Radio interference... 1/4 wavelength and 1/30 wavelength

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Hello,
I have a chart here that shows the relationship of cable length to 1/4
and 1/30th wavelengths of certain frequencies. Is the 1/4 wave
important because it will have a node on one end and an anti-node on
the other of one RF cycle? Why is 1/30th of a wave important in cable
design?

Thanks,
Peter

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In article <1110381913.649313.190980@l41g2000cwc.googlegroups.com> thecatspjamas@aol.com writes:

> I have a chart here that shows the relationship of cable length to 1/4
> and 1/30th wavelengths of certain frequencies. Is the 1/4 wave
> important because it will have a node on one end and an anti-node on
> the other of one RF cycle?

Yup.

> Why is 1/30th of a wave important in cable design?

Probably someone's idea of where the length of the cable is
insignificant and doesn't contribute to reflected power. It's why,
even though companies would rather sell you digital cables, you can
get away with using mic cable for AES/EBU connections and home video
cables for S/PDIF coax.

Applying a conservative velocity factor (how much slower than the
speed of light electricity travels in a cable rather than a vacuum), a
wavelength at 96 kHz is about 2,000 meters. Most of our digital
cable runs are considerably less than 1/30 of that, about 200 feet.


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Chris Hornbeck wrote:
> On 9 Mar 2005 07:25:13 -0800, thecatspjamas@aol.com wrote:
>
> > I have a chart here that shows the relationship of cable length to
1/4
> >and 1/30th wavelengths of certain frequencies. Is the 1/4 wave
> >important because it will have a node on one end and an anti-node on
> >the other of one RF cycle?
>
> A 1/4 wavelength line is special because if you stick a signal
> into one end and don't terminate the other, the signal travels
> to the far end, is reflected, travels back another 1/4 wavelength,
> and arrives back at the source in opposite polarity. That's
> how you make antennas.
>
> Interesting in acoustics, but not an issue in audio electronics
> for ordinary folks.
>
> Chris Hornbeck

True when the "unterminated" end is an open circuit.

When the "unterminated" end is a short circuit, then the wave is
inverted by the reflection and ends up back at the original polarity
(phase) back at the source.

Is a sound wave bouncing off a wall analogous to the electrical short
circuit case (inverted by the reflection) or the electrical open
circuit case? I would guess the short circuit case but I'm not sure.

Mark