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word length in PCM and its relation to the dB scale

post #1 of 5
Thread Starter 


/Long time lurker, registered to ask this question because the internet doesn't know.


I have been looking around for some information on how PCM actually works. I'm quite comfortable with the Nyquist criterion and all the frequency-related jazz, but I'm still a bit stuck on the word-size and dynamic range.


I know the decibel is dimensionless quantity, and is used as a relative scale. In some applications, dBm is used, where 0dBm=1mW. There are a bunch of others with different reference values, but as far as I know 'dynamic range' uses the dB unit as a relative measurement, where 0dB is the peak of the signal. Is this correct?


In this case, what exactly is 0dB? is 0dB simply the voltage roof of the ADC used in the recording studio? And sound engineers manipulate the gain such that the highest peaks reach just below 0dB? (I understand more mastering can be done after this, but theoretically that would reduce the dynamic range slightly. Anyhow, most listeners won't notice the difference anyways.)


After mastering (ideally), the highest peaks will reach 0dB without clipping, and the lowest -96dB represents the lowest 'volume' which can be encoded. Is this the goal?


Also, why does +1-bit = +6 dB dynamic range? An extra bit of word length will double the resolution, that should give +3 dB of dynamic range (the lowest 'volume' limit is halved --> the lowest point in dynamic range decreases by 3dB).


Thank you in advance, and Merry Christmas.

post #2 of 5

Most engineers work in 24 bit, so it has nothing to do with ADC conversion or anything, as far as I know. I'm going to link a lot of Wikipedia because I don't know specifics myself.


Technically the dB used here is dBFS. FS stands for full scale, so 0 dBFS is full scale volume.


Increased bit depth increases signal-to-noise ratio (basically dynamic range, difference between 0 dBFS and the noise level) because it reduces quantization errors, noise introduced due to rounding errors. The more bits, the less rounding is required to reproduce the analog signal, which I guess means fewer (quieter) quantization errors.


This appears to explain why each bit means 6 dB lower noise, but I'll be damned if I read something mathematical on my break.

post #3 of 5
Thread Starter 

Perfect! That answered pretty much everything.


My confusion was in the usage of dB to refer to both the voltage and power. Where the word-length is used to encode the voltage level, the dynamic range is a measurement of power (thus voltage^2). +1-bit word length = +3dB voltage = +6dB power (dynamic range, audio level, /etc).

post #4 of 5


The Art of Digital Audio 3rd ed, Watkinson, p. 222

post #5 of 5

As stated by ch41rmn, the term "dB" is "used as a relative scale".  Therefore, the question is what is the reference?  As Head Injury pointed out, "FS" is the reference.  If we were talking about volts, the dB would need to be followed by u or V.


For reference, rane.com's RaneNotes are very helpful.  http://www.rane.com/note169.html

In some applications, dBm is used, where 0dBm=1mW

Don't forget that "0dBm=1mW" requires stating the load impedance.  Is it 50 Ohm? 75 Ohm? 600 Ohm?  :)



+1-bit word length = +3dB voltage = +6dB power


A Voltage factor of 2 (multiply or divide by 2) results in a 6dB voltage change.  A Wattage factor of 2 results in a 3dB power change.  Doubling a digital signal will increase the output voltage by 6dB.


for example:

dBm = 10*log(measured Watts / 0.001Watts)

dBV = 20*log(measured Voltage / 1Volt)


If you start with 1 Volt (0dB V), and then double it, you will get 2 Volts (+6dB V)


note the "10" and "20", which result in the factor of 2 difference.


I know, it is confusing.  I still have to pull-out books.  :)


SPL is 20*log(Pressure2/Pressure1)


Reference: Handbook for Sound Engineers, 3rd Edition, Glen M. Ballou.  Chapter 1, pages 6-8.

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