13
\$\begingroup\$

How does a coaxial cable carry RF, audio and low freq. signals? I understand there must be a difference between all these, for example the return path being through the shield or not.

Can someone explain what happens in each case of the 3 types of signals? In general terms, how the shield is used, etc.

\$\endgroup\$
7
  • \$\begingroup\$ What cable? What signal? What pages on the internet? Are we talking about an audio signal? RF? \$\endgroup\$ – Phil Frost Oct 8 '13 at 19:10
  • \$\begingroup\$ @PhilFrost Coaxial cable with a general voltage signal. Does it matter if it's audio or RF? \$\endgroup\$ – 1p2r3k4t Oct 9 '13 at 7:47
  • 1
    \$\begingroup\$ Yes it matters. That's why I asked. \$\endgroup\$ – Phil Frost Oct 9 '13 at 12:37
  • \$\begingroup\$ @PhilFrost Can you tell me about both? If that would be too long, then let it be only RF. Thank you \$\endgroup\$ – 1p2r3k4t Oct 9 '13 at 12:44
  • 1
    \$\begingroup\$ It's still pretty unclear. The answer to how a cable (coax or not) carries signals in the general case encompasses the entire field of electrical engineering. You will have to narrow it down more. Your second paragraph presents a false dichotomy and is not answerable. \$\endgroup\$ – Phil Frost Oct 14 '13 at 17:31
5
\$\begingroup\$

For RF signals, the shield carries the signal just as much as the center conductor, and if it is broken anywhere, the cable performance will seriously degrade.

The shield does limit noise, by containing the desired "signal" EM field between the center conductor and the inside of the shield, acting essentially as a Faraday cage, keeping your signal inside, and other signals (noise) outside. The shield is able to completely (ideally) contain the EM field carrying the signal precisely because it carries equal and opposite currents to those in the center conductor at every point along the cable. Were this not the case, there would have to be an external field. Thus, the shield is also the return current path.

The geometry of the shield relative to the center conductor also defines the characteristic impedance of the cable. If there are any discontinuities in the shield, the signal will be distorted by reflections. In the case of the shield being entirely disconnected at one end, the distortions are likely quite horrible, and the power transfer from line driver to receiver will likely be quite poor, as most of the power will be reflected back at the line driver.

\$\endgroup\$
5
+50
\$\begingroup\$

A coax cable is what is known as a waveguide and the energy flows in the space between the central conductor and the inside of the outer (shield). The propagation is TEM00 (Transverse Electric Magnetic) mode which duplicates free-space propagation. Interestingly, the propagating wavefront creates electric currents on the surfaces of the conductors that supports the wavefront. The depth of penetration of these currents is dictated by the skin depth and hence is controlled by the frequency of the signal (the higher the frequency the less the penetration). It is this effect that essentially isolates the inner signal from any signal that is flowing on the outside of the coax (which also has a limited depth of influence).

Here is an excellent picture from York University enter image description here

It shows the very desirable TEM mode of operation, with the electric field lines being radial and the magnetic field lines being circumferential. The energy flows in the medium between the conductors. The localized currents in the conductors will support the adjacent fields with no net movement of charge along the length.

Conversely a DC signal will flow along the conductors.

And here is a picture from Microwaves101.com

enter image description here

Shows surface current at 40 GHZ.

\$\endgroup\$
2
  • \$\begingroup\$ This is for RF signals? How are the two cable conductors connected to this RF signal? \$\endgroup\$ – 1p2r3k4t Oct 17 '13 at 14:12
  • \$\begingroup\$ You get localized current loops that match the localized magnetic fields in the medium between the conductors. THere is no net energy flow in the metal (except for DC). This is proven because you can also make a waveguide that uses the same material and analysis that doesn't have a central conductor and indeed all that is needed is a dielectric slab and the wave will propagate down it, it slightly different configuration (TE and TM modes vs TEM mode). The coax because it has both a central and outside conductor supports TEM mode. \$\endgroup\$ – placeholder Oct 17 '13 at 14:22
4
\$\begingroup\$

What is a signal? It is a voltage, so it is a potential difference between two conductors. Therefore, both conductors are necessary to transmit information. When you say that one of the conductors is ground, all it means is that you measure all the voltages with respect to that potential, ie it is your reference.

\$\endgroup\$
2
  • \$\begingroup\$ So the shield carries the same current as the main conductor? And is shields from outside noise at the same time? \$\endgroup\$ – 1p2r3k4t Oct 9 '13 at 7:50
  • \$\begingroup\$ yes, the current is the same. As far as shielding, it simply does not allow almost any EM waves to penetrate the cable creating a 'noise' voltage signal \$\endgroup\$ – Yuriy Oct 10 '13 at 15:12

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.