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?

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 probably forgot that

power db = 10 * (P1/P2)

voltage db = 20 * (V1/V2)

which is way you are off by a factor of 2.

--
Jim Pennino

Remove -spam-sux to reply.

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. According to ohms law, whenever you
double the voltage, (and resistance remains the same) the current
must also double so your power quadruples. For example,

10 Volts / 5 Ohms = 2 Amps Power = 10 V x 2 A = 20 Watts

20 volts / 5 Ohms = 4 Amps Power = 20 V x 4 A = 80 Watts

So taking it in reverse, if you power quadruples from 7 Watts to 28
Watts, the voltage in you example doubles from 5 uV to 10 uV.

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.

Can someone help me out with the math here?

You probably forgot that
power db = 10 * (P1/P2)
voltage db = 20 * (V1/V2)

which is way you are off by a factor of 2.

Or to put it simply, if the voltage doubles, then the current also doubles
and since power is volts x current, and 2 x 2 = 4, then the power increase
is 4 x if the voltage increase is 2 x.

Hairy R. Beanz
(On second thoughts, forget that - probably never get away with it !:-)

And the really simple way to remember it:
2 times the power = 3 db.
2 times the voltage = 6 db.

73
Gary K4FMX

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?