Larry is correct. Mismatch loss does not necessarily involve a feed
line. Mismatch loss is the loss in dB of the power delivered to a load
that is not conjugately matched to the source compared to the power in
the load when there is a conjugate match. For example, a 2 volt voltage
source with a 50 ohm source impedance will deliver 20 mA of current to a
50 ohm load resulting in a load power of 20 mW. If the load is change
to 100 ohms, the current is now 13.3333 mA and the power in the load is
now 17.7777 mW. The ratio of those two powers is 10*LOG(17.7777/20)
which equals 0.512 dB. If the load is changed to 25 ohms, the mismatch
loss is exactly the same. Both of these cases represent a 2:1 SWR and
in this case no transmission line was present. This is also true for
all complex loads that fall on the 2:1 SWR circle on the Smith chart.
73 Tom W0IVJ
On 5/2/2012 10:45 AM, Jim Brown wrote:
> 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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