin 1 50meg 50meg
.print ac v(1,2) v(1) v(2) v(3)
.end
freq v(1,2) v(1) v(2) v(3)
5.000E+07 5.000E-01 1.000E+00 5.000E-01 1.000E+00
At a frequency of 50 MHz, our
1-volt signal source drops half of its voltage across the series 75 Ω
impedance (v(1,2))
and the other half of its voltage across the input terminals of the
transmission line (v(2)).
This means the source "thinks" it is powering a 75 Ω load. The actual
load impedance, however, receives a full 1 volt, as indicated by the
1.000 figure at v(3).
With 0.5 volt dropped across 75 Ω, the source is dissipating 3.333 mW of
power: the same as dissipated by 1 volt across the 300 Ω load,
indicating a perfect match of impedance, according to the Maximum Power
Transfer Theorem. The 1/4-wavelength, 150 Ω, transmission line segment
has successfully matched the 300 Ω load to the 75 Ω source.
Bear in mind, of course, that this only
works for 50 MHz and its odd-numbered harmonics. For any other signal
frequency to receive the same benefit of matched impedances, the 150 Ω
line would have to lengthened or shortened accordingly so that it was
exactly 1/4 wavelength long.
Strangely enough, the exact same line can
also match a 75 Ω load to a 300 Ω source, demonstrating how this
phenomenon of impedance transformation is fundamentally different in
principle from that of a conventional, two-winding transformer:
Transmission line
v1 1 0 ac 1 sin
rsource 1 2 300
t1 2 0 3 0 z0=150 td=5n
rload 3 0 75
.ac lin 1 50meg 50meg
.print ac v(1,2) v(1) v(2) v(3)
.end
freq v(1,2) v(1) v(2) v(3)
5.000E+07 5.000E-01 1.000E+00 5.000E-01 2.500E-01
Here, we see the 1-volt source
voltage equally split between the 300 Ω source impedance (v(1,2))
and the line's input (v(2)),
indicating that the load "appears" as a 300 Ω impedance from the
source's perspective where it connects to the transmission line. This
0.5 volt drop across the source's 300 Ω internal impedance yields a
power figure of 833.33 µW, the same as the 0.25 volts across the 75 Ω
load, as indicated by voltage figure v(3).
Once again, the impedance values of source and load have been matched by
the transmission line segment.
This technique of impedance matching is
often used to match the differing impedance values of transmission line
and antenna in radio transmitter systems, because the transmitter's
frequency is generally well-known and unchanging. The use of an
impedance "transformer" 1/4 wavelength in length provides impedance
matching using the shortest conductor length possible.
- REVIEW:
- A transmission line with standing
waves may be used to match different impedance values if operated at
the correct frequency(ies).
- When operated at a frequency
corresponding to a standing wave of 1/4-wavelength along the
transmission line, the line's characteristic impedance necessary for
impedance transformation must be equal to the square root of the
product of the source's impedance and the load's impedance.
Waveguides
A waveguide is a special form of
transmission line consisting of a hollow, metal tube. The tube wall
provides distributed inductance, while the empty space between the tube
walls provide distributed capacitance:
Waveguides are practical only for signals
of extremely high frequency, where the wavelength approaches the
cross-sectional dimensions of the waveguide. Below such frequencies,
waveguides are useless as electrical transmission lines.
When functioning as transmission lines,
though, waveguides are considerably simpler than two-conductor cables --
especially coaxial cables -- in their manufacture and maintenance. With
only a single conductor (the waveguide's "shell"), there are no concerns
with proper conductor-to-conductor spacing, or of the consistency of the
dielectric material, since the only dielectric in a waveguide is air.
Moisture is not as severe a problem in waveguides as it is within
coaxial cables, either, and so waveguides are often spared the necessity
of gas "filling."
Waveguides may be thought of as conduits
for electromagnetic energy, the waveguide itself acting as nothing more
than a "director" of the energy rather than as a signal conductor in the
normal sense of the word. In a sense, all transmission lines function as
conduits of electromagnetic energy when transporting pulses or
high-frequency waves, directing the waves as the banks of a river direct
a tidal wave. However, because waveguides are single-conductor elements,
the propagation of electrical energy down a waveguide is of a very
different nature than the propagation of electrical energy down a
two-conductor transmission line.
All electromagnetic waves consist of
electric and magnetic fields propagating in the same direction of
travel, but perpendicular to each other. Along the length of a normal
transmission line, both electric and magnetic fields are perpendicular
(transverse) to the direction of wave travel. This is known as the
principal mode, or TEM (Transverse Electric and
Magnetic) mode. This mode of wave propagation can exist only
where there are two conductors, and it is the dominant mode of wave
propagation where the cross-sectional dimensions of the transmission
line are small compared to the wavelength of the signal.
At microwave signal frequencies
(between 100 MHz and 300 GHz), two-conductor transmission lines of any
substantial length operating in standard TEM mode become impractical.
Lines small enough in cross-sectional dimension to maintain TEM mode
signal propagation for microwave signals tend to have low voltage
ratings, and suffer from large, parasitic power losses due to conductor
"skin" and dielectric effects. Fortunately, though, at these short
wavelengths there exist other modes of propagation that are not as "lossy,"
if a conductive tube is used rather than two parallel conductors. It is
at these high frequencies that waveguides become practical.
When an electromagnetic wave propagates
down a hollow tube, only one of the fields -- either electric or
magnetic -- will actually be transverse to the wave's direction of
travel. The other field will "loop" longitudinally to the direction of
travel, but still be perpendicular to the other field. Whichever field
remains transverse to the direction of travel determines whether the
wave propagates in TE mode (Transverse Electric) or
TM (Transverse Magnetic) mode.
Many variations of each mode exist for a
given waveguide, and a full discussion of this is subject well beyond
the scope of this book.
Signals are typically introduced to and
extracted from waveguides by means of small antenna-like coupling
devices inserted into the waveguide. Sometimes these coupling elements
take the form of a dipole, which is nothing more than two open-ended
stub wires of appropriate length. Other times, the coupler is a single
stub (a half-dipole, similar in principle to a "whip" antenna, 1/4λ in
physical length), or a short loop of wire terminated on the inside
surface of the waveguide:
In some cases, such as a class of vacuum
tube devices called inductive output tubes (the so-called
klystron tube falls into this category), a "cavity" formed of
conductive material may intercept electromagnetic energy from a
modulated beam of electrons, having no contact with the beam itself:
Just as transmission lines are able to
function as resonant elements in a circuit, especially when terminated
by a short-circuit or an open-circuit, a dead-ended waveguide may also
resonate at particular frequencies. When used as such, the device is
called a cavity resonator. Inductive output tubes use toroid-shaped
cavity resonators to maximize the power transfer efficiency between the
electron beam and the output cable.
A cavity's resonant frequency may be
altered by changing its physical dimensions. To this end, cavities with
movable plates, screws, and other mechanical elements for tuning are
manufactured to provide coarse resonant frequency adjustment.
If a resonant cavity is made open on one
end, it functions as a unidirectional antenna. The following photograph
shows a home-made waveguide formed from a tin can, used as an antenna
for a 2.4 GHz signal in an "802.11b" computer communication network. The
coupling element is a quarter-wave stub: nothing more than a piece of
solid copper wire about 1-1/4 inches in length extending from the center
of a coaxial cable connector penetrating the side of the can:
A few more tin-can antennae may be seen
in the background, one of them a "Pringles" potato chip can. Although
this can is of cardboard (paper) construction, its metallic inner lining
provides the necessary conductivity to function as a waveguide. Some of
the cans in the background still have their plastic lids in place. The
plastic, being nonconductive, does not interfere with the RF signal, but
functions as a physical barrier to prevent rain, snow, dust, and other
physical contaminants from entering the waveguide. "Real" waveguide
antennae use similar barriers to physically enclose the tube, yet allow
electromagnetic energy to pass unimpeded.
- REVIEW:
- Waveguides
are metal tubes functioning as "conduits" for carrying electromagnetic
waves. They are practical only for signals of extremely high
frequency, where the signal wavelength approaches the cross-sectional
dimensions of the waveguide.
- Wave propagation through a waveguide
may be classified into two broad categories: TE (Transverse
Electric), or TM (Transverse Magnetic), depending on which
field (electric or magnetic) is perpendicular (transverse) to the
direction of wave travel. Wave travel along a standard, two-conductor
transmission line is of the TEM (Transverse Electric and
Magnetic) mode, where both fields are oriented perpendicular to the
direction of travel. TEM mode is only possible with two conductors and
cannot exist in a waveguide.
- A dead-ended waveguide serving as a
resonant element in a microwave circuit is called a cavity
resonator.
- A cavity resonator with an open end
functions as a unidirectional antenna, sending or receiving RF energy
to/from the direction of the open end.
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