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.
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