The cut-off frequency of an RL circuit is described by the following equation: $$f = \frac{R}{2\pi L}$$ where R is the resistance of the circuit and L is the inductance. Applying this equation to a straight length of RG-58/U coax cable (no resistors or inductors in the circuit) with the following specs: $$Length = 5m$$ $$OD = 5mm$$ $$ID = 0.81mm$$ $$L = 0.00000159H$$ $$R = 0.163\Omega$$ yields a cut-off frequency of 16.32KHz. I know the cable is good for at least a few GHz. What's going on here? Does the formula not work for distributed inductances?

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    \$\begingroup\$ That formula does not work for distributed inductances, no. And I'm suspicious of your resistance measurement -- it seems too high. \$\endgroup\$ – TimWescott Jan 8 '19 at 20:49
  • \$\begingroup\$ That might apply to short circuit current BW but if you use matched R source and load, the RLC Telegrapher's equation prevails for distributed RLC \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 8 '19 at 20:50

That formula does not apply to distributed inductances.

Coax cable does have a cutoff frequency, when wavelength of the signal gets somewhere on the order of the diameter of the cable (it's some fraction, like 1/4, but I can't remember exactly). For most coax, that cutoff frequency is considerably higher than the useful frequency of the cable because of dielectric dissipation -- it is educational to find specifications for various sorts of coax and look at their attenuation/meter at various frequencies.



simulate this circuit – Schematic created using CircuitLab

If you short circuit the source and load impedance then you get what you calculated with 2pif=R/L


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