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16th April 2007, 06:21 AM  #1 
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Dynamic range confusion
Human hearing is 130 dB from lowest intensity to threshold of pain, which for sound intensity is ten trillion to one. For voltage, however, 100 dB is somewhat over three million.
A 24 bit digital system has 16.7 million levels of quantization, and they are equal. So you get 16.7 million different voltage levels. I fail to see how that can cover the 1:10 trillion dynamic range of human hearing. Since the quantization is made of even differences between consecutive binary digits and not weighted, the quietest level you can represent is only at 16.7 million times lower voltage than the loudest. How can that be sufficient? ??? 
16th April 2007, 06:39 AM  #2 
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Join Date: Apr 2002
Location: Munich

Even extreme live music presentation would not cover the whole 130dB ?

16th April 2007, 06:55 AM  #3 
Sin Bin
Join Date: Nov 2003
Location: Brighton UK

Hi,
Normal domestic listening levels do not exceed say ~ 100dB, and ambient background level is always above say ~ 35dB. /sreten. 
16th April 2007, 08:54 AM  #4 
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Join Date: Mar 2004
Location: Budapest, Hungary

We can represent 2^24 discrete levels which is 144.5 dB in voltage, but 72 dB in power ratio. The background noise in a very quiet listening room is say 35 dB as sreten remarked, so we can reproduce up to 107 dB range above the hearing threshold. The hearing threashold is considered as 20 uPa (0 dB). An average lodspeaker has 93 dB/1W/1m efficiency. It can produce 113 dB for 100 W drive at 1 meter, and 107 dB at 4 meters, which is extremely loud. I found some typical sound pressure levels at Wikipedia:
http://en.wikipedia.org/wiki/Sound_pressure_level More interesting is how the 48 dB power dynamic range of the 16bit DC audio can be sufficient. 
16th April 2007, 05:36 PM  #5  
Sin Bin
Join Date: Nov 2003
Location: Brighton UK

Quote:
You have your voltages and powers the wrong way round. 16 bit = 48dB voltage and 96dB power = signal to noise ratio. One reason vinyl sounds good is 70dB power range is adequate. /sreten. 

16th April 2007, 07:21 PM  #6 
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Join Date: Jun 2003
Location: Victoria, BC, Canada

You guys have your math mixed up. dB in voltage is the same dB in power, or more accurately, the measure of dB is always a scale relative to power!
The decibel scale is defined as: dB = 10 log(Watts) When calculating using voltage you use the relationship P = VxV/R > V squared divided by R mix the two equations together and dB = 10 log(VxV/R) now log (VxV) = 2 log(V) so you can simplify to dB = 20 log (V/R) since for the load R is gnerally a constant value, you can simplify to dB = 20 log (V) But this scale is still relating to power. Now, with that confusion sorted out, lets look at the original post. Crowbars confusion comes from the incorrect statement that 130dB equals a range of 10 trillion discrete voltage levels. Using our equations from above, V = 10 exp (130dB / 20) = 3.16 Million voltage levels. Crowbars next statement that 24 bit = 16.7 Million levels is correct. For reference: 21 bit = 2.10 Million = 126.4dB 22 bit = 4.19 Million = 132.5dB and 10 Trillion = 260dB = 44 bit Hope this helps. Terry 
17th April 2007, 12:57 AM  #7  
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Quote:


17th April 2007, 12:57 AM  #8 
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Continuing then.
I've sketched in the image below the dynamic range compression that the ear hasthe higher the intensity of a signal, the larger the difference has to be with a signal of similar intensity to be detectable. So differences on the order of the threshold of hearing at the bottom may be just around 10^12 W/m^2, but higher up are much larger to be detectable. So it would seem you don't need ten trillionlevel quantization to fully capture the detail of audible dynamic range. Maybe the 16.7 million levels of 24bit PCM are sufficientthat is, they might be sufficient if the quantization were nonlinear in the same way as actual hearing. The way it's now, either it can only capture a small portion of the dynamic range, or it is not sufficiently dense at the low end. I want to bring an analogy from the video world, specifically high dynamic range displays. Regular displays are usually between 300:1 and 600:1, whereas the human eye is about 60 dB (a million to one). Most HDR displays have an LED array behind the LCD screen, so the dynamic range is the LED multiplied by the LCD. Both are usually 8 bit (256) levels. In this case, the nonlinear range happens naturally due to the multiplication. At the bottom you have differences of one, say LED:LCD intensities go as 1x1=1, 1x2=2, 1x3=3, 2x2=4, etc., but near the top differences get very large: 255x254=64770, 255x255=62065. This fits the eye's nonlinearity with lower sensitivity to level differences when intensity is higher. However, with audio the linear dynamic range of PCM DACs is hugely suboptimal. Even as 32bit PCM has even more levels, it is still linear and if you try to cover threshold of hearing to threshold of pain (as you should), then you'll still have too granular quantization in the low end. If you have a nonlinear quantization (I'm sure DACs can be built, R2R design would be trivial to modify), then perhaps even 24bit is sufficient. 
17th April 2007, 12:57 AM  #9 
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17th April 2007, 05:11 AM  #10  
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Location: Victoria, BC, Canada

Quote:
But I think you missed some of what I was saying, which actually supports your view. PCM encoding uses discrete voltage levels that are converted to a wattage output by amplifier and speaker. Using P = V^2/R , roughly 3 million to 1 range of voltage, when squared produces the 10 Trillion to one intensity range you describe. This matches exactly with your description of an exponential relationship, which is already present when taking the square of voltae to produce intensity. I'm likely not explaining this well, but I think we are two breeds of dog barking at the same cat, if you will. 

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