# Need help with inverse laplace transform problem

I'm really stuck this problem.

This actually resulted because of equations for a circuit analysis problem, so in case it would help I'll list the equations here too. Although, feel free to ignore them. We are solving for currents $i_1$ and $i_2$,

$$(i_1 - i_2)s - 2s^{-1} + i_1 = 0 \\ (i_2 - i_1)s + i_2 + (s^{-1} + i_2)2s^{-1} = 0 \tag{1}.$$

I get that $i_2$ is given by,

$$i_2 = {2(-s^3 + s + 1)\over s((s^2 - s - 1)s^2 - s(s + 1) - 2(s + 1))} \tag{2}$$

I am supposed to calculate the inverse transform of,

$$\displaystyle V(s) = (s^{-1} + i_2)2s^{-1} \\ \ \ \ \ \ \ \ \ \ \ = \ {2(s^3 - 3s^2 - 2s - 1) \over s(s^4 - s^3 - 2s^2 - 3s - 2)} \tag{3}$$

The expression for $V$ was obtained with sage. I can't see any easy way to decompose $V$, and this is where I am stuck.

All help is greatly appreciated! Thanks!

• First, are you sure you wrote the systems equations properly and then solved them properly? I get different values for $i_1$ and $i_2$? – Amzoti Apr 28 '13 at 2:41
• @Amzoti Yeah, that could be it. I got two different expressions for $i_2$ by solving the system differently. I guess I'll just have to try another approach. – ThisIsNotAnId Apr 28 '13 at 20:43
• Glad it all worked out and you are very welcome! Regards – Amzoti May 4 '13 at 20:40

I get the following results (you can plug these into $(1)$ and see if they satisfy each of the system's equations):

$$i_1 = \frac{2(s^2 + 2)}{s(2s^2 + 3s + 2)}$$

and

$$i_2 = \frac{2(s^2 - s - 1)}{s(2s^2 + 3s + 2)}$$

You gave:

$$\displaystyle V(s) = (s^{-1} + i_2)2s^{-1} = \left(\frac{1}{s} + \frac{2(s^2 - s - 1)}{s(2s^2 + 3s + 2)}\right)\frac{2}{s} = \frac{8s + 2}{s(2s^2 + 3s + 2)}$$

The Inverse Laplace Transform is given by:

$$\displaystyle \mathcal{L}^{-1} (V(s)) = V(t) = 1-\frac{e^{-3 t/4} \left(\sqrt 7 \cos\left(\sqrt 7 t/4\right)-13 \sin(\sqrt 7 t/4)\right)}{\sqrt 7}$$

However, like I said, you had better validate my $i_1$ and $i_2$ since they do match yours.

Update

Note: I substitute $a = i_1$ and $b = i_2$ to get rid of any potential issues. Find a Wolfram Alpha solution to the simultaneous equation for $i_1$ and $i_2$.

Here is a WA solution to the simultaneous equations.

• Thanks for solving the system! Your solution seems to be on mark, also because of the type of circuit at hand; so I'll check and get back. – ThisIsNotAnId Apr 28 '13 at 20:53
• @ThisIsNotAnId: Please see the Wolfram Alpha link and solution to compare against what you are using. SAGE should certainly work! Regards – Amzoti Apr 28 '13 at 21:00
• Ok, I double checked and I finally have the proper expressions for $i_2 \text{ and } V(s).$ Thank you so much for all the help! – ThisIsNotAnId Apr 28 '13 at 23:00
• Can you let us know if Ron was correct or if the above was correct or if it turned out to be another answer? Thanks – Amzoti Apr 28 '13 at 23:02
• Well, Ron's was correct under the assumption that the expression for $V(s)$ in my op was correct, which it turns out it wasn't. You provided the correct expression for $i_2, V(s), \text{ and } \mathcal{L}^{-1}[V(s)].$ – ThisIsNotAnId Apr 28 '13 at 23:08

If your expression for $V(s)$ is correct, there really isn't much you can do about the roots: the expression is not factorable over the rationals, or even square roots. Rather, a plot reveals two real roots and a pair of complex roots; these you can find in terms of the horrifying quartic solution, or just numerically as I did.

Let the roots of the quartic be as follows:

$$s_1 \approx 2.45111 \quad s_2 \approx -0.788821 \quad s_{\pm} \approx -0.331144 \pm 0.931636$$

These are all simple roots, as is the root at $s_0=0$. We may determine the ILT from the sum of the residues of the poles of the function $V(s) e^{s t}$. The residue at $s_1$ is, for example:

$$2 \frac{2 s_1^3-3 s_1^2-2 s_1-1}{5 s_1^4-4 s_1^3-6 s_1^2-6 s_1-2} e^{s_1 t} \approx -0.26737 e^{2.45111 t}$$

where I used the fact that, for a simple pole $s_0$ of a function $p(s)/q(s)$ is $p(s_0)/q'(s_0)$ Repeat for all roots, and add; I get

$$f(t) = 1-0.26737 e^{2.45111 t}-1.22806 e^{-0.788821 t} + e^{-0.331144 t} [0.247714 \cos{0.961636 t} + 0.962263 \sin{0.961636 t} + i (-0.962263 \cos{0.961636 t} +0.247714 \sin{0.961636 t})]$$

• First, thank you for going through the trouble of finding the roots and the expression for $f$. Although, it now seems that I really did obtain an incorrect expression for $i_2$, because $f$ is supposed to be capacitor voltage which, here, will eventually go to negative infinity. Uhhhh, back to the drawing board for me. – ThisIsNotAnId Apr 28 '13 at 20:46