I hate to sound stupid, but at that risk here goes. When I was in school for psychology one course on IQ and testing said, "IQ is what an IQ test measures." I have transliterated that into, "SWR is w
Oh oh, now you started it Gene. :>) I will give the first answer. No measurable difference at the other end. 73 Gary K4FMX _______________________________________________ See: http://www.mscomputer.c
With a 2:1 mismatch at the far end the return loss is 10 dB. This is another way of saying that for every 1000 watts you send to the antenna, 100 watts is reflected back towards the transmitter. The
the receive end of that has a 2:1 has a measured antenna AND given dB. This is to the antenna, radiated power is gets reflected back. transmitters don't in the 2:1 case, <snip> Actually not. Reflect
I agree with Tom about reflected power not being dissipated in the transmitters final. For all practical purposes, it all gets returned to the antenna and radiated except for the small amount of loss
decrease when Losses are not evenly distributed between dielectric losses and conductor losses at HF. As a matter of fact dielectric losses are very small, almost all of the loss relates to I^2 R lo
Which is what the whole last part of my post was all about.. Hmmm... I haven't run the numbers, but I would think that this is only the case for lines <1/2 wavelength, and maybe <1/4 wavelength. Cert
<monster snip> For a tube amp, sure. And for a solid state amp with a tuner on the output. But not for a broadband SS amp, in general. While they're not 50 ohm resistive, they're also probably not to
-- Assuming the SWR is correct, you will lose 4% of the power transmitted. Say 40 watts at one KW out. 73 Ed _______________________________________________ See: http://www.mscomputer.com for "Self S
26.4 db at 1.2:1? Not to mention the 10 db. Ed _______________________________________________ See: http://www.mscomputer.com for "Self Supporting Towers", "Wireless Weather Stations", and lot's more
Regardless of the type of transmitter, the concept of "reflected power" and [it] "winds up 'inside' the TX" is a highly unprofitable way to look at this problem. A much easier way to think of it is:
Hi Ian, Nice to see you again! difference between go there... so It works perfectly, or as perfectly as I can measure, on my meters. calculate the suddenly requires that we need and loaded, etc previ
tuner on the output. not 50 ohm mismatch at the TX big mismatch, most dissipated (or fed that can do that). Not so Jim. As Ian and others have tried to point out the end result is just as if the loa
I'd rather say that "impedance at the transmission line" and "forward / reverse / circulating power" are just different ways of analyzing the same physical system. Sometimes one is more useful that t
hot. It typical SWR directions. Not really. The typical SWR bridge compares the voltage across the transmission line to current through the line. Every bridge we commonly use works that way. It does
Hi Tom! Nice to be back to this list. We're still commuting between two homes, in G and GM, and work levels and Internet access are causing some "QSB" on the newsgroups and mailing lists... I agree w
is back. OK, so I was inconsistent in my rounding and reading off the chart.. 10dB return loss is actually about 1.9:1 VSWR. 2:1 is more like 9.5 dB RL. ______________________________________________
with your don't RF voltage and get a vector form. current pushes a responding to RF power If you read my last response to Martin you'll see we agree even on that point!! I'm not saying the meters ac
Here is another way to look at it: With a circulator at the transmitter output and a load on the return port of the circulator, disconnect the antenna line from the circulator. Now all the power out
I hope there's more light than heat (forward or backward) in this discussion. Anyway, the common SWR meter samples forward and reverse power by adding a current sample and a voltage sample. This make