# Capacitance crosstalk voltage help

#### ham3388

Joined Jul 3, 2012
97
I have the following doubt only!
Shall I consider the 100 kHz or the 793 kHz as the cut-out frequency?

#### WBahn

Joined Mar 31, 2012
26,037
The question does NOT say that the cutoff frequency is 100 kHz. It merely says that the frequency of the signal source driving the circuit happens to be 100 kHz.

#### ham3388

Joined Jul 3, 2012
97
Thank Wbahn
So shall I asume that my method is correct?
And I believe that the source frequency 100 kHz should not have any influence on the calculation of the crosstalk voltages at the following frequencies 100 Hz, 1khz, 1 Mhz. Am I right?

#### bertus

Joined Apr 5, 2008
20,650
Hello,

There is asked to calculate the amount of crosstalk on the given frequencies.
They will all be different.

Bertus

#### WBahn

Joined Mar 31, 2012
26,037
Thank Wbahn
So shall I asume that my method is correct?
And I believe that the source frequency 100 kHz should not have any influence on the calculation of the crosstalk voltages at the following frequencies 100 Hz, 1khz, 1 Mhz. Am I right?
I didn't say anything about your method being correct or not, I only commented that the signal frequency does not determine the circuit cutoff frequency.

In general there will be some amount of crosstalk at all frequencies and the degree of crosstalk will be different at different frequencies. You need to analyze the circuits to determine how much crosstalk there is.

#### ham3388

Joined Jul 3, 2012
97
Thanks for both of you Wbahn and Bertus
Below is the original figure 1

I'm going to calculate the crosstalk voltage's for the different frequencies based on the below formula's keeping in mind the cut-outf frequency as the reference frequency.

#### WBahn

Joined Mar 31, 2012
26,037
I'm not a big fan of just using cookbook formulas. You have a circuit so analyze the circuit. Just find the Thevenin equivalent as seen by the load resistor in the second circuit when only the source in the other circuit is taken into account. Or, equivalently, find the transfer function for the voltage across the second circuit's load resistor as a result of the first circuit's voltage source. The analysis isn't hard -- you've only got one non-trivial node.