2
\$\begingroup\$

I'm expecting a midterm question about the circuit below that involves the equation t=RC and using a Laplace transform. We won't actually be using differential equations. Memorizing a certain equation is all we need, but we're not told what it is. I've found a lot of equations when looking up Laplace transforms. Which equations would be helpful for analyzing this circuit, and where in the circuit could they be applied?

enter image description here

\$\endgroup\$
3
  • \$\begingroup\$ So what good does it do you if we just tell you the answer? \$\endgroup\$
    – W5VO
    Mar 23, 2011 at 9:21
  • \$\begingroup\$ @W5VO Moot question because we won't do it. :P \$\endgroup\$
    – jpc
    Mar 23, 2011 at 12:23
  • \$\begingroup\$ Find the laplace form and matrix of the electrical circuit given above, provided that the starting condition is 0 \$\endgroup\$
    – Yusuf
    Jan 18, 2021 at 23:15

1 Answer 1

6
\$\begingroup\$

Using the Laplace transform in circuit analysis works just like normal complex analysis. You just plug \$s\$ instead of \$j\omega\$ everywhere.

Laplace circuit analysis rules from Wikipedia

The interesting part comes after you calculate one thing or another since you can use Laplace transfer function to draw a Bode (frequency and phase response) plot or to calculate the circuit's response to any stimulus.

\$\endgroup\$
3
  • \$\begingroup\$ Why are those voltage/current sources there? That makes no sense to me. \$\endgroup\$
    – Jason S
    Jun 11, 2011 at 0:10
  • \$\begingroup\$ @Jason the voltage/current sources are in the s-Domain models because a capacitor/inductor can act like a time dependent voltage or current source. Also, each have a frequency dependent resistance (impedance), which is why the resistor is in the model. \$\endgroup\$
    – gallamine
    Jun 13, 2011 at 16:04
  • 1
    \$\begingroup\$ @Jason They represent the current or voltage at t = 0 and are needed to account for the energy stored inside the reactive elements. \$\endgroup\$
    – jpc
    Jun 25, 2011 at 1:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.