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My family and I managed to break the aerial cable of our TV a few days ago.

The cable is of this kind of design:

enter image description here

And I believe its called a coaxial cable.

Fortunately, someone more practical than me has fixed the cable, but it made me wonder: how do these cables work?

The person in question was talking about the signal going from the outer of the inner to the inner of the outer. This made sense at the time, but I was doing a little more reading and that seemed to focus on the current flowing through the inner core, with the outer layer acting as a shield.

So, how does a signal get passed through a coaxial cable?

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  • \$\begingroup\$ "the outer of the inner to the inner of the outer" would be trying to express the effect of "skin depth" (or "skin effect") and high-frequency signals. \$\endgroup\$ – Ecnerwal Dec 26 '15 at 15:52
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    \$\begingroup\$ This is a very broad subject for someone with limited knowledge. \$\endgroup\$ – Andy aka Dec 26 '15 at 16:11
  • \$\begingroup\$ At a basic level... it's the same as any two-wire cable, isn't it? \$\endgroup\$ – user253751 Dec 27 '15 at 10:25
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Fortunately, someone more practical than me has fixed the cable, but it made me wonder: how do these cables work?

The very simplest and most basic of answers is there are two conductors and one conductor carries current in one direction while the other conductor carries current in the opposite direction.

But there is a lot more to the pair of wires that make up a cable to consider if you wanted to look into it in detail. For a coax: -

enter image description here

For all two-wire cables there are electric fields and magnetic fields set-up between the two conductors but the beauty about coaxial cables is that these fields, in a proper installation, do not extend outside the perimeter of the coax cable.

So, how does a signal get passed through a coaxial cable?

The signal's energy exists in the gap between the outer and inner conductor and it travels through the cable to the far-end (the load) as an electromagnetic wave. This EM wave carries the power of the signal and it carries the electric field and magnetic fields in a certain ratio. This ratio is known as the characteristic impedance of the cable.

There are also losses due to the resistance of the conductors and these can be significant. There are also losses in the dielectric (the material that seperates the inner and outer conductors) and at higher frequencies this loss can limit the use of a coax cable.

Giving a simple answer to the question is really problematic if all you might know is ohms law but if you are interested there are a lot of things you can look-up on google such as: -

  • Characteristic impedance
  • Speed of propagation of signals in cables
  • Reflections coefficient
  • Voltage standing wave ratio

All the above can contribute to signal reflections such as shown below: -

enter image description here

A signal travels from left to right along a perfect coax cable but that coax cable changes to a different characteristic impedance at a position shown by the vertical line. When the signal "hits" that point, some energy is reflected back up the cable whilst some energy continues down to the load.

This answer may be already more complex than you are currently able to cope with so I'll stop at this point.

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    \$\begingroup\$ What is the source of that reflection animation? That's something I'd love to show to my interns. \$\endgroup\$ – Krunal Desai Dec 26 '15 at 18:00
  • \$\begingroup\$ @KrunalDesai just right-click and "save image as" will do that. It came off the web somewhere but I can't remember where. \$\endgroup\$ – Andy aka Dec 26 '15 at 18:07
  • \$\begingroup\$ @Andy - Can the reflected signal distort the incoming signal? If yes, how is the signal integrity usually maintained in these cases? \$\endgroup\$ – Whiskeyjack Dec 26 '15 at 20:40
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    \$\begingroup\$ @Whiskeyjack yes it can - for instance, if the signal were "continual" like data, the reflected signal has a non detrimental effect initially - the data amplitude can be doubled in some cases but when it finally returns to the original sending end it can be a much delayed version of the data and completely screw up the data integrity IF it also reflects at the sending end because then, it will be like new data and old data travelling simultaneously together and that doesn't work too well! \$\endgroup\$ – Andy aka Dec 26 '15 at 20:44
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    \$\begingroup\$ @Whiskeyjack In addition to what Andy wrote above, if there is sufficient power involved, you can get high enough unintended voltages at the near end (transmitter side) of the cable that it can actually cause equipment damage to the transmitter, particularly in the final power amplifier stage. This is why some computer wiring like old-fashioned SCSI, coaxial cable networking and so on relied on terminators. It's less of a problem now that many types of computer cabling are strictly point-to-point by design rather than use a bus topology. \$\endgroup\$ – a CVn Dec 27 '15 at 0:14
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This cable contains two "wires", which are used to pass the signal.

The difference is that the two wires are concentric, one is completely around the other. That's why this is called a "co-axial" cable. Both wires are around the same axis.

The wires being coaxial has two important effects:

  1. The outer conductor acts as a shield for the inner. Any external electrical fields can only couple to the outer conductor. If this is grounded, then sensitive signals can be run on the inner conductor without them picking up noise due to these fields.

  2. The transmission line impedance can be well controlled. TV electronics tends to be designed for 75 Ω transmission lines.

    TV signals are high enough frequency so that transmission line effects are significant. To deal with that, the electronics is designed for a specific cable impedance, and cables are designed to have a well controlled impedance close to what the electronics expect. As I said above, that impedance is usually 75 Ω.

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    \$\begingroup\$ And if you're wondering what "impedance" is from this answer, an easy explanation is that it's a measure of the amount the wire suffers from the thing that makes coils of wire into electromagnets (a phenomenon /along/ the wire), vs the thing that makes +ve and -ve attract (/between/ the two wires). It turns out this is constant depending on what the crosssection of the wire looks like. The electrons at these high frequencies slosh to-and-fro being affected by these two things in ratio. If you get it wrong, the sloshing goes wrong, like the way water waves break on an underground obstruction. \$\endgroup\$ – Dan Sheppard Dec 26 '15 at 23:51

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