> So....If I happen to insert my Heath, Bird, or japanese SWR/Power meters
> into my transimssion line at a high impedance point, ,
> say 200 ohms or
> so, will the meter measure true SWR at the antenna, (minus the error
> created by the losses in the transmisson line) or will it indicate an swr
> roughly 200 divided by 50, or approximately 5:1? Can we use transmisson
> line transformer theory to match the feed point impedance a the
> transmitter to the transmitter output impedance?
>
> 73, de Pat AA6EG/N6IJ
Pat -
SWR is only a function of the line's characteristic impedance and the
load's impedance. If you had a lossline 50-ohm line and connected it to a
200-ohm load, the SWR is 4:1 - everywhere.
Assuming a 200-ohm purely resistive load - if you travel 1/8-wavelength
towards the transmitter, the complex impedance at that point in the line
will be about 180-ohms of pure inductive reactance. 1/8-wave further
(1/4-wave from the load), the impedance will be 12.5-ohms of pure
resistance. 1/8-wave further (3/8-wave from the load) and the impedance
will be about 180-ohms of pure capacitive reactance. 1/8-wave further
(1/2-wave from the load) and we're back to the original 200-ohms of pure
resistance. Notice that the SWR is 4:1 at all points and the impedance is
never 50 ohms of pure resistance.
If you look at a Smith Chart, this path describes a circle, centered on a
point which represents the characteristic impedance of the line. The
radius of the circle is constant and represents the SWR. Notice that the
path repeats around the Smith Chart once per half-wavelength. Notice also
that the central point, representing the line's characteristic impedance,
corresponds to an SWR of 1:1. The outer edge of the chart represents an
SWR of infinity.
Line losses cause the circle to spiral in towards the center point with
distance, gradually reducing SWR to 1:1. Note that a lossy line has no
effect on a perfectly matched load (except to reduce the power getting to
it). A lossy line will cause an antenna to look "broader" by flattening
its SWR curve. Imagine taking all of the SWR readings and reducing them
by a constant fraction...
So, to get back to your original question - the meter *should* read 4:1,
or 200/50, whereever you put it in the line, no matter how long the line
is, within reason. If you're only using 50-ohm line, you can't transform
200-ohms into 50-ohms through line length alone.
However, you could simulate the reactances of an L-network by adding open
or shorted stubs of the appropriate length, spaced an appropriate amount
on the line. Or, you could use lines of other characteristic impedances
(such as 75-ohms, 92-ohms, etc.) to shift the center of the circle such
that a certain length brings the path of the impedance in its travels
sufficiently close to 50-ohms of resistance.
The latter is the technique employed to transform a loop's driving-point
impedance to approx. 50-ohms by using a quarter-wave transformer of 75-ohm
or 92-ohm line. A G5RV also uses this technique - a section of 300-ohm
twinlead does the trick in this case. G5RV's genius was in finding a
combination of antenna length and 300-ohm line length that resulted in a
value of impedance at the terminals of acceptably close to 50-ohms on all
the popular ham bands. Its hard enough at one frequency - but SIX?? And
done by hand, too...
73, Ward N0AX
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