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The voltage across the power source equals the summed voltage across the resistor, capacitor, and inductor at any time (t). This is shown in the equation below:

$$Ri(t)+\frac{1}{C}\int_0^ti(t)+L\frac{d\:i(t)}{d\:t} = v_s(t)$$

for this example let: $$v_s(t) = 6$$

The laplace transform of this is:

$$RI(s)+\frac{1}{Cs}I(s)+LsI(s)=\frac{6}{s}$$

Rearrange it to make I(s) the subject:

$$I(s)=\frac{(\frac{6}{s})}{R+\frac{1}{Cs}+Ls}$$

Do some algebra to put it in a form that is easy to do an inverse laplace transform (ie. a form that represents an example in a laplace transform table)

$$I(s) = \frac{6}{Ls^2+Rs+\frac{1}{C}}$$

But I get stuck at this part. How would I do the algebra and inverse laplace transform so I can find what i(t) equals?

I expect that the answer will have the exponation e^-at in it, as the current will decay because of the energy dissipated in the resistor. I also expect it will have a sin, cos, both, or an imaginary number in it as the current will oscillate. According to wolfram alpha the answer is this. From this you can see that if 4L > CR^2 then the discriminant (the square root in the numerator of exponent) will be an imaginary number. You could then use eulers formula to get an equations with sin and cos in it.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I think you're missing the initial conditions in your equations \$\endgroup\$ – MightyPork Feb 8 '15 at 11:19
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Elaborating on Andy aka's answer ...

You need the transform pair

$$\frac{\omega_0}{(s+\alpha)^2+\omega_0^2}\Longleftrightarrow e^{-\alpha t}\sin(\omega_0t)u(t)$$

with

$$\omega_0^2=\frac{1}{LC}-\frac{R^2}{4L^2}$$

and

$$\alpha=\frac{R}{2L}$$

Of course you'll also get some multiplicative constant, but that's straightforward.

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  • \$\begingroup\$ Thanks for your answer. I just now have a slight confusion, shouldn't ω0 equal ((1/LC)-(R^2/4L^2))^0.5? On Mathematics Stack Exchange An answer has said that this is the frequency that it will ocilate at. \$\endgroup\$ – Blue7 Apr 13 '14 at 20:32
  • \$\begingroup\$ @Blue7 You're right, sorry for the mistake. I just corrected it in my answer! \$\endgroup\$ – Matt L. Apr 14 '14 at 6:57
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I suppose you need the TIME behaviour of the circuit after the circuit is excited with an input voltage Vo at t=0, correct?

Step 1: Start with the equation as given in your post. This equation is in the time domain and there is no need to step into the frequency domain.

Step2: Multiply the whole equation by C and differentiate the equation with respect to time. As a result, you have a homogenious differentiual equation of second order (right side of the equation is zero).

Step3: This diff. equation can be solved setting (using the "Ansatz")

i(t)=I*exp(st)

Step4: Introducing this expression into the diff. equation leads to

I*exp(st)*(1+sRC+s^2*LC)=0

and for t>0 we have

(1+sRC+s^2*LC)=0

Step5: This quadratic equation can be easily solved leading to

s1=sigma+jwo and s2=sigma-jwo

with sgma=R/2L and wo=SQRT(1/LC- sigma^2)

Step6: Introducing and adding both solutions into the equation in step3 and using EULER´s equation for sinusoidal expressions we arrive at

i(t)=Io*exp(sigma*t)*sin(wo*t) with Io=Vo/wo*L

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  • \$\begingroup\$ Yup i'm after the time behavior :) This answer is helpful, thank you. I'm just a bit confused as to what you mean by introducing the s1 and s2 into the equation in step 3. How would I do this. Also, I have a second question if you dont mind: What happens to the imaginary j in the solution? When you use eulers formula you will get a sinusoidal expression but there will still be a j (or i) in the expression, no? What does this j (or i) actually mean in terms of the time behavior of the current? \$\endgroup\$ – Blue7 Apr 14 '14 at 7:37
  • \$\begingroup\$ The equation in step3 contains the expression s*t. Introduce s1 into this equation as well as s2 and ADD both parts. Now you have two expressions with an exponential function and you can apply EULER´s formula. As the result, the imaginary term "j" disappears! \$\endgroup\$ – LvW Apr 14 '14 at 8:14
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I'm no expert on this but I'd say you need to do more work on the denominator to make it fit a standard transform. Maybe: -

\$I(s) = \dfrac{6}{Ls^2+Rs+\dfrac{1}{C}}\$ becomes...

\$I(s) = \dfrac{\dfrac{6}{L}}{s^2+\dfrac{R}{L}s+\dfrac{1}{LC}}\$

Then convert denominator to \$(s + \dfrac{R}{2L})^2+(\dfrac{1}{LC} - \dfrac{R^2}{4L^2})\$

I'm not going further because I'm unsure - not done this stuff in ages

EDIT

Drawing attached to help understand the relationships between \$\omega_N\$ and \$\omega_0\$: -

enter image description here

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  • \$\begingroup\$ Thanks for your answer. Looking at the second equation this looks like the standard form of a second order system. So does 2ζωn = R/L and +ωn^2 = 1/LC? \$\endgroup\$ – Blue7 Apr 13 '14 at 19:19
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    \$\begingroup\$ @Blue7 I'd have to check on the exact stuff but i suspect is does. In fact yes of course it does!! \$\endgroup\$ – Andy aka Apr 13 '14 at 20:06
  • \$\begingroup\$ It does seem like it would, but answers are disagreeing on different sites. If ωn^2 = 1/LC then the natural frequency equals (1/LC)^0.5. But I asked a similar question to this one on Mathematics Stack Exchange and the answer there says the function will have frequency ((1/LC)-(R/2L)^2)^0.5. Why are these answers disagreeing? \$\endgroup\$ – Blue7 Apr 13 '14 at 20:19
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    \$\begingroup\$ \$\omega_N\$ is the natural resonant frequency and is just based on LC but \$\omega_0\$ is the damped resonant frequency or frequency at which jw peaks at. the 2nd equation is the damped resonant frequency and if you did a pole zero diagram (or pythagorous) you'd see (I believe). \$\endgroup\$ – Andy aka Apr 13 '14 at 20:22

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