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Ash View Drop Down
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Ash Quote  Post ReplyReply Direct Link To This Post Posted: 13 Jul 2020 at 8:46pm
Thanks for taking the time to rewrite this. I find this topic interesting and would like to understand it in more detail.

I think:
Anti-node = region of peak amplitude
Node= region of zero displacement
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Post Options Post Options   Thanks (1) Thanks(1)   Quote Fatmangolf Quote  Post ReplyReply Direct Link To This Post Posted: 13 Jul 2020 at 10:09pm
It will be the anti-nodes, my notes from October 2012 have:

velocity factor, VF = speed of light in dielectric / speed of light in vacuum = 1/sqrt(e) where e is the dielectric constant

Reception is zero at beginning of cable so quarter wavelength (antinode) out from shielded source is the key position, that is the first peak in the wavelength or lamda /4

For 100 MHz in vaccuum would get 75cm for lambda = 3m, adjust for the VF of PE cable and you get 50cm for antinode/ferrite position (using PE which has VF 66%) - put the ferrite bead there

Halve this for 200MHz, it has half the wavelength so 25cm. This would be at lambda/8 of 100MHz so one bead at 25cm and one bead at 75cm (i.e. 50cm further on from the first one)

Foamed PE has VF 82% so increase the spacing in proportion. Nylon and PVC (in some mains cables) is around 50% so closer than solid PE.

Graham was right and it worked! The theory made sense to me back then and still does. Science and good engineering put a ferrite in the right place on an audio or SPDIF cable for known radio interference, better than randomly chucking a ferrite on a cable as seen elsewhere which makes no sense if your aim is to protect an unbalanced signal by stopping a RF signal imposing itself on the ground/return so it appears on the signal too and overloads the input of the next piece of equipment.

Hope that helps Ash and my acknowledgement to Graham for repeating what he worked out and shared.

Jon

Open mind and ears whilst owning GSP Genera, Accession M, Accession MC, Elevator EXP, Solo ULDE, Proprius amps, Cusat50 cables, Lautus digital cable, Spatia cables and links, and a Majestic DAC.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote jupiterboy Quote  Post ReplyReply Direct Link To This Post Posted: 14 Jul 2020 at 1:19am
Can you only tune the ferrite core by position for one frequency, and how do you choose that frequency?
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Ash Quote  Post ReplyReply Direct Link To This Post Posted: 14 Jul 2020 at 2:16am
A particular frequency is attenuated based on its wavelength. The attenuation curve is a phase-shifted sine wave?? A range of frequencies in the region are reduced in amplitude, with the frequency at 90 degrees receiving the highest reduction.

100MHz region of the radio band is chosen as likely has the best potential for attenuation as it is the typical frequency area used in FM radio transmission so contains the most radio interference to suppress.

If you wish for signal attenuation at an alternative frequency region, the ferrite position is moved to reflect this. The cable has to be connected the "correct" way around (marked arrows on sheath) for the signal attenuation to be in the "culprit" region chosen. It is not a "palindromic" distance.


Edited by Ash - 14 Jul 2020 at 2:23am
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Graham Slee Quote  Post ReplyReply Direct Link To This Post Posted: 14 Jul 2020 at 3:05am
The reason for knowing where the nodes and anti-nodes are so the ferrite can be positioned where it counts (I am grateful to Ash and Jon for reminding me which is what).

A node is where the waveform passes through zero. Positioning for that will do nothing because the signal is already zero.

The anti-node is where the voltage rises to maximum, and it needs to be in the first half cycle at the peak of the quadrant, for maximum attenuation.

The frequencies either side, near to that point, will receive some attenuation.

As I've said, I can offer no proof, just the theory, but the circuit is shown in fig 3C on this page (http://www.compliance-club.com/archive/old_archive/990609.htm). It is the left-hand circuit of the 2-circuit CM choke. The "short" is because the two conductors transform to each other. Its nickname is a half-turn transformer because it isn't a loop-around. Obviously, its efficiency isn't high, but subjectively, it works.

The ferrite bead or tube must be chosen to have maximum impedance for the range of frequencies you wish the filter to work on.

The USB cable I made was based on 200MHz. If made for a lower frequency, it might affect the rise and fall times of the signal. You can get a clue from the clock frequency. A square wave is made up of all the whole number odd harmonics of its fundamental (3, 5, 7, 9, etc.), and the idea is to keep it as square as possible while filtering out higher frequency noise.

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Post Options Post Options   Thanks (0) Thanks(0)   Quote gwebster Quote  Post ReplyReply Direct Link To This Post Posted: 14 Jul 2020 at 2:01pm
As Graham found (even though the ASA thought otherwise), the appropriate use of ferrites on cables is definitely a worthwhile thing. Along similar lines, my son made definite and demonstrable improvements to RFI noise picked up by his remotely mounted radio receiver stick being controlled by a Raspberry Pi over his home network. If it's of interest, have a look here.


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Post Options Post Options   Thanks (0) Thanks(0)   Quote Ash Quote  Post ReplyReply Direct Link To This Post Posted: 14 Jul 2020 at 3:25pm
Finally getting my head around Jon's mental arithmetic there. Calculating the distances for ferrite placement is one thing; what about the qualities of the ferrite itself? Does it have to be a certain size or have a particular inductance or "crystal" resonance?

Why a 200MHz calculation? For the second harmonic of the fundamental wave?? If you had a 10m cable and the signal amplitude were not significantly attenuated by that length, you could have a series of ferrites along the cable that gradually get closer together by a factor of VF%, as long as they fit on the cable?

The physical contact that the sheath/insulation makes with the EM shield is also important. I assume that is why users were encouraged to gently and carefully bend the cable slightly along different parts of its length to ensure that the sheath is as separated from the shield as possible. If not, could the filter be bypassed to some extent? I understand that I may be asking too many questions...


One more thing, is velocity factor a similar principle to refractive index? A constant between 0 and 1 that considers the speed of light in the medium compared to the speed of light in free space (vacuum).


Edited by Ash - 14 Jul 2020 at 4:05pm
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