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I know this is a very basic question but the answers returned by google are way too complicated for me to understand. I am not asking about modulation here. What I want to know is what exactly is carrying the data.

Please let me explain my doubts:

Suppose from my PC, if I want to transmit the number ten. It will be converted to binary and become 00001010. Then it will be sent to the modem which will convert to an analog signal. This analog signal will then travel over the wire and reach its destination where it will again be converted to binary and the user will receive the number.

Now if it were a digital signal, the value would be transmitted as a combination of high and low voltage.

What flows through the wire is current.

How does this current carry the data? Current is basically flowing electrons.

The speed of the electrons depends on the voltage applied (that's what I remember from school). But my data is received almost instantly.

So if it were current carrying my data it would not travel this fast.

I read somewhere that wires transmit data almost at the speed of light. How?

What is carrying my data? Only EM waves travel this fast.

Please help me. I may be missing a lot many basic points here. I have not studied communication modes.

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    \$\begingroup\$ The speed of the electrons depends on the voltage applied (that's what I remember from school). But my data is received almost instantly. If you push a rigid rod on one side, the other side will move (almost) immediately regardless(if we are reasonable enough about it) of the rod length. The electron on the transmitting side is not the same electron on the receiving side.. \$\endgroup\$ – Eugene Sh. Jul 28 '16 at 17:01
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    \$\begingroup\$ To add on to that analogy, the motion will travel through the rod at a specific speed: the speed of sound in that medium. The analogy carries through to electricity, there is a specific quantifiable speed the signals travel down the wire, which is related to the wires permittivity. \$\endgroup\$ – whatsisname Jul 28 '16 at 17:18
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How does this current carry the data?

Current and voltage are inseperable. The current is flowing because there is a voltage on the wire, and there is a conductive path from that voltage to a lower voltage.

So we can say the data is encoded as voltage pulses or current pulses, it doesn't really matter. Often a high voltage (5 V) indicates a "1" and a low voltage (0 V) indicates a "0". But you could choose any two voltages you like. 3.3 and 0 V. 0 and 3.3 V. -0.8 and -1.2 V. According to what works best in your design.

I read somewhere that wires transmit data almost at the speed of light. How? What is carrying my data? Only EM waves travel this fast.

Another way to look at things is that the voltage at a location on the wire is just a simpler way of looking at the fact that there is an electric field between the wire and everything around it.

When a signal propagates along a wire, it's actually the electromagnetic field between the wire and a nearby "ground" or "return" conductor that is propagating. So it is actually an EM wave, not a massive object (like an electron) that is carrying the signal along the wire.

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  • \$\begingroup\$ So when the data leaves the modem, it is carried by EM Waves that are generated because of the flowing current? \$\endgroup\$ – Sunshine Jul 28 '16 at 17:17
  • \$\begingroup\$ Also from what I remember, the carrier wave (in this case the EM wave) is modified (or modulated). For example, the amplitude, frequency or phase is changed according to the data signal. So one of these attributes of the EM waves must be changing right? \$\endgroup\$ – Sunshine Jul 28 '16 at 17:24
  • \$\begingroup\$ @Sunshine Well, it's more complicated than that. A modulated signal (that's what a signal that carries information is) can be very different according to the modulation scheme it is used to "put the information" onto the carrier signal (i.e. the unmodulated signal). You can vary any property of the carrier signal (amplitude, phase, frequency - even polarization of an EM wave can be varied to transmit information) and this only in the basic analog modulation schemes. In complex modulation schemes (especially digital modulations) more than one property is varied at once. \$\endgroup\$ – Lorenzo Donati supports Monica Jul 28 '16 at 18:09
  • \$\begingroup\$ It is indeed modulated by the MOdulator-DEModulator. The exact form of the modulation is described by various 'V' standards: en.wikipedia.org/wiki/List_of_ITU-T_V-series_recommendations - start at V21 and work upwards. \$\endgroup\$ – pjc50 Jul 28 '16 at 18:18
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    \$\begingroup\$ @Shamtam, that's more or less what my last paragraph was trying to say, without turning into a full textbook chapter. \$\endgroup\$ – The Photon Dec 22 '17 at 2:26
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I read somewhere that wires transmit data almost at the speed of light. How? What is carrying my data? Only EM waves travel this fast.

Ohms law is great. It tells you that if you put 1 volt across a 1 ohm resistor, then 1 amp will flow. However it hides a darker truth that is best uncovered if you imagine that the 1 ohm resistor is several miles distant from the 1 volt source and connected by cable.

So, you apply 1 volt and some time later you will see that 1 volt across the 1 ohm load - well that's what you think might happen but it's more complex than that in the microseconds it takes to get down the cable.

In reality, the cable "informs" the 1 volt power source that it's taking 20 mA (this is for cable with 50 ohm characteristic impedance i.e. a lot of coax cables have this impedance). Clearly 1 volt / 50 ohm = 20 mA. So current is initially determined not by the load (too far away) but by the medium of the cable.

So, the 20 mA AND the 1 volt go hurtling down the cable as an EM wave - the cable ensures this and, there is an E field and a H field just like a real radio wave transmitted into the air/atmosphere/vacuum/medium. A vacuum has a characteristic impedance too - it's approximately 377 ohms; meaning that the ratio of E field to H field is 377.

The E and H fields journey to the far end of the cable to be greeted with a 1 ohm load and then strange things start to happen. If the load at the far end was 50 ohms it would be "end of story" but, because the load doesn't match the EM wave "characteristics" you get a reflection sent back to the power source and, after many times too-ing and fro-ing eventually the right current is sent down the cable to suit the load. It's all over in a few microseconds though.

So, it is an EM wave travelling down the cable. And, for that reason, it is always a good idea to consider the use of matching impedances to prevent reflections causing data corruptions.

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Since you are asking this question within the context of a PC and a modem, the answers I present are confined to the telephone domain.

You are correct in your explanation of sending the value "10" from your PC up to the point of the modem converting the 1's and 0's which make up the binary value 00001010. In general the modem actually is converting the 1's and 0's into two different audio tones. This is basically because the telephone system was designed to transmit and receive audio waveforms as a varying electric current. These two discrete values of audio tones ( two distinct frequencies) pass thru the local telephone system as a time-varying current. Once these signals are received at your local telephone company's central office ("CO") (i.e. the place where the telephone wire from your house connects to), they are generally converted to digital data right there and sent over the national trunk lines digitally. At the receiving end of the phone call the CO there reconverts these digital signals back to a time varying current for presentation to the copper telephone lines that run to the subscriber with the receiving modem.

The receiving modem recognizing these two specific audio tones (one tone is a "zero", the other is a "one") and converts them back to a binary string of 1's and 0's. Then, it's up to the PC connected to the receiving modem to convert these 0's and 1's back into 8-bit values.

So that to answer your question about what actually carries the data, it is really a multi-tiered mechanism. The modem converts the 0's and 1's to different time varying signals (the two tones, represented by an analogous time varying voltage) and then pushes these time varying signals thru the copper telephone wires to the CO as time varying currents. The modem converts the time varying signals to time varying currents because the connection to the CO is what is known as a "current loop". The local, copper wire telephone loop to your CO carries electrically-encoded audio signals as currents, not voltages. These electric currents flow very swiftly, so your "data" (which the time-varying current represents), flows very swiftly. Maybe not at the speed of light, but some significant fraction thereof depending on various conditions in the connecting lines.

You see? There are two mechanisms at play here: The binary data is represented as to audio frequency tones and the tones are transmitted in the form of electric currents. At least that's how it works between the modem and the telephone company's CO at both ends of the connection. In between the two participating CO's a whole other set of mechanisms come into play.

Also to correct your thinking, binary data is indeed commonly encoded as two voltage levels in electronic systems, but not always. Some systems encode data as frequencies, like the modem. Others encode data as the phase of a constant frequency signal. And there a few other methods as well.

And leave all that electric wave & E-field propagation stuff to the physicists. It will only confuse you when you are dealing with practical electronic equipment. In this world of EE it's all about voltages and currents. You don't have to understand the phenomena beyond these two parameters to understand much of what goes on in most common electronic devices.

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