function form and partial fraction expansion form remain the same as Case 1:
The unknown coefficients
are calculated in the same manner as Case 1, as well:
[
]
The primary difference from Case 1 comes in the calculated values for
. In particular, with
a pair of complex conjugate poles,
and
, then the coefficients
and
will also be
complex conjugates. This means that we really only need to calculate one of
and
. It
will also mean that we can likely combine the resultant complex terms using Euler‟s
formulas to create sine and cosine terms in the resultant function
:
̅
 
{ }
2
 
 
3
 
 
 
(
)
 
Once again, an example is in order to clarify the above procedure. Consider the following
function,
, for which we want to find
:
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39
Using partial fraction decomposition, we can rewrite
as:
Note that as mentioned above,
̅̅̅
; we really only need to calculate one of these
coefficients. Using the method for calculation above:




̅̅̅
̅̅̅
2
̅̅̅
3
{
}
2
3
Case 4
–
Repeated Complex Poles
The last case we must deal with involves the potential for repeated complex poles. However,
aside from some slight differences in the combination of conjugate terms, the procedure is
otherwise identical to that of Case 2. To begin, the expansion can be written for a function,
, with a single repeated (
times, where
) complex pole,
, as follows:
[
] [
]
0
̅̅̅̅
̅̅̅
̅̅̅̅
̅̅̅
̅̅̅̅
̅̅̅
1
One can notice that we have effectively doubled the number of terms related to the repeated
pole,
. However, the coefficients
all have complex conjugate pairs, meaning we do
not actually require any additional calculations over a repeated real pole problem.
40
Lastly, since we have pairs of complex conjugate poles, they will combine to create cosine
and sine terms, as per the following:
̅
 
2
̅
3
2
 
 
3
 
 
 
(
)
 
To finish off this section, let us examine one last example. Consider a function with
repeated complex poles,
, for which we want to determine
.
Using the above expansion, we obtain:
̅̅̅̅
̅̅̅̅
Calculating the coefficients yields the following:



The remaining unknown coefficients are simply complex conjugates of
and
. We can
then rewrite the partial fraction expansion and complete the inverse:
0
1 0
1
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41
{
}
2
3
2
3
This completes our examination of the inverse Laplace Transform for circuits. We will
assume from now on that any particular function for which we need the inverse transform
will be easily calculated using either this method, or through direct table lookup. Most
importantly, now that we have completed a review of the basics of the Laplace Transform in
general, we are now able to apply what we have learned to useful realworld tasks involving
actual circuits.
As with the Laplace Transform, the software package Maple is able to directly calculate the
inverse Laplace Transform (after including the „inttrans‟ package, as before). For example,
if we enter the function from the first example in this section, we get the correct
:
>
invlaplace( 96*(s+5)*(s+12)/(s*(s+8)*(s+6)), s, t);
It should be noted that for more complex functions, the result obtained by Maple may not
look exactly like those calculated by hand. However, the result will be mathematically
 Spring '17