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A ~100kHz square wave through a 100m open end coax (RG-58) produces a slightly sloped shoulder during the time from initial pulse until reflected pulse (A-B), and from initial pulse drops until reflected pulse ends (C-D).

What are the correct explanations for these slopes [not being flat]?

(Scaling: 2.0V/ 2us/) enter image description here enter image description here EDIT: enter image description here

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  • \$\begingroup\$ Maybe the driver impedance is changing due to temperature sensitive impedance. What is the driving source? What is the nature of the source resistance? \$\endgroup\$
    – user57037
    Feb 23, 2017 at 21:49
  • \$\begingroup\$ It could also be source capacitance. \$\endgroup\$
    – user57037
    Feb 23, 2017 at 21:50
  • \$\begingroup\$ impedance, capacitance or inductance, depending on where it is and how the whole setup is done, which we know nothing about \$\endgroup\$
    – PlasmaHH
    Feb 23, 2017 at 21:50
  • \$\begingroup\$ See edit: function generator. \$\endgroup\$
    – bretddog
    Feb 23, 2017 at 21:54
  • \$\begingroup\$ What kind of connection do you have to the scope? What is the input impedance? 50 Ohms? \$\endgroup\$ Feb 23, 2017 at 23:59

1 Answer 1

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The slope on a shoulder (first plateau) is defined by interplay of resistive and dielectric losses in the transmission line (cable), see the iconic source, by Howard Johnson. In your case the resistive loss effect seems to dominate.

There are also 13,000 hits on Google for [tdr lossy transmission line], from Tektronix and Agilent and other good places.

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