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\begin{document}
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{\Large\bf The Method of Partial Fractions for Inverting Laplace
Transforms}

\vspace{.3in}

Consider the Laplace transform:

$$
Y(s)=\frac{2s-8}{s^2-5s+6}
$$

Some manipulations must be done before Y(s) can be inverted since
it does not appear directly in our table of Laplace transforms.

As we will show below:

$$
Y(s)=\frac{4}{s-2}-\frac{2}{s-3}
$$

Now, we can invert Y(s). From the table, we see that the inverse
of 1/(s-2) is exp(2t) and that inverse of 1/(s-3) is exp(3t). 
Using the linearity of the inverse transform, we have 

$$
y(t)=L^{-1}[Y(s)](t)=4e^{2t}-2e^{3t}
$$

The method of partial fractions is a technique for decomposing
functions like Y(s) above so that the inverse transform can be
determined in a straightforward manner. It is applicable to
functions of the form

$$
Y(s)=\frac{Q(s)}{P(s)}
$$

where Q(s) and P(s) are polynomials and {\it the degree of
Q is less than the degree of P}. For simplicity we assume that
Q and P have real coefficients. We consider the following cases:

\begin{itemize}
\item P(s) is a quadratic with 2 real roots
\item P(s) is a quadratic with a double root
\item P(s) is a quadratic with complex conjugate roots
\item P(s) is of degree 3 and has 1 real root and 2 complex conjugate
roots
\end{itemize}

\vspace{.3in}

{\bf P(s) is a Quadratic with 2 Real Roots}

\vspace{.2in}

Consider the initial example on this page:

$$
Y(s)=\frac{2s-8}{s^2-5s+6}
$$

The denominator can be factored: s^2-5s+6=(s-2)(s-3). The
denominator has 2 real roots. In this case we can write

$$
\frac{2s-8}{s^2-5s+6}=\frac{A}{s-2}+\frac{B}{s-3}
$$

Here A and B are numbers. This is always possible if the
numerator is of degree 1 or is a constant. The goal now is
to find A and B. Once we find A and B, we can invert
the Laplace transform:

$$
y(t)=L^{-1}[Y(s)](t)=Ae^{2t}+Be^{3t}
$$

If we add the two terms on the RHS, we get 

$$
\frac{A}{s-2}+\frac{B}{s-3}=\frac{A(s-3)+B(s-2)}{(s-2)(s-3)}
=\frac{(A+B)s+(-3A-2B)}{s^2-5s+6}
$$

Comparing the LHS and RHS, we have:

$$
\frac{2s-8}{s^2-5s+6}=\frac{(A+B)s+(-3A-2B)}{s^2-5s+6}
$$

The denominators are equal. The numerators are equal for
all s if the the coefficients of s and the constant term
are equal. Hence, we get the following equations:

$$
2=A+B \hspace{.2in} -8=-3A -2B
$$

This is a system of 2 linear equations with 2 unknowns. It
can be solved by standard methods. We obtain A=4 and B=-2.
Hence

$$
Y(s)=\frac{4}{s-2}-\frac{2}{s-3}
$$

If the denominator has n distinct roots:

$$
P(s)=(s-a_1)(s-a_2)\cdots(s-a_n),
$$

then the decomposition has the form

$$
\frac{Q(s)}{P(s)}=\frac{A_1}{s-a_1}+\frac{A_2}{s-a_2}
+\cdots+\frac{A_n}{s-a_n}
$$

where A_1, A_2, ..., A_n are unknown numbers. The inverse
transform is

$$
y(t)=L^{-1}[\frac{Q(s)}{P(s)}](t)=A_1e^{a_1t}+A_2e^{a_2t}+\cdots
A_ne^{a_nt}
$$

The unknown coefficients can be determined using the
same technique as in the case of only 2 factors, as shown
above.

\vspace{.3in}

{\bf P(s) is a Quadratic with a Double Root}

\vspace{.2in}

Consider the example:

$$
Y(s)=\frac{s+1}{(s+2)^2}
$$

The denominator has double root -2. The appropriate decomposition
in this case is 

$$
Y(s)=\frac{s+1}{(s+2)^2}=\frac{A}{s+2}+\frac{B}{(s+2)^2}
$$

Here A and B are numbers. From the table the inverse transform
of 1/(s+2) is exp(-2t) and the inverse transform of 
1/(s+2)^2 is texp(-2t). Hence, the inverse transform of 
Y(s) is

$$
y(t)=L^{-1}[Y(s)](t)=Ae^{-2t}+Bte^{-2t}
$$

Using the same technique as in the case of distinct roots
above, it can be shown that A=1 and B=-2. 

\vspace{.3in}

{\bf P(s) is a Quadratic and has Complex Conjugate Roots}

\vspace{.2in}

Consider the example:

$$
Y(s)=\frac{3s+9}{s^2+4s+5}
$$

The roots of the denominator are -2+/-i. We can complete the
square for the denominator. We have

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

Hence, we have

$$
Y(s)=\frac{3s+9}{(s+2)^2+1}
$$

Note the denominator (s+2)^2+1 is similar to that for
Laplace transforms of exp(-2t)cos(t) and exp(-2t)sin(t).
We need to manipulate the numerator. Note that in the
formula in the table, we have a=-2 and b=1. We look
for a decomposition of the form

$$
\frac{3s+9}{(s+2)^2+1}=\frac{A(s+2)+B}{(s+2)^2+1}
$$

If we can find A and B, then: 

$$
\frac{3s+9}{(s+2)^2+1}=A\frac{s+2}{(s+2)^2+1}+B\frac{1}{(s+2)^2+1}
$$

The inverse transform is

$$
y(t)=L^{-1}[Y(s)](t)=Ae^{-2t}\cos t+Be^{-2t}\sin t
$$

We can determine A and B by equating numerators in the
expression

$$
\frac{3s+9}{(s+2)^2+1}=\frac{A(s+2)+B}{(s+2)^2+1}
=\frac{As+2A+B}{(s+2)^2+1}
$$

Comparing coefficients of s in the numerator we
conclude 3=A. Comparing the constant terms we
conclude 2A+B=9. Hence A=3 and B=3.

\vspace{.3in}

{\bf P(s) is of Degree 3 and has 1 Real Root and 2 Complex Conjugate
Roots}

\vspace{.2in}

Consider the example:

$$
Y(s)=\frac{s^2+9s+3}{(s-1)(s^2+4s+5)}
$$

The roots of the denominator are 1 and -2+/-i. In this case
we look for a decomposition of the form:

$$
\frac{s^2+9s+3}{(s-1)(s^2+4s+5)}=\frac{A}{s-1}+{Bs+C}{s^2+4s+5}
$$

where A, B and C are unknowns. The inverse transform
of A/(s-1) is Aexp(t). Once B and C are determined the
second term can be manipulated as in the previous example.

\end{document}
