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Starting with the the equation:

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

I need to find what $i(t)$ is by doing the inverse Laplace transform.

I need to do some algebra to put it in a form that is easy to do an inverse Laplace transform (i.e. a form that represents an example in a Laplace transform table). 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?

According to Wolfram Alpha the answer is this. But how was this calculated?

From the answer you can see that if $4L > CR^2$ then the discriminant (the square root in the numerator of exponent) will be an imaginary number. If this is the case, how would I use Euler's formula to get an equations with sine and cosine in it?

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Use partial fraction decompostion. This requires factoring the quadratic first.

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

The roots of that quadratic are $$ s_{1,2} = -\frac{R}{2L}\pm\sqrt{\left(\frac{R}{2L}\right)^2-\frac{1}{LC}} $$

If $s_{1,2}$ are distinct, your function can be represented as $$ I(s)=\frac{6}{L}\frac{1}{(s-s_1)(s-s_2)} = \frac{A}{s-s_1} + \frac{B}{s-s_2} $$

$A$ can be found by multiplying the equation through by $s-s_1$ and then taking $\lim_{s\rightarrow s_1}$; $B$ can be found in the analogous way with $s_2$. The form of each term is in the Laplace transform tables; each term will correspond to an exponential.

If $s_{1,2}$ happen to be identical (i.e. the discriminant is exactly zero), you'll use the representation $$ I(s)=\frac{6}{L}\frac{1}{(s-s_1)^2} = \frac{A}{s-s_1} + \frac{B}{(s-s_1)^2} $$

and again, find constants $A$ and $B$ that satisfy the above. Each of that kind of term will also be found in the transform tables.

The top case is more likely, where the roots are distinct. In that case you'll either get two real roots or a complex conjugate pair. If there are two real roots, you'll get two decaying exponentials after the inverse Laplace transform. If you have a complex conjugate pair, you'll get an exponentially decaying sinusoid, where the decay will look like $e^{-\frac{R}{2L}}$ and the sinusoidal part will have real frequency $\sqrt{\frac{1}{LC}-\left(\frac{R}{2L}\right)^2}$

In the second case, you should get something like $(C_1t+C_2)e^{-\frac{R}{2L} t}$ for some constants $C_1,C_2$.

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

Having fixed $R$, $C$ and $L$ (for example $R=6$, $C=1/5$, $L=1$), we have:

$$I(s) = \frac{6}{s^2+6s+5}$$

First of all, find the zeros of denominator:

$$s^2 + 4s + 1= 0 \Rightarrow s = -5 \vee s = -1$$

Then $s^2+4s+1= (s+5)(s+1)$.

We can write $$I(s) = \frac{a}{s+5} + \frac{b}{s+1} = \frac{a(s+1)+b(s+5)}{(s+1)(s+5)} = \frac{s(a+b) + (a + 5b)}{s^2 + 6s + 5} = \frac{6}{s^2+6s+5}$$

Then:

$$\left\{\begin{array}{lcl}a+b & = & 0\\a + 5b & = & 6 \end{array}\right. \Rightarrow \left\{\begin{array}{lcl}a & = & -b\\-b + 5b & = & 6 \end{array}\right. \Rightarrow \left\{\begin{array}{lcl}a & = & -\frac{3}{2}\\b & = & \frac{3}{2} \end{array}\right.$$

Then:

$$I(s) = -\frac{\frac{3}{2}}{s+5} + \frac{\frac{3}{2}}{s+1}$$

The antitransformation yield to:

$$i(t) = -\frac{3}{2}e^{-5t} + \frac{3}{2}e^{-t}$$

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  • $\begingroup$ Thanks for your answer. But wouldn't the antitransformation be: (-3/2)e^(-5t) + (3/2)e^(-5t), which equals 0? $\endgroup$ – Blue7 Apr 13 '14 at 19:23
  • $\begingroup$ why it should be equal to $0$?? $\endgroup$ – the_candyman Apr 13 '14 at 19:40
  • $\begingroup$ I might be missing something, but isn't the inverse laplace transform of (3/2)/(s+5) equal to (3/2)e^-5t? You have made it equal (3/2)e^-t instead. Where did the 5 go? Another way to look at it is your equation I(s)=-(3/2)/(s+5)+(3/2)/(s+5). This equals $0$ $\endgroup$ – Blue7 Apr 13 '14 at 19:46
  • $\begingroup$ I fixed it, sorry. $\endgroup$ – the_candyman Apr 14 '14 at 5:26

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