One of my work colleagues told me that a cable he is sending a signal through is losing power at high frequencies. So he recommends the signal should be amplified before being sent. The explanation was given for the power loss is that higher frequency signals are lossier.

As a newcomer to signal processing, I'd like to understand more about why or how this effect occurs. How is the frequency of the signal causing the power loss to occur?

From Wikipedia's article on Coaxial cable I found this, which seems promising:

If an ordinary wire is used to carry high frequency currents, the wire acts as an antenna, and the high frequency currents radiate off the wire as radio waves, causing power losses.

Is understanding how antennae work key to understanding why the high frequency results in power loss?

  • $\begingroup$ "Is understanding how antennae work key to understanding why the high frequency results in power loss?" Yep. :) $\endgroup$ – Lagerbaer Jun 9 '11 at 21:29

Basically a high frequency means a high change in the electric field sourrounding the conductor. That results in a higher emittance of electromagnetic waves.

If you want an easy to understand analogon: Suppose you are wiggling a rope. That creates a standing wave in the rope. The faster you wiggle, the more wavelength you will see in the rope and the more you will sweat, while you do it :D

  • $\begingroup$ So (to continue with your analogy), my playing the part of the wire, my wiggling is affecting the rope, which is attached to me? I could wiggle a lot more if the rope weren't attached to me, but it is, so I've got to share my potential for wiggling with the rope... Is that about the gist of it? $\endgroup$ – Atreys Jun 9 '11 at 23:50
  • $\begingroup$ Well as model picture yes. But there is a lot of other more complicated effects inherent to the conductor too. Someone mentioned the skin effect and resistance, but radiation is a big contributer and more general. But basically changing E/M-Field will always dissipate energy - be it to internal energy (heat) or an E/M-wave. The higher the change rate (frequency), the higher the total amount of power lost. You can see that mathematically too: $\endgroup$ – con-f-use Feb 23 '12 at 12:49

Losses in coaxial cable are resistive. For low frequencies, one uses the full thickness of the coaxial cable and resistance is low. As frequencies increase, the signal is unable to penetrate as deeply into the conductor. This is called the skin effect.

So as frequencies increase, the amount of metal that is used to carry the signal decreases. The result is an increase in the resistance and hence higher losses.


Your wiggle analogy contains too many poorly defined terms to be absolutely sure, but I think it is correct and may well help you understand what is going on.

Coaxial cables have, as does every other circuit element in the universe, a high frequency cutoff. It comes gradually as the frequency increases but eventually you can't get any power through the coax. The main reason is that each type of coax has a characteristic impedence that is a function of frequency. Another reason is that most losses in the system are frequency dependent. There are dielectric losses inside the plastic that separates the central conductor from the outer conductive shield. With some plastics, high frequencies can make it through the plastic and reach the center conductor. There are stray currents, parasitics, leaks, losses, radiating elements and pretty much all of them get worse at higher frequencies.

Solutions include bigger coax, swapping aluminum for copper, upgrading to teflon wherever possible.


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