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 �W, 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.
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REVIEW:
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A transmission line with standing waves
may be used to match different impedance values if
operated at the correct frequency(ies).
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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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