# Determine $i_L(t)$of the RLC circuit

I have an RLC circuit that I am supposed to solve using s-domain analysis and I have gotten stuck when trying to transform back into time domain.

I utilized node voltage analysis to determine the voltage at node B since it is needed to solve for $$\i_L\$$. However, I got stuck here because I noticed that the denominator is not able to be simplified so I cannot perform partial fraction decomposition to solve for $$\i_L\$$.

Could someone explain where I went wrong or explain a potential next step? My work is shown below:

• Why is there a voltage source of $3/s$? The initial capacitor voltage is zero as you showed. Commented Dec 3, 2021 at 22:33
• – jonk
Commented Dec 4, 2021 at 2:00
• Why can't do decomposition? What's wrong with $x^2+2x+2=(x+1+\mathrm{j})(x+1-\mathrm{j})$ ? Commented Dec 4, 2021 at 15:05
• 55a, If you already have what you need then it's considerate to select an answer. If not and if you don't have anything to add here then you may not get the help you want. You may need to write something if you haven't received the help you wanted.
– jonk
Commented Dec 5, 2021 at 4:09

However, I got stuck here because I noticed that the denominator is not able to be simplified so I cannot perform partial fraction decomposition to solve

Quite right: $$\\dfrac{1}{s^2 + 2s +2}\$$ isn't reducible by partial fractions.

But, it is reducible to this: -

$$\dfrac{1}{(s+1)^2+1}$$

And that is easily inverse Laplaced to this: -

$$e^{-t} \sin(t)$$

Of course, if you used the full fraction ($$\\dfrac{4s}{s^2 + 2s + 2}\$$) then, it inverse Laplaces to this: -

$$4e^{-t}\cos(t)- 4e^{-t}\sin(t)$$

This inverse Laplace calculator may be of use to you in the future.

• Very minor comment, it is reducible by partial fractions but just with complex roots and you can use Euler to massage it into forms that inverse transform. Agree? Commented Dec 4, 2021 at 19:48
• @relayman yes, I see what you mean. Good point, it could be solved that way. Commented Dec 4, 2021 at 20:28

Well, the current through the inductor can be found using the current divider formula:

$$\text{i}_\text{L}\left(t\right)=\mathscr{L}_\text{s}^{-1}\left[\frac{\frac{1}{\text{sC}}}{\frac{1}{\text{sC}}+\text{sL}}\cdot\text{I}_\text{i}\left(\text{s}\right)\right]_{\left(t\right)}\tag1$$

Where $$\\mathscr{L}_\text{s}^{-1}\left[\cdot\right]_{\left(t\right)}\$$ is the inverse Laplace transform.

And it is not hard to see that:

$$\text{I}_\text{i}\left(\text{s}\right)=\frac{\text{V}_\text{i}\left(\text{s}\right)}{\text{R}_1+\text{R}_2+\left(\frac{1}{\text{sC}}\space\text{||}\space\text{sL}\right)}\tag2$$

Where $$\\text{V}_\text{i}\left(\text{s}\right)=\frac{\hat{\text{u}}}{\text{s}}\$$.

So, we get:

$$\text{i}_\text{L}\left(t\right)=\mathscr{L}_\text{s}^{-1}\left[\frac{\frac{1}{\text{sC}}}{\frac{1}{\text{sC}}+\text{sL}}\cdot\frac{\hat{\text{u}}}{\text{s}}\cdot\frac{1}{\text{R}_1+\text{R}_2+\left(\frac{1}{\text{sC}}\space\text{||}\space\text{sL}\right)}\right]_{\left(t\right)}=$$ $$\mathscr{L}_\text{s}^{-1}\left[\frac{\hat{\text{u}}}{\text{s}}\cdot\frac{1}{\text{CL}\left(\text{R}_1+\text{R}_2\right)\text{s}^2+\text{Ls}+\text{R}_1+\text{R}_2}\right]_{\left(t\right)}\tag3$$

Using your values, you will (must) find:

$$\text{i}_\text{L}\left(t\right)=\frac{2}{7}-\frac{2\exp\left(-\frac{t}{7}\right)}{679}\cdot\left(\sqrt{97} \sin \left(\frac{\sqrt{97} t}{7}\right)+97 \cos \left(\frac{\sqrt{97} t}{7}\right)\right)\tag4$$

• Is there a reason on why you type real variables in non-italics? Commented Dec 4, 2021 at 4:49
• Ah, I got the numbers wrong, my apologies. I'll delete the comments. Commented Dec 4, 2021 at 8:55
• @aconcernedcitizen it's okay, mate! No worries Commented Dec 4, 2021 at 8:57
• @Jan Your answer is sterile, though I agree with results. Straight math is usually fine for me, these days. But I certainly remember times when I'm in deep water and wish the answer text included more discussion to help me align my poor brain. Still, I don't yet see the OP writing anything since the first posted question. It may be because they are looking a free answer without any real respect for those trying to help (Elliot's hot button.) It may be for other reasons. If the OP says anything I may jump in and write some context. Either way, +1. But the OP may need a little more hand-holding.
– jonk
Commented Dec 5, 2021 at 1:47