I often see papers that average dB SNRs. Obviously this does not make sense since dB is a log function and you should average and then take dBs. However, can you get away with such a method? Suppose we have 4 SNRs of value 1,10,100 and 1000. Convert to dB and we get 0,10,20,30dB. Averaging we get (0+10+20+30)/4 = 60/4 = 15dB. Averaging the SNRs gives (1+10+100+1000)/4 = 277. Now convert to dB and the answer ain't 15dB but what if the dB variation is say within a few dBs - might this be ok?
Average dB
Started by ●June 13, 2007
Reply by ●June 13, 20072007-06-13
On 6 13 , 3 27 , gyansor...@gmail.com wrote:> I often see papers that average dB SNRs. Obviously this does not make > sense since dB is a log function and you should average and then take > dBs. However, can you get away with such a method? Suppose we have 4 > SNRs of value 1,10,100 and 1000. Convert to dB and we get > 0,10,20,30dB. Averaging we get (0+10+20+30)/4 = 60/4 = 15dB. > > Averaging the SNRs gives (1+10+100+1000)/4 = 277. Now convert to dB > and the answer ain't 15dB but what if the dB variation is say within a > few dBs - might this be ok?I preffer the latter one. since your method of averaging is a linear weighted one, which is not suitable for a log function.
Reply by ●June 13, 20072007-06-13
On 6 13 , 3 27 , gyansor...@gmail.com wrote:> I often see papers that average dB SNRs. Obviously this does not make > sense since dB is a log function and you should average and then take > dBs. However, can you get away with such a method? Suppose we have 4 > SNRs of value 1,10,100 and 1000. Convert to dB and we get > 0,10,20,30dB. Averaging we get (0+10+20+30)/4 = 60/4 = 15dB. > > Averaging the SNRs gives (1+10+100+1000)/4 = 277. Now convert to dB > and the answer ain't 15dB but what if the dB variation is say within a > few dBs - might this be ok?humm, about the problem of variation, I think you could treat it as just a pure numeric problem. to solve it, what you need to do is just provide your method of calcaluting.
Reply by ●June 13, 20072007-06-13
gyansorova@gmail.com wrote:> I often see papers that average dB SNRs. Obviously this does not make > sense since dB is a log function and you should average and then take > dBs. However, can you get away with such a method? Suppose we have 4 > SNRs of value 1,10,100 and 1000. Convert to dB and we get > 0,10,20,30dB. Averaging we get (0+10+20+30)/4 = 60/4 = 15dB. > > Averaging the SNRs gives (1+10+100+1000)/4 = 277. Now convert to dB > and the answer ain't 15dB but what if the dB variation is say within a > few dBs - might this be ok? >If you average the dB values you generate the geometric mean of the SNR values, expressed in dB. (Whether you have a need the geometric mean of the SNRs is another matter entirely.) Example: Geometric mean of the SNR values you quoted is: (1*10*100*1000)^(1/4) = 31.62 10log(31.62) = 15 dB, as you found by averaging the dB values. Regards, John
Reply by ●June 13, 20072007-06-13
gyansorova@gmail.com writes:> I often see papers that average dB SNRs.Can you cite an example? I have not seen that. -- % Randy Yates % "I met someone who looks alot like you, %% Fuquay-Varina, NC % she does the things you do, %%% 919-577-9882 % but she is an IBM." %%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO http://home.earthlink.net/~yatescr
Reply by ●June 13, 20072007-06-13
gyansorova@gmail.com wrote:> I often see papers that average dB SNRs. Obviously this does not make > sense since dB is a log function and you should average and then take > dBs.It depends. In many cases, the SNR in dB is taken per a segment of a signal. The nature of the signal and noise can be different at the different segments. Thus the averaged SNR in dB per all segments is the better measure of the quality then All_signal/All_noise. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
Reply by ●June 13, 20072007-06-13
gyansorova@gmail.com wrote:> I often see papers that average dB SNRs. Obviously this does not make > sense since dB is a log function and you should average and then take > dBs. However, can you get away with such a method? Suppose we have 4 > SNRs of value 1,10,100 and 1000. Convert to dB and we get > 0,10,20,30dB. Averaging we get (0+10+20+30)/4 = 60/4 = 15dB. > > Averaging the SNRs gives (1+10+100+1000)/4 = 277. Now convert to dB > and the answer ain't 15dB but what if the dB variation is say within a > few dBs - might this be ok?If 47 dB is "within a few dBs" of 15, then it's OK. It's also OK if the number has a special significance, say for statutory purposes. Suppose I have two loudspeakers. One is twice as efficient as the other. The more efficient one is connected to an amplifier establishes the reference level in the room. How much is the sound level (in dB) increased by connecting the second one also? How much is the level decreased by using the second alone? One way on a 60-mile trip, a car goes 60 mph. On the way back, it goes 40 mph. What is its average speed? Not only can't you average dBs, you can't average averages. Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by ●June 13, 20072007-06-13
On Jun 13, 3:27 am, gyansor...@gmail.com wrote:> I often see papers that average dB SNRs. Obviously this does not make > sense since dB is a log function and you should average and then take > dBs. However, can you get away with such a method? Suppose we have 4 > SNRs of value 1,10,100 and 1000. Convert to dB and we get > 0,10,20,30dB. Averaging we get (0+10+20+30)/4 = 60/4 = 15dB. > > Averaging the SNRs gives (1+10+100+1000)/4 = 277. Now convert to dB > and the answer ain't 15dB but what if the dB variation is say within a > few dBs - might this be ok?What you should do is what gives a useful result for a given application. IIRC (it has been >20 years) in the Jayant and Noll waveform coding book, it was reported that the average of the SNRs in dB was more correlated with either speech quality or intelligibility (don't remember which) than was the average of the SNRs, not in dB (all calculated over intervals). Th former calculation would be useful if you were trying to evaluate/improve performace (whichever it was, quality or intelligibility, don't remember). So, sometimes 'what makes sense' is 'what is useful'. Dirk
Reply by ●June 13, 20072007-06-13
On Wed, 13 Jun 2007 07:07:19 -0400, Randy Yates <yates@ieee.org> wrote:>gyansorova@gmail.com writes: > >> I often see papers that average dB SNRs. > >Can you cite an example? I have not seen that.It's commonly done for evaluating bursty systems in fading. There's more than one way to do it and if the paper didn't say how they were doing the averaging then that's an oversight IMHO. Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org
Reply by ●June 13, 20072007-06-13
On Jun 13, 12:27 am, gyansor...@gmail.com wrote:> I often see papers that average dB SNRs. Obviously this does not make > sense since dB is a log function and you should average and then take > dBs. However, can you get away with such a method? Suppose we have 4 > SNRs of value 1,10,100 and 1000. Convert to dB and we get > 0,10,20,30dB. Averaging we get (0+10+20+30)/4 = 60/4 = 15dB. > > Averaging the SNRs gives (1+10+100+1000)/4 = 277. Now convert to dB > and the answer ain't 15dB but what if the dB variation is say within a > few dBs - might this be ok?Well, it all depends on what you are trying to do. By finding an average you are trying to find an agregate statastic. If your original distribution is lognormal, then finding the average of the logs makes since. RF signals that have underqone many diffractions where the main straight path to the receiver is occluded tend to have a lognormal distribution. In this case averaging dB makes since. Even peoples' salaries tend to have a lognormal distribution. Clay
