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In Reply to: Re: No cable warm up: proven by a Korean hi-end critique posted by Thorsten on February 13, 2002 at 12:32:09:
I do how things are derived and so on. However, where I disagree with all the neat textbook math is in the reality. The 96db claimed dynamic range for CD is not as such directly comparable to a 96db dynamic range in analogue system.Of course it's comparable. Unless you're using some strange definition of "dynamic range" which otherwise is the ratio of the maximum signal level versus the noise floor. In other words, it is the signal to noise ratio expressed in decibels rather than percent.
Simple proof (which you conveniently avoided) is the ability of analogue system having 96db of dynamic range to store a signal with a level of -93db (both 0db and -93db being defined as RMS measurements).
Because it's not proof as a 16 bit system can encode a signal 93 dB down from its full scale. It can encode a signal WORST CASE at 96 dB below full scale. And best case 98 dB. And when you decorrelate the signal from the quantization error using techniques such as dither, even more.
Please encode (purely as mathematical excercise) a 1KHz sinewave having an RMS level of 0.045mV for a given 16 Bit PCM Digital System having an input/output level of 2000mV RMS for a 1kHz sinewave having a peak level equal to 0dbfs and 44.1KHz sample rate. (BTW, purely for fun repeat at the same level for a 10KHz sinewave.
Why should I have to go throuh all that rigamarole? Any digital system can encode a signal down to ±1/2 LSB. And for a 16 bit system, the ratio between ±full scale and ±1/2 LSB will always be 65,574. Isn't that obvious enough for you? If not let me know and I'll go through the rigamarole.
Please present the data generated by your sampling algorythm here. Sow me that a reliable encoding is possible and we will agree on 96db dynamic range.
Here, let me 'splain this in a way that might cause you to understand this issue of the dynamic range of digital systems.
Again, dynamic range is the ratio between a system's peak signal and the noise floor. For analog systems, the noise floor is typically the 1/f thermal noise of the system. In the digital domain, such noise doesn't exist. Instead the noise is the system's quantization error.
You seem to be in agreement that your "15 bit plus sign" gives two domains of ±215. So let's just look at the postivie domain. 215 gives 32,768 discrete states. If we scale this to represent a full scale signal of 1 volt peak, then each quantization interval will represent a voltage change of 1/(215 - 1) or 30.5 microvolts.
As I said previously, the quanitzation error is a maximum of ±1/2 LSB as a signal which falls directly between sample intervals and will be quantized to one or the other intervals. That being the case, the maximum peak error voltage will be 1/2 LSB or 15.25 microvolts, which is 1/2 of 30.5 microvolts.
For a peak signal of 1 volt and a peak error voltage of 15.25 microvolts, we can then determine the worst case dynamic range. Our peak signal is 1 volt. Our peak error voltage is 15.25 microvolts. Therefore 1/0.00001525 = 65,574. And 20log(10)65,574 = 96 dB.
This is absolutely no different than an analog system which has a 1 volt peak signal and a peak noise floor of 15.25 microvolts. Both will have a dynamic range of 96 dB.
Get it now?
se
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- Re: No cable warm up: proven by a Korean hi-end critique - Steve Eddy 17:21:37 02/13/02 (0)
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