 RMS Value  RF Cafe Forums 

Because of the high maintenance needed to monitor and filter spammers from the RF Cafe Forums, I decided that it would
be best to just archive the pages to make all the good information posted in the past available for review. It is unfortunate
that the scumbags of the world ruin an otherwise useful venue for people wanting to exchanged useful ideas and views.
It seems that the more formal social media like Facebook pretty much dominate this kind of venue anymore anyway, so if
you would like to post something on RF Cafe's
Facebook page, please do.
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.
Top
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.
Top
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



