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For a 100 meters RG58 coaxial cable, according to the datasheet, the capacitance of the cable is 10 nF and the inductance is of 25 µH approximately. A sinusoidal voltage is applied at the input of the line with a 50 Ohm source, and the end of the line is connected to an oscilloscope, HZ mode (1 MOhm//10 pF), AC coupling. This give, according to me, the following schematic:

enter image description here

The RG58 coaxial cable is replaced by the lumped element R,L,G,C, G is neglected. The simulation of this circuit with LTSpice, and theoretical calculation give an upper cutoff frequency of 300 kHz approximately. However, I personally made the measurement with a 100 meters RG58 coaxial cable and the measured -3 dB bandwidth is of 5 MHz approximately. (I don't recall the exact measurement value, sorry) The coax cable model(picture below) gives a value in perfect agreement with measurement enter image description here

Once again, I don't have a screenshot of the simulation result, sorry.

I do not understand why the measurement and the simulation of the coaxial cable give a cutoff frequency greater than the one given by taking into account only theoretical effect of the parasitic capacitance. Every physical phenomenon I neglected such as skin effect, or conductance of the dielectric should lower the upper cutoff frequency.

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  • \$\begingroup\$ SHow exactly how you. physically connected the cable to 100 pf 1M scope. You realizethis is mismatched? \$\endgroup\$
    – D.A.S.
    Commented May 27, 2020 at 22:22

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You need to model your lumped line in much smaller quantities like "per metre". At the moment you are modeling it as "per 100 metres" and that will give you severe problems above a fairly low frequency.

Consider this 25 uH and 10 nF have a resonant frequency of: -

$$\dfrac{1}{2\pi\sqrt{LC}} = 318 \text{ kHz}$$

And clearly that is the problem with using a lump of line equal to 100 metres. If you were only interested in frequencies of a few tens of kHz then a 100 metre model would be fine. So, take the maximum frequency you are interested in and convert to a wavelength hence, 5 MHz has a wavelength of 60 metres. Now make sure that the lumped model length is no longer than one twentieth this distance i.e. 3 metres.

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  • \$\begingroup\$ As I recall it is the sheath and dielectric that determines the transfer impedance losses. The LC determines the delay time and fractional wavelength Return Loss with a mismatched load which adds to the loss. \$\endgroup\$
    – D.A.S.
    Commented May 27, 2020 at 21:49
  • \$\begingroup\$ I think your right but are missing the point for this question. \$\endgroup\$
    – Andy aka
    Commented May 27, 2020 at 21:54
  • \$\begingroup\$ Even if you say the cable is 250 nH/mm and 1 pF/mm, It is not affecting the amplitude at -3dB but does affect the phase. More likely Probe effects of a poor 1x probe ground. I.e. measurement error. \$\endgroup\$
    – D.A.S.
    Commented May 27, 2020 at 22:04
  • \$\begingroup\$ I think my point is on target but making some assumptions. \$\endgroup\$
    – D.A.S.
    Commented May 27, 2020 at 22:20
  • \$\begingroup\$ Thank you a lot Andy, that was the problem. By simulating 100 cells of 1 meter plugged all together, I manage to get the same cutoff frequency than ADS and measurement. I will try to explain this with periodical structure. \$\endgroup\$
    – moumi Lelion
    Commented May 28, 2020 at 14:26

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