Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Laplace Transforms and Phasors in Circuit Analysis

  1. Jan 11, 2012 #1
    So last semester, I had a Circuit Analysis course where I learned about phasors. Basically, when dealing with AC circuits, I should convert everything to the frequency domain where [itex]X = j\omega L[/itex] and [itex]X = \frac{1}{j\omega C}[/itex]. I feel like I understood this part really well.

    However, in Circuits II, my prof was used Laplace transforms when dealing with DC circuits (at least I think?) involving capacitors and inductors and derived the same result for their reactance. I'm really confused at how he did this because aren't inductors just treated as short and capacitors as open circuits?

    Also, are Laplace transforms and phasors related? They produced the same impedences, but they feel too different to be the same...

    I hope I made my questions clear because I am currently very confused, and thanks in advance.
     
  2. jcsd
  3. Jan 11, 2012 #2
    I think you studied circuit analysis related to transient processes. Am I right?

    If so, no, in DC analysis when you are studying transient processes, capacitor doesn't act like an open, and inductor doesn't act like a short circuit.

    It has a transient process that LEADS to that.
     
  4. Jan 11, 2012 #3
    In beginning circuit theory, you learn a natural response and a forced response when you work with phasors.

    The forced response is what you see at DC and this is when dv/dt or di/dt are 0 and you can consider the circuit in DC, and the natural response is when you have transients. You learn these ideas in the time domain.

    The laplace transform changes your time information in complex number phasors into frequency information using the e^(-jwt+sigma) kernel

    This is tied to the laplace transform, which is in the frequency domain and is in terms of a transfer function. When you let s = 0, this means you are setting the frequency to 0, and this corresponds to the forced, DC, steady-state response, or at t = infinity. Same goes when you set s = infinity as in infinite frequency, you approach t = 0 and are looking at the transients.

    So to answer your question, laplace transforms and phasors are representing the same information. However, laplace transforms reveal information more easily and are easier to work with, since convolution becomes multiplication in the frequency domain. Also, in the laplace domain, s = jw, and so the impedance of a capacitor is 1/sC which is like you wrote. The best way to see the relation is when you look at euler's identity and see that a time domain sine wave is composed of complex exponentials much like you see in the laplace kernel.

    As to your other question, when your professor is looking at these circuits, you must consider their initial conditions. A DC current source in the laplace domain sort of has a transient built into it, and that is that the voltage is a step from 0 to its DC value at time t = 0 in the time domain. If a capacitor's initial condition is such that it was already charged to the DC voltage of the voltage source before time t = 0, then the capacitor will look like an open. If the initial condition is that the capacitor was at 0V when the DC supply is applied, then there is a transient component that must charge the capacitor before it could ever reach the DC voltage, and this is where the impedance plays its factor and how he can derive it in the laplace domain.
     
    Last edited: Jan 11, 2012
  5. Jan 12, 2012 #4
    I just wanted to also add that although they are related, I don't know how to prove the exact relationship mathematically and if I had the time I would try to help you show that.
     
  6. Jan 12, 2012 #5
    Yes, when you reduce s=jw you actually get Fourier transform and those phasors you are working with.

    DragonPetter, very nice post.
     
  7. Jan 13, 2012 #6

    jim hardy

    User Avatar
    Science Advisor
    Gold Member
    2016 Award

    i echo sentiment above , very nice posts ...
    here's what helped me:
    in beginning circuit analysis courses, we use phasors and assume sinewaves.

    To the point that in the beginning of our studies we forget that in nature, a sinewave is a special case. (electric company goes to a lot of trouble to make theirs fairly pure.)
    but in EE we do not limit our studies to sinewaves.



    I think of the phasor as representing steady state operation for sinewave input after all transients have died out.
    you can think of s = jw as true for the long haul when input is a sinewave..
    but that won't solve for the initial turn-on transient.


    The Laplace notation gives you the initial transient response and it handles non-sinewave inputs.
    so Laplace is a way more powerful tool.
    like any sophisticated tool it takes practice to become adept with it.

    learn to use Laplace notation even if you can't quite tie anything physical to that integral that goes out to infinity.
     
  8. Jan 14, 2012 #7
    Thanks for the response you guys. They really helped me out!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Laplace Transforms and Phasors in Circuit Analysis
  1. Laplace transforms (Replies: 1)

Loading...