Average value
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From: Vedran on 6 Jul 2010 04:16 Hello! If the measuring is performed in decibels. What is the average value of the measurements? Do I have to convert the measurements to their absolute values before calculating the average? Measurements in dB are distributed according to the normal distribution. My problem is that: AVG(mesurements_{dB}) = \sum_n {x_i} / n y_i is the absolute value of x_i, therefore: y_i = 10^(x_i/10) AVG(measurements_abs) = \sum_n{y_i}/n 10*log_{10} (AVG(measurements_abs)) != AVG(measurements_dB) Which is the correct way to calculate the average value?
From: Henry on 6 Jul 2010 05:52 On 6 July, 09:16, Vedran <not_my_em...(a)sadf.com> wrote: > Hello! > > If the measuring is performed in decibels. What is the average value of > the measurements? Do I have to convert the measurements to their absolute > values before calculating the average? Measurements in dB are distributed > according to the normal distribution. > > My problem is that: > > AVG(mesurements_{dB}) = \sum_n {x_i} / n > y_i is the absolute value of x_i, therefore: y_i = 10^(x_i/10) > AVG(measurements_abs) = \sum_n{y_i}/n > > 10*log_{10} (AVG(measurements_abs)) != AVG(measurements_dB) > > Which is the correct way to calculate the average value? As you implicitly say, the geometric mean is less than the arithmetic mean. It all depends on what you are trying to average and why.
From: alainverghote on 6 Jul 2010 06:41 On 6 juil, 11:52, Henry <s...(a)btinternet.com> wrote: > On 6 July, 09:16, Vedran <not_my_em...(a)sadf.com> wrote: > > > > > > > Hello! > > > If the measuring is performed in decibels. What is the average value of > > the measurements? Do I have to convert the measurements to their absolute > > values before calculating the average? Measurements in dB are distributed > > according to the normal distribution. > > > My problem is that: > > > AVG(mesurements_{dB}) = \sum_n {x_i} / n > > y_i is the absolute value of x_i, therefore: y_i = 10^(x_i/10) > > AVG(measurements_abs) = \sum_n{y_i}/n > > > 10*log_{10} (AVG(measurements_abs)) != AVG(measurements_dB) > > > Which is the correct way to calculate the average value? > > As you implicitly say, the geometric mean is less than the arithmetic > mean. > > It all depends on what you are trying to average and why.- Masquer le texte des messages précédents - > > - Afficher le texte des messages précédents - Good morning, "Which is the correct way to calculate the average value?" It seems the question is about physical "measurements" , in some way about measuring real facts. I mean here 'averaging' terms of power (watt) and later on converting into dB, Alain
From: Vedran on 6 Jul 2010 11:40 On Tue, 06 Jul 2010 02:52:28 -0700, Henry wrote: > On 6 July, 09:16, Vedran <not_my_em...(a)sadf.com> wrote: >> Hello! >> >> If the measuring is performed in decibels. What is the average value of >> the measurements? Do I have to convert the measurements to their >> absolute values before calculating the average? Measurements in dB are >> distributed according to the normal distribution. >> >> My problem is that: >> >> AVG(mesurements_{dB}) = \sum_n {x_i} / n y_i is the absolute value of >> x_i, therefore: y_i = 10^(x_i/10) AVG(measurements_abs) = \sum_n{y_i}/n >> >> 10*log_{10} (AVG(measurements_abs)) != AVG(measurements_dB) >> >> Which is the correct way to calculate the average value? > > As you implicitly say, the geometric mean is less than the arithmetic > mean. > > It all depends on what you are trying to average and why. I'm measuring transfer functions in the communication cables. After that I calculated average and standard deviations to determine which value is worst than 99% of the cables in general. That is easily done with the data that is normally distributed (average + 2,33*standard deviation). But the normal distribution of the measured data is obtained by applying a function (abs -> dB conversion) to the real "physical" ratios of the input and output values.
From: Tim Little on 6 Jul 2010 20:15
On 2010-07-06, Vedran <not_my_email(a)sadf.com> wrote: > I'm measuring transfer functions in the communication cables. After > that I calculated average and standard deviations to determine which > value is worst than 99% of the cables in general. The mean and standard deviation suffice for that only if you are *certain* that the distribution of the population is some form of normal distribution. That is a rather unlikely scenario. For example, the normal distribution has indefinite tails, which would correspond to power *gain* in the cable. So from a theoretical point of the view the distribution can't possibly be normal. The question then remains of whether it is "close enough" to normal that the first percentile is very close to -2.33 standard deviations. That is something that you can only verify experimentally, by taking a large sample and measuring both the 1st %ile and comparing with the -2.33 sd value. If you have been explicitly told by a credible source that the population is definitely close enough to normally distributed in dB for your approach to make sense, you should take the average of dB values of the transfer function, and not the raw amplitude or power values. - Tim |