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How does network interface card work physically?

Hi,

This may be more a question of physics, so pardon me if there's any inconvenience.

When I study computer networks, I often read something like this

in order to represent a signal, we place some voltage on one end of the wire and the other end will detect the voltage and thus the signal.

So I am wondering how a signal exactly passes through wire?

Here's my current understanding based on my formal knowledge about electronics:

First we need a close circuit to constrain/hold the electronic field. When we place a voltage at somewhere A of the circuit, electronic field will start to build up within the circuit medium, this process should be as fast as light speed. And as the electronic field is being built up, the electrons within the circuit medium are moved, and thus electronic current occurs, and once the electronic current is strong enough to be detected at somewhere else B on the complete circuit, then B knows about what has happend at A and thus communication between A and B is achieved.

The above is only talking about the process of sending a single voltage through wire. If there's a bitstream and we need to send a series of voltages, I am not sure which of the following is true:

  • The 2nd voltage should only be sent from A after the 1st voltage has been detected at B, the time interval is time needed to stimulate the electronic field in the medium and form a detectable electronic current at B.

  • Several different voltages could be sent on wire one by one, different electronic current values will exists along the wire simutaneously and arrive at B successively.

I hope I made myself clear and someone else has ever pondered this question.

(I tag this question with network cause I don't know if there's a better option.)

Thanks,

Sam

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marked as duplicate by Dennis Williamson, Randolph Potter, Linker3000, Ivo Flipse Feb 11 '11 at 0:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Isn't this just a reworking of your previous question "How does network interface card work physically?" superuser.com/questions/238617/… –  Linker3000 Feb 10 '11 at 7:10
    
@Linker3000 this one has a different focus to his previous one –  barlop Feb 10 '11 at 7:16
    
Agreed - different focus (hence 'reworking'), but still asking how the physical layer works. –  Linker3000 Feb 10 '11 at 7:23
    
Just FYI, the speed of a signal through a cable is not at light speed (c) - it's typically about 0.6c to 0.7c –  Linker3000 Feb 10 '11 at 7:25
    
The speed of an electromagnetic wave in free space is c, whether the wave is at light frequencies or radio frequencies. If you could build a transmission line tiny enough to carry waves at light frequencies, those waves would travel at about 0.6c, too. The speed of light in an optical fiber is also about 0.6c. –  garyjohn Feb 10 '11 at 8:53

3 Answers 3

You have two questions

Do a succession of electrical signals pass along a wire like multiple small railway carriages simultaneously moving along a railway track? Or is it like two men pushing on the ends of a wooden pole?

Make a simple thought experiment - lets reinvent gigabit ethernet.

Our ethernet cable will be 300 metres long. It will be a perfect conductor and we will ignore any laws of physics that prevent a signal propagating at the speed of light. So when I connect one end of the wire to my PP3 battery, the signal will travel to the other end, 300m away, in a microsecond (300/(3*10^8)). Gigabit means 10^9 bits a second, so in a microsecond I need to send 1000 bits, so those 1000 voltage changes must all be present somewhere in the 300m wire just as the first bit reaches the other end.

How does an electrical signal pass through a wire

An electric field is applied at the ends of the wire. This attracts or repels nearby charge carriers and makes them slowly move a tiny distance. For metallic wires the charge carriers are electrons, for other conductors they might be positively charged ions or a mix of positively and negatively charged ions moving in opposite directions simultaneously. The charge carriers a bit further along the wire are affected by the tiny movements of their neighbours. Although the physical movement of charge carriers is very slow, the effect on their neighbours occurs extremely quickly (think of Newton's cradle). In this way a tiny disturbance in the position of charge carriers propagates along the wire until it reaches the other end.

This part of the question might be better asked on http://physics.stackexchange.com/

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Fairly logical. But it is more like a proof by contradiction. –  smwikipedia Feb 10 '11 at 14:21
    
I have expanded the answer –  RedGrittyBrick Feb 10 '11 at 16:14
    
Thanks RedGrittyBrick. I really appreciated it. And thanks for show me the physics site. –  smwikipedia Feb 16 '11 at 14:10

I'll take a stab at it. First off, the concepts of open and closed circuits work well only at DC and at frequencies where the wavelength of the signal is significantly longer than the size of the circuit. At higher frequencies, the behavior of the voltages and currents becomes more complicated.

As for sending a bitstream through a pair of wires, both of your alternatives are true. The IEEE 488 bus (a.k.a. GP-IB or previously, HP-IB) uses your first method. The data transfer process uses 8 data lines and 3 handshake lines. The sender puts voltages on the 8 data lines, waits long enough for the voltages to reach the receiver, then puts a voltage on one of the handshake lines telling the receiver that the data is there. When the receiver sees the voltage on that handshake line, it measures the voltages on the data lines and determines the 8 binary values that those voltages represent. The receiver then puts a voltage on another handshake line to tell the sender that it has received the data and that the sender can send the next 8 bits. (The handshaking is a little more complicated than that, but that's close enough for this discussion.)

All that waiting takes time and limits the data rate on the bus. Also, the maximum data rate becomes lower as the distance between sender and receiver increases because it takes longer for the voltage changes to travel between sender and receiver.

Long-distance communications links and modern computer networks work more like your second method. The sender sends many bits together as a sequence of voltages on a pair of wires. The rate at which the sender can change the voltage on the wires is limited by the bandwidths of the wire pair, the sender's transmitting circuits and the receiver's receiving circuits. When the bandwidths of the components are such that the voltage can be changed rapidly, and the distance between sender and receiver is large, the sender can send many bits in sequence before the receiver has seen the first bit.

There are many ways to send binary values on a pair of wires besides just having one voltage represent a "1" and another voltage represent a "0". For example, you could use 4 different voltages, representing the binary values "00", "01", "10" and "11". On a pair of wires that allow the voltage to be changed only a certain number of times a second, using 4 voltages instead of just 2 allows you to send twice as many bit values in a given interval of time.

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I find it very hard to believe that the understanding he showed in his question was correct. Yet you haven't countered anything he said. He said "different electronic current values will exists along the wire simutaneously" are you sure about this? He talked about electronic field building up as fast as light speed. Are you sure about that? If you know better, then rather than write a generic essay leaving his mistakes unresponded to, and not dealing directly with any misunderstandings he has, why not correct his understanding. I find it hard to believe that his understanding is correct. –  barlop Feb 10 '11 at 7:22
    
@barlop: He was looking for understanding. I thought I could help him with understanding some of what he was asking about, and I addressed the simpler issues. Electromagnetic fields and the propagation of waves on transmission lines are complicated subjects that I don't know how to explain adequately in a few paragraphs and without drawing pictures. Yes, the voltage and current can vary along a transmission line. Light is an electromagnetic wave and travels at a speed depending on the medium, just as waves on a transmission line do, so saying that fields travel at light speed is reasonable. –  garyjohn Feb 10 '11 at 8:33

There are various ways of sending an information signal along a wire medium. Three ways are a) varying the voltage, b) varying the current, and c) varying the phase.

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AFAIK, what we could measure/detect ultimately is the current, things like voltage or else are just deduction of current. Am I right? –  smwikipedia Feb 10 '11 at 14:22
    
Voltage is the potential difference across two lines, i.e. between the signal line and ground. Current, if I recall correctly, is the flow of electrons over time, or better stated coulombs per second. However, I think what you measure will ultimately depend on your instrument or detecting device. If you'd like to get into the theory of electronics I would suggest starting to explore Coulomb's Law, Ohm's Law, and Kirchoff's Law. You might look through MIT's OpenCourseWare (OCW) for lectures, learning materials, etc on Electricity. This one might be helpful link –  Mike Feb 10 '11 at 16:17
    
It should be noted that voltage and current cannot be varied independently. Their values are coupled by the impedance across and through which they are induced or measured. –  garyjohn Feb 10 '11 at 20:07
    
@garyjohn - I think in the context of the question, we are looking at the variations of a signal, whether it be voltage, current or phase so yes, in this aspect, the hardware is going to control one independently of the other. For instance the simplest example is flipping between 0V and 5V to represent 0s and 1s. We are not concerned with the current other than there is enough to overcome the resistance/impedance on the line and make it to the other side where the receiver detects the voltage level change (theoretical model). –  Mike Feb 10 '11 at 21:03
    
@garyjohn - In the past I have used constant voltages sources and varied the current output in order to reach the lasing threshold of laser diodes. So in this example, current was varied independently of voltage. The same can be said for constant current sources as well where the voltage is varied instead of the current. Have I misunderstood your statement? –  Mike Feb 10 '11 at 21:10

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