> I don't think the cable cares whether the source and load
> impedances are 50 ohms
> resistive. I think the cable is indifferent to whether the
> load and source values are
> resistive or whether they present a complex impedance
> involving +/- J. as long as the
> composite value looks like 50 ohms.
The cable only acts as a transformer if the *load* Z is not the same as the
cable's characteristic Z. It doesn't care about the source Z; the mismatch
that occurs at the source end only affects power transfer into the cable at
that point.
> The conventional wisdom generally expressed is that as long
> as the cavities are properly
> tuned, that the interconnect length from the TX is
> immaterial. I question that:
>
> Properly tuned? When what's properly tuned and for what
> parameter? Is the pass section of the cavity(s)
> being tuned for maximum output or is it tuned for minimum
> reflection back to the TX source?
Reflection (S11). Always always always tune pass or pass/reject cavity
filters for best match (pounding on the desk as I type). Too many
manufacturers' instructions say to tune for maximum power transfer or least
insertion loss, probably because they assume the field techs don't have
equipment for measuring return loss properly. That's just not good advice
in my book. If the filter is designed and built right, the insertion loss
minima and return loss maxima should conincide very closely. Even if
they're off a tad and you can't get them to line up perfectly, you're always
better off taking 10 dB more return loss in exchange for 0.1 dB of insertion
loss degradation. That primarily applies to our little narrowband two-way
radio world. In wideband applications, there are other things involved,
such as group delay, that come into play, but for what we're talking about,
return loss is the key.
> Only if the TX output impedance exactly matches the cavity
> impedance and the impedance of the interconnecting cable will
> the cavity tuning point be the same for either parameter. To
> assume that the
> TX output impedance is 50 ohms is optimistic and as you point
> out, altering the power level of the
> TX can affect TX output Z, the amount dependant on what TX
> stages are used to control TX output.
Let's straighten something out here before we get off track. Most
transmitters don't HAVE a 50 ohm source Z. They are designed to work INTO a
50 ohm Z. They have internal matching transformers (stripline or otherwise)
to convert the very-low-Z output of the bipolar transistors to something
approaching 50 ohms so that when it is connected to our external 50 ohm
world that the devices are able to transfer power.
> Considering how nit-picky forum members are about designing
> and building their systems,
> (and I mean that in the best sense of the word), it seems
> inconsistent to be indifferent to
> how the duplexers might be affected by inserting what is
> potentially a radical impedance
> transformer between the TX and the cavities. In the absence
> of any way to measure any
> source and load mismatch, using a 1/2 wave (or half wave
> repeating) cable length will at
> least keep any existing mismatch status quo. It won't improve
> the match but at least it won't
> increase a mismatch because the 1/2 wave length simply
> repeats the TX output Z and does
> not act as a line transformer. But as the cable length
> departs from a 1/2 wave and approaches
> a 1/4 wave, the game changes and a 1/4 wave interconnect
> between a mismatched source
> and load can produce some eye opening shifts in the impedance
> reflected to the load and
> back to the source.
Whether you have a half-wave or a quarter-wave cable terminated by a
mismatched load, the VSWR remains the same. As as a simple example, assume
the Z of the duplexer is 100 ohms. If you use a half-wave cable, the PA
sees 100 ohms, a 2:1 VSWR. If the cable is a quarter wave, it transforms
the 100 ohms to 25 ohms, again a 2:1 VSWR. Yes, the Z is not the same, but
the VSWR is. You don't know whether the PA will be better off looking into
the 25 ohm load versus the 100 ohm load, so why would you hold fast to the
half-wave rule?
For a given load Z, the VSWR remains constant no matter what cable length
you use. A 50 ohm cable can't transform a non-50 ohm load to 50 ohms; it
can only roll you around the Smith Chart at a constant VSWR, that being
something other than 1:1. Round and round the Smith Chart we go, where she
stops, nobody knows.
Point being, if there is a mismatch, using a half wave cable does nothing to
improve your chances of making your PA happy any more than would a quarter
wave cable or any other random length. Without knowing the actual
impedences involved, your odds of making an improvement using an X-length
cable (pick your favorite value for X) are 50/50, nothing more, nothing
less.
Also keep in mind that the transformation the cable does in the case of a
load mismatch is, for all practical purposes, random as you sweep across a
range of frequencies both due to its electrical length not being uniform
(the electrical length of the cable remains fixed, while the wavelength of
the signals vary), and also due to the varying load Z presented by the
duplexer changing with frequency. So again, if your PA has issues with the
off-channel reactances posed by your duplexer, cable lengths and Z-matchers
aren't going to give you a flat Z across a fairly wide range like an
isolator will.
--- Jeff
