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.