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
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