RLC Circuit - 2. Order differential equation solution

[santais]

Homework Statement

Hey guys.

So we've been working on a RLC circuit, where the subject is oscillations. We accomplished to work out a differential equation, for this specific circuit, however solving it is a completely other part of the problem. We're completely stuck here and hopefully, there are some of you, who could help us move on :)

The equation is as following, using Kirchoffs law of the sum of the currents in a closed circuit:

(d^2v/dt) L + (dv/dt)R + (1/C)v = 0

Homework Equations

We tried looking at the general solution, where we did find a complex solution. Looking at the real part, we get equation which looks like:

k * e^-(α/2βt)sinw0(t).

where k, α and β are constants, here our R, L and C values. However, we just can't seem to find the solution, where our constants are included. In other words, how to determine k, α and β as R, L and C.

Also, searching at the internet, we found that w0 is:

w0 = 1/Sqrt(LC)

But how do you solve w0? Or simply just find the expression for it?

 

[ehild]

The equation is for the current in a series resonant circuit, set up according to Kirchhoff's Loop Rule.
Anyway, you have a second order differential equation,and can find the solution in exponential form: V=e^λt.
Taking the derivatives and substituting back into the equation, you get the equation for λ: Lλ²+rλ+1/C=0. The solution is
λ=-R/(2L)±√((R/2L)²-1/(LC)) In case 1/(LC)>R/(2L), λ is complex,

λ=-R/(2L)±i√(1/(LC)-(R/2L)²) ,

and the real solutions are V₁=e^-R/(2L)tcos(ωt) and V₂=e^-R/(2L)tsin(ωt), with ω=sqrt(1/(LC)-(R/(2L))²)

ω₀=1/sqrt(LC) would be the angular frequency in case of a lossless circuit, with no resistor in it. Your parameters α and β correspond to α =R and β =L.

ehild

 

[Boorglar]

You have a second-order differential equation of the form

{ay(t)'' + by(t)' + cy(t) = 0}

There are a few different methods to find the general solution.
Plug in {y = e^{λt}} in the equation. You'll get:

{aλ^{2} + bλ + c = 0} as the characteristic equation.

If you solve for {λ}, you have:

{λ_{1} = \frac{-b+\sqrt{b^{2}-4ac}}{2a}} and {λ_{2} = \frac{-b-\sqrt{b^{2}-4ac}}{2a}}


\textbf{1.} {b^{2}-4ac > 0:}

{y(t) = c_{1}e^{λ_{1}t} + c_{2}e^{λ_{2}t}}



\textbf{2.} {b^{2}-4ac = 0:}

Then {λ_{1} = λ_{2} = λ}

{y(t) = c_{1}e^{λt} + c_{2}te^{λt}}



\textbf{3.} {b^{2}-4ac < 0:}

Then {λ_{1} = α + βi} and {λ_{2} = α - βi}, where {α = \frac{-b}{2a}} and {β = \frac{\sqrt{4ac-b^{2}}}{2a}}

{y(t) = e^{αt}[c_{1}cos(βt) + c_{2}sin(βt)]}


In the case when it oscillates, it is a case 3. Then its angular frequency is {ω_{0} = \sqrt{\frac{1}{LC}-\frac{R^{2}}{4L^{2}}}}.

 

[santais]

You have a second-order differential equation of the form

{ay(t)'' + by(t)' + cy(t) = 0}

There are a few different methods to find the general solution.
Plug in {y = e^{λt}} in the equation. You'll get:

{aλ^{2} + bλ + c = 0} as the characteristic equation.

If you solve for {λ}, you have:

{λ_{1} = \frac{-b+\sqrt{b^{2}-4ac}}{2a}} and {λ_{2} = \frac{-b-\sqrt{b^{2}-4ac}}{2a}}


\textbf{1.} {b^{2}-4ac > 0:}

{y(t) = c_{1}e^{λ_{1}t} + c_{2}e^{λ_{2}t}}



\textbf{2.} {b^{2}-4ac = 0:}

Then {λ_{1} = λ_{2} = λ}

{y(t) = c_{1}e^{λt} + c_{2}te^{λt}}



\textbf{3.} {b^{2}-4ac < 0:}

Then {λ_{1} = α + βi} and {λ_{2} = α - βi}, where {α = \frac{-b}{2a}} and {β = \frac{\sqrt{4ac-b^{2}}}{2a}}

{y(t) = e^{αt}[c_{1}cos(βt) + c_{2}sin(βt)]}


In the case when it oscillates, it is a case 3. Then its angular frequency is {ω_{0} = \sqrt{\frac{1}{LC}-\frac{R^{2}}{4L^{2}}}}.

Great thanks for the answer. Anyway the resonance, w0, how to get to that answer ? {ω_{0} = \sqrt{\frac{1}{LC}-\frac{R^{2}}{4L^{2}}}} specifically. I get the solution for the differential equation, but the equation for the resonance simply doesn't make sense.

 

[Boorglar]

Well, I simply replaced

a with L
b with R
c with 1/C

And I plugged them in the formula for beta (which is the angular frequency)

If you know how to solve the diff. equation, then this is simply a direct consequence of plugging in these variables.

EDIT:

This is not the formula for resonance, but rather the angular frequency. I'm not sure we can talk about resonance when there is resistance because the oscillations eventually die out (inverse exponentially) (but I might be wrong). If you set R = 0, then the system will oscillate forever with the frequency ω0 = 1/√(LC) as you mentioned. This is the resonance frequency of the system. If you induce oscillations (for example with alternating voltage) at this precise frequency w0, then the system will enter in resonance and the amplitude will tend to infinity.
I think this article might interest you: http://en.wikipedia.org/wiki/Harmonic_oscillator#Damped_harmonic_oscillator

 

[ehild]

We speak about resonance when there is a sinusoidal driving force of angular frequency ω. The own vibration of the circuit dies out in a short time and the voltage or current oscillates with the driving frequency. The amplitude of the forced oscillation depends on the parameters of the circuit, and it is maximum at the resonant frequency, 1/sqrt(LC).

ehild

 

[santais]

Thanks :D that did it make it a lot clearer. Tried to solve it myself, as you said, with replacing the parameters in the beta notation, where I ended up with the same result. Now it's acutally just to put the equations in the general answer, of a 2nd order differential equation and it should start to look like something :)