3 ms·

A wired phone (POTS) doesn't have "bit rate", it has bandwidth. It could very well be that VOIP uses a lower bitrate than Skype, or Skype uses a better algorith

by tremon 6y ago

A wired phone (POTS) doesn't have "bit rate", it has bandwidth. It could very well be that VOIP uses a lower bitrate than Skype, or Skype uses a better algorithm, but I would lump VOIP in the "Skype" category, not the "wired phone" category.

WalterBright 6y ago
I do know that POTS is analog (at least the wire to my house is, I presume at some point it becomes digital), not digital, but the larger point stands.
mnw21cam 6y ago
A wire phone (POTS) definitely does have a "bit rate". The entire POTS back-end system is digital, and each circuit gets 8000 8-bit samples per second, or 64kbps, uncompressed.
msla 6y ago
Anything with a bandwidth has bitrate: https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampli...
- tremon 6y ago
  Well, to put it simply, no. Nyquist's theories (and Shannon's sampling theorem based on it) only concerned themselves with the time-based discretization of analog signals. As Shannon showed, to accurately reconstruct a signal of bandwidth x, you need to take samples at a constant frequency of 2x. But when speaking of digital signals, discretization is only half the story. The other part of the story is quantization, i.e. how many bits to use to represent each sample. And that amplitude quantization is a lossy operation, regardless of Shannon's theorem about discretization in the time domain. You can't simply say that an analog signal of 40kHz therefore has a bitrate of 80kbps -- using just one bit to represent each sample will not accurately reproduce the original analog waveform, unless the original waveform was a square wave (and you would still need that bit of information to accurately reconstruct it). So what, then, is the bitrate of a discretized, analog signal? It is still infinite, because the amplitude of every sample can be any element of R! Only once you decide on a quantization resolution, i.e. once you determine your acceptable level of signal loss, can you speak of "the bitrate of an analog signal". But that bitrate is only an approximation of the original signal, Shannon notwithstanding.
  - msla 6y ago
    Your flat dismissals do nothing to disprove Shannon's math. Come back when you have equations.

by tremon 6y ago

WalterBright 6y ago

I do know that POTS is analog (at least the wire to my house is, I presume at some point it becomes digital), not digital, but the larger point stands.

mnw21cam 6y ago

A wire phone (POTS) definitely does have a "bit rate". The entire POTS back-end system is digital, and each circuit gets 8000 8-bit samples per second, or 64kbps, uncompressed.

msla 6y ago

Anything with a bandwidth has bitrate: https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampli...

tremon 6y ago
Well, to put it simply, no. Nyquist's theories (and Shannon's sampling theorem based on it) only concerned themselves with the time-based discretization of analog signals. As Shannon showed, to accurately reconstruct a signal of bandwidth x, you need to take samples at a constant frequency of 2x. But when speaking of digital signals, discretization is only half the story. The other part of the story is quantization, i.e. how many bits to use to represent each sample. And that amplitude quantization is a lossy operation, regardless of Shannon's theorem about discretization in the time domain. You can't simply say that an analog signal of 40kHz therefore has a bitrate of 80kbps -- using just one bit to represent each sample will not accurately reproduce the original analog waveform, unless the original waveform was a square wave (and you would still need that bit of information to accurately reconstruct it). So what, then, is the bitrate of a discretized, analog signal? It is still infinite, because the amplitude of every sample can be any element of R! Only once you decide on a quantization resolution, i.e. once you determine your acceptable level of signal loss, can you speak of "the bitrate of an analog signal". But that bitrate is only an approximation of the original signal, Shannon notwithstanding.
- msla 6y ago
  Your flat dismissals do nothing to disprove Shannon's math. Come back when you have equations.