Hi to all, I would like to compute the so called RMS Power but it's not simple how seems to be. RMS= square root ( sum of square (signal[k])/N) (1) First thing is :some programs like Cool Edit calculate the RMS with a negative number ( or zero) and since the square root (and the square) are always greater or equal to zero I think I must change something here. The " unit " for RMS is decibel FS (the maximum value possible for RMS is 0 dB FS). My signal is a sine wave in the range -32767,32767 with theoric RMS equal to zero ( min,max and average RMS are zero,too). Cool Edit analysis obtains exactly these values for this signal. To compute the RMS by myself,I transform the signal normalizing it,that is for each sample k compute signal[k]=signal[k]/32767 and then taking the logarithm of the absolute values : signal[k] =20*log10(abs(signal[k]))-0.7458 (the last one is a reference costant). Now I apply the equation (1) but I'm not able to find what to subctrat to the calculate RMS to match Cool Edit's results . Someone know where I'm wrong?
RMS Power of audio digital signal
Started by ●February 25, 2004
Reply by ●February 25, 20042004-02-25
You should be doing equation (1) before you do the log conversion. The order should be 1) normalize signal, 2) take RMS, 3) convert to dB. CoolEdit's RMS numbers are in dB relative to full scale, calibrated for a sinusoidal input. If you create a full-scale sine wave, you will see an RMS value of 0dB (or very nearly so). If you create a full scale square wave, you will see a +3dBFS RMS. Because of this, you will need to add a correction factor based on the RMS value of a sine wave to match CoolEdit. I think just adding 3dB to your final result is all that is required. "Martin" <martindegaio@lycos.com> wrote in message news:OZ8%b.33204$FJ6.1232515@twister1.libero.it...> Hi to all, I would like to compute the so called RMS Power but it's not > simple > how seems to be. > RMS= square root ( sum of square (signal[k])/N) (1) > First thing is :some programs like Cool Edit calculate the RMS with a > negative number ( or zero) and since the square root (and the square)are> always greater or equal to zero > I think I must change something here. > The " unit " for RMS is decibel FS (the maximum value possible for RMS is0> dB FS). > My signal is a sine wave in the range -32767,32767 with theoric RMS equal > to zero > ( min,max and average RMS are zero,too). > Cool Edit analysis obtains exactly these values for this signal. > To compute the RMS by myself,I transform the signal > normalizing it,that is for each sample k compute > signal[k]=signal[k]/32767 > and then taking the logarithm of the absolute values : > signal[k] =20*log10(abs(signal[k]))-0.7458 (the last one is areference> costant). > Now I apply the equation (1) but I'm not able to find what to subctrat to > the calculate RMS to match Cool Edit's results . > Someone know where I'm wrong? > > >
Reply by ●February 25, 20042004-02-25
Yes! It works fine! Thank you for your help. "Jon Harris" <goldentully@hotmail.com> ha scritto nel messaggio news:c1j6mo$1jq139$1@ID-210375.news.uni-berlin.de...> You should be doing equation (1) before you do the log conversion. The > order should be 1) normalize signal, 2) take RMS, 3) convert to dB. > > CoolEdit's RMS numbers are in dB relative to full scale, calibrated for a > sinusoidal input. If you create a full-scale sine wave, you will see anRMS> value of 0dB (or very nearly so). If you create a full scale square wave, > you will see a +3dBFS RMS. Because of this, you will need to add a > correction factor based on the RMS value of a sine wave to match CoolEdit. > I think just adding 3dB to your final result is all that is required.
Reply by ●October 21, 20152015-10-21
what would be the most efficient way in terms of area to perform the above equation (1) - calculating RMS of signal in digital domain in FPGA? Seems to me like dividing with N and square root can take a lot of resources? At how many samples is this done? Last "n" samples or all samples as long as your program is running? --------------------------------------- Posted through http://www.DSPRelated.com
Reply by ●October 21, 20152015-10-21
there is no mathematical meaning to RMS power there is RMS voltage and RMS current and V RMS x I RMS x phase = AVERAGE power. RMS Power is a misnomer and not mathematically correct however it is often (mis) used Mark
Reply by ●October 21, 20152015-10-21
On Thursday, February 26, 2004 at 11:04:30 AM UTC+13, Martin wrote:> Hi to all, I would like to compute the so called RMS Power but it's not > simple > how seems to be. > RMS= square root ( sum of square (signal[k])/N) (1) > First thing is :some programs like Cool Edit calculate the RMS with a > negative number ( or zero) and since the square root (and the square) are > always greater or equal to zero > I think I must change something here. > The " unit " for RMS is decibel FS (the maximum value possible for RMS is 0 > dB FS). > My signal is a sine wave in the range -32767,32767 with theoric RMS equal > to zero > ( min,max and average RMS are zero,too). > Cool Edit analysis obtains exactly these values for this signal. > To compute the RMS by myself,I transform the signal > normalizing it,that is for each sample k compute > signal[k]=signal[k]/32767 > and then taking the logarithm of the absolute values : > signal[k] =20*log10(abs(signal[k]))-0.7458 (the last one is a reference > costant). > Now I apply the equation (1) but I'm not able to find what to subctrat to > the calculate RMS to match Cool Edit's results . > Someone know where I'm wrong?No. The RMS is the same as standard deviation when the mean is zero. So just calculate the sum of squares, divide by N and take the square root. That is once the numbers have been converted to amplitude levels of course.
Reply by ●October 21, 20152015-10-21
On Thursday, October 22, 2015 at 3:29:45 AM UTC+13, mako...@yahoo.com wrote:> there is no mathematical meaning to RMS power > > there is RMS voltage > and RMS current > > and V RMS x I RMS x phase = AVERAGE power. > > RMS Power is a misnomer and not mathematically correct > however it is often (mis) used > > MarkIndeed it is by hi-fi idiots who talk of peak power.
Reply by ●October 21, 20152015-10-21
<makolber@yahoo.com> wrote:>RMS Power is a misnomer and not mathematically correct >however it is often (mis) usedI agree. What it means, as jargon is usually something like "Power measured by a true RMS calculation of voltage or current, instead of a less accurate approximation". Steve
Reply by ●October 21, 20152015-10-21
gyansorova@gmail.com writes:> On Thursday, October 22, 2015 at 3:29:45 AM UTC+13, mako...@yahoo.com wrote: >> there is no mathematical meaning to RMS power >> >> there is RMS voltage >> and RMS current >> >> and V RMS x I RMS x phase = AVERAGE power. >> >> RMS Power is a misnomer and not mathematically correct >> however it is often (mis) used >> >> Mark > > Indeed it is by hi-fi idiots who talk of peak power.RMS is an average, theoretically over infinity, which can be defined as xrms = sqrt{lim_{T \approaches\infty} \frac{1}{T}\int_{-T/2}^{T/2} |x|^2 dx} Now there are certainly useful concepts of power over the shorter term. I wouldn't call them total idiots! -- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com
Reply by ●October 21, 20152015-10-21
Randy Yates <yates@digitalsignallabs.com> writes:> gyansorova@gmail.com writes: > >> On Thursday, October 22, 2015 at 3:29:45 AM UTC+13, mako...@yahoo.com wrote: >>> there is no mathematical meaning to RMS power >>> >>> there is RMS voltage >>> and RMS current >>> >>> and V RMS x I RMS x phase = AVERAGE power. >>> >>> RMS Power is a misnomer and not mathematically correct >>> however it is often (mis) used >>> >>> Mark >> >> Indeed it is by hi-fi idiots who talk of peak power. > > RMS is an average, theoretically over infinity, which can be defined as > > xrms = sqrt{lim_{T \approaches\infty} \frac{1}{T}\int_{-T/2}^{T/2} |x|^2 dx} > > Now there are certainly useful concepts of power over the shorter term. > I wouldn't call them total idiots!PS: I'm not trying to change the nomenclature: I agree V RMS x I RMS is by convention "average power." I'm just saying this method of computing power is based on an average over a long time, and that averages performed over shorter intervals have their uses. -- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com
