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Below are all of the forum threads, including all
the responses to the original posts.
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
Top
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.
Top
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