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Signal cable does not work as optical

The characteristic impedance of a coaxial cable used for SPDIF will matter very little for short distances—ten to fifteen feet. If you take a look at the signal output from just about any consumer device with a coaxial SPDIF output, the signal level will likely far exceed the specification which is 0.5 - 0.6 Vpp. Many consumer devices have output signal levels as high as 7 Vpp. This is probably intentional to mitigate signal loss over long distances or bad cables—even though the copper spec for SPDIF is thirty-three feet (10m).

A good app note from ST Micro:

Receiving S/PDIF audio stream with the STM32F4/F7/H7 Series
 
One works, the other doesn't- easy to hear.

OP said “I have measured the frequency response of both cables and there are no differences but my mind may be playing against me because I could feel more definition in the high part”

How can you feel more/less definition in the high part from a cable that makes no sound. That’s my point.
 
The characteristic impedance of a coaxial cable used for SPDIF will matter very little for short distances—ten to fifteen feet. If you take a look at the signal output from just about any consumer device with a coaxial SPDIF output, the signal level will likely far exceed the specification which is 0.5 - 0.6 Vpp. Many consumer devices have output signal levels as high as 7 Vpp. This is probably intentional to mitigate signal loss over long distances or bad cables—even though the copper spec for SPDIF is thirty-three feet (10m).

A good app note from ST Micro:

Receiving S/PDIF audio stream with the STM32F4/F7/H7 Series
It is primarily not about signal loss, but about reflections distorting the wave form.
 
OP said “I have measured the frequency response of both cables and there are no differences but my mind may be playing against me because I could feel more definition in the high part”

How can you feel more/less definition in the high part from a cable that makes no sound. That’s my point.
The cable works at lower bit rates. So perceptive/cognitive (etc) bias.
 
The characteristic impedance of a coaxial cable used for SPDIF will matter very little for short distances—ten to fifteen feet. If you take a look at the signal output from just about any consumer device with a coaxial SPDIF output, the signal level will likely far exceed the specification which is 0.5 - 0.6 Vpp. Many consumer devices have output signal levels as high as 7 Vpp.
Are you sure? That is out of spec and might be dangerous for the receiving device. The manual of the KH750DSP for example staates explicitely that one must not connect an AES/EBU Output to its SPDIF Input due to the higher voltage.
This is probably intentional to mitigate signal loss over long distances or bad cables—even though the copper spec for SPDIF is thirty-three feet (10m).
At 13 MHz the signal loss is not that high. When I designed my DIY digital patchbay I used a 16 m RG59 cable for Tests @ 6 MHz, and with a 75 Ohm satellite cable you could go even longer.
I used a line receiver chip for both SPDIF and AES/EBU. The Arcam Black Box 3 uses an inverter for SPDIF.
 
Are you sure? That is out of spec and might be dangerous for the receiving device.

Years ago, when testing some older DVD players, I was surprised to see this myself. I chalked it up to the fact that most of the consumer devices of that time were being delivered with the cheapest cables possible that still worked. I don't know if it is still the case. That would be something for someone who has more time and interest at this point to test.

Legacy Rane app note (PDF):

RaneNote 149 - Interfacing AES3 & SPDIF
 
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The characteristic impedance of a coaxial cable used for SPDIF will matter very little for short distances—ten to fifteen feet.
The note you shared also suggests the opposite:

1714836706553.png
 
The note you shared also suggests the opposite:

The specification shows the limitations of consumer SPDIF. No fixed data rate. Supports only 32KHZ, 44.1KHz or 48KHz. Only supports a max of 20 bits. Only supports stereo. But we know that it does more than that. Those 'unspecified' capabilities can largely be attributed to the designers of the chipsets that are used in modern implementations.
 
Well, yes, but normally in exactly the way you have experienced here... either the signal passes or it doesn't. :)

If you need more reassurance, consider how the digital signal moves through the cable.

One sample of a 16-bit signal looks like this:

1101001010001001

The next sample might be:

1101001010001011

There are 44,100 of these per second moving through the cable. They are represented by high and low voltages moving through the cable, high voltage =1, low voltage =0. I am oversimplifying it since there is a protocol around the raw data which I frankly don't fully understand, but this is the basic idea.

To alter the frequency response of an analog signal, sometimes just a certain value of capacitance in the circuit is enough. But to alter the frequency response of a DIGITAL signal, you need to operate on multiple samples at once, and have a feedback loop that adds/subtracts/multiplies the samples together. This is how digital filters are constructed.

But if there is something wrong with the cable, it will only affect individual 1s and 0s going by. There is not even a way for full samples to be affected in a consistent way, let alone entire groups of samples being mathematically compared to each other.

So if there is digital interference, generally you will only hear an interruption in the whole signal (as you heard here) or random noise. If you randomly "flip bits" in digital signals, the audible result is simply noise. Because of the protocols I mentioned earlier, often there is some protection against noise also. (Consider that your bank transactions or video downloads don't get any "random noise"... ever, and we can see why digital audio is so robust, too. "Bits are bits". )

So don't worry about losing high frequencies or anything like that over a digital cable, that kind of audible effect is essentially impossible.
@kemmler3D

I think you have over simplified this. I might agree with your conclusion but not your reasoning.
Suppose you have a 16 bit sample which looks like:

0000000000000000 and it is followed by other samples which look like:
0000000000000000
0000000000000000
0000000000000000

etc etc

As the transmitting & receiving devices don't share a clock signal it would be impossible to maintain synchronisation if voltage levels were used to denote logic levels.
Instead a 'self clocking' protocol is used, wherein the clock is embedded within the data. It is completely wrong to say that 'high voltage =1, low voltage =0'

AFAIK SP/DIF uses biphase encoding. I think there is a link in this thread to an STM paper which explains this? A similar protocol is used by ethernet which uses Manchester encoding. In essence, a logic level is denoted by a transition in the signal level. Not by the signal level itself.

Sorry to be pedantic. I used to teach this stuff ;)

Cheers

Ian
 
@kemmler3D

I think you have over simplified this. I might agree with your conclusion but not your reasoning.
Suppose you have a 16 bit sample which looks like:

0000000000000000 and it is followed by other samples which look like:
0000000000000000
0000000000000000
0000000000000000

etc etc

As the transmitting & receiving devices don't share a clock signal it would be impossible to maintain synchronisation if voltage levels were used to denote logic levels.
Instead a 'self clocking' protocol is used, wherein the clock is embedded within the data. It is completely wrong to say that 'high voltage =1, low voltage =0'

AFAIK SP/DIF uses biphase encoding. I think there is a link in this thread to an STM paper which explains this? A similar protocol is used by ethernet which uses Manchester encoding. In essence, a logic level is denoted by a transition in the signal level. Not by the signal level itself.

Sorry to be pedantic. I used to teach this stuff ;)

Cheers

Ian
This is great, thanks for the clarification! I know enough to know I don't know the protocols, and I am not surprised that someone here actually does know. Cool!

e: The wikipedia page does make refernence to this, I have no excuse. https://en.wikipedia.org/wiki/S/PDIF

The point I really wanted to make was just "digital signals don't work like that" but I got carried away saying things that were only vaguely related to the reality of it, instead. :D
 
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I think you have over simplified this. I might agree with your conclusion but not your reasoning. [..]
In essence, a logic level is denoted by a transition in the signal level. Not by the signal level itself.

Sorry to be pedantic. I used to teach this stuff ;)
Welcome to ASR, Ian. One more expert, thanks for joining.
 
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