5 ms·

What is the theoretical limit (for comparison)?

by 3rd3 12y ago

quink 12y ago
Whatever the theoretical limit is, they're pretty much right at it at this point. Similar to how ADSL2+ was near enough at the theoretical limit of what the frequency spectrum it used could do given the attenuation it was asked to do it at. For the same frequency ranges as ADSL2+, VDSL2 delivered pretty much no improvement. What VDSL2 did do is expand the frequency range up to 17 MHz/30 MHz, depending on the profile. So the question you might want to ask is "How high a frequency can we use?" and in that case the real question you're looking for is "How short a piece of copper do you have?". And the answer to that is very quickly becoming "a very short piece of copper indeed". Either that or a lot of copper that's ideally also very thick... which, considering most copper was rolled out with the aim of providing a low attenuation at 3000 Hz over at most a low number of kilometres isn't often the case. tl;dr: How long is a piece of string?
ajb 12y ago
Depends on a number of factors: the major one is the length of the wire, but also what gauge of wire, how well twisted it is (phone wires aren't very twisted due to not expecting to use the high frequencies), the radio environment, local radio interference standards (telephone wires are leaky at these frequencies), how many of your neighbours also have DSL (although this is improving due to crosstalk cancellation mechanisms in the newer standards), how many watts the local operator is prepared to put down the wire...
rayiner 12y ago
The bandwidth of an electrical link is a function of bandwidth (the range of frequencies present on the cable) and the encoding (bits per second per hertz). The first on is limited by attenuation on the cable. As you increase the frequency, the cable does a less good job of propagating the signal, and your signal level decreases at the other end. This is limited by power and cable quality. The efficiency of the encoding is limited by the sensitivity of the receiver and its internal noise along with external noise. Depending on these parameters, power, bandwidth, and signal to noise, the bandwidth can be arbitrarily high (subject I suppose to quantum effects). See: http://en.m.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem http://en.m.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theor... This system uses a bandwidth of 500 MHz over two pairs. That gives roughly 10 bits per second per hertz. Both are pretty close to the state of the art over unshielded copper cable. You coaxial cable is usually about 600 MHz. But if you use waveguides and the like, you can get into the GHz range.

What is the theoretical limit (for comparison)?

by 3rd3 12y ago

What is the theoretical limit (for comparison)?

quink 12y ago

Whatever the theoretical limit is, they're pretty much right at it at this point. Similar to how ADSL2+ was near enough at the theoretical limit of what the frequency spectrum it used could do given the attenuation it was asked to do it at. For the same frequency ranges as ADSL2+, VDSL2 delivered pretty much no improvement. What VDSL2 did do is expand the frequency range up to 17 MHz/30 MHz, depending on the profile. So the question you might want to ask is "How high a frequency can we use?" and in that case the real question you're looking for is "How short a piece of copper do you have?". And the answer to that is very quickly becoming "a very short piece of copper indeed". Either that or a lot of copper that's ideally also very thick... which, considering most copper was rolled out with the aim of providing a low attenuation at 3000 Hz over at most a low number of kilometres isn't often the case. tl;dr: How long is a piece of string?

ajb 12y ago

Depends on a number of factors: the major one is the length of the wire, but also what gauge of wire, how well twisted it is (phone wires aren't very twisted due to not expecting to use the high frequencies), the radio environment, local radio interference standards (telephone wires are leaky at these frequencies), how many of your neighbours also have DSL (although this is improving due to crosstalk cancellation mechanisms in the newer standards), how many watts the local operator is prepared to put down the wire...

rayiner 12y ago

The bandwidth of an electrical link is a function of bandwidth (the range of frequencies present on the cable) and the encoding (bits per second per hertz). The first on is limited by attenuation on the cable. As you increase the frequency, the cable does a less good job of propagating the signal, and your signal level decreases at the other end. This is limited by power and cable quality. The efficiency of the encoding is limited by the sensitivity of the receiver and its internal noise along with external noise. Depending on these parameters, power, bandwidth, and signal to noise, the bandwidth can be arbitrarily high (subject I suppose to quantum effects). See: http://en.m.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem http://en.m.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theor... This system uses a bandwidth of 500 MHz over two pairs. That gives roughly 10 bits per second per hertz. Both are pretty close to the state of the art over unshielded copper cable. You coaxial cable is usually about 600 MHz. But if you use waveguides and the like, you can get into the GHz range.