I would like to simulate a short length transmission line in MATLAB/Simulink. In the library, there is a block that simulates a transmission line. The parameters that are required are the resistance R, the impedance L and the capacitance C of the line. However, the capacitance effect is not present in a short transmission line but Simulink does not accept zero (0) as a capacitance value. So, I am thinking of putting a very small capacitance value (1e-09). Do you think that I am going to have an accurate simulation or I should put another capacitance value? What do you propose?

  • 2
    \$\begingroup\$ Capacitance is a fundamental property of a t-line and cannot be guessed at. \$\endgroup\$
    – Andy aka
    Dec 3, 2018 at 10:33
  • \$\begingroup\$ Start with 100 picoFarad per meter, if you need a guess. Better is to read the data sheet or the catalog. \$\endgroup\$ Dec 3, 2018 at 18:11
  • \$\begingroup\$ Thanks all of you for your comments and answers. As I mentioned in my question in shorts lines modeling the capacitance is neglected. Thus, it is hard to find the capacitance value of short transmission lines. But when I use the Simulink block of transmission lines I have to enter a value of the capacitance. In some researches, a typical value of the capacitance is given around 8 nF/km. So, in other words, I am asking for a typical capacitance value that is gonna have almost zero impact in my simulation since zero, according to Simulink, it is not a valid capacitance value. \$\endgroup\$
    – Yiannis S.
    Dec 3, 2018 at 19:43
  • \$\begingroup\$ Why can't you just use a simple RL block instead of a transmission line block \$\endgroup\$ Jul 20, 2019 at 21:12
  • \$\begingroup\$ You could use 0.1 pF to 1 pF for your model. This relates to what you'll find for wiring below a meter of length down to a few centimeters. You will hardly find less in practical wiring. \$\endgroup\$
    – datenheim
    Feb 13, 2023 at 7:49

1 Answer 1


If you want to simulate a transmission line, you must choose its impedance.

From there, via the Telegrapher's Equation, you can get an equivalent L per unit length, and C per unit length, which together will give you that impedance.

For a short length of line, length << wavelength, the open circuit capacitance will be given by this C per length.

  • \$\begingroup\$ For a long length of line, isn't the open-circuit capacitance still given by C per length (multiplied by the length)? \$\endgroup\$ Dec 3, 2018 at 15:50
  • \$\begingroup\$ @davidmneedham, if the length is long relative to the frequencies present in the circuit, you'd need to do some Smith chart math to figure out the input impedance of the line. Depending on the length of the line, it could end up looking like a pure open circuit, pure capacitance, pure resistance, a short circuit or something in between. \$\endgroup\$
    – The Photon
    Dec 3, 2018 at 16:06
  • \$\begingroup\$ @ThePhoton , Thanks. I believe I understand. For example, if the length of the line is a quarter-wavelength, then the line looks like a pure resistance, and the capacitance is zero at that frequency? \$\endgroup\$ Dec 3, 2018 at 16:14
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    \$\begingroup\$ @davidmneedham, actually (as I think about it more carefully) if it's a lossless line, it'll never look like any resistance other than pure short or open, but it can look like different inductive or capacitive values. \$\endgroup\$
    – The Photon
    Dec 3, 2018 at 16:23

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