1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Cable-to-cable crosstalk (Capacitative Coupling)

  1. Mar 16, 2016 #1
    1. The problem statement, all variables and given/known data
    Compute the cable-to-cable crosstalk due to capacitive coupling in a harness between two cable pairs having an average separation distance of 3 mm and a 10 m in a cable tray. The cable diameters are 1 mm and both cables are operating at a 100 ohm impedance level. Assume h = 5 mm, source logic = 3.5 V with rise/fall times of 500 ns.

    The lecturer showed the following diagram when he was talking about the capacitative coupling, and the equivalent (part of) circuit representation.


    2. Relevant equations
    The Capacitive Cross-talk Coupling (CCC) is defined as

    CCC(dB) = 20 log (VV/VC)

    where VC is the culprit source line voltage and VV is the victim load line voltage.

    3. The attempt at a solution

    The solution given is as follows.

    The second corner frequency, fc, in the spectrum of a pulse (or > pulses) corresponding to rise/fall time of 500 ns is: fc = 1/π τ = 637 kHz

    CCV = 12 pF/m (read from a graph), and [ωCCVl]-1 = 2 kΩ
    Thus CCC dB = -26.4 dB
    Source logic = 3.5 V = 11 dBV, then
    Noise = 11 dB – 26.4 dB = -15.4 dBV = 0.17 V < 0.4 V
    Thus, switching OK.

    I have several questions.

    1) I don't know how to get CCC dB = -26.4 dB. I know I can use the voltage divider on the circuit representation to find VV/VC, which means I need to find ZV. But to do that, I need to find CV, which is the capacitance per unit length of the victim wire, and I don't know how to get this value.

    2) What is "source logic"?

    3) Why does noise < 0.4V imply "switching OK", and why "0.4V"?

    Thank you.
  2. jcsd
  3. Mar 16, 2016 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    You just calculated fc and you know CCV and l, so what's the problem? The given formula is kind of obvious and the answer is indeed close to 2K ohms.
    The source voltage. By "logic" we mean the source from a logic integrated circuit.
    Because if the "low" voltage limit is 0.4V then anything below that looks like a "0" and that's what you want here.
  4. Mar 17, 2016 #3
    ZV is close to 2K ohms? Why is that?

    I thought the combined resistance ZV is calculated as ZV = [1/100 + 2000 + 1/100]-1 = 0.0005 if CV = CCV. This gives me the voltage divider ZV / (ZV + [jwCCV]-1) = 0.005/(0.005 + 2000) = 2.5 * 10-7, which doesn't give me the CCC value of -26.4dB.
    Last edited: Mar 17, 2016
  5. Mar 17, 2016 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    [ωCCV]l-1 = 2 kΩ.

    You're adding impedances to admittances! What is really your voltage divider?
    BTW I don't get -26.4 dB. I get a somewhat more negative number than that.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted