# Laplace analysis of simple LC tank (no resistance)

1. Jul 16, 2014

### jrive

I am stumped by an exercise in using Laplace transforms to analyze the voltage and current in simple LC tank. My issue is with the correct sign of the voltage across the capacitor ...let me pose the problem.

A circuit consists of a voltage source V, 2 switches, a cap C and an inductor L. The switch from the source to the cap has been on for a long time (cap is fully charged), while the switch connecting the cap to the inductor is open. Then at time t(0+), the switches toggle, and the voltage source is disconnected and the cap is now connected to the inductor.

v(0-)=V,
il(0-)=0,

The laplace circuit models are:
Cap:
i(t)=Cdv(t)/dt
I(s)=CsV(s)-Cv(0-)
V(s)=I(s)/sC+ v(0-)/S

Ind.
v(t)=Ldi(t)/dt
V(s)=LsI(s)-Li(0-)
I(s)=V(s)/sL+i(0-)/s

So....I can get the answer for the current fairly easily...Since the current into the cap is defined as positive when the switch from the source to the cap is on , then when the current flows from the cap to the inductor at t(0+), it is negative, or -Ic. So,
-Ic=IL

Cv(0-)-CsV(s)=V(s)/sL
solving for V(s),
V(s)=Cv(0-)sL/(s^2LC+1)
invLaplace(V(s))=V cos[t/sqrt(LC)] --> this is fine....

my problem is when I try to solve for the voltage directly...(keep in mind that if I obtain voltage by using this current across the cap or inductor, I do get the correct answer, but not when I try to do it directly), I have a sign problem that I can't figure out....

Since the voltage across the cap = the voltage across the inductor at time t(0+), then
I(s)/sC+v(0-)/s=LsI(s) (i(0-)==0)

this is the problem....the sign is incorrect, and this will lead to a transfer function where I have s^2LC - 1 in the denominator, and not s^2LC + 1 to get an oscillatory response. What am I missing here?

The cap model during the charging phase is as shown in figure in file cap_t(0-).bmp....and for the math to work, I need to change it to the model in figure cap_t(0+).bmp at t(0+). I just can't convince myself as to why....

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• ###### cap_t(0+).bmp
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Last edited: Jul 16, 2014
2. Jul 16, 2014

### Staff: Mentor

Hi. You're a bit long-winded, but I think I get the gist of your angst.

You say the sign is wrong, and I agree that it's wrong....so fix it!

Have you drawn the circuit of a capacitor parallel with an inductor? And marked in the current? And written the voltage across each element in terms of that current direction you drew?

Try it again.

3. Jul 17, 2014

### jrive

Thanks for the response (and the criticism)....

4. Jul 17, 2014

### Staff: Mentor

So you discovered what you'd been doing wrong?

5. Jul 22, 2014

### jrive

Yep....stupid mistake!!