Primer on dacs

Atmasphere

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the 44.1KHz sample rate of a CD would faithfully reproduce frequencies of 22.05 KHz which is already above the threshold of even young humans.
This is false just FWIW; when I worked at Allied Radio Shack service department back in the 1970s it was always super annoying when they left the ultrasonic motion detector on. Fortunately it had a big switch so you could just turn it off. I could hear it up the stairs before I even entered the space. It ran at 25KHz. I sure can't hear anywhere near that now!
 

microstrip

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This is false just FWIW; when I worked at Allied Radio Shack service department back in the 1970s it was always super annoying when they left the ultrasonic motion detector on. Fortunately it had a big switch so you could just turn it off. I could hear it up the stairs before I even entered the space. It ran at 25KHz. I sure can't hear anywhere near that now!

IMHO this type of comments must be considered statistically - the very few people that could listen to ultrasonics are not relevant to the high-end industry.

Anyway CD is now obsolete, many excellent modern recordings are now available in 96/24, that does not suffer from bandwidth limitations.
 

DonH50

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Minor correction: The Nyquist sampling criteria (not the stability criteria) is that the sampling rate must be >2x the signal (information) bandwidth. No equality. You cannot recover a signal precisely at fs/2; the signal bandwidth must be a little lower in frequency.

Bandwidth is essentially DC to 20 kHz or whatever for audio, but the "bandwidth" part is important for systems that capture or produce signals at much higher frequencies. An example is an ADC running at 100 kS/s can recover a 40 kHz signal band centered at 1 GHz, assuming the ADC has the front-end bandwidth to handle it. That eliminates a whole bunch of components otherwise needed to convert the signal all the way down to DC-40 kHz. A lot of radios, cell phones, and the like take advantage of that property.

FWIWFM - Don
 
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Gregadd

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Thanks Don. That start the debate how much greater than twice. One guy just does it three times. He claims it just makes life easier
 

Atmasphere

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Minor correction: The Nyquist sampling criteria (not the stability criteria) is that the sampling rate must be >2x the signal (information) bandwidth. No equality. You cannot recover a signal precisely at fs/2; the signal bandwidth must be a little lower in frequency.

Bandwidth is essentially DC to 20 kHz or whatever for audio, but the "bandwidth" part is important for systems that capture or produce signals at much higher frequencies. An example is an ADC running at 100 kS/s can recover a 40 kHz signal band centered at 1 GHz, assuming the ADC has the front-end bandwidth to handle it. That eliminates a whole bunch of components otherwise needed to convert the signal all the way down to DC-40 kHz. A lot of radios, cell phones, and the like take advantage of that property.

FWIWFM - Don
Isn't the sample also supposed to be an analog sample?
 

DonH50

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Isn't the sample also supposed to be an analog sample?
I don't know what that means... Signal, whatever the signal looks like, must not have any frequency component equal to or greater than Nyquist if you want to prevent aliasing.
 

Atmasphere

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I don't know what that means... Signal, whatever the signal looks like, must not have any frequency component equal to or greater than Nyquist if you want to prevent aliasing.
Right. That's not the question.
When a sample is taken, a 16 bit word is used to represent the voltage value at that instant. I don't see anything in the Theorem that suggests it be that way; the Theorem seems to suggest that a 'sample' in this case a voltage, is the actual value of the voltage. If I have this right, that is why there are 24- and 32-bit codices now. The voltage of any given sample will be more accurately portrayed.
 

microstrip

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Right. That's not the question.
When a sample is taken, a 16 bit word is used to represent the voltage value at that instant. I don't see anything in the Theorem that suggests it be that way; the Theorem seems to suggest that a 'sample' in this case a voltage, is the actual value of the voltage. If I have this right, that is why there are 24- and 32-bit codices now. The voltage of any given sample will be more accurately portrayed.

No way the Theorem refers to digitizing - it just says take the amplitude of the signal with a certain time interval and keep it! The quantification error is a different subject.

The way we get the values is not part of the Theorem - modern high resolution ADC's are complex devices. Some people are very upset with the DAC's - if they knew about the ADC's they would not sleep at night! :oops:

BTW, the old bucket brigade delay lines where the signal amplitude was stored in capacitors is an excellent example of using the Nyquist theorem in the analogue domain.
 
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DonH50

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Right. That's not the question.
When a sample is taken, a 16 bit word is used to represent the voltage value at that instant. I don't see anything in the Theorem that suggests it be that way; the Theorem seems to suggest that a 'sample' in this case a voltage, is the actual value of the voltage. If I have this right, that is why there are 24- and 32-bit codices now. The voltage of any given sample will be more accurately portrayed.

Ah, correct, digitization is separate. The Theorem is about information bandwidth. In an ADC with a track-and-hold, technically we go from analog, to sampled-analog with discrete time and continuous voltage, then to a digital word. The resolution (bit depth) impacts amplitude and timing resolution but not bandwidth.

More on sampling: https://www.whatsbestforum.com/threads/sampling-101.1209/
More about the time resolution: https://www.whatsbestforum.com/threads/jitter-101.1322/

My signature link has other articles (yes, I know Ralph knows, posting to the thread in case it helps others understand).

HTH - Don
 

sbnx

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Right. That's not the question.
When a sample is taken, a 16 bit word is used to represent the voltage value at that instant. I don't see anything in the Theorem that suggests it be that way; the Theorem seems to suggest that a 'sample' in this case a voltage, is the actual value of the voltage. If I have this right, that is why there are 24- and 32-bit codices now. The voltage of any given sample will be more accurately portrayed.

When talking digital, the input signal -- meaning the recording of a band in a studio (with an ADC) -- has to be separated from the reproduction of that signal -- through a DAC -- in our listening room.

The input signal (from the microphone) can be sampled at 44.1 kHz or higher. But once it is put onto a CD it will be a 44.1 kHz sample rate. Of course "hi resolution" files one downloads could be 96 kHz, 192 kHz or higher. Of course what is being sampled is the amplitude of the voltage coming from the microphone. This sampled amplitude data will then be encoded (usually PCM) when it put onto the CD.

On the output side the CD transport or streamer sends this digital stream (along with the clock signal) to the DAC which will decode the digital amplitude data and map that from 0 to 2 Volt output (usually, but this is DAC dependent).

What cooks my brain is the realization of how small a voltage the least significant bits are. For standard CD (16 bits) the LSB is only about 30 micro volts. Going up to 24 bits the LSB is only about 0.1 microvolts. Getting that 24th bit from the DAC to the preamp poses a big challenge. Of course a higher output voltage of the DAC helps with this some. But the stability of the power supply can't be underestimated here.
 

Atmasphere

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On the output side the CD transport or streamer sends this digital stream (along with the clock signal) to the DAC which will decode the digital amplitude data and map that from 0 to 2 Volt output (usually, but this is DAC dependent).
Isn't it the case that at -45dB, you only have 8 bits to express the signal if 16 bits expresses 0VU?
 

sbnx

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Isn't it the case that at -45dB, you only have 8 bits to express the signal if 16 bits expresses 0VU?
Not sure I 100% understand the question but ... Each bit of resolution provides 6 dB of dynamic range. So 16 bits gives 96 dB of dynamic range while 8 bits of resolution would give 48 dB of dynamic range.

If you turn the volume down on your preamp that does not affect the dynamic range of the recording. It is still 96 dB (assuming it is CD). However, let's say you listen in your living room with the AC going, refrigerator in the kitchen, outside traffic etc. Then the noise floor of your living room may be 50 dB. In this case if you listen to a CD at 80 dB then you really only have 30 dB of dynamic headroom because the noise floor of the room is obscuring a lot of the low level detail. As you turn up the volume you will be able to hear more of the detail in the recording.
 

Atmasphere

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Each bit of resolution provides 6 dB of dynamic range. So 16 bits gives 96 dB of dynamic range while 8 bits of resolution would give 48 dB of dynamic range.
Yes- this is what I was getting at. So isn't it true then that at -48dB you'd have half of the bits in a 16 bit word turned off? If I have this right, if any of the bits to the left of the remaining 8 were on, the word would be representing something greater than -48dB?
 

DonH50

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Couple of (OK, three) quick comments:

The SNR from quantization noise (conversion from continuous analog to discrete digital amplitude values) is about 6N dB for N bits. That does not include noise or distortion from any other sources. That is also for a sinusoidal signal.

The SNR is basically the integrated noise across the converter's bandwidth. The spurious-free dynamic range is ~9N dB for N bits. If you think of the noise floor as the "grass", then from your head to the grass is 9N dB. If you cut the grass and bundle it all into a bag, that bundle represents the noise in the SNR equation, ~6N dB. This post discusses it a bit: https://www.whatsbestforum.com/threads/sampling-101.1209/

Finally, we can hear, and processors can extract, signals from below the noise floor. How far below varies, but anywhere from a few dB to 10 dB or more is possible.

HTH - Don
 

Atmasphere

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Finally, we can hear, and processors can extract, signals from below the noise floor. How far below varies, but anywhere from a few dB to 10 dB or more is possible.
We can hear into certain forms of hiss and 10dB is reasonable. But oddly, if the hiss doesn't have a 'natural' quality (like white noise) our ability to hear into the noise floor is curtailed. For example in power amplifiers, if feedback is poorly applied, it can result in a noise floor composed of harmonic and inharmonic (intermodulations) information. The ear has less ability to hear into this sort of noise floor (which sounds like 'hiss' when no signal; is present).
 

audiobomber

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My infatuation with the IFi hip dac has prompted me to learn more about dacs. I think it is the hardware that has improved. I don't know. Can anyone refer me to a good primer to a straightforward primer.
Thanking you in advance.
Greg
This is not new but it's the best explanation of jitter and digital sound I've seen on the net.
 

sbnx

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This is not new but it's the best explanation of jitter and digital sound I've seen on the net.
Good point bringing up jitter. Everything discussed above about bit depth, digital noise, sampling etc. is all assuming everything works perfectly. Of course, in the real world, nothing is ever perfect. Thus clock jitter rears its ugly head and screws up all of the beautiful theory.
 

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