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I've seen a number of coaxial connectors claimed to be rated for "mode-free operation up to (whatever) GHz"--but after much searching, I can't seem to find any definition of "mode-free".

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What does "mode-free" mean in this context?

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    \$\begingroup\$ I think it's because coax cable can have higher-order propagation modes -- like a waveguide, except the lowest "mode" of a coax goes to 0Hz. When your coax starts exhibiting these higher-order modes, Bad Things happen (at least weird dispersion, maybe heating? I dunno). I assume that "mode free" means the highest frequency you can trust the connector not to have such modes within itself. \$\endgroup\$
    – TimWescott
    Jul 14, 2021 at 3:03

3 Answers 3

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EM energy can travel through a waveguide in different modes, i.e. different ways for the electrical and magnetic standing waves to fit inside of the confining cavity. Normally propagation in coax is in the TEM00 fundamental mode, which means that there are no standing waves perpendicular to the direction of propagation. However, if the wavelength used is small enough compared to the diameter of the coax or connector, other modes will also be able to propagate. This is generally a bad thing, since the different modes will have different propagation velocities, leading to intersymbol interference and loss. Arguably a better term would be "single-mode" instead of "mode-free", but I guess people in the coax world aren't used to thinking of TEM00 as a mode at all.

In any case, the mode-free limit is the frequency below which an insignificant amount of energy couples into the higher modes, so they don't cause any trouble.

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    \$\begingroup\$ This is what I was confused about--I thought about waveguide modes, but as far as I could figure, "mode-free" would mean "no signal at all", which didn't make any sense at all. \$\endgroup\$
    – Hearth
    Jul 14, 2021 at 3:12
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    \$\begingroup\$ Bah why don't they just use the same terminology as fiber optics. \$\endgroup\$
    – DKNguyen
    Jul 14, 2021 at 3:13
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    \$\begingroup\$ @Hearth look ma no modes! Apologies if I was writing beneath your level, but I think this still stands as a decent general-purpose answer. \$\endgroup\$
    – hobbs
    Jul 14, 2021 at 3:15
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    \$\begingroup\$ No, I'm not offended or anything! This site is here to give answers to people other than the person who asked the question in the first place, after all. \$\endgroup\$
    – Hearth
    Jul 14, 2021 at 3:16
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    \$\begingroup\$ @DKNguyen Because coax cable was practical engineering technology more than a century before fiber optics. Coaxial cables were used for long distance telephony back in the 1850s. Fiber optics only became a practical solution in the 1980s when it was discovered how to produce them quickly. Before then, the maximum speed for manufacturing fiber optic cable was only 2 meters / second. \$\endgroup\$
    – alephzero
    Jul 14, 2021 at 13:59
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In a waveguide consisting of a thin hollow tube, there's a cutoff frequency below which it will not work, it will not propagate energy. Above that frequency, various different patterns of E and H field can propagate, each pattern being called a mode.

If you run a wire down the middle of the tube, you add an extra mode, the coaxial or TEM mode, which conveniently goes right down to DC. Adding this extra wire changes the higher frequency waveguide modes in detail, but they still exist.

As long as you operate your coax below the waveguide-without-centre-wire-cutoff frequency, then only the TEM mode propagates, and you'll have very predictable performance, even if you bend the cable.

If you operate above this cutoff frequency, then discontinuities in the waveguide, like those that happen at connectors, or bends, can move energy from the TEM mode to other propagating waveguide modes, and back again to the TEM mode. As these other modes have different propagation speeds to the TEM mode, this will make the performance of your coax unpredictable, changing every time you bend it.

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    \$\begingroup\$ Still doesn't make sense to me that they'd call it "mode-free" instead of "single-mode", personally. \$\endgroup\$
    – Hearth
    Jul 14, 2021 at 14:32
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Waveguide modes depend on the direction of E and H Field relative to current.

Most DC, AC and RF fields are all Transverse or TEM electromagnet field directions which are orthogonal to the direction of current.

EM Mode-Free

It is called mode-free in microwave for absence of abnormal directions for E & H fields. (according to Maxwell's boundary conditions for wavelengths)

Higher than mode-free transmission frequencies are possible with a resonant cavity operation with harmonic products that change the direction of the E or H field.

Transverse magnetic (TM) and electric (TE) modes

  • No magnetic (TM) or electric (TE) field in the direction of propagation.
    TM are sometimes called E modes because there is only an electric field along the direction of propagation or "longitudinal waves" ( not the Bearden hypothetical ones)

Hybrid modes = Non-zero E & H fields in the direction of propagation.

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    \$\begingroup\$ "No magnetic (TM) or electric (TE) field in the direction of propagation." "sometimes called E modes because there is only an electric field along the direction of propagation." I think these two sentences contradict each other, no? \$\endgroup\$
    – Hearth
    Jul 14, 2021 at 3:42
  • \$\begingroup\$ E mode fields are longitudinal \$\endgroup\$ Jul 14, 2021 at 3:47

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