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

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1. Mar 16, 2016

### wu_weidong

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. Mar 16, 2016

### rude man

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.

3. Mar 17, 2016

### wu_weidong

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
4. Mar 17, 2016

### rude man

[ωCCV]l-1 = 2 kΩ.