In message , You
writes
In article ,
Ian Jackson wrote:

In message , Roy Lewallen
writes
Sal M. Onella wrote:
I've heard the same thing, but the reality is that the S-meter
usually
follows the AGC and the signal for S9 could be different for every design.
But Flex radio and others define S9 as 50 uV. . .

Adjusting the S-meter to read a specific value for one signal strength
is simple, and there's often an adjustment for doing it. I think it's
fairly common to find S9 to be around 50 uV -- on one band at least --
although I wouldn't be surprised to see a fair amount of variation from
band to band. But that has nothing to do with what the difference is
between S7 and S8, S8 and S9, etc.

Isn't there a school of thought which says that, above 1000MHz, an
S-point is 3dB? [Don't ask me why!]

there are "Many Schools of thought".... but unless someone rewrites the
Laws of Physics, and increase in Tx Output will show and Increase of Rx
Input, on the same path, period.

I'm not sure what statement you are arguing against. Has anyone said
otherwise?

S-meters are NOT necessarily actually
showing Rx Receive Signal Strength

So what DO they 'show'? OK, they may not 'show' it very accurately, and
the way they work is usually a very indirect way of 'measuring' the
signal power or voltage at the receiver input, but I'm sure that this is
what the do 'show'.

For convenience, most S-meters use the AGC voltage to provide an
indication of incoming signal level. However, some FM-only CB sets have
no AGC, and have to pick off an input level dependent voltage from
somewhere else in the circuit. These are usually VERY inaccurate. What
alternative methods do you suggest?

, and they are rarely, either linear

No one is disagreeing. S-meters are renowned liars. Mind you, I have a
CB set (which has an AGC line) which is converted to 10m, and this has
an S-meter which has remarkably consistent 6dB S-points between S2 and
S9 +20dB.
,
or even Logrythmnic

No one's disagreeing.

in presentation of what they do represent. In the
Microwave Field if you measure a path, and also calculate that path,
comparing the two will show that they are very close, if you did it
right....

You are undoubtedly correct. But did anyone say that this was not the
case?
--
Ian