# Laplace Relation to Capacitors and Inductors

Can anyone recommend a good resource to understand how a capacitor works as an integrator and an inductor works as differentiator? I'm looking for something to show me how the laplace transform comes into play, resulting in 1/(sC) and (sL) for impedances. Thanks!

• You can (and at some point should) look at the physics and the math, but there's something to be said for a function generator, a scope, and an assortment of inductors, capacitors, and resistors. – Chris Stratton Mar 17 '13 at 18:23
• Use SPICE. Most versions include Laplace stuff. – Leon Heller Mar 17 '13 at 19:22
• There are too many things to consider to answer this question properly. For instance caps can be differentiators and inductors can be integrators. It all depends on the circuit and what you are looking at i.e. current or voltage. Laplace transformation is quite deep and I wouldn't like to attempt a general answer given that I don't know how much you know on the subject. I look forward to a decent answer on this - don't forget to tell me!!! – Andy aka Mar 17 '13 at 19:57
• I know a decent amount about the laplace transform, and I was referring to the how (physically) a capacitor operates as an integrator and how an inductor acts as a differentiator for voltage. – anonymousfox Mar 17 '13 at 21:27

Integration is a continuous form of summation. We take infinitesimal pieces of a function and, well, integrate them to make a sum.

A capacitor physically integrates by packing electrons and developing charges across its plates. The current charge on a capacitor is the result of all previous current flow: the integral of the current function from minus infinity to the current time.

Think of the capacitor's charge as a bank account balance. The bank account current balance is the result of the opening balance (initial charge) plus the sum of all of the transactions since opening (all tiny pieces of the current flow this way and that way, added together).

An inductor works differentially because whereas it develops a magnetic field proportional to the current flowing through it, only changes in the magnetic field generate a voltage. An inductor with a steady DC current flowing through it is bathed in a magnetic field, but that field does not change and so the inductor does not develop a voltage. To obtain changes in the magnetic field, we must change the current. The current is a function of time, and a derivative of any such a function gives us another function which informs us of the rate of change in time. For instance, derivative of position is velocity, and the derivative of that is acceleration. If you visualize current flow as a speed, then the inductor voltage indicates its acceleration, so to speak.

• Excellent answer! Thanks so much for the reply - you really clarified the topic for me. – anonymousfox Mar 18 '13 at 4:32

Most simply 1/s is the Laplace operator for integration, and s, for differentiation. Start with the differential equations describing the reactive circuit elements, like so.

$i=C\frac{dV}{dt}$

$V=\frac{1}{C} \int i dt$

$V = L \frac {di}{dt}$

Now, try to replace things with the Laplace operators and a simple Ohms Law relationship V=iR and see what the complex impedances must be.

For example, $V(s) = \frac{1}{C} \frac{1}{s} I(s)$

yields an effective impedence of $\frac{1}{sC}$, and it would be a similar derivation for inductance.

It all comes from the relationships between the Laplace operators and the differential equations. This used to be routinely taught in differential equations classes, and engineering students would eventually fill in the blanks to understand how it applied to what they learn in circuits. Unfortunately, diffeqs are just not taught with the same rigor anymore.

• Thanks for the reply. I know these equations for the current and voltage of capacitors and inductors, respectively. I was actually hoping for a resource that would explain more of the physical aspect that creates these relationships. – anonymousfox Mar 17 '13 at 21:25
• Energy storage in magnetic and electric fields - as illustrated by Maxwell's equations... – Chris Stratton Mar 17 '13 at 21:29
• That's a little more complex, but since you understand how firmly rooted the relationship is in the differential equations, any basic signals book can show you how to derive, say, the integral property from the definition of the Laplace transform. Actually, the derivations might be easier to follow in Fourier space, but they'll be very similar. – Scott Seidman Mar 17 '13 at 21:30
• That was a decent answer – Andy aka Mar 17 '13 at 22:06