I still am not sure if I've learned anything. Maybe I have.
Jon's somewhat rhetorical question about how come a quarter-wave
transformer works brought back why I initially entered into this
discussion in the first place. Years ago, I had personal success at
matching funky loads (phased wire arrays) using a noise bridge, a Smith
Chart and a hunk of 75 ohm coax (series section transformer, actually).
Worked great, nice matches - 50ohm SWR meter said 1:1 or thereabouts.
I had successfully created systems whose inputs I figured were (close
to) 50 ohms because the VSWR (as measured) at the inputs were good.
So I posted:
(Once again, ignore line losses, common mode crap, bumps and other
imperfections, please!)
************
In a system where (at the far end) the load (antenna) and
cable have different impedances, the impedance seen looking into the
near end of the cable will change depending on cable length. Correct?
Therefore, does that not imply that the SWR at, say, the amp/cable
interface (50 ohm source, changing load), also change depending on the
length of said cable?
************
Several answers came back like this:
************
No, that's not correct. That is the same as saying that SWR varies
along
the length of the cable. SWR is constant along the cable regardless of
length.
************
True or false?
CASE 1. If you assume the source, cable, and SWR meter have matched Z
(load Z is different), this is true; any forward wave on the cable
creates standing waves that are unaffected by line length. Sticking the
SWR meter at the output of the source (input of the cable) and messing
with the cable length should show that. This is because the load gets
transformed 'round the VSWR circle to an impedance that is still
creating the same mismatch regardless of where on the circle it is.
(The reflected waves, once created are already in a cable (of Z-ohms),
so when they see the Z-ohm source, they are "absorbed" completely.)
Any mistakes yet? :)
CASE 2. My experience with the phased array was a situation where the
cable and SWR bridge were not matched and things change a bit.
Rich's example: 50-ohm load, 93 ohm 1/2 wave coax, 50-ohm SWR meter.
Here, if the SWR meter were 93 ohms, then it would fit CASE 1 regardless
of line length (system is 93ohms with a mismatched load) and it would
read 1.86:1. OK so far?
But, if a 50ohm SWR meter is used instead as per Rich's example, what
happens? Does anyone think it would read the same as a 93ohm SWR meter?
I don't, so...
This got me thinking - what would happen to the standing waves in the
93ohm cable created by the mismatch at the (50ohm) load end? They must
exist, right? Any forward waves in the 93ohm coax will dump some power
into the 50ohm load and there will be some (smaller magnitude)
reflections from it. Now, any waves "traveling backwards" in the 93ohm
coax will again dump some power into the SWR meter end (the load in this
case) and again have smaller reflections and so on. Unless all these
reflections cancel each other out (I don't think they can) this tells me
that there are standing waves "on the cable" and therefore the SWR on it
is not 1:1.
However, this 50ohm SWR meter is not "on the cable". It's in its own
world, on the near side of the 50-to-93ohm impedance step. So it
doesn't see the standing waves that are on the 93ohm cable, does it?
Rich contends that the SWR meter will READ 1:1. Maybe it will. But
that is not the REAL SWR "on the cable"; it's just what the meter
reads.
Now, assuming I've gotten anything correct, what happens to the
backwards-traveling reflections from the 50 ohm load that make it
through this same impedance step? They should cause some standing waves
in the 50ohm system, where the SWR meter is, no? And...
73
Mike
N2MG
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