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?
Best Answer
$$I(s)=\frac{6}{Ls^2 + Rs + \frac{1}{C}}$$
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:
$$i(t) = -\frac{3}{2}e^{-5t} + \frac{3}{2}e^{-t}$$