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An interesting read from someone from Apogee

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#1
https://www.headphone-earphone.reviews/2018/12/05/apogee-unfiltered-and-straight-from-the-source/

I haven't finished the whole thing yet, as I am reading it from work. Which my work involves riding motorcycle...

The interview come across wide topics from digital audio, MQA, practicality of using 96k...

The thing caught my eyes was the explanation to apogee groove have a higher than normal impedance. Is it true that variable voltage output can overcome the impedance matching problem?
 

amirm

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#2
No. He is assuming that your headphone has hundreds of ohm impedance so what they have is very good. He doesn't realize there are consumer headphones with very low impedances.
 

restorer-john

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#3
Amir, he's basically saying the amplifier output stage is configured a current source instead of a voltage source (hence the high output impedance you measured), but in reality, your testing showed it clearly is limited by the internal rails wouldn't you say?

NP did some experimentation with full range speaker drivers (like headphone design) and transconductance amps:

http://www.firstwatt.com/pdf/art_cs_amps.pdf

"The most precise way to develop that specific current is with a current-source amplifier. Such
an amplifier ignores the impedances in series with the circuit, the resistance and inductance
of the wire and voice coil and the back electromotive force (EMF) produced by the cone
motion. As I said, most speakers are designed around voltage sources but there are few
instances where a current source can be used to advantage. One of the best ones is the
category of full-range high-efficiency drivers.
Why is that? First, such drivers are able to take advantage of acoustic and suspension
resistance to achieve some or all of the damping that they need to prevent excessive
overhang because their moving mass is very light. With their efficient motors, even a high
source impedance is often enough to give critical damping. Second, their impedance curve
tends to reflect their needs – more current both at low frequency resonance and in the treble,
two areas where frequency response has fallen off with increased speaker impedance. If you
want, the current through the voice coil can be made constant regardless of the variations in
the acoustic environment. The voice coil force is invariant whether the cone is loaded into a
horn, sealed box, bass reflex or whatever else you care to mount it in."








The above linked Apogee article has a few issues IMO...

"Roger: Of course. Let’s take a 48 kHz 24 bit file. 24 bit means you should have several hundred thousand different voltage levels. You have 24 ‘levels’ to define a voltage level between 0 dBFS and -144 dBFS if you’re in the digital domain. I think you have around 600,000 individual bit levels for a 24 bit file (recording format), and for a 16 bit file (CD audio) you have about 16,000 bit levels. The total amount of levels are your dynamic range. In the signal path, you are asking a computer or converter chip to look at the analogue signal 48,000 times per second (assuming a 48kHz file), look at the incoming voltage and finally associate that voltage with one of 600,000 different levels. For a 96kHz file, the converter has to do this 96,000 times per second and so on."
 
Last edited:
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#4
Amir, he's basically saying the amplifier output stage is configured a current source instead of a voltage source (hence the high output impedance you measured), but in reality, your testing showed it clearly is limited by the internal rails wouldn't you say?

NP did some experimentation with full range speaker drivers (like headphone design) and transconductance amps:

http://www.firstwatt.com/pdf/art_cs_amps.pdf

"The most precise way to develop that specific current is with a current-source amplifier. Such
an amplifier ignores the impedances in series with the circuit, the resistance and inductance
of the wire and voice coil and the back electromotive force (EMF) produced by the cone
motion. As I said, most speakers are designed around voltage sources but there are few
instances where a current source can be used to advantage. One of the best ones is the
category of full-range high-efficiency drivers.
Why is that? First, such drivers are able to take advantage of acoustic and suspension
resistance to achieve some or all of the damping that they need to prevent excessive
overhang because their moving mass is very light. With their efficient motors, even a high
source impedance is often enough to give critical damping. Second, their impedance curve
tends to reflect their needs – more current both at low frequency resonance and in the treble,
two areas where frequency response has fallen off with increased speaker impedance. If you
want, the current through the voice coil can be made constant regardless of the variations in
the acoustic environment. The voice coil force is invariant whether the cone is loaded into a
horn, sealed box, bass reflex or whatever else you care to mount it in."








The above linked Apogee article has a few issues IMO...

"Roger: Of course. Let’s take a 48 kHz 24 bit file. 24 bit means you should have several hundred thousand different voltage levels. You have 24 ‘levels’ to define a voltage level between 0 dBFS and -144 dBFS if you’re in the digital domain. I think you have around 600,000 individual bit levels for a 24 bit file (recording format), and for a 16 bit file (CD audio) you have about 16,000 bit levels. The total amount of levels are your dynamic range. In the signal path, you are asking a computer or converter chip to look at the analogue signal 48,000 times per second (assuming a 48kHz file), look at the incoming voltage and finally associate that voltage with one of 600,000 different levels. For a 96kHz file, the converter has to do this 96,000 times per second and so on."
If I' m correct (!) 2**16=65,536, and 2**24=16,777,216
This is the voltage resolution of 16bit and 24bit samples respectively.
 

RayDunzl

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#5
If I'm correct (!) 2**16=65,536, and 2**24=16,777,216
16 bit is +/- 32,767

The 16th Big Bit flips the polarity, it's the sign bit.

The increments (value of the LSB), assuming 2.818Vpk volts = +/-0.00008600115V at the DAC output

24 bit is +/- 8,388,607

The LSB, assuming 2.818Vpk volts = +/-0.000000335931818 V

(If I'm correct (!))

Amplification (or attenuation, as the case may be) expands or conracts those voltage increments.
 
Last edited:

RayDunzl

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#6
Ok, I'm a little bit wrong... That's normal.

WAV Files

"For the sample data, it’s important to note that each element of the data array is a signed value – it can be negative, positive or zero. The range of each of these elements is important too.

Since each sample represents amplitude, and we are working with signed values, we have to consider what is our minimum and maximum amplitude given the data type we’ve chosen. Due to some crazy business involving Endianness and 2’s complement, 16-bit samples range from -32760 to 32760 instead of -32768 to 32768 (2^16 / 2). We don’t worry about this with 8-bit data, because 8 bits is only one byte and Endianness is not an issue, nor with 32-bit data because it is represented as a proper float from -1.0f to 1.0f."

I don't remember taking any of that into account when I composed some "music" directly calculating sample values and writing to WAV (to avoid figuring out how to tickle the synthesizer in Windows).
 
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#7
Ok, I'm a little bit wrong... That's normal.

WAV Files

"For the sample data, it’s important to note that each element of the data array is a signed value – it can be negative, positive or zero. The range of each of these elements is important too.

Since each sample represents amplitude, and we are working with signed values, we have to consider what is our minimum and maximum amplitude given the data type we’ve chosen. Due to some crazy business involving Endianness and 2’s complement, 16-bit samples range from -32760 to 32760 instead of -32768 to 32768 (2^16 / 2). We don’t worry about this with 8-bit data, because 8 bits is only one byte and Endianness is not an issue, nor with 32-bit data because it is represented as a proper float from -1.0f to 1.0f."

I don't remember taking any of that into account when I composed some "music" directly calculating sample values and writing to WAV (to avoid figuring out how to tickle the synthesizer in Windows).
You are right!
 

KSTR

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#8
Microsoft Blog said:
Due to some crazy business involving Endianness and 2’s complement, 16-bit samples range from -32760 to 32760 instead of -32768 to 32768 (2^16 / 2).
Utter nonsense. I'm very disappointed to see this kind of misinformation and blatant error in a "Microsoft developer blog".

16bit integers *always* range from -32768 to +32767, period. No matter in what Endianess the two bytes comprising it are stored. And of course no matter what the data actually represents.

Generally, with n=number of bits, n-bit signed integers span values from -(2^(n-1)) to +(2^(n-1)-1).
The "two's complement thing" is that the max positive value is one less than max negative value and this has to be handled when eg when we want to invert the signal. We cannot simply use x := -x, rather we must additionally check for -32768 and convert that special case to +32767.
 
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#10
Read the comments: that article is riddled with errors. You were almost right, barring the slight asymmetry of signed numbers...
I'm awfully sorry that my comment about the values of 2**16 and 2**24 elicited this discussion (which may nonetheless be useful to some).
I only meant to correct Roger's statement that 16 bit samples give about 16,000 values and 24 bit about 600,000 values.
I couldn't figure out where those gross approximations came from.
But of course, I had forgotten about the signed integer business. My bad (and apologies).
 
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