When solving circuit analysis problems in the s-Domain with Laplace transforms, you are generally left with a ratio of polynomials of s. To convert these back into functions in the time domain, the ratio may need to be converted into the addition of simpler polynomials which can then be looked up in a table to find their inverse.
The first step in expanding the ration is dividing if the ratio is improper. The ration is improper if the order (the largest power) of the numerator is greater than the order of the denominator. If this is the case then the numerator should be divided by the denominator in a way which is very similar to standard long division.
The two terms which are not part of a fraction now can be looked up in a table to find the inverse Laplace transform. The rest of this reference section will deal with how to reduce the remaining fractional part to something that can be converted easily.
After polynomial division, we are left with a proper rational function. At this point, our objective is to convert this function into the sum of simple fractions. For example we have
and we want to find A and B such that
The factors s + 3 and s + 4 result in the solutions of s = -3 and s = -4. These solutions are called simple poles. To find the numerators we do the following
likewise
In general, a simple pole s = p will result in a simple fraction of the form A / (s - p) where A can be found from the function F(s) with the following formula
When there is a simple pole that is a complex number, then its conjugate is also a pole. Additionally, the numerator of the simple fractions of these two poles are also conjugates, meaning that it is only necessary to find one of the numerators. Here is an example.
The two poles are s = -2 + j4 and s = -2 - j4, so we are looking for A and B such that
The thing we do know though is that B = A*, so once we find A, we will know B as well. We can find A in the same way we found a real simple pole
so the partial fraction expansion of F(s) is
A whole new situation arises when one of the poles occurs more than once in the denominator. In this case, there are as many terms as the order of the pole. For example, the following function has a pole of order 3 and a pole of order 1, this means there will be 4 total terms in the expansion
Basically, there is a term in the expansion for the highest order of the pole, and then each decreasing order down to one. The following example will illustrate how to solve the expansion when there are repeated poles.
A and C can be solved by using the same method we did for simple poles.
B cannot be solved for by this method since multiplying F(s) by (s + 2) and evaluating at s=2 would lead to an undefined answer. Instead, we will use the following formula
The final answer is then
In general, when there is a repeated pole, it has r terms where r is the order of the pole. The numerators can be found from the following formula where r is the order of the pole and k is the number of the term being solved for (k = 1 . . . r)
This formula will work to find all the numerators of the terms for both real and complex repeated poles.