Hi, 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 (VV/VC)
where VC is the culprit source line voltage and VV is the victim load line voltage.
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/πτ = 637 kHz
CCV = 12 pF/m (read from a graph), and [ωCCVl]−1=>2 kΩ, thus CCC dB = -26.4 dB
I don't know how to get CCC dB = -26.4 dB.
I assume CV=CCV, and I thought I should use the voltage divider on the circuit representation to find VV/VC, which means I need to find ZV, the combined resistance of the 2 wires and CV.
ZV = 1/[1/ZV1 + 1/(jwCVl) + 1/ZV2] = 1/[1/100+2000+1/100] = 0.0005
This gives me
VV/VC = ZV/(ZV+1/[jwCCVl]) = 0.0005/(0.0005+2000 )= 2.5×10^(−7)
Where did I go wrong in my calculations?