2
\$\begingroup\$

I am not an electric engineer so sorry for questions that might miss some fundamentals.

I am trying to understand the ADSL modem. The ADSL modem sends the data over different frequencies in parallel. However I have learned that faster frequencies means higher data rate. So I assume the data is transferred over the "bins" with different speed.

Which algorithm exactly is deployed to ensure that data is restored in the correct order? The link with simple explanation will be fine.

UPD: As smart people mentioned, I didn't meant actual speed of signal but bitrate per channel.

Improve this question
\$\endgroup\$
3
\$\begingroup\$

No, the speed (distance covered per unit of time) is always the same. A higher (not 'faster') frequency channel can accommodate more data per unit of time.

Think of it as sending a sequence of small letters and a sequence of fat books by mail. They will both travel at the same speed (let's say 100 kilometers in one day), but the 'book stream' conveys much more information per day.

Improve this answer
\$\endgroup\$
4
  • \$\begingroup\$ What you mean is that the data is distributed in different sizes per channel across the same time slots? Suppose I have two channels 1 and 2 with speed X and 10X per timeslot. If for simplicity I have to transmit 11X, what I will do is allocate the data of sizes X and 10X respectively and send it over both channels simultaneously. Both ends will agree that channel 1 contains the first bytes. After both channel completed the transfer over the same time of virtual timeslot, receiver will add X (from channel 1) and 10X to restore the sequence? \$\endgroup\$ – Boris Jul 27 '14 at 9:36
  • 2
    \$\begingroup\$ Actually, the speed isn't necessarily the same on each channel. The number of symbols per second is the same for each channel, but the number is bit per symbol (using QAM encoding) is dynamically altered based on the amount of noise on each channel. \$\endgroup\$ – RJR Jul 27 '14 at 10:22
  • \$\begingroup\$ That is data speed though @RJR, rather than signal propagation speed. Hopefully my answer covers this, feel free to suggest if I've made any mistakes. \$\endgroup\$ – David Jul 27 '14 at 10:40
  • \$\begingroup\$ @Boris I don't know how data is split and combined, but I can thing of a lot of ways I could do it. Data is typically divided in packets, so I would for instance start transmitting a packet on a currently idle channel. That would automatically divide the packets over the channels according to their data rate. \$\endgroup\$ – Wouter van Ooijen Jul 27 '14 at 19:01
4
\$\begingroup\$

I think you are possibly confusing two different issues. Throughout I'll be quoting from this page, which is superb and worth a read.

Firstly we see this:

The whole available frequency band is split into a distinct number of sub-channels.

So we split our entire available frequency up, the page also says that sub-channels may be referred to as "carriers, bins or buckets" if used for data (instead of signalling).

Not all possibly channels are used and not all are used for data. Importantly, as Wouter says, the different frequencies used will not affect speed.

However, if the sub-channel is used for data, it's a "bucket" or "bin", then the following is relevant:

The amount of data bits that can be carried in the bin can vary depending upon the quality of the signal at the particular frequency range for that particular bin.

So if you examine two sub-channels that have been allocated for carrying data you might find that one can carry data at a "faster speed". This means more data at a time rather than "quicker" data that arrives faster.

This is because the signal quality is higher (less noise, less interference, etc.) - all the standard reasons in the black art of RF engineering.

Because the data doesn't arrive "quicker" or "slower", or before or after data sent at the same time, reconstructing it is not a particular challenge. It is the same as reconstructing data if all bins carried the same amount of data per time slot, there is just slightly more data from some bins.

Improve this answer
\$\endgroup\$
3
  • \$\begingroup\$ Thanks, so the piece I was missing is this equal "timeslot" allocation for all channels. In this case it is of cause no problem to restore the original sequence. \$\endgroup\$ – Boris Jul 27 '14 at 10:55
  • \$\begingroup\$ From the page I linked in the answer: "The QAM rate is said to be 4,000 symbols per second". These symbols carry a varying amount of bits, depending on SNR. This is where it gets quite complex and the page is worth reading. \$\endgroup\$ – David Jul 27 '14 at 11:03
  • \$\begingroup\$ I appreciate the answer, I think I get it now. I am assigning answer though, to Wouter as he was first to point out my missing" link. \$\endgroup\$ – Boris Jul 27 '14 at 12:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.