Consider the Laplace transform:

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:

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

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

where Q(s) and P(s) are polynomials and *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:

- P(s) is a quadratic with 2 real roots
- P(s) is a quadratic with a double root
- P(s) is a quadratic with complex conjugate roots
- P(s) is of degree 3 and has 1 real root and 2 complex
conjugate
roots

**P(s) is a Quadratic with 2 Real Roots**

Consider the initial example on this page:

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

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:

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

Comparing the LHS and RHS, we have:

The denominators are equal. The two sides will be equal if the
numerators

are equal. The numerators are equal for
all s if the coefficients of s and the

constant term
are equal. Hence, we get the following equations:

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

If the denominator has n distinct roots:

then the decomposition has the form

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

transform is

The unknown coefficients can be determined using the
same

technique as in the case of only 2 factors, as shown
above.

**P(s) is a Quadratic with a Double Root**

Consider the example:

The denominator has double root -2. The appropriate decomposition

in this case is

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

Using the same technique as in the case of distinct roots
above,

it can be shown that A=1 and B=-2.

**P(s) is a Quadratic and has Complex Conjugate Roots**

Consider the example:

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

square for the denominator. We have

Hence, we have

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

If we can find A and B, then:

The inverse transform is

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

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.

**P(s) is of Degree 3 and has 1 Real Root and 2 Complex
Conjugate
Roots**

Consider the example:

The roots of the denominator are 1 and -2+/-i. In this case
we

look for a decomposition of the form:

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.

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**Copyright **© **1996
Department
of Mathematics, Oregon State
University**

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