Re: [TenTec] Tennessee Dreamin'

["James Duffer" <dufferjames@hotmail.com>]

You didn't mention any change in impedance so assuming an impedance of 10 
Ohms, a voltage of 10 volts would then result of power of 10 watts.  Now 
doubling the power to 20 watts, and assuming no change in resistance the 
resulting voltage would be 14.14 volts so the answer to your question, if it 
was a question, is no, the voltage did not quadruple, it only increased by 
the square route of the increase in power ratio.
powe ratios result in a 3 dB change for doubleing, voltage ratios (assuming 
equal impedance) would be 6 dB for a doubling.  Now when power is doubled 
measured at the 50 ohm output, what would be the change field in strength 
(volts/meter) measured at the impedance of free space (377 Ohms) My guess 
would be 3 dB. What is your guess?

> From: "Mike Hyder  -N4NT-" <mike_n4nt@charter.net>
> Reply-To: tentec@contesting.com
> To: <tentec@contesting.com>
> Subject: Re: [TenTec] Tennessee Dreamin'
> Date: Mon, 2 Aug 2004 19:47:08 -0400
>
> 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 N4NT
>
> ____________________________
>
> It's even better than that. One S unit is 6db so 25 watts is one S unit 
> less
> than 100 watts.  So an S9 signal at 100 watts is just a frog's hair under 
> S8
> at 20 watts.
>
> Carl Moreschi N4PY
> Franklinton, NC
>
> _______________________________________________
> TenTec mailing list
> TenTec@contesting.com
> http://lists.contesting.com/mailman/listinfo/tentec