Laplace analysis of simple LC tank (no resistance)

In summary, the conversation discusses a problem with analyzing the voltage and current in a simple LC tank using Laplace transforms. The issue is with the correct sign of the voltage across the capacitor, which leads to a transfer function with the wrong denominator. The solution is to fix the sign and redraw the circuit to properly represent the voltage and current directions.
  • #1
jrive
58
1
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...
 

Attachments

  • lc_tank.bmp
    277.2 KB · Views: 559
  • cap_t(0-).bmp
    341 KB · Views: 556
  • cap_t(0+).bmp
    329.7 KB · Views: 547
Last edited:
Engineering news on Phys.org
  • #2
jrive said:
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
Hi. You're a bit long-winded, but I think I get the gist of your angst. :wink:

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. :smile:
 
  • #3
Thanks for the response (and the criticism)...
 
  • #4
jrive said:
Thanks for the response ...
So you discovered what you'd been doing wrong?
 
  • #5
Yep...stupid mistake!
 

1. What is a simple LC tank circuit?

A simple LC tank circuit is an electronic circuit that consists of a capacitor (C) and an inductor (L) connected in parallel. The capacitor and inductor together create a resonant circuit that can store and release energy at a specific frequency. This type of circuit is commonly used in radio frequency (RF) applications.

2. What is the purpose of Laplace analysis in a simple LC tank circuit?

Laplace analysis is used to analyze the behavior and response of a simple LC tank circuit over time. It allows for the calculation of the circuit's transfer function, which describes the relationship between the input and output signals. This is useful in determining the circuit's frequency response and stability.

3. How is Laplace analysis performed on a simple LC tank circuit?

Laplace analysis involves transforming the circuit's differential equations into algebraic equations using Laplace transforms. This allows for the use of standard algebraic techniques to solve for the circuit's response. The resulting equations can then be inverted back to the time domain to obtain the response of the circuit over time.

4. What are the advantages of using Laplace analysis in a simple LC tank circuit?

One of the main advantages of Laplace analysis is that it allows for the analysis of complex circuits using simple algebraic equations. This makes it easier to solve and understand the behavior of the circuit. Additionally, Laplace analysis can provide insight into the stability and frequency response of the circuit, which can be useful in circuit design and optimization.

5. Are there any limitations to using Laplace analysis in a simple LC tank circuit?

While Laplace analysis is a powerful tool for analyzing circuits, it does have some limitations. One limitation is that it assumes the circuit is linear, which may not always be the case in practical circuits. Additionally, Laplace analysis does not take into account any parasitic elements or non-idealities in the circuit, which can affect its performance. Therefore, it is important to validate the results of Laplace analysis with real-world measurements.

Similar threads

  • Electrical Engineering
Replies
3
Views
938
Replies
4
Views
894
  • Electrical Engineering
2
Replies
38
Views
3K
  • Electrical Engineering
Replies
20
Views
1K
Replies
5
Views
9K
  • Electrical Engineering
Replies
7
Views
3K
Replies
12
Views
1K
  • Electrical Engineering
Replies
15
Views
2K
Replies
20
Views
2K
Replies
6
Views
2K
Back
Top