0
\$\begingroup\$

Assuming there is one node that is connected to four wires/transmission lines, and a perfect step function(no rise time) is injected into the node, how does it show up on the four wires due to the cross talk(parasitic capacitors) between them ?

My guess is that four wires will each have some resistance and capacitance so it acts like low pass filter, the signal should be rising slowly with different time constant on four different wires

\$\endgroup\$
  • \$\begingroup\$ You could simulate this in Falstead to find out. Use the transmission line template to start out. \$\endgroup\$ – DKNguyen Feb 7 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.