And I fear that you UNDERcomplicated them a little. Jetfixr properly
calculated, using the formula

dB = 10 log P1/P2 = 10 log 28/7 = 10 (0.60) = 6 dB

that the gain in transmitter power was 6 dB.

What he failed to calculate was the concomitant voltage gain for 6 dB. If

dB = 20 log V1/V2, then V1/V2 = 10 ^ (dB/20) = 10 ^ (0.3) = 2.0

So a 6 dB gain in power gives you a 2:1 gain in voltage.

Jim


"B.R. Smith" wrote in message
...
On 9 Jan 2005 14:06:22 -0800, "jetfixr" wrote:

I recently took a test and I failed to understand how the answered was
derived.. Heres the question:

A malfunctioning transmitter designed to transmit at 28 watts of power
has a power output of 7 watts. In its current state, its signal is
being received by a base station at 5uV. If the transmitter were to be
repaired and had its proper output of 28 watts, what would the receive
signal signal at the base station be?

I figured if the transmitter were repaired, that would be a improvement
of 6dB on the transmit side. At the receiver I would figure the signal
should be 20uV, but somehow the the correct answer was a receive signal
of 10uV. I have no idea how they derived 10 uV and no explaination was
given. Can someone help me out with the math here?

You are over complicating things.