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The Inverse Laplace Transform (example #1)

Determine f(t) given:

$$ F(s) = \frac{48(s+2)}{(s+1)(s+3)(s+4)} $$ Using partial fraction decomposition, we expand F(s) to get the following: $$ F(s) = \frac{A}{s+1} + \frac{B}{s+3} + \frac{C}{s+4} \qquad,(Eqn\;1)$$ We will now examine two different approaches that can be used to determine the expansion coefficients (A, B, and C).

Determine coefficients, Method #1 (use Heaviside theorem)

Recall the definition of the Heaviside theorem (from the previous page): $$ k_i = (s+p_i) F(s) \Big|_{s=-p_i} $$

Expansion coefficient "A":

To determine "A" via the Heaviside theorem, we have the following: $$ \begin{align} A &= (s+1)F(s) \Big|_{s=-1} \\\\ &= \frac{48(s+2)}{(s+3)(s+4)} \Big|_{s=-1} \\\\ &= \frac{48(-1+2)}{(-1+3)(-1+4)} \\\\ &= \frac{48(1)}{(2)(3)} \\\\ A &= 8 \end{align} $$

Expansion coefficient "B":

To determine "B" via the Heaviside theorem, we have the following: $$ \begin{align} B &= (s+3)F(s) \Big|_{s=-3} \\ \\ &= \frac{48(s+2)}{(s+1)(s+4)} \Big|_{s=-3} \\\\ &= \frac{48(-3+2)}{(-3+1)(-3+4)} \\\\ &= \frac{48(-1)}{(-2)(1)} \\\\ B &= 24 \end{align} $$

Expansion coefficient "C":

To determine "C" via the Heaviside theorem, we have the following: $$ \begin{align} C &= (s+4)F(s) \Big|_{s=-4} \\ \\ &= \frac{48(s+2)}{(s+1)(s+3)} \Big|_{s=-4} \\\\ &= \frac{48(-4+2)}{(-4+1)(-4+3)} \\\\ &= \frac{48(-2)}{(-3)(-1)} \\\\ C &= -32 \end{align} $$ We can now substitute these coefficients into equation #1:

$$ F(s) = \frac{8}{s+1} + \frac{24}{s+3} - \frac{32}{s+4} \qquad,(Eqn\;2) $$

Determine coefficients, Method #2 (Algebraic method/system of equations)

Once again, recall equation #1:

$$ F(s) = \frac{48(s+2)}{(s+1)(s+3)(s+4)} = \frac{A}{s+1} + \frac{B}{s+3} + \frac{C}{s+4} \qquad,(Eqn\;1)$$

We begin by multiplying both sides by the common denominator: $$ 48(s+2) = A(s+3)(s+4) + B(s+1)(s+4) + C(s+1)(s+3) $$ Now multiply through: $$ 48(s+2) = A(s^2+7s+12) + B(s^2+5s+4) + C(s^2+4s+3) $$ $$ 48s+96 = As^2+7As+12A+Bs^2+5Bs+4B+Cs^2+4Cs+3C $$ $$ 48s+96 = s^2(A+B+C) + s(7A+5B+4C) + (12A+4B+3C) $$ Lastly, we equate coefficients: $$ for \; s^2: \;\;\; A+B+C=0 $$ $$ for \; s: \;\;\; 7A+5B+4C = 48 $$ $$ for \; constants: \;\;\; 12A+4B+3C=96 $$ Now create an augmented matrix of coefficients containing the above three equations: $$ \begin{pmatrix} 1&1&1&0\\ 7&5&4&48\\ 12&4&3&96\\ \end{pmatrix} $$ We will now use Cramer's Method for solving the system of three equations.

1) Calculate the determinant of the matrix of coeffificients:

$$ D = \begin{vmatrix} 1&1&1\\ 7&5&4\\ 12&4&3\\ \end{vmatrix} $$ $$ = 1 \begin{vmatrix} 5&4\\ 4&3\\ \end{vmatrix} - 1 \begin{vmatrix} 7&4\\ 12&3\\ \end{vmatrix} + 1 \begin{vmatrix} 7&5\\ 12&4\\ \end{vmatrix} $$ $$ = [3(5)-4(4)] - [7(3)-4(12)] + [7(4)-12(5)] $$ $$ -1-(-27)+(-32) $$

$$ D = -6 $$

2) Create a matrix that consists of the matrix of coefficients with its 1st column replaced by the 4th column of the augmented matrix of coefficients (and calculate its determinant).

$$ D_A = \begin{vmatrix} 0&1&1\\ 48&5&4\\ 96&4&3\\ \end{vmatrix} $$ $$ = 0[5(3)-4(4)]-1[48(3)-4(96)]+1[48(4)-5(96)]$$ $$ = 0+240-288 $$

$$ D_A = -48 $$

3) Create a matrix that consists of the matrix of coefficients with its 2nd column replaced by the 4th column of the augmented matrix of coefficients (and calculate its determinant).

$$ D_B = \begin{vmatrix} 1&0&1\\ 7&48&4\\ 12&96&3\\ \end{vmatrix} $$ $$ = 1[48(3)-96(4)]-0[7(3)-12(4)]+1[7(96)-12(48)]$$ $$ = -240-0+96 $$

$$ D_B = -144 $$

4) Create a matrix that consists of the matrix of coefficients with its 3rd column replaced by the 4th column of the augmented matrix of coefficients (and calculate its determinant).

$$ D_C = \begin{vmatrix} 1&1&0\\ 7&5&48\\ 12&4&96\\ \end{vmatrix} $$ $$ = 1[5(96)-4(48)]-1[7(96)-12(48)]+0[7(4)-12(5)]$$ $$ = 288-96 $$

$$ D_C = 192 $$

5) Use Cramer's Rule to solve the system of equations:

$$ A = \frac{D_A}{D} = \frac{-48}{-6} $$ $$ A = 8 $$ $$ $$ $$ B = \frac{D_B}{D} = \frac{-144}{-6} $$ $$ B = 24 $$ $$ $$ $$ C = \frac{D_C}{D} = \frac{192}{-6} $$ $$ C = -32 $$ We can now substitute these coefficients into equation #1 and obtain the same result as we did using the Heaviside Theorem:

$$ F(s) = \frac{8}{s+1} + \frac{24}{s+3} - \frac{32}{s+4} \qquad,(Eqn\;2) $$

Determine the Inverse Laplace Transform

The inverse Laplace transform is defined as: $$ f(t) = \mathcal{L^{-1}}[F(s)] $$ Substituting equation #2 into the above expression we have: $$ f(t) = 8\mathcal{L^{-1}}\Big[ \frac{1}{s+1} \Big] + 24\mathcal{L^{-1}}\Big[ \frac{1}{s+3} \Big] - 32\mathcal{L^{-1}}\Big[ \frac{1}{s+4} \Big] $$ We can now use property #2 in our Laplace transform table on each of the three terms in the expression above. Doing so gives us the final answer of:

$$ f(t) = 8e^{-t} + 24e^{-3t} - 32e^{-4t} $$

Continue on to example problem #2 ...