If qq is a rational function with
poles
λ
j
j=1…h
λ
j
j
1
…
h
then the inverse Laplace transform of
qq is

ℒ-1qt≡12πi∫qzeztd
z
ℒ
q
t
1
2
z
C
q
z
z
t

(1)
where

CC is a curve that encloses
each of the poles of

qq. As a
result

ℒ-1qt=∑
j
=1hres
λ
j
ℒ
q
t
j
1
h
res
λ
j

(2)
Let us put this lovely formula to the test. We take our
examples from discussion of

the Laplace Transform and

the inverse Laplace
Transform. Let us first compute the inverse Laplace
Transform of

qz=1z+12
q
z
1
z
1
2
According to

Equation 2 it is simply
the residue of

qzezt
q
z
z
t
at

z=-1
z
-1
,

i.e.,

res-1=limit
z
→
-1
deztd
z
=te−t
res
-1
z
-1
z
z
t
t
t
This closes the circle on the example begun in the discussion
of

the Laplace
Transform and continued in exercise one for chapter 6.

For our next example we recall
ℒ
x
1
s=0.19(s2+1.5s+0.27)s+1/64(s3+1.655s2+0.4978s+0.0039)
ℒ
x
1
s
0.19
s
2
1.5
s
0.27
s
16
4
s
3
1.655
s
2
0.4978
s
0.0039
from the Inverse Laplace
Transform. Using `numde`

,
`sym2poly`

and
`residue`

, see
`fib4.m`

for details, returns
r
1
=(
0.0029
262.8394
-474.1929
-1.0857
-9.0930
-0.3326
211.3507
)
r
1
0.0029
262.8394
-474.1929
-1.0857
-9.0930
-0.3326
211.3507
and
p
1
=(
-1.3565
-0.2885
-0.1667
-0.1667
-0.1667
-0.1667
-0.0100
)
p
1
-1.3565
-0.2885
-0.1667
-0.1667
-0.1667
-0.1667
-0.0100
You will be asked in the exercises to show that this indeed
jibes with the
x
1
t=211.35e−t100−(0.0554t3+4.5464t2+1.085t+474.19)e−t6+e-329t400(262.842cosh73t16+262.836sinh73t16)
x
1
t
211.35
t
100
0.0554
t
3
4.5464
t
2
1.085
t
474.19
t
6
-329
t
400
262.842
73
t
16
262.836
73
t
16
achieved in the Laplace
Transform via `ilaplace`

.