# Cable-to-cable crosstalk (Capacitative Coupling)

I'm having some difficulty understanding the solution to this question.

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.

The Capacitive Cross-talk Coupling (CCC) is defined as

CCC(dB) = 20 log ($V_V/V_C$)

where $V_C$ is the culprit source line voltage and $V_V$ is the victim load line voltage.

The solution given is as follows.

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

$C_{CV}$ = 12 pF/m (read from a graph), and $[ωC_{CV}l]^{-1} = > \text{2 k}Ω$, 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 $V_v/V_c$, which means I need to find $Z_v$. But to do that, I need to find $C_v$, 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"?

• Ok, so because $Z_{V1}, C_V and Z_{V2}$ are in parallel, the potential difference at each branch across them is the same at $V_V$, right? That's why the voltage divider is as you stated, 100/(2000+100), i.e. take $Z_{V1}/([jwC_{CV}]^{-1} + Z_{V1})$. But my question is, doesn't this value change if instead of taking $Z_{V1}$, I take $[jwC_{V}]^{-1}$ instead, which then gives 2000/(2000+2000) = 0.5? – Rayne Mar 17 '16 at 5:52
• That's why I thought we needed to consider $Z_V$, the combined resistance, for the voltage divider? I thought in parallel, we have $Z_V = [1/100 + 2000 + 1/100]^{-1} = 0.0005$, then voltage divider is $Z_V / (Z_V + [jwC_{CV}]^{-1}) = 2.5 \times 10^{-7}$? Where did I go wrong in my calculations? – Rayne Mar 17 '16 at 5:53