[TenTec] Tennessee Dreamin'

[James Duffer]

Thanks Ken, I think I was confusing the issue for Mike.  Your explanation 
was lucid.


>From: "Mike Hyder  -N4NT-" <mike_n4nt at charter.net>
>Reply-To: tentec at contesting.com
>To: <tentec at contesting.com>
>Subject: Re: [TenTec] Tennessee Dreamin'
>Date: Tue, 3 Aug 2004 01:41:58 -0400
>
>Thanks, Ken--
>
>This one I understand.  Thank you for your patience and your explanation.
>
>73, Mike N4NT
>
>----- Original Message -----
>From: "Ken Brown" <ken.d.brown at verizon.net>
>To: <tentec at contesting.com>
>Sent: Tuesday, August 03, 2004 1:25 AM
>Subject: Re: [TenTec] Tennessee Dreamin'
>
>
>Mike Hyder -N4NT- wrote:
>
> >Since we measure signal strength in volts and power in watts and since
>watts
> >vary with the square of the voltage, are your 6dB per S-unit dB of 
>voltage
> >difference or of power difference?  In other words, if I double my power 
>am
> >I quadrupling my volts?  Inquiring minds want to know.
> >
> >
> >
>Mike,
>
>If the impedance remains the same doubling voltage quadruples power.
>Either way it is a 6 dB increase. That explains why there are two
>formulas for dB change, one for power ratios and one for voltage ratios:
>10 log P1/P2 and 20 log V1/V2. We measure signal strength in a lot of
>different ways. Probably a better way would be dBm, which is decibles
>relative to 1 milliwatt. This is they way receiver sensitivities are
>usually expressed, these days. Sometimes you will see specifications in
>microvolts and dBm. Even when we are using 50 uV as our point of
>reference (when S9 = 50 uV) the ratios relative to that reference point
>are still in dB or S units, from which either power or voltage ratios
>can easily be derived. S units are generally agreed to equal 6 dB. A one
>S unit change would be a power ratio of 4 (or 1/4) and that would be a
>voltage ratio of 2 (or 1/2). Either way it is 6 dB.
>
>DE N6KB
>
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