Laplace Transforms of Kirchhoff's Laws

1. Apr 6, 2012

rdfloyd

Back again with another problem. These transforms are difficult...

1. The problem statement, all variables and given/known data
http://i.imgur.com/LM38A.png

3. The attempt at a solution

His instructions are to take the Laplace Transform of each of Kirchhoff's laws. However, when we do, nothing useful comes out of them. I'm completely stuck as to what the problem is asking for.

The question (I think) asks: "Find the voltage across each element as a function of time." But he gives us 3 equations where the voltage is already in respect to time. Can anyone help decode this?

EDIT: Sorry mods. Didn't realize this was Advanced physics. >.<

Last edited: Apr 6, 2012
2. Apr 7, 2012

vela

Staff Emeritus
From the wording of the question, it looks like this is from a differential equations course. Am I right?

You're looking for an explicit expression of each current as a function of time, like IL(t) = I0e-t/τ cos ωt where you say what the constants τ and ω are equal to in terms of R, L, and C.

You have three unknowns: IL, IR, and IC. You have two equations, e.g., ΔVC=ΔVL and ΔVR=ΔVL, which will allow you to write two of the currents in terms of the third one. (The third condition, ΔVC=ΔVR, automatically follows from the other two, so it's not independent.) You can therefore write IR+IL+IC=0 in terms of just one of the currents and solve for it.

3. Apr 7, 2012

rude man

He wants an explicit expression for the voltage across R. Not the integro-differential equations he gave you.

4. Apr 7, 2012

rdfloyd

It's a math methods course, but we are studying differential equations now. Finished partial diff. eq. on Thursday.

As far as the solution, this is what we did. I was trying to understand exactly what was going on, but then it clicked and I finished it in about half an hour. The answer wasn't very pretty, but it looked right.

But didn't we have to use those to get to Vr?

5. Apr 8, 2012

rude man

Of course. That's why he gave them to you ....
Did he cover transforming differential equations (linear, constant-coefficient ones) to Laplace-transformed algebraic equations, I hope?

6. Apr 9, 2012

rdfloyd

This just clicked for me. He doesn't like to "spoil the surprise" by giving away the whole idea of a concept (forces us to choose whether to use something he taught or not), so it wasn't directly implied.

The whole idea behind Laplace/Fourier Transforms is to take a differential equation, put it in algebraic terms (which I have many more tricks to use), then bring them back to differential terms. It makes so much sense now!

I guess the only question I have now is why don't we do this to every differential equation?

Also, why would we use Fourier Transforms? To my understanding, Fourier Transforms are much more restrictive (it's limits at +/-∞ → 0).

7. Apr 9, 2012

RoshanBBQ

The Fourier transform has a more intuitive connection with response of a system based on the frequency components of the input function. The system, for example, could be a bunch of connected capacitors, inductors, and resistors, modeled by a differential equation. The input could be a voltage applied to that circuit at a certain point with the output being voltage developed elsewhere. The Fourier transform would give easier insight into the characteristics of the developed voltage relative to the applied voltage if the applied voltage had dominant frequencies -- such as an applied sinusoidal voltage that has only a single frequency at which it interacts with the system.

8. Apr 10, 2012

rude man

1. Laplace only works for linear, constant-coefficient diff. eq's, basically.
2. You don't learn the 'innards' of diff eq's by looking up the answer in a table of transforms! (There is of course a way to compute the inverse transforms but it involves contour integration in the complex plane, which nobody does if tables are available!)

The Fourier transform is in particular isn't suitable for transient problems, like a step voltage into an R-C network. The Fourier integral is used for single pulses but all you get from that is the frequency spectrum of the output.

9. Apr 10, 2012

rude man

Pure sinusoidal excitations are handled simply by replacing s → jω.

10. Apr 10, 2012

RoshanBBQ

Yeah. Then you have turned your Laplace transform into a Fourier transform. The idea is simple: If you deal with analyzing frequency response all the time, you do not want to do the simplifications that follow from letting s be jω. Instead, you just use a Fourier transform whose tables are already in a simplified fashion in terms of ω.

11. Apr 10, 2012

rude man

Can't think of anything much simpler than s → jω and computing the magnitude ....

Give me an example of what you mean. Take a simple R-C lo-pass with Laplace 1/(Ts + 1), and tell me what your table of Fourier transforms does for you?