2
\$\begingroup\$

This is regarding wired connections. I have heard that higher frequencies mean higher data rates since there are more cycles per second you can fit more data in per second. That makes sense but I don't understand why we need them in the first place. Couldn't we have a data scheme that just relies on the presence of voltage being a 1 and the absence being a 0. Just imagine a very long wire and a battery. When I touch the wire to complete the circut it's a 1 when the circut shows no voltage it's a 0. Now imagine I alternate this very very very fast. Other binary schemes can be:

"Low frequency pulse" = 0 "high frequency pulse" = 1, no sound = wait or next bit (imagine pressing 1 vs 9 on a touch tone phone as your coding scheme)

+Voltage = 1, - Voltage = 0, No voltage = no bits yet (To send 1001 you would touch positive, unapply voltage and switch wires, unapply voltage and touch them again, unapply voltage and switch back to the original position).  Once again imagine a machine doing this very very fast.
\$\endgroup\$
  • 3
    \$\begingroup\$ Changing from 1 to 0 very fast is a signal with frequency! \$\endgroup\$ – pjc50 Oct 3 '13 at 20:24
  • \$\begingroup\$ I hear of circuts using a "carrier frequency" and modulating it and it seems as if the speed is dependent on this frequency (for example ethernet is several hundred MHz). What I mean is why can't one skip all the modulation business and have a circut where they simply flip the voltage high\low super super fast and achieve speeds of 1 THz (= 1 Tb/sec) \$\endgroup\$ – John Smith Oct 3 '13 at 20:29
  • \$\begingroup\$ Don't feed the trolls. \$\endgroup\$ – Samuel Oct 3 '13 at 21:05
  • \$\begingroup\$ Not trolling. Sorry if I didn't explain it well. I meant "square wave pulses" as was answered in another comment. \$\endgroup\$ – John Smith Oct 3 '13 at 21:26
  • 1
    \$\begingroup\$ How are there already so many answers to this? I don't understand the question at all... \$\endgroup\$ – travisbartley Oct 4 '13 at 1:58
5
\$\begingroup\$

I don't know what do you mean by "presence/absence" of voltage, but in general, what you describe is exactly how it works:

  • High voltage on the data line is (usually) defined as logical 1
  • Low voltage is logical 0

The "frequency of the circuit" is just the maximal number of times the transition between voltage levels can happen in a second.

Please note that if you send a long string of logical 1's, there is no need for these transitions at all - you can keep the data line constantly at high voltage. However, the question arises "how many 1's were sent during some period of time"? How do you count them if the data line is at constant voltage? For the purpose of resolving the above ambiguity, there is usually a "control" signal which determines the "pace" of the circuitry - the Clock. On each rising/falling edge of the clock signal, the voltage is sampled from the data bus and its value is checked for determining whether it is logical 1 or 0.

In the case of a circuit where Clock is present, the frequency of the circuit is just the frequency of the clock signal.

There are schemes when the circuit is not synchronized with Clock, but these schemes are much more complex and they are (usually) employed in special application which require very high bit rates.

\$\endgroup\$
  • \$\begingroup\$ I hear of circuts using a "carrier frequency" and modulating it and it seems as if the speed is dependent on this frequency (for example ethernet is several hundred MHz). What I mean is why can't one skip all the modulation business and have a circut where they simply flip the voltage high\low super super fast and achieve speeds of 1 THz (= 1 Tb/sec) \$\endgroup\$ – John Smith Oct 3 '13 at 20:27
  • \$\begingroup\$ @John Because the transmission medium has unavoidable intrinsic frequency filters, which means that this super fast signal will be extremely attenuated and distorted, to the point the original message is unrecoverable. \$\endgroup\$ – apalopohapa Oct 3 '13 at 20:36
  • \$\begingroup\$ The common Ethernet data links (there are many) use baseband == no modulation. \$\endgroup\$ – Wouter van Ooijen Oct 3 '13 at 20:47
  • \$\begingroup\$ To further elaborate - the wires themselves have an intrinsic inductance and capacitance. The inductance comes because the wire is carrying a current, and all current-carrying wires have an inductance. The capacitance comes because the wire is a metal conductor at a different potential to its surroundings, i.e. a capacitor. The inductance and capacitance of the wire limit the maximum rate at which the signal can change. \$\endgroup\$ – Li-aung Yip Oct 3 '13 at 20:48
  • 1
    \$\begingroup\$ @JohnSmith, I forgot the main advantage of modulation schemes - using modulation you can use a single physical media and transfer few channels of information over it. \$\endgroup\$ – Vasiliy Oct 3 '13 at 22:16
2
\$\begingroup\$

I hear of circuts using a "carrier frequency" and modulating it and it seems as if the speed is dependent on this frequency (for example ethernet is several hundred MHz). What I mean is why can't one skip all the modulation business and have a circut where they simply flip the voltage high\low super super fast and achieve speeds of 1 THz (= 1 Tb/sec)

The quote above is a comment from the OP but I think it gets closer to the misunderstanding than what he puts in the question.

Pure data transmissions (1s and 0s) don't make efficient use of bandwidth. That's not a problem - the simplicity of transmitting 1 and 0 makes reception of those bits very easy. Between 5V and 0V (5V logic signalling) noise and glitches can come long and make the signal somewhat "different" to the original 0V and 5V but providing there is still a distinction to be made by the receiver, data can be faithfully recovered.

The trouble with normal data is that it can't share a line with another normal data system. The two lots of 1s and 0s end up being additive and sometimes you'll get 0V, sometimes you'll get 5V and sometimes you'll get 10V i.e. rubbish.

A carrier wave, when (say 1kbps) data is applied to it has a bandwidth that is a few thousand Hertz centred on a frequency that might be 20MHz. Below (say) 19.97MHz and above 20.03MHz the side frequencies can be largely removed and not transmitted AND importantly the "modulated" carrier can be received and turned back into the original data.

You could have another data system whose carrier is at 19.9MHz - if modulated with different data (still 1kbps for this example), the useful bandwidth it might occupy is from 19.87MHz to 19.93MHz. The receiver tuned into this carrier won't be interfered with by the transmission at 20MHz.

You could repeat this system and have a multitude of exclusive data systems all sharing the same wire (different carriers of course) and all the data streams are perfectly recoverable by their own receivers.

This is why modulation schemes of one sort or another are used. This one is called frequency division multiplexing. It can use a wire or radio.

It doesn't stop there - you can have time division multiplexing - this type of system allocates a time slot for each data stream. Let's say there are ten data streams each at 1kbps. If the data was sent ten times as fast it only needs to occupy 10% of the usable "capacity" of the cable. Ten systems, each with their own time slot can share one cable.

There are a few other schemes as well but this is beyond the scope of the question.

\$\endgroup\$
1
\$\begingroup\$
  • Transmitting data by completing a switch wouldn't be as effective as you'd think.

    1. You could have a ground loop

      • The ground of the transmiter circuit and the ground of the receiver circuit could be at different potentials, especially true over longer distances in cables. This is generally more true for transmission lines, so you may not have this issue with a PCB.
    2. Transients from turning ON and OFF

      • Turning on anything is similar to introducing a new current source to circuit or sub-circuit. This could mean spikes in your data-stream which may effect your ICs or other PCB components.
    3. Data isn't transmitted as ON/OFF, but as HI/LO

      • HI is above a certain threshold and LO is below that threshold. A signal that is only turning a switch ON and OFF makes it more difficult to differentiate
    4. There is less disambiguation in a data stream instead of single bit pulses.

      • Consider "1001". How would a receiver view this word? "101" is just as easy to interpret as "1001". But how would you know for sure it's "1001"? Maybe by pre-defining a predetermined signal length, i.e. frequency?
  • Using low-frequency and high-frequency signals to denote 0 and 1, respectively, is some sort of frequency modulation method. This would be modulating a carrier wave, similar to your standard FM radio transmission. This method is called frequency shift keying.

  • The third method you described is more similar to RS-232 data transmission scheme. It actually uses +3 to + 15V as logical 1 and -3 to -15V as logical 0. Once again, you want to avoid a 0 level so that you have a clear, definable signals.

\$\endgroup\$
  • 1
    \$\begingroup\$ "I don't know exactly which names denote this type of encoding..." - I think that's Frequency Shift Keying, or FSK. \$\endgroup\$ – Li-aung Yip Oct 3 '13 at 20:45
  • \$\begingroup\$ @Li-aungYip Thanks! I know I'd heard about it some point in the past, but the name escaped me earlier. I'm editing my answer and putting your definition in. \$\endgroup\$ – Shabab Oct 3 '13 at 20:47
  • \$\begingroup\$ What about instead of the clock you used + voltage as 1, - voltage as 0, and no voltage as wait for the next bit? 1001 would be recoverable as +,wait,-,wait,-,wait,+,wait.... Assume the waits are sufficiently small \$\endgroup\$ – John Smith Oct 3 '13 at 21:32
  • \$\begingroup\$ @JohnSmith Yes, your data transfer is correct. Though I don't think I've seen anyone account for the "waiting period" before in the middle of a signal transmission. However, if 0 is wait, then let's consider two words "1001" and "1101". By this new definition, your received word will be "10011101" since a 0-level signal is telling it to get the next bit. \$\endgroup\$ – Shabab Oct 3 '13 at 21:40
  • \$\begingroup\$ To be blunt, it seems like you're trying really hard to avoid clocks and synchronous signals. Is there a particular reason behind this? \$\endgroup\$ – Shabab Oct 3 '13 at 21:41

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.