Determine f(t) given:
$$ F(s) = \frac{s+1}{(s+2)(s^2+2s+5)} $$ Begin by expanding F(s) using partial fraction decomposition:
$$ F(s) = \frac{s+1}{(s+2)(s^2+2s+5)} = \frac{A}{s+2} + \frac{Bs+C}{s^2+2s+5} \qquad,(Eqn\;1) $$
Determine expansion coefficients A, B and C
Now multiply both sides of equation #1 by the common denominator: $$ \begin{align} s+1 &= A(s^2+2s+5) + (Bs+C)(s+2) \\ &= As^2 + 2As + 5A + Bs^2 + 2Bs + Cs + 2C \\ s+1 &= s^2(A+B) + s(2A+2B+C) + 5A + 2C \end{align} $$ We now equate coefficients: $$ for \; s^2: \;\;\; A+B=0 $$ $$ for \; s: \;\;\; 2A+2B+C = 1 $$ $$ for \; constants: \;\;\; 5A+2C=1 $$ Now create an augmented matrix of coefficients containing the above three equations: $$ \begin{pmatrix} 1&1&0&0\\ 2&2&1&1\\ 5&0&2&1\\ \end{pmatrix} $$ We will now perform row operations to solve the system of equations. We begin by multiplying row #1 by -2 and adding the results to row #2. Doing so gives the following result: $$ \begin{pmatrix} 1&1&0&0\\ 0&0&1&1\\ 5&0&2&1\\ \end{pmatrix} $$ From this operation, we now know that: $$ C = 1 $$ Plugging this into row #3 we have: $$ 5A + 2(1) = 1 $$ $$ 5A = -1 $$ $$ A = \frac{-1}{5} $$ Plugging our value for "A" into row #1, we get: $$ \frac{-1}{5} + B = 0 $$ $$ B = \frac{1}{5} $$ We now plug the expansion coefficients into equation #1: $$ F(s) = \frac{-1}{5(s+2)} + \frac{\frac{1}{5}s+1}{s^2+2s+5} $$ $$ F(s) = \frac{-0.2}{s+2} + \frac{0.2s+1}{s^2+2s+5} \qquad, (Eqn\;2)$$
Rewrite the expression to match the format of a property in the table:
By inspection, you may recognize that for the second term of equation #2, we may be able to use one of the properties 19-22 in our table to find the inverse. The problem is that the format is a bit off. In such a case, we use the technique of "completing the square" so that the denominator of the second term can be rewritten in a form that resembles one of those properties in the Laplace transform table. This isn't a tutorial on completing the square, but the steps are as follows: $$ \begin{align} s^2+2s+5 \;&=\; s^2 + 2s + (\frac{2}{2})^2 + 5 - (\frac{2}{2})^2 \\\\ \;&=\; (s^2+2s+1) + 4 \\\\ \;&=\; (s+1)(s+1) + 4 \\\\ s^2+2s+5 \;&=\; (s+1)^2 + 4 \qquad, (Eqn\;2) \end{align} $$ We now substitute equation #2 into equation #1 and get the following: $$ F(s) = \frac{-0.2}{s+2} + \frac{0.2s+1}{(s+1)^2 + 4} \qquad, (Eqn\;3)$$
Determine the inverse Laplace transform:
By definition, the inverse is defined as: $$ f(t) = \mathcal{L^{-1}}[F(s)] $$ Substituting in equation #3:
$$ f(t) = \mathcal{L^{-1}}\Big[\frac{-0.2}{s+2} + \frac{0.2s+1}{(s+1)^2 + 4}\Big] $$ We will now continue to manipulate the expression until it resembles the properties in our table:
$$ \begin{align} &= -0.2\mathcal{L^{-1}}\Big[ \frac{1}{s+2} \Big] + 0.2\mathcal{L^{-1}}\Big[ \frac{s}{(s+1)^2+4} \Big] + \mathcal{L^{-1}}\Big[ \frac{1}{(s+1)^2+4} \Big] \\\\ &= -0.2\mathcal{L^{-1}}\Big[ \frac{1}{s+2} \Big] + 0.2\mathcal{L^{-1}}\Big[ \frac{(s+1)-1}{(s+1)^2+4} \Big] + \frac{1}{2}\mathcal{L^{-1}}\Big[ \frac{2}{(s+1)^2+4} \Big] \\\\ \end{align} $$
$$ \begin{align} &= -0.2\mathcal{L^{-1}}\Big[ \frac{1}{s+2} \Big] + 0.2\mathcal{L^{-1}}\Big[ \frac{(s+1)}{(s+1)^2+4} \Big] -0.2\mathcal{L^{-1}}\Big[ \frac{1}{(s+1)^2+4} \Big]+ \frac{1}{2}\mathcal{L^{-1}}\Big[ \frac{2}{(s+1)^2+4} \Big] \\\\ &= -0.2\mathcal{L^{-1}}\Big[ \frac{1}{s+2} \Big] + 0.2\mathcal{L^{-1}}\Big[ \frac{(s+1)}{(s+1)^2+4} \Big] \frac{-0.2}{2}\mathcal{L^{-1}}\Big[ \frac{2}{(s+1)^2+4} \Big]+ \frac{1}{2}\mathcal{L^{-1}}\Big[ \frac{2}{(s+1)^2+4} \Big] \\\\ \end{align} $$
The expression is now in a form that allows us to use properties 2, 19 and 20 of our table to find the inverse Laplace transform:
$$ f(t) = -0.2e^{-2t} + 0.2e^{-t} \cos(2t) - 0.1e^{-t} \sin(2t) + 0.5e^{-t} \sin(2t) $$
$$ f(t) = -0.2e^{-2t} + 0.2e^{-t} \cos(2t) + 0.4e^{-t} \sin(2t) $$
Continue on to example problem #3 ...