

RMS Value  RF Cafe Forums

ljoseph Post subject: RMS Value Posted: Wed Apr 29, 2009 6:34
pm
Captain
Joined: Tue Sep 04, 2007 7:54 am
Posts: 13 Location: Dallas This is a simple one. Just
consider two sine waves each with 1V peak value. Assume 1ohm resistor.
The RMS value of each sine is 1/sqrt(2) and the average power is
0.5W.
Now, consider the signal which sum of the above sine
waves. This will be a sinewave with 2V paek and average power of
2W. But the individual signls have only 0.5W so the sum should be
1W.... where am I missing? Thanks, Leyo
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VSWR Post subject: Re: RMS ValuePosted: Wed Apr 29, 2009
11:29 pm
Captain
Joined: Sun Jan 11, 2004 4:44 pm
Posts: 8 Hi ljoseph
The power formula is as you know
P=V^2/R
So you see the power is a function of the square
of the voltage. A 2x factor in voltage yields a 4x (2^2) increase
in power. That's why you go from 1/2 W to 2 W (1/2 x 4 = 2).
You are thinking of the two independent 1 Vpk sinewaves across
two independent 1 ohm resistors. It might not seem intuitive that
impressing 2 Vpk across a single 1 ohm resistor would dissipate
2 W when individually (1 Vpk and 1 ohm, twice) they would dissipate
only 1 W total, but that's Ohm's law (and real life). The two scenarios
simply are not equivalent.
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ljoseph Post
subject: Re: RMS ValuePosted: Thu Apr 30, 2009 8:50 am
Captain
Joined: Tue Sep 04, 2007 7:54 am Posts: 13 Location:
Dallas Hi Thanks for the answer. One morer question on
this. I have a two signal generator set to 1Ghz and are connected
through a combiner to the power meter. Each of the signal generator
levels are adjusted so that each signal alone measures 0dBm on the
power meter. Now, I turn on both signal generators. As per your
analysis, the power meter should 6dBm.... it reads only 3dBm.
Looking forward for your comments. Thanks.
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VSWR Post subject: Re: RMS ValuePosted: Thu Apr 30, 2009
9:26 am
Captain
Joined: Sun Jan 11, 2004 4:44 pm
Posts: 8 Hi again ljoseph,
The 3 dB power increase is
correct when two equal amplitude, equal phase signals are added
linearly. It is due to the way decibels work for power and voltage.
For a factor of 2 in power with units of power:
10
log (2) = 3.01 dB
For a factor of 2 in power with units of
volts:
10 log (2^2) = 20 log (2) = 6.02 dB
In your
previous case from the first post, you were doubling the voltage
(two, 1 Vpk signals). In this one, you are doubling the power.
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Peter Raynald Post subject: Re: RMS ValuePosted:
Fri May 01, 2009 9:03 am
Captain
Joined: Tue
Sep 07, 2004 3:09 pm Posts: 11 When strictly looking at adding
voltage waveforms at a voltage summation point in a circuit, it
is right to assume that you will get 6dB increase in power if the
two waveforms are coherent.
But don't forget that your combiner
performs an impedance transformation, some voltage from each waveform
get's transfered to current by a squarerooth of 2 factor, and therefore
you fall again on 3dB increase.
Posted 11/12/2012



