Speaking of connector loss, the mismatch loss that a connector may provide
is only valid when the system generator is a fixed value and there is no
re-reflection of reflected power at the generator.
In a typical HF system we usually can readjust the transmitter/tuner
matching network to compensate for any mismatch so that reflected power gets
re-reflected to the antenna.
Also you can not just add together multiple connector losses in a system and
say that the sum is the total loss, as each has a different impedance
transformation that may or may not add directly to that of other losses.
73
Gary K4FMX
> -----Original Message-----
> From: amps-bounces@contesting.com [mailto:amps-bounces@contesting.com]
> On Behalf Of Jim Brown
> Sent: Wednesday, May 02, 2012 12:45 PM
> To: amps@contesting.com
> Subject: Re: [Amps] every db lost re Tubes vs. Solid State
>
> On 5/1/2012 2:05 PM, Larry Benko wrote:
> > Jim,
> >
> > I have made the exact measurement I mentioned below as an exercise.
> > Are you questioning the 2:1 SWR or the fact that a 2"1 SWR is a
> > mismatch loss of .55dB?
>
> Both. See the family of curves in the ARRL Handbook for excess
> attenuation due to mismatch, which has been in every edition of the
> handbook since I've been buying it (the 50s), and which I've confirmed
> by cranking the equations. The curve for a 2:1 SWR shows an excess
> attenuation of 0.1 dB if the matched loss is 0.4dB, 0.2dB if the matched
> loss is 1dB, 0.4dB if the matched loss is 10dB. And remember, the
> connector is quite small as a fraction of a wavelengh at HF, which is
> what this discussion is about, and not much more in the lower half of
> the VHF spectrum. If you've measured more than that, I suspect you are
> simply seeing the standing waves at the point of the measurement.
>
> Very small values of attenuation can be VERY difficult to measure.
> Several years ago, I tried to measure the attenuation of Commscope 3227
> from 1 MHz to 1 GHz Getting good data below a few MHz requires a VERY
> long sample -- even with 1,000 ft, there is some mismatch between the 50
> ohm resistive terminations of the HP gear and the complex Zo of the
> cable at those frequencies, which is NOT 50+j0.
>
> Another point. While we like to ASSUME that input stages have an ideal
> resistive impedance that matches the cable we're using, many do not.
> Ditto for output stages -- indeed, the Zo of most output stages is
> nowhere near an ideal match for the cable.
>
> 73, Jim K9YC
>
>
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