- #1

wu_weidong

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## Homework Statement

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.

## Homework Equations

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 Attempt at a Solution

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/π τ = 637 kHz

C

_{CV}= 12 pF/m (read from a graph), and [ωC

_{CV}l]

^{-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 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"?

Thank you.