We know that in terms of copper or metal conductors, when one part is given heat the other part also gets hot. This is some kind of transmission of electrons.

Signal transmission in terms of data communication is not any different than that, I believe. But I can't get a clear picture of how the digital or analog signal that is passed through a coaxial or twisted pair copper cable is handled by the conductor. What happens to the voltage of the signal? Or why don't the collisions of electrons inside the conductor attenuate the signal, distort the signal, or don't add any noise to it?

You can point me to any resource book if suitable or answer it here.


closed as unclear what you're asking by John Rennie, ACuriousMind, Rob Jeffries, Gert, heather Aug 3 '16 at 11:48

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.


You seem to have a good instinct about the issue. The resistance of conductors causes signal degradation for high frequency components and it adds noise. You can separate the two issues in most cases, but they both play a major role in the design of communications systems.

The most important step in understanding signal conduction on wires is to realize that it is not actually the electrons that get transmitted, but it's the electromagnetic field that surrounds the conductors. The most obvious sign of this is that the electrons in a wire are moving at a fraction of a mm/s, but the signals on cables are being transmitted at roughly one half to two thirds of the speed of light.

One way of thinking about this is that the main function of the electrons in the conductor is to keep the electromagnetic field from "escaping" into free space. You probably know that in the static case the potential on the surface of a conductor is constant. A wire is therefor a "guide" for an electrostatic potential. Wherever the wire leads, the potential follows. While the field of a free charge has a $1/r$ potential and decays very quickly with distance, the potential of the same charge on a long wire will be exactly the same, independent of the distance (for the static case without current flow). This is the reason why we can generate a GW of electric power in a large power plant and transmit it hundreds of miles to where the energy is needed with reasonable losses. The potential along the wired does, of course, give rise to a radial electric field $\vec E$ surrounding the wire.

In addition, when there is a current flowing trough the wire, there will also be a cylindrical magnetic field $\vec B$ surrounding the wire and it will also be the strongest right around the wire along its entire length, so, again, the wire acts as just the right boundary conditions for the magnetic field to stay localized right where we want it.

In electrodynamics you will learn that there is also a quantity called the Poynting-vector $\vec S$, which is a cross product of electric and magnetic field: $\vec S=\vec E \times \vec B$. The Poynting vector describes the flow of energy due to the electromagnetic field. Because the electric field surrounding a straight wire is radial and the magnetic field has cylinder symmetry, the Poynting vector happens to be aligned parallel to the wire. Wikipedia has a nice drawing showing the fields and the Poynting vector in a simple circuit with energy flow: https://en.wikipedia.org/wiki/Poynting_vector

This is how energy flows trough space around the wire from the source to the load. It is not being transported by the electrons in the wire, but it is being transported by the field on the outside! So while the guiding electrons can only move very slowly, the energy can flow almost at the speed of light.

Now, if we want to understand the transmission of dynamic signals around pairs of conductors (we always need two because there has to be a current return path), then we have to solve Maxwell's equations or a simplified model of them called the Telegrapher's equations: https://en.wikipedia.org/wiki/Telegrapher%27s_equations

To derive the Telegrapher's equations we imagine two long parallel conductors and we look at the electric and magnetic fields along a small section. One can then either use symmetries and simplifications to turn the Maxwell equations into a one dimensional wave equation or one can use a simplified circuit model for short segments. The most common way of thinking about it is that the wire segment is an inductor in series with a small resistance (that's the resistance of the wires on that length of wire) and a small capacitance between the two wires in parallel with another resistance which models the finite isolation resistance of the dielectric (the latter is usually not a big problem, though).

Both approaches lead to the same equations and when we solve them for the lossless case then we get traveling wave solutions in both directions. The signal transmission can be characterized by an effective velocity and a cable impedance. When there are losses in the cable due to the finite conductivity of the wires, then there will be an exponential dampening and some signal dispersion, i.e. a wave packet will slowly spread out as it travels down the cable. Both effects can be minimized by choosing proper materials and cable geometries and today they are being carefully managed by digital signal equalization technologies, which mathematically model the dispersion and correct for it.

And this bring us to the question of noise. As we said, our conductors have a finite resistance and there is an exponential loss of signal along the length of a cable. This loss is frequency dependent and greatly increases with frequency. In practice these signal losses will limit cable runs to around 100m for frequencies above 1GHz, which you will easily be able to identify as typical limits for e.g. GBit ethernet connections.

But we are not only losing signal, we are also picking up noise along the way. A simple thermal spectral noise density formula is given by https://en.wikipedia.org/wiki/Johnson%E2%80%93Nyquist_noise: $v_n=\sqrt{4kTR}$.

For a typical $50\Omega$ coaxial system this means that the irreducible thermal noise floor is set by the terminating resistance (either on the driver or the receiver side) and it is limited to a approx. $0.9nV/\sqrt{Hz}$. For a 1GHz bandwidth system this give us a noise voltage of roughly $0.9\times\sqrt{10^9}nV\approx 28.5\mu V$. In practice this will probably be more like 50-100% larger due to the additional noise in our electronic circuits. So if we want to have, at least, a $5\sigma$ signal to noise ratio, then we need to inject something on the order of $200\mu V$ times the signal attenuation along the cable, which can range from 30-50dB. In practice this means that we need tens to hundreds of mV for said GBit/s connections over 100m distances. One can get a little better than that with error correction codes, but that's an entire lecture on digital communication. :-)

Hope this gives you an idea where all of this leads...

  • $\begingroup$ Thanks a lot. Its a Very detailed answer. There is a lots of Interesting physics involved here. I am a Computer Science student and it seems that I need to study some Physics and Electronics books now to understand Data Communication Better. $\endgroup$ – Alex Vega Aug 2 '16 at 10:21
  • $\begingroup$ It's a complex topic all the way from the physics of electromagnetic fields to the construction of optimal error correction codes. I used to have a textbook about digital communications and I didn't understand half of it (the half that was related to the mathematical issues :-)). The best way to get a feel for these things is if you can get your hands on lab equipment, i.e. fast oscilloscopes and, even better, a vector impedance analyzer/spectrum analyzer. The second best approach beyond learning the math is by playing with a circuit simulator. $\endgroup$ – CuriousOne Aug 2 '16 at 10:34

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