0
\$\begingroup\$

I am reading this doc

I see this magic happening on the third line of equations...

enter image description here

How in the name of Math, does this \$ Ae^{st} \$ appeared there?

Magic?

\$\endgroup\$
  • 1
    \$\begingroup\$ Literally the very first bullet point on that document lol. \$\endgroup\$ – KingDuken Aug 1 '18 at 20:07
  • 1
    \$\begingroup\$ It's weird I know. Whoever wrote this paper was probably just thinking, "Yeah it's usually not like this but let's assume that the components in the circuit take form in exponential characteristics." \$\endgroup\$ – KingDuken Aug 1 '18 at 20:18
  • 1
    \$\begingroup\$ SpaceDog, a basic text for first year calculus will cover the details when they get to teaching you about the standard form for 1st order differential equations and how to develop solutions. Your equation is 2nd order, but most of what you need to learn is already covered by the time you get through 1st order. Have you gone to Khan Academy, yet? It's all pretty much there in simple videos. \$\endgroup\$ – jonk Aug 1 '18 at 21:43
  • 1
    \$\begingroup\$ A nice link (not the best, I'm sure, but it seems good enough as a quick catch) is here. See if reading through that helps. (You might also go through this answer on EESE to see how one specific example plays out.) \$\endgroup\$ – jonk Aug 1 '18 at 21:59
  • 1
    \$\begingroup\$ If multiply both sides by C you get the classic relationships for linear current 1st order exponential RC=dI/dt and 2nd order decay. These expand the partial derivatives into time dependent equivalents known from fundamentals. \$\endgroup\$ – Sunnyskyguy EE75 Aug 1 '18 at 22:11
0
\$\begingroup\$

This is really a question about solving differential equations. If you massage the first equation a bit you will see that the shape (the form of the function) for \$i\$, \$di/dt\$, and \$d^2i/dt^2\$ must all be the same. There are two such functions that we run into quite a bit in electronics: exponentials and sinusoids. For transient behavior the solution is often in the form of an exponential, so the author has substituted an exponential function for \$i\$. The remaining steps will be for solving for the constants \$A\$ and \$s\$.

\$\endgroup\$
  • \$\begingroup\$ ok, but suppose this was the first time these equations were derived. How someone would guess that. This is my problem. \$\endgroup\$ – SpaceDog Aug 1 '18 at 20:12
  • \$\begingroup\$ That's a math question. Electrical engineers are usually happy to just use the result without deriving it from first principles. \$\endgroup\$ – Elliot Alderson Aug 1 '18 at 20:15

Your Answer

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

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