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

Applied example of Laplace Transform?

  1. Jan 16, 2008 #1
    I'm looking some applied example of Laplace transform.

    For instance:

    If I have function f(t) = Asin(ωt + φ), where;
    A is peak deviation from center
    ω is Angular velocity
    t is time
    φ is phase

    and now I want to laplace transform it and get some function like
    A*(cos(φ)*ω+sin(φ)*s)/(s^2+ω^2) where is variable s.

    What s represents in this case??
  2. jcsd
  3. Jan 16, 2008 #2

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Laplace transforms are used in electric circuit theory to transform functions of [itex]t[/itex] to functions of the complex variable [itex]s[/itex]. Here [itex]s=\sigma+j\omega[/itex] ([itex]\sigma[/itex], [itex]\omega[/itex] both real), and it is referred to as the complex frequency.

    Is that what you're looking for?
  4. Jan 16, 2008 #3
    Yes, this is exactly what I was wondering. Thank you. Hence s is complex frequency it is not sum of phase and angular velocity but complex number which both parts are real? Hmm, what this actually means and what kind of unit function A*(cos(φ)*ω+sin(φ)*s)/(s^2+ω^2) returns?
    Last edited: Jan 16, 2008
  5. Jan 16, 2008 #4
    To specify my question; Is variable s always complex number which both real numbers parts are somehow original (not transformed) function other parameters like this example case it was?
  6. Jan 17, 2008 #5

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No, the complex frequency has both real and imaginary parts. Since I'm talking circuits, I used the electrical engineer's notation for the imaginary unit: [itex]j=\sqrt{-1}[/itex].

    What it means in the context of electric circuit theory is that if you start with a function (say, for electric current) in the time domain, upon taking the Laplace transform you end up with a function in the (complex) frequency domain. Look for "frequency domain analysis" in any book on introductory circuit analysis.

    Now, as for the units of a laplace transformed function, let [itex]i(t)[/itex] equal the electric current through a circuit element. It has units of amperes. Now look at the laplace transform [itex]I(s)=\mathcal{L}[i(t)][/itex].


    The units of [itex]I(s)[/itex] are the units of the integrand. [itex]i(t)[/itex] has dimensions of electric current, [itex]exp(-st)[/itex] is dimensionless, and [itex]dt[/itex] has dimensions of time. So apparently, [itex]I(s)[/itex] has dimensions of current*time, although I've never seen this mentioned in the literature. We usually don't bother talking about the units of [itex]I(s)[/itex] since in the end we always invert the transform to recover [itex]i(t)[/itex] anyway.

    As for the question in your next post, I'm afraid I don't understand what you're asking.
  7. Jan 17, 2008 #6
    Thank you Tom!

    Things getting less blurry...

    Is the whole idea of Laplace transform to be a tool that help us solve equations?
    We have so hard equation in our time domain that we have to transform function to another domain and solve some equation there (wonderland to me). And then invert solved equation back to our time domain? Does it make any sense to calculate and observe some values in another domain or is it just a wonderland to solve equations?
  8. Jan 18, 2008 #7

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I don't know what originally motivated the Laplace transform, but one nice application of them is that they turn some differential equations into algebraic equations.

    Ultimately, we want to get back to the time domain because that is the world we see. However, there is a good reason to spend a little more time in wonderland :rofl:. It turns out that stability analysis is most easily done in the frequency domain. In a linear system (with zero initial conditions) the laplace transform of the output divided by the laplace transform of the input is called the transfer function [itex]H(s)[/itex]. It turns out that the zeros of [itex]H(s)[/itex] give you information about the system's transient response, and the poles of [itex]H(s)[/itex] give you information about the stability of the system.

    Rather than type out all of the details, I'll refer you to the following.

    http://planetmath.org/encyclopedia/StabilityOfTransferFunctionsInTheLaplaceDomain.html [Broken]

    And here is even a little quiz on transfer functions.
    Last edited by a moderator: May 3, 2017
  9. Jan 19, 2008 #8
    This is great!

    My eyes opened, I just enter in a new world of mathematics...the wonderland. Thank you for your great and helpful answers.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook